JEE Main 2014 12th April Morning Shift Question Paper Question Paper with Solutions | JEE Main Mock Test | JeeMate

Q 1PhysicsSection A • MCQunits and measurements

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From the following combinations of physical constants (expressed through their usual symbols) the only combination, that would have the same value in different systems of units, is:

A

$\frac{ch}{2\pi\varepsilon_0^2}$

B

$\frac{e^2}{2\pi\varepsilon_0 Gm_e^2}$

C

$\frac{\mu_0\varepsilon_0}{c^2} \cdot \frac{G}{he^2}$

D

$\frac{2\pi\sqrt{\mu_0\varepsilon_0}}{ce^2} \cdot \frac{h}{G}$

Official Correct Answer:Option (B)

Step-by-Step Solution:

To determine which combination of physical constants has the same value in different systems of units, we need to find the dimensionless quantity. A dimensionless quantity has dimensions of mass, length, time, and current all raised to the power zero, i.e., $[M^0 L^0 T^0 I^0]$ . We will perform dimensional analysis on each option using the following dimensions of the fundamental constants:

Speed of light, $c$ : $[L T^{-1}]$

Planck's constant, $h$ : $[M L^2 T^{-1}]$ (since energy $E = h\nu$ and $\nu$ has $[T^{-1}]$ )

Elementary charge, $e$ : $[I T]$ (charge = current $\times$ time)

Gravitational constant, $G$ : $[M^{-1} L^3 T^{-2}]$ (from Newton's law of gravitation)

Permittivity of free space, $\varepsilon_0$ : $[M^{-1} L^{-3} T^4 I^2]$ (from Coulomb's law: $F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$ )

Permeability of free space, $\mu_0$ : $[M L T^{-2} I^{-2}]$ (from the force between parallel currents)

Note that $\mu_0 \varepsilon_0 = \frac{1}{c^2}$ , so the dimensions of $\mu_0 \varepsilon_0$ are $[T^2 L^{-2}]$ .

Also, constants like $2\pi$ are dimensionless and do not affect the dimensional analysis.

Now, we evaluate each option:

Option A: $\frac{ch}{2\pi\varepsilon_0^2}$

Dimensions of numerator $c h$ : $[L T^{-1}] \times [M L^2 T^{-1}] = [M L^3 T^{-2}]$

Dimensions of denominator $\varepsilon_0^2$ : $[M^{-1} L^{-3} T^4 I^2]^2 = [M^{-2} L^{-6} T^8 I^4]$

Dimensions of fraction: $\frac{[M L^3 T^{-2}]}{[M^{-2} L^{-6} T^8 I^4]} = M^{1 - (-2)} L^{3 - (-6)} T^{-2 - 8} I^{-4} = [M^3 L^9 T^{-10} I^{-4}]$

This is not dimensionless.

Option B: $\frac{e^2}{2\pi\varepsilon_0 G m_e^2}$

Dimensions of numerator $e^2$ : $[I T]^2 = [I^2 T^2]$

Dimensions of denominator $\varepsilon_0 G m_e^2$ : $\varepsilon_0: [M^{-1} L^{-3} T^4 I^2]$ , $G: [M^{-1} L^3 T^{-2}]$ , $m_e^2: [M]^2 = [M^2]$ , so $\varepsilon_0 \times G \times m_e^2 = [M^{-1} L^{-3} T^4 I^2] \times [M^{-1} L^3 T^{-2}] \times [M^2] = M^{-1-1+2} L^{-3+3} T^{4-2} I^2 = [M^0 L^0 T^2 I^2]$

Dimensions of fraction: $\frac{[I^2 T^2]}{[T^2 I^2]} = [I^{2-2} T^{2-2}] = [M^0 L^0 T^0 I^0]$

This is dimensionless.

Option C: $\frac{\mu_0\varepsilon_0}{c^2} \cdot \frac{G}{h e^2}$

First part: $\frac{\mu_0\varepsilon_0}{c^2}$ , dimensions of $\mu_0\varepsilon_0$ : $[T^2 L^{-2}]$ , dimensions of $c^2$ : $[L^2 T^{-2}]$ , so fraction: $\frac{[T^2 L^{-2}]}{[L^2 T^{-2}]} = T^{2 - (-2)} L^{-2 - 2} = [T^4 L^{-4}]$

Second part: $\frac{G}{h e^2}$ , dimensions of $G$ : $[M^{-1} L^3 T^{-2}]$ , dimensions of $h e^2$ : $h: [M L^2 T^{-1}]$ , $e^2: [I^2 T^2]$ , so $h e^2 = [M L^2 T^{-1}] \times [I^2 T^2] = [M L^2 T I^2]$ , fraction: $\frac{[M^{-1} L^3 T^{-2}]}{[M L^2 T I^2]} = M^{-1-1} L^{3-2} T^{-2-1} I^{-2} = [M^{-2} L T^{-3} I^{-2}]$

Product: $[T^4 L^{-4}] \times [M^{-2} L T^{-3} I^{-2}] = M^{-2} L^{-4+1} T^{4-3} I^{-2} = [M^{-2} L^{-3} T I^{-2}]$

This is not dimensionless.

Option D: $\frac{2\pi\sqrt{\mu_0\varepsilon_0}}{c e^2} \cdot \frac{h}{G}$

Note that $\sqrt{\mu_0\varepsilon_0} = \sqrt{\frac{1}{c^2}} = \frac{1}{c}$ , so the expression simplifies to $\frac{2\pi \cdot \frac{1}{c}}{c e^2} \cdot \frac{h}{G} = \frac{2\pi}{c^2 e^2} \cdot \frac{h}{G}$

Dimensions of denominator $c^2 e^2$ : $c^2: [L^2 T^{-2}]$ , $e^2: [I^2 T^2]$ , so $c^2 e^2 = [L^2 T^{-2}] \times [I^2 T^2] = [L^2 I^2]$

Dimensions of $\frac{h}{G}$ : $h: [M L^2 T^{-1}]$ , $G: [M^{-1} L^3 T^{-2}]$ , so fraction: $\frac{[M L^2 T^{-1}]}{[M^{-1} L^3 T^{-2}]} = M^{1-(-1)} L^{2-3} T^{-1-(-2)} = [M^2 L^{-1} T]$

Product: $\frac{1}{[L^2 I^2]} \times [M^2 L^{-1} T] = [M^2 L^{-3} T I^{-2}]$

This is not dimensionless.

Only Option B is dimensionless and thus has the same value in different systems of units. Hence, the correct answer is Option B.

Q 2PhysicsSection A • MCQmotion in a straight line

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A person climbs up a stalled escalator in 60 s. If standing on the same but escalator running with constant velocity he takes 40 s. How much time is taken by the person to walk up the moving escalator?

A

37 s

B

27 s

C

24 s

D

45 s

Q 3PhysicsSection A • MCQwork energy and power

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A bullet of mass 4 g is fired horizontally with a speed of 300 m/s into 0.8 kg block of wood at rest on a table. If the coefficient of friction between the block and the table is 0.3, how far will the block slide approximately?

A

0.19 m

B

0.379 m

C

0.569 m

D

0.758 m

Official Correct Answer:Option (B)

Step-by-Step Solution:

First, we have a bullet of mass 4 g fired horizontally at 300 m/s into a wooden block of mass 0.8 kg at rest. The bullet embeds into the block, making this an inelastic collision. We use conservation of momentum to find the velocity of the block-bullet system right after the collision.

Convert the bullet's mass to kilograms: mass of bullet, $m_b = 4$ g = $0.004$ kg. Mass of block, $m_{bl} = 0.8$ kg. Initial velocity of bullet, $u_b = 300$ m/s. Initial velocity of block, $u_{bl} = 0$ m/s. Let $v$ be the common velocity after collision.

Conservation of momentum:

$m_b u_b + m_{bl} u_{bl} = (m_b + m_{bl}) v$

Substitute the values:

$(0.004 \times 300) + (0.8 \times 0) = (0.004 + 0.8) v$

$1.2 + 0 = 0.804 v$

$1.2 = 0.804 v$

Solve for $v$ :

$v = \frac{1.2}{0.804} = \frac{1200}{804}$

Simplify the fraction by dividing numerator and denominator by 12:

$\frac{1200 \div 12}{804 \div 12} = \frac{100}{67}$

So, $v = \frac{100}{67}$ m/s ≈ 1.4925 m/s.

Now, the block-bullet system slides on the table with initial velocity $v = \frac{100}{67}$ m/s and comes to a stop due to friction. The coefficient of friction is $\mu = 0.3$ . The friction force opposes the motion.

The normal force $N$ equals the weight of the system since the table is horizontal. Total mass, $m = m_b + m_{bl} = 0.004 + 0.8 = 0.804$ kg. Using $g = 9.8$ m/s², weight = $m g = 0.804 \times 9.8$ N.

Friction force, $f = \mu N = \mu m g = 0.3 \times 0.804 \times 9.8$ .

By Newton's second law, friction causes deceleration $a$ :

$f = m a$

$0.3 \times 0.804 \times 9.8 = 0.804 a$

The mass $0.804$ cancels out:

$a = 0.3 \times 9.8 = 2.94 \text{ m/s}^2$

So, deceleration $a = -2.94$ m/s² (negative indicates opposition to motion).

We need to find the distance $s$ the block slides before stopping. Initial velocity $u = v = \frac{100}{67}$ m/s, final velocity $v_f = 0$ m/s, acceleration $a = -2.94$ m/s².

Use the equation of motion:

$v_f^2 = u^2 + 2 a s$

Substitute the values:

$0^2 = \left( \frac{100}{67} \right)^2 + 2 \times (-2.94) \times s$

$0 = \frac{10000}{4489} - 5.88 s$

Solve for $s$ :

$5.88 s = \frac{10000}{4489}$

$s = \frac{10000}{4489 \times 5.88}$

Convert $5.88$ to a fraction: $5.88 = \frac{588}{100} = \frac{147}{25}$ .

$s = \frac{10000}{4489} \times \frac{1}{5.88} = \frac{10000}{4489} \times \frac{25}{147}$

$s = \frac{10000 \times 25}{4489 \times 147}$

Calculate numerator: $10000 \times 25 = 250000$ .

Calculate denominator: $4489 \times 147$ . Break it down:

$4489 \times 100 = 448900$

$4489 \times 40 = 179560$

$4489 \times 7 = 31423$

Add them: $448900 + 179560 = 628460$ , then $628460 + 31423 = 659883$ .

So,

$s = \frac{250000}{659883} \approx 0.37878 \text{ m}$

Approximately, $s \approx 0.379$ m.

Comparing with the options, 0.379 m corresponds to Option B.

Hence, the correct answer is Option B.

Q 4PhysicsSection A • MCQwork energy and power

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A spring of unstretched length l has a mass m with one end fixed to a rigid support. Assuming spring to be made of a uniform wire, the kinetic energy possessed by it if its free end is pulled with uniform velocity v is:

A

$\frac{1}{2}mv^2$

B

$mv^2$

C

$\frac{1}{3}mv^2$

D

$\frac{1}{6}mv^2$

Official Correct Answer:Option (D)

Step-by-Step Solution:

A spring of unstretched length $l$ and mass $m$ is fixed at one end and pulled at the free end with a uniform velocity $v$ . The spring is made of a uniform wire, so its mass per unit length is constant. We need to find the kinetic energy of the entire spring.

Since the spring is uniform, its mass distribution is linear along its unstretched length. Let the mass per unit length be $\mu = \frac{m}{l}$ . Consider a small element of the spring at a distance $s$ from the fixed end in the unstretched state. The mass of this element is $dm = \mu ds = \frac{m}{l} ds$ .

When the free end is pulled with uniform velocity $v$ , the spring stretches uniformly. The displacement of a point originally at distance $s$ from the fixed end is proportional to $s$ . If the displacement of the free end (at $s = l$ ) is $\delta(t) = v t$ (since velocity is constant), then the displacement of a point at $s$ is $u(s, t) = \frac{s}{l} \delta(t) = \frac{s}{l} v t$ .

The velocity of this point is the time derivative of displacement: $v(s) = \frac{\partial u}{\partial t} = \frac{s}{l} v$ .

The kinetic energy of the small element is $dK = \frac{1}{2} dm [v(s)]^2$ . Substituting $dm$ and $v(s)$ :

$dK = \frac{1}{2} \left( \frac{m}{l} ds \right) \left( \frac{s}{l} v \right)^2$

Simplify the expression:

$dK = \frac{1}{2} \cdot \frac{m}{l} \cdot \frac{s^2}{l^2} v^2 ds = \frac{1}{2} \cdot \frac{m v^2}{l^3} s^2 ds$

To find the total kinetic energy, integrate $dK$ over the entire length of the spring from $s = 0$ to $s = l$ :

$K = \int_{0}^{l} dK = \int_{0}^{l} \frac{1}{2} \cdot \frac{m v^2}{l^3} s^2 ds$

Factor out the constants:

$K = \frac{1}{2} \cdot \frac{m v^2}{l^3} \int_{0}^{l} s^2 ds$

Evaluate the integral:

$\int_{0}^{l} s^2 ds = \left[ \frac{s^3}{3} \right]_{0}^{l} = \frac{l^3}{3}$

Substitute back:

$K = \frac{1}{2} \cdot \frac{m v^2}{l^3} \cdot \frac{l^3}{3} = \frac{1}{2} \cdot \frac{m v^2}{3} = \frac{1}{6} m v^2$

Hence, the kinetic energy possessed by the spring is $\frac{1}{6} m v^2$ . Comparing with the options, this corresponds to Option D.

So, the answer is Option D.

Q 5PhysicsSection A • MCQlaws of motion

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Three masses m, 2m and 3m are moving in the x-y plane with speeds 3u, 2u and u respectively as shown in the figure. The three masses collide at the same point at P and stick together. The velocity of the resulting mass will be:

A

$\frac{u}{12}(\hat{i} + \sqrt{3}\hat{j})$

B

$\frac{u}{12}(\hat{i} - \sqrt{3}\hat{j})$

C

$\frac{u}{12}(-\hat{i} + \sqrt{3}\hat{j})$

D

$\frac{u}{12}(-\hat{i} - \sqrt{3}\hat{j})$

Q 6PhysicsSection A • MCQrotational motion

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A particle of mass 1 unit is moving in a circular path of radius a, with a constant velocity v as shown in the figure. The centre of circle is marked by 'C'. The angular momentum from the origin O can be written as:

A

$va(1 + \cos 2\theta)$

B

$va(1 + \cos\theta)$

C

$va\cos 2\theta$

D

$va$

Q 7PhysicsSection A • MCQgravitation

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Two hypothetical planets of masses m $_1$ and m $_2$ are at rest when they are infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centres. What is their speed when their separation is 'd'? (Speed of m $_1$ is v $_1$ and that of m $_2$ is v $_2$ )

A

$v_1 = v_2$

B

$v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)}}$ , $v_2 = m_1\sqrt{\frac{2G}{d(m_1+m_2)}}$

C

$v_1 = m_1\sqrt{\frac{2G}{d(m_1+m_2)}}$ , $v_2 = m_2\sqrt{\frac{2G}{d(m_1+m_2)}}$

D

$v_1 = m_2\sqrt{\frac{2G}{m_1}}$ , $v_2 = m_2\sqrt{\frac{2G}{m_2}}$

Q 8PhysicsSection A • MCQproperties of matter

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Steel ruptures when a shear of $3.5 \times 10^8$ N m $^{-2}$ is applied. The force needed to punch a 1 cm diameter hole in a steel sheet 0.3 cm thick is nearly:

A

$1.4 \times 10^4$ N

B

$2.7 \times 10^4$ N

C

$3.3 \times 10^4$ N

D

$1.1 \times 10^4$ N

Official Correct Answer:Option (C)

Step-by-Step Solution:

The steel ruptures when the shear stress reaches $3.5 \times 10^8$ N m $^{-2}$ . To punch a hole, we need to apply a force that causes this shear stress over the area being sheared.

When punching a hole of diameter 1 cm in a sheet of thickness 0.3 cm, the sheared area is the lateral surface area of the cylindrical hole. This area is calculated as the circumference of the hole multiplied by the thickness of the sheet.

First, convert all measurements to meters for consistency. The diameter is 1 cm, which is 0.01 m, so the radius $r$ is half of that: $r = \frac{0.01}{2} = 0.005$ m. The thickness $t$ is 0.3 cm, which is 0.003 m.

The circumference of the hole is $2 \pi r$ . Therefore, the sheared area $A$ is:

$A = 2 \pi r t$

Substituting the values:

$A = 2 \times \pi \times 0.005 \times 0.003$

First, multiply the numerical parts:

$0.005 \times 0.003 = 0.000015$

Then:

$A = 2 \times \pi \times 0.000015 = 0.00003 \pi$

Using $\pi \approx 3.1416$ ,

$A \approx 0.00003 \times 3.1416 = 0.000094248 \text{ m}^2$

In scientific notation, $A \approx 9.4248 \times 10^{-5}$ m $^2$ .

The shear stress $\tau$ is related to the force $F$ and area $A$ by:

$\tau = \frac{F}{A}$

At rupture, $\tau = 3.5 \times 10^8$ N m $^{-2}$ . Solving for $F$ :

$F = \tau \times A$

Substituting the values:

$F = (3.5 \times 10^8) \times (9.4248 \times 10^{-5})$

First, multiply the coefficients:

$3.5 \times 9.4248 = 32.9868$

Then multiply the powers of 10:

$10^8 \times 10^{-5} = 10^{3}$

So:

$F \approx 32.9868 \times 10^3 = 32986.8 \text{ N}$

Which is $3.29868 \times 10^4$ N. Rounding to two significant figures, as in the options, gives approximately $3.3 \times 10^4$ N.

Alternatively, using exact values:

$A = 2 \pi r t = 2 \times \pi \times 0.005 \times 0.003 = 2 \times \pi \times 1.5 \times 10^{-5} = 3\pi \times 10^{-5} \text{ m}^2$

Then:

$F = (3.5 \times 10^8) \times (3\pi \times 10^{-5}) = 3.5 \times 3 \times \pi \times 10^{8-5} = 10.5 \pi \times 10^3 = 10500\pi \text{ N}$

With $\pi \approx 3.1416$ ,

$F \approx 10500 \times 3.1416 = 32986.8 \text{ N} \approx 3.3 \times 10^4 \text{ N}$

Comparing with the options, $3.3 \times 10^4$ N corresponds to option C.

Hence, the correct answer is Option C.

Q 9PhysicsSection A • MCQproperties of matter

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A cylindrical vessel of cross-section A contains water to a height h. There is a hole in the bottom of radius 'a'. The time in which it will be emptied is:

A

$\frac{2A}{\pi a^2}\sqrt{\frac{h}{g}}$

B

$\frac{\sqrt{2}A}{\pi a^2}\sqrt{\frac{h}{g}}$

C

$\frac{2\sqrt{2}A}{\pi a^2}\sqrt{\frac{h}{g}}$

D

$\frac{A}{\sqrt{2}\pi a^2}\sqrt{\frac{h}{g}}$

Official Correct Answer:Option (B)

Step-by-Step Solution:

We are given a cylindrical vessel with cross-sectional area $A$ filled with water to a height $h$ . There is a hole at the bottom with radius $a$ , so the area of the hole is $\pi a^2$ . We need to find the time taken for the vessel to empty completely.

According to Torricelli's theorem, the speed of efflux $v$ of water through the hole is given by $v = \sqrt{2gh}$ , where $g$ is the acceleration due to gravity and $h$ is the height of water above the hole at any instant.

The volume flow rate of water through the hole is the product of the area of the hole and the speed of efflux. Since the volume is decreasing, we write:

$\frac{dV}{dt} = - (\pi a^2) \sqrt{2gh}$

Here, $V$ is the volume of water in the vessel. For a cylinder, $V = A h$ . Differentiating both sides with respect to time $t$ :

$\frac{dV}{dt} = A \frac{dh}{dt}$

Setting the two expressions for $\frac{dV}{dt}$ equal:

$A \frac{dh}{dt} = - \pi a^2 \sqrt{2gh}$

We can rearrange this equation to separate the variables $h$ and $t$ . First, express $\sqrt{2gh}$ as $\sqrt{2g} \sqrt{h}$ :

$A \frac{dh}{dt} = - \pi a^2 \sqrt{2g} \sqrt{h}$

Bring terms involving $h$ to one side and $t$ to the other:

$A dh = - \pi a^2 \sqrt{2g} \sqrt{h} dt$

Divide both sides by $\sqrt{h}$ and rearrange:

$dt = - \frac{A}{\pi a^2 \sqrt{2g}} \frac{dh}{\sqrt{h}}$

Simplify the right side:

$dt = - \frac{A}{\pi a^2 \sqrt{2g}} h^{-1/2} dh$

To find the total time $T$ for the vessel to empty, integrate both sides. At $t = 0$ , $h = h$ (initial height). At $t = T$ , $h = 0$ (vessel empty). So:

$\int_{0}^{T} dt = - \frac{A}{\pi a^2 \sqrt{2g}} \int_{h}^{0} h^{-1/2} dh$

The left side integrates to $T$ . For the right side, reversing the limits of integration removes the negative sign:

$T = \frac{A}{\pi a^2 \sqrt{2g}} \int_{0}^{h} h^{-1/2} dh$

Now, integrate $h^{-1/2}$ :

$\int h^{-1/2} dh = \frac{h^{1/2}}{1/2} = 2 h^{1/2}$

Evaluate the definite integral:

$\int_{0}^{h} h^{-1/2} dh = \left[ 2 \sqrt{h} \right]_{0}^{h} = 2\sqrt{h} - 2\sqrt{0} = 2\sqrt{h}$

Substitute back:

$T = \frac{A}{\pi a^2 \sqrt{2g}} \times 2 \sqrt{h}$

Simplify:

$T = \frac{2A \sqrt{h}}{\pi a^2 \sqrt{2g}}$

Express $\sqrt{h}$ and $\sqrt{2g}$ in terms of $\sqrt{\frac{h}{g}}$ :

$T = \frac{2A}{\pi a^2} \cdot \frac{\sqrt{h}}{\sqrt{2g}} = \frac{2A}{\pi a^2} \sqrt{\frac{h}{2g}}$

Now, $\sqrt{\frac{h}{2g}} = \frac{1}{\sqrt{2}} \sqrt{\frac{h}{g}}$ , so:

$T = \frac{2A}{\pi a^2} \cdot \frac{1}{\sqrt{2}} \sqrt{\frac{h}{g}} = \frac{2A}{\sqrt{2} \pi a^2} \sqrt{\frac{h}{g}}$

Simplify $\frac{2}{\sqrt{2}}$ :

$\frac{2}{\sqrt{2}} = \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

Thus:

$T = \frac{\sqrt{2} A}{\pi a^2} \sqrt{\frac{h}{g}}$

Comparing with the options, this matches option B.

Hence, the correct answer is Option B.

Q 10PhysicsSection A • MCQproperties of matter

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Two soap bubbles coalesce to form a single bubble. If V is the subsequent change in volume of contained air and S change in total surface area, T is the surface tension and P atmospheric pressure, then which of the following relation is correct?

A

4PV + 3ST = 0

B

3PV + 4ST = 0

C

2PV + 3ST = 0

D

3PV + 2ST = 0

Official Correct Answer:Option (B)

Step-by-Step Solution:

To solve this problem, we need to find the correct relation between the change in volume (V), change in surface area (S), surface tension (T), and atmospheric pressure (P) when two soap bubbles coalesce into one. We start by recalling the physics of soap bubbles.

A soap bubble has an excess pressure inside due to surface tension. Since a soap bubble has two surfaces (inner and outer), the excess pressure is given by $\Delta P = \frac{4T}{r}$ , where $r$ is the radius. Therefore, the pressure inside a bubble is $P_{\text{inside}} = P + \frac{4T}{r}$ , with $P$ being atmospheric pressure.

Let the two initial bubbles have radii $r_1$ and $r_2$ . After coalescing, they form a single bubble of radius $r$ . The air inside is conserved, and assuming constant temperature, we use the ideal gas law. The number of moles of air is conserved.

For the first bubble, the number of moles is:

$n_1 = \frac{\left(P + \frac{4T}{r_1}\right) \cdot \frac{4}{3}\pi r_1^3}{RT}$

For the second bubble:

$n_2 = \frac{\left(P + \frac{4T}{r_2}\right) \cdot \frac{4}{3}\pi r_2^3}{RT}$

The total initial moles are:

$n_1 + n_2 = \frac{\frac{4}{3}\pi}{RT} \left[ \left(P + \frac{4T}{r_1}\right) r_1^3 + \left(P + \frac{4T}{r_2}\right) r_2^3 \right]$

Simplifying the terms inside the brackets:

$\left(P + \frac{4T}{r_1}\right) r_1^3 = P r_1^3 + 4T r_1^2$

$\left(P + \frac{4T}{r_2}\right) r_2^3 = P r_2^3 + 4T r_2^2$

So:

$n_1 + n_2 = \frac{\frac{4}{3}\pi}{RT} \left[ P r_1^3 + 4T r_1^2 + P r_2^3 + 4T r_2^2 \right]$

For the final single bubble:

$n = \frac{\left(P + \frac{4T}{r}\right) \cdot \frac{4}{3}\pi r^3}{RT} = \frac{\frac{4}{3}\pi}{RT} \left[ P r^3 + 4T r^2 \right]$

Since moles are conserved, $n_1 + n_2 = n$ :

$P r_1^3 + 4T r_1^2 + P r_2^3 + 4T r_2^2 = P r^3 + 4T r^2$

Rearranging:

$P r_1^3 + P r_2^3 - P r^3 + 4T r_1^2 + 4T r_2^2 - 4T r^2 = 0$

$P (r_1^3 + r_2^3 - r^3) + 4T (r_1^2 + r_2^2 - r^2) = 0 \quad ...(1)$

Now, define the changes. The change in volume of contained air, $V$ , is the final volume minus the initial total volume. The volume of a bubble is $\frac{4}{3}\pi r^3$ , so:

$V = \frac{4}{3}\pi r^3 - \left( \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 \right) = \frac{4}{3}\pi (r^3 - r_1^3 - r_2^3)$

Solving for $r^3 - r_1^3 - r_2^3$ :

$r^3 - r_1^3 - r_2^3 = \frac{3V}{4\pi} \quad ...(2)$

The change in total surface area, $S$ , is defined as the final surface area minus the initial total surface area. For a soap bubble, the surface area is often considered as the area of one surface (e.g., the outer surface), which is $4\pi r^2$ . Thus:

Initial total surface area = $4\pi r_1^2 + 4\pi r_2^2 = 4\pi (r_1^2 + r_2^2)$

Final surface area = $4\pi r^2$

So:

$S = 4\pi r^2 - 4\pi (r_1^2 + r_2^2) = 4\pi (r^2 - r_1^2 - r_2^2)$

Solving for $r^2 - r_1^2 - r_2^2$ :

$r^2 - r_1^2 - r_2^2 = \frac{S}{4\pi} \quad ...(3)$

Note that equation (1) contains $r_1^3 + r_2^3 - r^3$ and $r_1^2 + r_2^2 - r^2$ . Using equations (2) and (3):

$r_1^3 + r_2^3 - r^3 = - (r^3 - r_1^3 - r_2^3) = - \frac{3V}{4\pi}$

$r_1^2 + r_2^2 - r^2 = - (r^2 - r_1^2 - r_2^2) = - \frac{S}{4\pi}$

Substitute these into equation (1):

$P \left( - \frac{3V}{4\pi} \right) + 4T \left( - \frac{S}{4\pi} \right) = 0$

$- \frac{3PV}{4\pi} - \frac{4T S}{4\pi} = 0$

Simplify the second term:

$- \frac{3PV}{4\pi} - \frac{T S}{\pi} = 0$

Multiply both sides by $-1$ :

$\frac{3PV}{4\pi} + \frac{T S}{\pi} = 0$

To eliminate denominators, multiply both sides by $4\pi$ :

$4\pi \cdot \frac{3PV}{4\pi} + 4\pi \cdot \frac{T S}{\pi} = 0$

$3PV + 4 T S = 0$

Since $T S = ST$ , we write:

$3PV + 4ST = 0$

This matches option B.

Hence, the correct answer is Option B.

Q 11PhysicsSection A • MCQthermal properties of matter

+4 / -1

Hot water cools from 60°C to 50°C in the first 10 minutes and to 42°C in the next 10 minutes. The temperature of the surroundings is:

A

25°C

B

10°C

C

15°C

D

20°C

Official Correct Answer:Option (B)

Step-by-Step Solution:

According to Newton's law of cooling, the rate of cooling is proportional to the difference between the temperature of the body and the surroundings. The formula for the average rate of cooling over a time interval is given by:

$\frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_s \right)$

where $T_1$ is the initial temperature, $T_2$ is the final temperature, $t$ is the time interval, $T_s$ is the temperature of the surroundings, and $k$ is a constant.

For the first 10 minutes:

Initial temperature $T_1 = 60^\circ \text{C}$
Final temperature $T_2 = 50^\circ \text{C}$
Time $t = 10$ minutes
Rate of cooling $\frac{60 - 50}{10} = 1^\circ \text{C per minute}$
Average temperature $\frac{60 + 50}{2} = 55^\circ \text{C}$

So, the equation becomes:

$1 = k (55 - T_s) \quad \text{(Equation 1)}$

For the next 10 minutes:

Initial temperature $T_1 = 50^\circ \text{C}$
Final temperature $T_2 = 42^\circ \text{C}$
Time $t = 10$ minutes
Rate of cooling $\frac{50 - 42}{10} = 0.8^\circ \text{C per minute}$
Average temperature $\frac{50 + 42}{2} = 46^\circ \text{C}$

So, the equation becomes:

$0.8 = k (46 - T_s) \quad \text{(Equation 2)}$

We now have two equations:

$1 = k (55 - T_s)$

$0.8 = k (46 - T_s)$

To solve for $T_s$ , divide Equation 1 by Equation 2:

$\frac{1}{0.8} = \frac{k (55 - T_s)}{k (46 - T_s)}$

Simplify the left side:

$\frac{1}{0.8} = 1.25$

So:

$1.25 = \frac{55 - T_s}{46 - T_s}$

Cross-multiply:

$1.25 \times (46 - T_s) = 55 - T_s$

Calculate $1.25 \times 46$ :

$1.25 \times 46 = 1.25 \times (40 + 6) = (1.25 \times 40) + (1.25 \times 6) = 50 + 7.5 = 57.5$

So:

$57.5 - 1.25 T_s = 55 - T_s$

Bring all terms involving $T_s$ to one side and constants to the other. Add $1.25 T_s$ to both sides:

$57.5 = 55 - T_s + 1.25 T_s$

$57.5 = 55 + 0.25 T_s$

Subtract 55 from both sides:

$57.5 - 55 = 0.25 T_s$

$2.5 = 0.25 T_s$

Solve for $T_s$ :

$T_s = \frac{2.5}{0.25} = \frac{2.5 \times 100}{0.25 \times 100} = \frac{250}{25} = 10$

So, the temperature of the surroundings is $10^\circ \text{C}$ .

Verification: Substitute $T_s = 10$ into Equation 1:

$1 = k (55 - 10) = k \times 45 \implies k = \frac{1}{45}$

Now substitute into Equation 2:

$0.8 = \frac{1}{45} \times (46 - 10) = \frac{1}{45} \times 36 = \frac{36}{45} = \frac{4}{5} = 0.8$

This matches, confirming the solution.

Hence, the correct answer is Option B.

Q 12PhysicsSection A • MCQheat and thermodynamics

+4 / -1

A Carnot engine absorbs 1000 J of heat energy from a reservoir at 127°C and rejects 600 J of heat energy during each cycle. The efficiency of engine and temperature of sink will be:

A

20% and $-43$ °C

B

40% and $-33$ °C

C

50% and $-20$ °C

D

70% and $-10$ °C

Official Correct Answer:Option (B)

Step-by-Step Solution:

A Carnot engine absorbs 1000 J of heat energy from a reservoir at 127°C and rejects 600 J of heat energy during each cycle. We need to find the efficiency of the engine and the temperature of the sink.

First, recall that the efficiency (η) of any heat engine is given by the ratio of the work done to the heat absorbed. The work done by the engine is the difference between the heat absorbed and the heat rejected. So:

Heat absorbed, $Q_1 = 1000 \text{J}$

Heat rejected, $Q_2 = 600 \text{J}$

Work done, $W = Q_1 - Q_2 = 1000 - 600 = 400 \text{J}$

Efficiency, $\eta = \frac{W}{Q_1} = \frac{400}{1000}$

Simplifying the fraction: $\frac{400}{1000} = 0.4$

To express as a percentage: $0.4 \times 100 = 40\%$

So, the efficiency is 40%.

For a Carnot engine, efficiency is also related to the absolute temperatures of the source and sink. The formula is:

$\eta = 1 - \frac{T_2}{T_1}$

where $T_1$ is the absolute temperature of the source (in Kelvin) and $T_2$ is the absolute temperature of the sink (in Kelvin).

The source temperature is given as 127°C. Convert this to Kelvin by adding 273:

$T_1 = 127 + 273 = 400 \text{K}$

We know $\eta = 40\% = 0.4$ (in decimal form). Substitute into the formula:

$0.4 = 1 - \frac{T_2}{400}$

Solve for $T_2$ :

Rearrange: $\frac{T_2}{400} = 1 - 0.4$

$\frac{T_2}{400} = 0.6$

Multiply both sides by 400: $T_2 = 0.6 \times 400$

$T_2 = 240 \text{K}$

Convert $T_2$ from Kelvin to Celsius by subtracting 273:

$T_2 (\text{in °C}) = 240 - 273 = -33°C$

Therefore, the efficiency is 40% and the temperature of the sink is -33°C.

Comparing with the options:

A. 20% and -43°C

B. 40% and -33°C

C. 50% and -20°C

D. 70% and -10°C

Option B matches our calculated values.

Hence, the correct answer is Option B.

Q 13PhysicsSection A • MCQkinetic theory of gases

+4 / -1

At room temperature a diatomic gas is found to have an r.m.s. speed of 1930 ms $^{-1}$ . The gas is:

A

H $_2$

B

Cl $_2$

C

O $_2$

D

F $_2$

Official Correct Answer:Option (A)

Step-by-Step Solution:

The root mean square (r.m.s.) speed of a gas is given by the formula:

v_{rms} = \sqrt{\frac{3RT}{M}}

where $R$ is the universal gas constant, $T$ is the temperature in Kelvin, and $M$ is the molar mass of the gas in kg/mol. The r.m.s. speed is provided as 1930 m/s, and room temperature is taken as 300 K. The gas constant $R$ is 8.314 J/(mol·K), which is equivalent to 8.314 kg·m²/(s²·mol·K) since 1 J = 1 kg·m²/s².

To identify the gas, we rearrange the formula to solve for the molar mass $M$ :

v_{rms}^2 = \frac{3RT}{M} \implies M = \frac{3RT}{v_{rms}^2}

Substituting the given values:

M = \frac{3 \times 8.314 \times 300}{(1930)^2}

First, compute the numerator:

3 \times 8.314 = 24.942

24.942 \times 300 = 7482.6

So the numerator is 7482.6 kg·m²/(s²·mol).

Next, compute the denominator, which is $v_{rms}^2$ :

1930^2 = 1930 \times 1930

Calculating this:

1930 \times 1930 = (2000 - 70)^2 = 2000^2 - 2 \times 2000 \times 70 + 70^2 = 4,000,000 - 280,000 + 4,900 = 3,724,900

So $v_{rms}^2 = 3,724,900$ m²/s².

Now, substitute back into the formula for $M$ :

M = \frac{7482.6}{3,724,900} \text{ kg/mol}

Performing the division:

M = \frac{7482.6}{3724900} \approx 0.0020088 \text{ kg/mol}

To convert to grams per mole (g/mol), multiply by 1000:

M = 0.0020088 \times 1000 = 2.0088 \text{ g/mol}

The molar mass is approximately 2.009 g/mol. Comparing this with the molar masses of the given diatomic gases:

Hydrogen (H₂): 2 g/mol
Chlorine (Cl₂): 71 g/mol (since Cl is 35.5 g/mol)
Oxygen (O₂): 32 g/mol
Fluorine (F₂): 38 g/mol

The calculated molar mass of 2.009 g/mol is very close to the molar mass of hydrogen gas (H₂), which is 2 g/mol. The slight difference can be attributed to approximations in room temperature or the gas constant, but it is evident that hydrogen is the gas.

To verify, the r.m.s. speed for hydrogen at 300 K can be calculated:

v_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{0.002}} = \sqrt{\frac{7482.6}{0.002}} = \sqrt{3,741,300} \approx 1934.5 \text{ m/s}

This is close to the given 1930 m/s, considering typical approximations. The other gases have much lower r.m.s. speeds at the same temperature, confirming that hydrogen is the correct gas.

Hence, the correct answer is Option A.

Q 14PhysicsSection A • MCQoscillations

+4 / -1

Which of the following expressions corresponds to simple harmonic motion along a straight line, where x is the displacement and a, b, c are positive constants?

A

$a + bx - cx^2$

B

$bx^2$

C

$a - bx + cx^2$

D

$-bx$

Official Correct Answer:Option (D)

Step-by-Step Solution:

To determine which expression corresponds to simple harmonic motion (SHM) along a straight line, we must recall the defining characteristic of SHM. In SHM, the acceleration of the particle is directly proportional to its displacement from a fixed point and is always directed opposite to the displacement. Mathematically, this is expressed as $a = -\omega^2 x$ , where $\omega$ is a positive constant and $x$ is the displacement. Here, $a$ represents acceleration, which is the second derivative of displacement with respect to time, $a = \frac{d^2 x}{dt^2}$ .

The given expressions are functions of $x$ , and we interpret them as representing acceleration $a$ in terms of displacement $x$ . Therefore, for SHM, the expression for acceleration must be linear in $x$ and have a negative sign, ensuring the restoring force property. The constants $a$ , $b$ , and $c$ are positive as per the question.

Now, let's evaluate each option:

Option A: $a + bx - cx^2$

This expression is quadratic due to the $-cx^2$ term. At $x = 0$ , the acceleration is $a$ (a positive constant), which is not zero. However, in SHM, at the equilibrium position ( $x = 0$ ), acceleration must be zero. Additionally, the quadratic term makes the acceleration nonlinear, violating the proportionality condition $a \propto -x$ . Hence, this does not represent SHM.

Option B: $bx^2$

This expression is quadratic and always non-negative since $b > 0$ and $x^2 \geq 0$ . At $x = 0$ , acceleration is zero, but for $x \neq 0$ , acceleration is positive. If $x$ is positive, positive acceleration increases $x$ further away from zero. If $x$ is negative, positive acceleration pushes the particle towards more negative values initially but eventually towards zero, but the dependence is quadratic, not linear. The equation $\frac{d^2 x}{dt^2} = bx^2$ is nonlinear and does not yield sinusoidal solutions characteristic of SHM. Thus, this is not SHM.

Option C: $a - bx + cx^2$

This expression is also quadratic. At $x = 0$ , acceleration is $a$ (positive), not zero, which contradicts SHM. The presence of a constant term and a quadratic term means acceleration is not proportional to $-x$ . Therefore, this does not correspond to SHM.

Option D: $-bx$

This expression is linear in $x$ and has a negative sign. Since $b > 0$ , we can write $a = -bx$ . Comparing with $a = -\omega^2 x$ , we see that $\omega^2 = b$ , a positive constant. This satisfies the condition for SHM: acceleration is proportional to displacement and opposite in direction. The resulting differential equation is $\frac{d^2 x}{dt^2} = -bx$ , or $\frac{d^2 x}{dt^2} + bx = 0$ , which has sinusoidal solutions like $x = A \sin(\sqrt{b} t + \phi)$ , confirming SHM.

Hence, the correct answer is Option D.

Q 15PhysicsSection A • MCQwaves

+4 / -1

A source of sound A emitting waves of frequency 1800 Hz is falling towards ground with a terminal speed v. The observer B on the ground directly beneath the source receives waves of frequency 2150 Hz. The source A receives waves, reflected from ground of frequency nearly: (Speed of sound = 343 m/s)

A

2150 Hz

B

2500 Hz

C

1800 Hz

D

2400 Hz

Official Correct Answer:Option (B)

Step-by-Step Solution:

First, we are given that a source of sound A emits waves at a frequency of 1800 Hz and is falling towards the ground with a terminal speed $v$ . An observer B on the ground directly beneath the source receives waves at a frequency of 2150 Hz. The speed of sound is 343 m/s. We need to find the frequency of the waves that source A receives after they are reflected from the ground.

To solve this, we use the Doppler effect formula for sound. The general formula when the source and observer are moving relative to each other is:

$f' = f \left( \frac{v_{\text{sound}} \pm v_{\text{observer}}}{v_{\text{sound}} \mp v_{\text{source}}} \right)$

The signs depend on the directions of motion. For the first part, source A is moving towards the stationary observer B on the ground. Since the source is moving towards the observer, the observed frequency increases. The formula becomes:

$f_B = f \left( \frac{v_{\text{sound}}}{v_{\text{sound}} - v_{\text{source}}} \right)$

Here, $f = 1800$ Hz, $f_B = 2150$ Hz, and $v_{\text{sound}} = 343$ m/s. Plugging in the values:

$2150 = 1800 \left( \frac{343}{343 - v} \right)$

Now, solve for $v$ . Rearrange the equation:

$\frac{2150}{1800} = \frac{343}{343 - v}$

Simplify the left side:

$\frac{2150}{1800} = \frac{2150 \div 50}{1800 \div 50} = \frac{43}{36}$

So:

$\frac{43}{36} = \frac{343}{343 - v}$

Cross-multiply:

$43 \times (343 - v) = 36 \times 343$

Calculate $36 \times 343$ :

$343 \times 30 = 10290, \quad 343 \times 6 = 2058, \quad 10290 + 2058 = 12348$

Calculate $43 \times 343$ :

$343 \times 40 = 13720, \quad 343 \times 3 = 1029, \quad 13720 + 1029 = 14749$

So:

$14749 - 43v = 12348$

Rearrange to solve for $v$ :

$14749 - 12348 = 43v$

$2401 = 43v$

$v = \frac{2401}{43}$

Simplify:

$2401 \div 43 = 55.8372, \quad \text{but we keep it as a fraction:} \quad v = \frac{2401}{43} \text{m/s}$

Note that $2401 = 343 \times 7$ because $343 \times 7 = 2401$ . So:

$v = \frac{343 \times 7}{43} \text{m/s}$

Now, we need to find the frequency that source A receives from the waves reflected from the ground. The ground reflects the sound waves, and since the ground is stationary, the frequency of the reflected wave is the same as the frequency received by the ground. Observer B on the ground receives 2150 Hz, so the reflected wave has a frequency of 2150 Hz and is emitted by the stationary ground.

Source A is moving towards the ground (downwards) while the reflected wave is traveling upwards. So, for the reflected wave:

The source (ground) is stationary, so $v_{\text{source}} = 0$ .
The observer (source A) is moving towards the source (ground), so it is moving towards the sound wave.

The Doppler effect formula for a stationary source and an observer moving towards the source is:

$f'' = f_{\text{reflected}} \left( \frac{v_{\text{sound}} + v_{\text{observer}}}{v_{\text{sound}}} \right)$

Here, $f_{\text{reflected}} = 2150$ Hz, $v_{\text{sound}} = 343$ m/s, and $v_{\text{observer}} = v = \frac{2401}{43}$ m/s. Plugging in:

$f'' = 2150 \left( \frac{343 + v}{343} \right) = 2150 \left(1 + \frac{v}{343}\right)$

Substitute $v = \frac{343 \times 7}{43}$ :

$f'' = 2150 \left(1 + \frac{\frac{343 \times 7}{43}}{343}\right) = 2150 \left(1 + \frac{7}{43}\right)$

Simplify inside the parentheses:

$1 + \frac{7}{43} = \frac{43}{43} + \frac{7}{43} = \frac{50}{43}$

So:

$f'' = 2150 \times \frac{50}{43}$

Note that $2150 = 43 \times 50$ because $43 \times 50 = 2150$ . Therefore:

$f'' = 43 \times 50 \times \frac{50}{43} = 50 \times 50 = 2500$

Thus, the frequency that source A receives from the reflected waves is 2500 Hz.

Hence, the correct answer is Option B.

Q 16PhysicsSection A • MCQelectrostatics

+4 / -1

A spherically symmetric charge distribution is characterised by the following charge density variations: $\rho(r) = \rho_o\left(1 - \frac{r}{R}\right)$ for $r < R$ , $\rho(r) = 0$ for $r \geq R$ . Where r is the distance from the centre of the charge distribution and $\rho_o$ is a constant. The electric field at an internal point ( $r < R$ ) is:

A

$\frac{\rho_o}{4\varepsilon_o}\left(\frac{r}{3} - \frac{r^2}{4R}\right)$

B

$\frac{\rho_o}{\varepsilon_o}\left(\frac{r}{3} - \frac{r^2}{4R}\right)$

C

$\frac{\rho_o}{3\varepsilon_o}\left(\frac{r}{3} - \frac{r^2}{4R}\right)$

D

$\frac{\rho_o}{12\varepsilon_o}\left(\frac{r}{3} - \frac{r^2}{4R}\right)$

Official Correct Answer:Option (B)

Step-by-Step Solution:

To find the electric field at an internal point where $r < R$ , we use Gauss's law due to the spherical symmetry. Gauss's law states that the flux through a closed surface is equal to the charge enclosed divided by $\varepsilon_0$ . We choose a spherical Gaussian surface of radius $r$ centered at the same point as the charge distribution.

The electric field is radial and constant in magnitude on this surface, so the flux is $E \times 4\pi r^2$ . By Gauss's law:

$E \times 4\pi r^2 = \frac{Q_{\text{enc}}}{\varepsilon_0}$

Solving for $E$ :

$E = \frac{Q_{\text{enc}}}{4\pi \varepsilon_0 r^2}$

Now, we need to find $Q_{\text{enc}}$ , the charge enclosed within the sphere of radius $r$ . The charge density is given by $\rho(r') = \rho_o \left(1 - \frac{r'}{R}\right)$ for $r' < R$ . Since $r < R$ , we integrate the charge density over the volume inside radius $r$ . Using spherical coordinates, the volume element is $dV' = 4\pi r'^2 dr'$ , so:

$Q_{\text{enc}} = \int_0^r \rho(r') dV' = \int_0^r \rho_o \left(1 - \frac{r'}{R}\right) \times 4\pi r'^2 dr'$

Factor out the constants:

$Q_{\text{enc}} = 4\pi \rho_o \int_0^r \left(1 - \frac{r'}{R}\right) r'^2 dr'$

Expand the integrand:

$Q_{\text{enc}} = 4\pi \rho_o \int_0^r \left( r'^2 - \frac{r'^3}{R} \right) dr'$

Integrate term by term:

$\int_0^r r'^2 dr' = \left[ \frac{r'^3}{3} \right]_0^r = \frac{r^3}{3}$

$\int_0^r \frac{r'^3}{R} dr' = \frac{1}{R} \left[ \frac{r'^4}{4} \right]_0^r = \frac{r^4}{4R}$

So the integral becomes:

$\int_0^r \left( r'^2 - \frac{r'^3}{R} \right) dr' = \frac{r^3}{3} - \frac{r^4}{4R}$

Therefore:

$Q_{\text{enc}} = 4\pi \rho_o \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)$

Substitute $Q_{\text{enc}}$ back into the expression for $E$ :

$E = \frac{4\pi \rho_o \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)}{4\pi \varepsilon_0 r^2} = \frac{\rho_o \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)}{\varepsilon_0 r^2}$

Simplify by dividing each term by $r^2$ :

$E = \frac{\rho_o}{\varepsilon_0} \left( \frac{r^3}{3r^2} - \frac{r^4}{4R r^2} \right) = \frac{\rho_o}{\varepsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right)$

Comparing with the options, this matches option B:

$\frac{\rho_o}{\varepsilon_o}\left(\frac{r}{3} - \frac{r^2}{4R}\right)$

Hence, the correct answer is Option B.

Q 17PhysicsSection A • MCQcapacitor

+4 / -1

The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation: $K(x) = K_o + \lambda x$ ( $\lambda$ = a constant). The capacitance C, of the capacitor, would be related to its vacuum capacitance C $_o$ for the relation:

A

$C = \frac{\lambda d}{\ln(1+K_o\lambda d)} C_o$

B

$C = \frac{\lambda}{d\ln(1+K_o\lambda d)} C_o$

C

$C = \frac{\lambda d}{\ln(1+\lambda d/K_o)} C_o$

D

$C = \frac{\lambda}{d\ln(1+K_o/\lambda d)} C_o$

Official Correct Answer:Option (C)

Step-by-Step Solution:

The capacitance of a parallel plate capacitor with a varying dielectric constant can be found by considering the capacitor as composed of infinitesimal capacitors in series. The dielectric constant is given by $K(x) = K_o + \lambda x$ , where $x$ is the distance from one plate, and the plates are separated by a distance $d$ . The vacuum capacitance is $C_o = \frac{\varepsilon_o A}{d}$ , where $A$ is the plate area and $\varepsilon_o$ is the permittivity of free space.

For an infinitesimal capacitor of thickness $dx$ at position $x$ , the capacitance is $dC = \frac{K(x) \varepsilon_o A}{dx}$ . Since these infinitesimal capacitors are in series, the reciprocal of the total capacitance $C$ is the integral of the reciprocals of the infinitesimal capacitances:

$\frac{1}{C} = \int_0^d \frac{1}{dC} dx = \int_0^d \frac{dx}{K(x) \varepsilon_o A}$

Substituting $K(x) = K_o + \lambda x$ :

$\frac{1}{C} = \int_0^d \frac{dx}{(K_o + \lambda x) \varepsilon_o A} = \frac{1}{\varepsilon_o A} \int_0^d \frac{dx}{K_o + \lambda x}$

To solve the integral, use the substitution $u = K_o + \lambda x$ . Then $du = \lambda dx$ , so $dx = \frac{du}{\lambda}$ . The limits change: when $x = 0$ , $u = K_o$ ; when $x = d$ , $u = K_o + \lambda d$ . The integral becomes:

$\int_0^d \frac{dx}{K_o + \lambda x} = \int_{K_o}^{K_o + \lambda d} \frac{1}{u} \cdot \frac{du}{\lambda} = \frac{1}{\lambda} \int_{K_o}^{K_o + \lambda d} \frac{du}{u} = \frac{1}{\lambda} \left[ \ln|u| \right]_{K_o}^{K_o + \lambda d} = \frac{1}{\lambda} \left( \ln(K_o + \lambda d) - \ln(K_o) \right) = \frac{1}{\lambda} \ln\left(\frac{K_o + \lambda d}{K_o}\right) = \frac{1}{\lambda} \ln\left(1 + \frac{\lambda d}{K_o}\right)$

Substituting back:

$\frac{1}{C} = \frac{1}{\varepsilon_o A} \cdot \frac{1}{\lambda} \ln\left(1 + \frac{\lambda d}{K_o}\right)$

Recall that $C_o = \frac{\varepsilon_o A}{d}$ , so $\varepsilon_o A = C_o d$ . Substituting this in:

$\frac{1}{C} = \frac{1}{C_o d} \cdot \frac{1}{\lambda} \ln\left(1 + \frac{\lambda d}{K_o}\right) = \frac{\ln\left(1 + \frac{\lambda d}{K_o}\right)}{\lambda C_o d}$

Solving for $C$ :

$C = \frac{\lambda C_o d}{\ln\left(1 + \frac{\lambda d}{K_o}\right)} = \frac{\lambda d}{\ln\left(1 + \frac{\lambda d}{K_o}\right)} C_o$

Comparing with the options, this matches Option C.

Hence, the correct answer is Option C.

Q 18PhysicsSection A • MCQcurrent electricity

+4 / -1

The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3 $\Omega$ , 9 $\Omega$ and 9 $\Omega$ and a capacitor of 5.0 $\mu$ F. How much is the current I in the circuit in steady state?

A

1.6 A

B

0.67 A

C

2.5 A

D

0.25 A

Q 19PhysicsSection A • MCQcurrent electricity

+4 / -1

In the experiment of calibration of voltmeter, a standard cell of e.m.f. 1.1 volt is balanced against 440 cm of potential wire. The potential difference across the ends of resistance is found to balance against 220 cm of the wire. The corresponding reading of voltmeter is 0.5 volt. The error in the reading of volmeter will be:

A

$-0.15$ volt

B

0.15 volt

C

0.5 volt

D

$-0.05$ volt

Q 20PhysicsSection A • MCQmoving charges and magnetism

+4 / -1

A positive charge 'q' of mass 'm' is moving along the +x axis. We wish to apply a uniform magnetic field B for time $\Delta t$ so that the charge reverses its direction crossing the y axis at a distance d. Then:

A

$B = \frac{mv}{qd}$ and $\Delta t = \frac{\pi d}{v}$

B

$B = \frac{mv}{2qd}$ and $\Delta t = \frac{\pi d}{2v}$

C

$B = \frac{2mv}{qd}$ and $\Delta t = \frac{\pi d}{2v}$

D

$B = \frac{2mv}{qd}$ and $\Delta t = \frac{\pi d}{v}$

Official Correct Answer:Option (C)

Step-by-Step Solution:

A positive charge $q$ of mass $m$ is moving along the positive x-axis with velocity $v$ . To reverse its direction and make it cross the y-axis at a distance $d$ from the origin, a uniform magnetic field $B$ is applied for a time $\Delta t$ . The force on a moving charge in a magnetic field is given by $\vec{F} = q(\vec{v} \times \vec{B})$ , which acts as a centripetal force, causing circular motion. Since the charge is positive and initially moving in the $+x$ direction, to curve it upwards and cross the y-axis at a positive distance $d$ , the magnetic field must be directed along the negative z-axis (into the plane).

The magnetic force provides the centripetal force: $q v B = \frac{m v^2}{r}$ , where $r$ is the radius of the circular path. Solving for $r$ :

$q v B = \frac{m v^2}{r}$

$r = \frac{m v}{q B}$

Reversing direction requires the charge to traverse a semicircular path. After half a circle, the charge is diametrically opposite its starting point. Assuming it starts at the origin $(0,0)$ , after a semicircle, it crosses the y-axis at $(0, d)$ . The distance from the origin to this point is $d$ , and since the center of the circle is at $(0, r)$ (as the initial velocity is horizontal and force is vertical), the y-coordinate after semicircle is $2r$ . Thus:

$d = 2r$

Substituting $r = \frac{m v}{q B}$ :

$d = 2 \times \frac{m v}{q B}$

Solving for $B$ :

$d = \frac{2 m v}{q B}$

$q B d = 2 m v$

$B = \frac{2 m v}{q d}$

The time $\Delta t$ to traverse a semicircle is the distance divided by speed. The arc length of a semicircle is $\pi r$ , and speed is constant $v$ , so:

$\Delta t = \frac{\pi r}{v}$

Substituting $r = \frac{d}{2}$ from $d = 2r$ :

$\Delta t = \frac{\pi \times \frac{d}{2}}{v} = \frac{\pi d}{2 v}$

Thus, $B = \frac{2 m v}{q d}$ and $\Delta t = \frac{\pi d}{2 v}$ . Comparing with the options:

- Option A: $B = \frac{m v}{q d}$ , $\Delta t = \frac{\pi d}{v}$ — Incorrect.
- Option B: $B = \frac{m v}{2 q d}$ , $\Delta t = \frac{\pi d}{2 v}$ — Incorrect.
- Option C: $B = \frac{2 m v}{q d}$ , $\Delta t = \frac{\pi d}{2 v}$ — Correct.
- Option D: $B = \frac{2 m v}{q d}$ , $\Delta t = \frac{\pi d}{v}$ — Incorrect.

Hence, the correct answer is Option C.

Q 21PhysicsSection A • MCQmoving charges and magnetism

+4 / -1

Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field B $_1$ , at its centre when a current I passes through it. The ratio B $_1$ : B $_2$ is:

A

1:1

B

1 : 3

C

1 : 9

D

9 : 1

Q 22PhysicsSection A • MCQalternating current

+4 / -1

A sinusoidal voltage V(t) = 100 sin(500t) is applied across a pure inductance of L = 0.02H. The current through the coil is:

A

10 cos(500t)

B

$-10$ cos(500t)

C

10 sin(500t)

D

$-10$ sin(500t)

Official Correct Answer:Option (B)

Step-by-Step Solution:

We are given a sinusoidal voltage $V(t) = 100 \sin(500t)$ applied across a pure inductance of $L = 0.02 \, \text{H}$ . We need to find the current through the coil.

For a pure inductor, the relationship between voltage and current is given by:

$V(t) = L \frac{di}{dt}$

Substituting the given values:

$100 \sin(500t) = 0.02 \cdot \frac{di}{dt}$

Now, solve for $\frac{di}{dt}$ :

$\frac{di}{dt} = \frac{100 \sin(500t)}{0.02}$

Calculate the numerical value:

$\frac{100}{0.02} = 5000$

So,

$\frac{di}{dt} = 5000 \sin(500t)$

To find the current $i(t)$ , integrate both sides with respect to $t$ :

$i(t) = \int 5000 \sin(500t) \, dt$

The integral of $\sin(500t)$ is $-\frac{\cos(500t)}{500}$ , because the derivative of $\cos(500t)$ is $-500 \sin(500t)$ , so the antiderivative of $\sin(500t)$ is $-\frac{\cos(500t)}{500}$ . Therefore:

$i(t) = 5000 \cdot \left( -\frac{\cos(500t)}{500} \right) + C$

where $C$ is the constant of integration.

Simplify the expression:

$i(t) = -\frac{5000}{500} \cos(500t) + C$

$i(t) = -10 \cos(500t) + C$

In AC circuits with pure inductance, the average current over a full cycle is zero, and there is no DC component. Therefore, we can assume the constant $C = 0$ . Additionally, the current in an inductor lags the voltage by $90^\circ$ (or $\frac{\pi}{2}$ radians). Since the voltage is a sine function, the current should be a negative cosine function, which matches the form above without a constant term.

Thus, the current is:

$i(t) = -10 \cos(500t)$

Comparing with the options:

A. $10 \cos(500t)$

B. $-10 \cos(500t)$

C. $10 \sin(500t)$

D. $-10 \sin(500t)$

Our solution matches option B.

Hence, the correct answer is Option B.

Q 23PhysicsSection A • MCQelectromagnetic waves

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A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100 W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 5 m from the lamp will be nearly:

A

1.34 V/m

B

2.68 V/m

C

4.02 V/m

D

5.36 V/m

Official Correct Answer:Option (B)

Step-by-Step Solution:

The lamp consumes 100 W of electrical power and is 3% efficient in converting this power to electromagnetic waves. Therefore, the radiant power emitted as light is calculated as follows:

Radiant power = Efficiency × Consumed power = (3/100) × 100 W = 3 W.

This power is emitted uniformly in all directions, so at a distance of 5 m, it spreads over the surface of a sphere with radius 5 m. The surface area of a sphere is given by $4\pi r^2$ . Substituting $r = 5$ m:

Surface area = $4\pi (5)^2 = 4\pi \times 25 = 100\pi$ m².

The intensity $I$ is the power per unit area, so:

$I = \frac{\text{Radiant power}}{\text{Surface area}} = \frac{3}{100\pi} \text{ W/m}^2.$

For an electromagnetic wave, the intensity $I$ is related to the amplitude of the electric field $E_0$ by the formula:

$I = \frac{1}{2} \epsilon_0 c E_0^2,$

where $\epsilon_0$ is the permittivity of free space ( $8.85 \times 10^{-12}$ C²/N·m²) and $c$ is the speed of light ( $3 \times 10^8$ m/s). Rearranging for $E_0^2$ :

$E_0^2 = \frac{2I}{\epsilon_0 c}.$

Substituting the expression for $I$ :

$E_0^2 = \frac{2 \times \frac{3}{100\pi}}{\epsilon_0 c} = \frac{6}{100\pi \epsilon_0 c} = \frac{3}{50\pi \epsilon_0 c}.$

Now, compute $\epsilon_0 c$ :

$\epsilon_0 c = (8.85 \times 10^{-12}) \times (3 \times 10^8) = 8.85 \times 3 \times 10^{-12+8} = 26.55 \times 10^{-4} = 2.655 \times 10^{-3}.$

Using $\pi \approx 3.1416$ , compute $50\pi$ :

$50\pi \approx 50 \times 3.1416 = 157.08.$

Now, compute the denominator $50\pi \epsilon_0 c$ :

$50\pi \epsilon_0 c \approx 157.08 \times 2.655 \times 10^{-3}.$

First, $157.08 \times 2.655$ :

$157.08 \times 2 = 314.16,$

$157.08 \times 0.6 = 94.248,$

$157.08 \times 0.055 = 8.6394,$

Adding these: $314.16 + 94.248 = 408.408$ , then $408.408 + 8.6394 = 417.0474$ .

Now multiply by $10^{-3}$ :

$417.0474 \times 10^{-3} = 0.4170474.$

So,

$E_0^2 = \frac{3}{0.4170474} \approx 7.193.$

Taking the square root:

$E_0 \approx \sqrt{7.193} \approx 2.6815 \text{ V/m}.$

Rounding to two decimal places, $E_0 \approx 2.68$ V/m.

Hence, the amplitude of the electric field is nearly 2.68 V/m, which corresponds to option B.

Q 24PhysicsSection A • MCQray optics

+4 / -1

The refractive index of the material of a concave lens is $\mu$ . It is immersed in a medium of refractive index $\mu_1$ . A parallel beam of light is incident on the lens. The path of the emergent rays when $\mu_1 > \mu$ is:

A

qImage----Jeemate

B

qImage----Jeemate

C

qImage----Jeemate

D

qImage----Jeemate

Q 25PhysicsSection A • MCQwave optics

+4 / -1

Interference pattern is observed at 'P' due to superimposition of two rays coming out from a source 'S' as shown in the figure. The value of 'l' for which maxima is obtained at 'P' is: (R is perfect reflecting surface)

A

$l = \frac{2n\lambda}{\sqrt{3}-1}$

B

$l = \frac{(2n-1)\lambda}{2(\sqrt{3}-1)}$

C

$l = \frac{(2n-1\lambda)\sqrt{3}}{4(2-\sqrt{3})}$

D

$l = \frac{(2n-1)\lambda}{\sqrt{3}-1}$

Q 26PhysicsSection A • MCQwave optics

+4 / -1

In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength. If wavelength of red light is 6600 $\text{Å}$ , then wavelength of first maximum will be:

A

3300 $\text{Å}$

B

4400 $\text{Å}$

C

5500 $\text{Å}$

D

6600 $\text{Å}$

Official Correct Answer:Option (B)

Step-by-Step Solution:

In a single slit diffraction experiment, the position of minima and maxima are determined by specific conditions. For the first minimum of red light, the condition is given by:

$a \sin \theta = n \lambda$

where $a$ is the slit width, $\theta$ is the angle from the central maximum, $\lambda$ is the wavelength, and $n$ is the order of the minimum. For the first minimum, $n = 1$ . Given the wavelength of red light $\lambda_r = 6600 \text{Å}$ , we have:

$a \sin \theta = 1 \cdot 6600 = 6600 \text{Å} \quad \text{(Equation 1)}$

For the first maximum of another light with wavelength $\lambda$ , the condition is approximately:

$a \sin \theta = \left( m + \frac{1}{2} \right) \lambda$

where $m$ is the order of the maximum. For the first maximum, $m = 1$ , so:

$a \sin \theta = \left( 1 + \frac{1}{2} \right) \lambda = \frac{3}{2} \lambda \quad \text{(Equation 2)}$

The problem states that the first minimum for red light coincides with the first maximum of the other light. This means the angular position $\theta$ is the same for both, so $\sin \theta$ is identical. Therefore, the expressions for $a \sin \theta$ from Equation 1 and Equation 2 must be equal:

$6600 = \frac{3}{2} \lambda$

Solving for $\lambda$ :

$\lambda = 6600 \times \frac{2}{3}$

Performing the multiplication:

$\lambda = \frac{6600 \times 2}{3} = \frac{13200}{3} = 4400 \text{Å}$

Thus, the wavelength of the other light for which the first maximum coincides with the first minimum of red light is 4400 Å.

Comparing with the options:

A. 3300 Å

B. 4400 Å

C. 5500 Å

D. 6600 Å

Hence, the correct answer is Option B.

Q 27PhysicsSection A • MCQdual nature of matter and radiation

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A beam of light has two wavelengths of 4972 $\text{Å}$ and 6216 $\text{Å}$ with a total intensity of $3.6 \times 10^{-3}$ Wm $^{-2}$ equally distributed among the two wavelengths. The beam falls normally on an area of 1 cm $^2$ of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in 2 s is approximately:

A

$6 \times 10^{11}$

B

$9 \times 10^{11}$

C

$11 \times 10^{11}$

D

$15 \times 10^{11}$

Official Correct Answer:Option (B)

Step-by-Step Solution:

A beam of light has two wavelengths: 4972 Å and 6216 Å. The total intensity is $3.6 \times 10^{-3}$ W/m², equally distributed between the two wavelengths. This means each wavelength has an intensity of half the total, so intensity for 4972 Å is $1.8 \times 10^{-3}$ W/m² and for 6216 Å is also $1.8 \times 10^{-3}$ W/m².

The beam falls normally on an area of 1 cm². First, convert area to square meters: 1 cm² = $1 \times 10^{-4}$ m². The power (energy per second) for each wavelength is intensity multiplied by area.

For 4972 Å: power = $(1.8 \times 10^{-3}) \times (1 \times 10^{-4}) = 1.8 \times 10^{-7}$ W.

For 6216 Å: power = $(1.8 \times 10^{-3}) \times (1 \times 10^{-4}) = 1.8 \times 10^{-7}$ W.

We need the number of photoelectrons liberated in 2 seconds. The energy delivered by each wavelength in 2 seconds is power multiplied by time.

For 4972 Å: energy = $(1.8 \times 10^{-7}) \times 2 = 3.6 \times 10^{-7}$ J.

For 6216 Å: energy = $(1.8 \times 10^{-7}) \times 2 = 3.6 \times 10^{-7}$ J.

Each photon has energy $E_{\text{photon}} = \frac{hc}{\lambda}$ , where $h$ is Planck's constant ( $6.626 \times 10^{-34}$ J s), $c$ is the speed of light ( $3 \times 10^8$ m/s), and $\lambda$ is the wavelength in meters.

Convert wavelengths to meters:

4972 Å = $4972 \times 10^{-10}$ m = $4.972 \times 10^{-7}$ m.

6216 Å = $6216 \times 10^{-10}$ m = $6.216 \times 10^{-7}$ m.

Calculate photon energy for 4972 Å:

$E_{\text{ph1}} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{4.972 \times 10^{-7}} = \frac{1.9878 \times 10^{-25}}{4.972 \times 10^{-7}} = \frac{1.9878 \times 10^{-25}}{4.972 \times 10^{-7}} = 3.9976 \times 10^{-19} \text{ J}.$

Calculate photon energy for 6216 Å:

$E_{\text{ph2}} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{6.216 \times 10^{-7}} = \frac{1.9878 \times 10^{-25}}{6.216 \times 10^{-7}} = \frac{1.9878 \times 10^{-25}}{6.216 \times 10^{-7}} = 3.1987 \times 10^{-19} \text{ J}.$

The work function of the metal is 2.3 eV. Convert to joules using $1 \text{ eV} = 1.602 \times 10^{-19}$ J:

Work function $\phi = 2.3 \times 1.602 \times 10^{-19} = 3.6846 \times 10^{-19}$ J.

Compare photon energies with work function:

For 4972 Å: $E_{\text{ph1}} = 3.9976 \times 10^{-19}$ J > $3.6846 \times 10^{-19}$ J, so photons can eject electrons.

For 6216 Å: $E_{\text{ph2}} = 3.1987 \times 10^{-19}$ J < $3.6846 \times 10^{-19}$ J, so photons cannot eject electrons.

Only photons of 4972 Å contribute to photoelectrons. The number of photons for 4972 Å is the energy delivered divided by energy per photon:

$N_{\text{ph1}} = \frac{3.6 \times 10^{-7}}{3.9976 \times 10^{-19}} = \frac{3.6 \times 10^{-7}}{3.9976 \times 10^{-19}} = 9.003 \times 10^{11}.$

Since each capable photon ejects one electron, the number of photoelectrons is $9.003 \times 10^{11}$ , approximately $9 \times 10^{11}$ .

Hence, the correct answer is Option B.

Q 28PhysicsSection A • MCQsemiconductor devices and logic gates

+4 / -1

For LED's to emit light in visible region of electromagnetic light, it should have energy band gap in the range of:

A

0.1eV to 0.4eV

B

0.5eV to 0.8eV

C

0.9eV to 1.6eV

D

1.7eV to 3.0eV

Official Correct Answer:Option (D)

Step-by-Step Solution:

To determine the energy band gap required for LEDs to emit light in the visible region, we need to recall that the visible spectrum ranges from approximately 400 nm (violet) to 700 nm (red). The energy $E$ of a photon is related to its wavelength $\lambda$ by the formula:

$E = \frac{hc}{\lambda}$

where $h$ is Planck's constant ( $4.135667662 \times 10^{-15}$ eV·s), $c$ is the speed of light ( $3 \times 10^8$ m/s), and $\lambda$ is the wavelength in meters. However, it is convenient to use nanometers for wavelength and the approximation $hc \approx 1240$ eV·nm, so the formula simplifies to:

$E \text{ (in eV)} = \frac{1240}{\lambda \text{ (in nm)}}$

Now, calculate the energy for the shortest visible wavelength (400 nm, violet light, highest energy):

$E_{\text{violet}} = \frac{1240}{400} = 3.1 \text{ eV}$

Next, calculate the energy for the longest visible wavelength (700 nm, red light, lowest energy):

$E_{\text{red}} = \frac{1240}{700} \approx 1.771 \text{ eV}$

Thus, the energy range for visible light photons is approximately 1.771 eV to 3.1 eV. For an LED to emit visible light, its energy band gap must match the energy of the photons in this range because the emitted photon energy equals the band gap energy.

Comparing this range with the given options:

Option A: 0.1 eV to 0.4 eV → Too low (corresponds to infrared, not visible)
Option B: 0.5 eV to 0.8 eV → Too low (corresponds to infrared)
Option C: 0.9 eV to 1.6 eV → Too low (1.6 eV is about 775 nm, which is in the infrared; 1.6 eV is less than the 1.771 eV required for red light)
Option D: 1.7 eV to 3.0 eV → This covers energies from 1.7 eV (about 729 nm, near red) to 3.0 eV (about 413 nm, violet). While 3.0 eV is slightly less than 3.1 eV (400 nm), it still covers most of the visible spectrum, including blue and violet regions (since 413 nm is violet). The lower limit of 1.7 eV is very close to the 1.771 eV for red light.

Therefore, the band gap range for visible LEDs is 1.7 eV to 3.0 eV, which corresponds to Option D.

Hence, the correct answer is Option D.

Q 29PhysicsSection A • MCQatoms and nuclei

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A piece of bone of an animal from a ruin is found to have $^{14}$ C activity of 12 disintegrations per minute per gm of its carbon content. The $^{14}$ C activity of a living animal is 16 disintegrations per minute per gm. How long ago nearly did the animal die? (Given half life of $^{14}$ C is $t_{1/2}$ = 5760 years)

A

1672 years

B

2391 years

C

3291 years

D

4453 years

Official Correct Answer:Option (B)

Step-by-Step Solution:

The activity of radioactive carbon-14 decreases over time due to decay. The initial activity when the animal was alive is given as $A_0 = 16$ disintegrations per minute per gram, and the current activity of the bone is $A = 12$ disintegrations per minute per gram. The half-life of carbon-14 is $t_{1/2} = 5760$ years.

The decay follows the exponential law: $A = A_0 e^{-\lambda t}$ , where $\lambda$ is the decay constant and $t$ is the time elapsed since death. Substituting the given values:

$12 = 16 e^{-\lambda t}$

Divide both sides by 16:

$\frac{12}{16} = e^{-\lambda t}$

Simplify the fraction:

$\frac{3}{4} = e^{-\lambda t}$

Take the natural logarithm of both sides:

$\ln\left(\frac{3}{4}\right) = \ln\left(e^{-\lambda t}\right)$

The right side simplifies to $-\lambda t$ :

$\ln\left(\frac{3}{4}\right) = -\lambda t$

Multiply both sides by -1:

$-\ln\left(\frac{3}{4}\right) = \lambda t$

Using the property $-\ln(a) = \ln(1/a)$ , rewrite as:

$\ln\left(\frac{4}{3}\right) = \lambda t$

Therefore:

$t = \frac{\ln\left(\frac{4}{3}\right)}{\lambda}$

Now, find the decay constant $\lambda$ using the half-life formula $t_{1/2} = \frac{\ln(2)}{\lambda}$ . Solving for $\lambda$ :

$\lambda = \frac{\ln(2)}{t_{1/2}} = \frac{\ln(2)}{5760}$

Substitute this into the equation for $t$ :

$t = \frac{\ln\left(\frac{4}{3}\right)}{\frac{\ln(2)}{5760}} = \ln\left(\frac{4}{3}\right) \times \frac{5760}{\ln(2)}$

So:

$t = 5760 \times \frac{\ln\left(\frac{4}{3}\right)}{\ln(2)}$

Compute the numerical value. First, calculate $\ln(4/3)$ :

$\frac{4}{3} \approx 1.3333, \quad \ln(1.3333) \approx 0.287682$

Next, $\ln(2) \approx 0.693147$ :

$\frac{\ln(4/3)}{\ln(2)} = \frac{0.287682}{0.693147} \approx 0.415037$

Now multiply by 5760:

$t = 5760 \times 0.415037 \approx 2390.613$

Rounding to the nearest whole number gives approximately 2391 years.

Comparing with the options, 2391 years corresponds to option B.

Hence, the correct answer is Option B.

Q 30PhysicsSection A • MCQcommunication systems

+4 / -1

For sky wave propagation, the radio waves must have a frequency range in between:

A

1MHz to 2MHz

B

5MHz to 25MHz

C

35MHz to 40MHz

D

45MHz to 50MHz

Official Correct Answer:Option (B)

Step-by-Step Solution:

Sky wave propagation involves radio waves being reflected back to Earth by the ionosphere. For this reflection to occur effectively, the frequency of the radio waves must lie within a specific range. If the frequency is too low, the waves get absorbed by the ionosphere, especially in the D layer during the day. If the frequency is too high, the waves penetrate the ionosphere and do not return to Earth.

The critical frequency, denoted as $f_c$ , is the maximum frequency at which a radio wave can be reflected back to Earth when sent vertically upwards. It is calculated using the formula:

f_c = 9 \sqrt{N_{\text{max}}}

where $N_{\text{max}}$ is the maximum electron density in the ionosphere. A typical value for $N_{\text{max}}$ is $10^{12}$ electrons per cubic meter. Substituting this value:

f_c = 9 \times \sqrt{10^{12}}

Since $\sqrt{10^{12}} = (10^{12})^{1/2} = 10^{6}$ (because $12 \times \frac{1}{2} = 6$ ), we get:

f_c = 9 \times 10^6 \text{ Hz}

Converting Hz to MHz (where $10^6$ Hz = 1 MHz), this becomes:

f_c = 9 \text{ MHz}

However, for oblique incidence (waves sent at an angle), the maximum usable frequency (MUF) is higher than $f_c$ and is given by:

\text{MUF} = \frac{f_c}{\cos \theta}

where $\theta$ is the angle of incidence. For long-distance communication, $\theta$ is large, so $\cos \theta$ is small, allowing MUF to be significantly higher than $f_c$ . Typically, the MUF can reach up to about 25-30 MHz under favorable conditions.

The effective frequency range for sky wave propagation is generally from 3 MHz to 30 MHz. Frequencies below 3 MHz are prone to absorption, while frequencies above 30 MHz tend to penetrate the ionosphere. Now, examining the options:

Option A: 1 MHz to 2 MHz - This is too low; absorption occurs in the D layer.
Option B: 5 MHz to 25 MHz - This falls within the typical 3 MHz to 30 MHz range.
Option C: 35 MHz to 40 MHz - This is too high; waves penetrate the ionosphere.
Option D: 45 MHz to 50 MHz - This is also too high; penetration occurs.

Therefore, the correct frequency range for sky wave propagation is given in Option B.

Hence, the correct answer is Option B.

Q 31ChemistrySection A • MCQsome basic concepts of chemistry

+4 / -1

The amount of BaSO $_4$ formed upon mixing 100 mL of 20.8% BaCl $_2$ solution with 50 mL of 9.8% H $_2$ SO $_4$ solution will be: (Ba = 137, Cl = 35.5, S = 32, H = 1 and O = 16)

A

23.3 g

B

11.65 g

C

30.6 g

D

33.2 g

Q 32ChemistrySection A • MCQatomic structure

+4 / -1

If $m$ and $e$ are the mass and charge of the revolving electron in the orbit of radius $r$ for hydrogen atom, the total energy of the revolving electron will be:

A

$\frac{1}{2}\frac{e^2}{r}$

B

$-\frac{e^2}{r}$

C

$\frac{me^2}{r}$

D

$-\frac{1}{2}\frac{e^2}{r}$

Official Correct Answer:Option (D)

Step-by-Step Solution:

To find the total energy of the electron revolving in a hydrogen atom, we consider the Bohr model where the electron moves in a circular orbit of radius $r$ around the proton. The electron has mass $m$ and charge $e$ , while the proton has charge $+e$ . The electrostatic force of attraction provides the necessary centripetal force for circular motion.

The electrostatic force between the electron and proton is given by Coulomb's law. Since the charges are opposite, the magnitude is $\frac{1}{4\pi\epsilon_0} \frac{e \cdot e}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$ . This force acts as the centripetal force, which is $\frac{mv^2}{r}$ , where $v$ is the orbital speed of the electron.

Setting these equal:

$\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$

Multiply both sides by $r$ to simplify:

$mv^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$

Solve for $v^2$ :

$v^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{m r}$

The kinetic energy $K$ of the electron is $\frac{1}{2} m v^2$ . Substitute the expression for $v^2$ :

$K = \frac{1}{2} m \left( \frac{1}{4\pi\epsilon_0} \frac{e^2}{m r} \right)$

The $m$ cancels out:

$K = \frac{1}{2} \cdot \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$

The potential energy $U$ for the electrostatic attraction is negative and given by $-\frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$ .

The total energy $E$ is the sum of kinetic and potential energy:

$E = K + U = \left( \frac{1}{2} \cdot \frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \right) + \left( -\frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \right)$

Factor out $\frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$ :

$E = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \left( \frac{1}{2} - 1 \right)$

Simplify the term in parentheses:

$\frac{1}{2} - 1 = -\frac{1}{2}$

So:

$E = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \times \left( -\frac{1}{2} \right) = -\frac{1}{2} \cdot \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$

In atomic physics, it is common to use units where the constant $\frac{1}{4\pi\epsilon_0}$ is set to 1 (as in cgs units). Thus, the expression simplifies to:

$E = -\frac{1}{2} \frac{e^2}{r}$

Comparing with the options, this matches option D.

Hence, the correct answer is Option D.

Q 33ChemistrySection A • MCQatomic structure

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The de-Broglie wavelength of a particle of mass 6.63 g moving with a velocity of 100 ms $^{-1}$ is:

A

$10^{-33}$ m

B

$10^{-35}$ m

C

$10^{-31}$ m

D

$10^{-25}$ m

Official Correct Answer:Option (A)

Step-by-Step Solution:

The de-Broglie wavelength $\lambda$ of a particle is given by the formula $\lambda = \frac{h}{p}$ , where $h$ is Planck's constant and $p$ is the momentum of the particle. Momentum $p$ is calculated as $p = m \times v$ , where $m$ is the mass and $v$ is the velocity. So, the formula becomes $\lambda = \frac{h}{m v}$ .

We are given the mass $m = 6.63$ grams and velocity $v = 100$ meters per second. However, the SI unit for mass is kilograms, so we must convert grams to kilograms. Since 1 kilogram = 1000 grams, we have:

$m = \frac{6.63}{1000} = 0.00663 \text{ kg}$

This can also be written as $m = 6.63 \times 10^{-3}$ kg. The velocity $v = 100$ m/s remains as is. Planck's constant $h$ is $6.63 \times 10^{-34}$ Joule seconds (J s), and since Joule is kg m² s⁻², the units are consistent.

Now substitute the values into the formula:

$\lambda = \frac{6.63 \times 10^{-34}}{m \times v} = \frac{6.63 \times 10^{-34}}{(6.63 \times 10^{-3}) \times 100}$

First, compute the denominator: $m \times v = (6.63 \times 10^{-3}) \times 100$ . Since $100 = 10^2$ , we have:

$m \times v = 6.63 \times 10^{-3} \times 10^2 = 6.63 \times 10^{-3 + 2} = 6.63 \times 10^{-1}$

Now the expression is:

$\lambda = \frac{6.63 \times 10^{-34}}{6.63 \times 10^{-1}}$

Simplify by dividing the numerical coefficients and the powers of 10 separately. The numerical part is $\frac{6.63}{6.63} = 1$ . The exponent part is $\frac{10^{-34}}{10^{-1}} = 10^{-34 - (-1)} = 10^{-34 + 1} = 10^{-33}$ . Therefore:

$\lambda = 1 \times 10^{-33} \text{ m}$

So, the de-Broglie wavelength is $10^{-33}$ meters. Comparing with the options, this matches option A.

Hence, the correct answer is Option A.

Q 34ChemistrySection A • MCQatomic structure

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Excited hydrogen atom emits light in the ultraviolet region at $2.47 \times 10^{15}$ Hz. With this frequency, the energy of a single photon is: (h = $6.63 \times 10^{-34}$ Js)

A

$8.041 \times 10^{-40}$ J

B

$2.680 \times 10^{-19}$ J

C

$1.640 \times 10^{-18}$ J

D

$6.111 \times 10^{-17}$ J

Official Correct Answer:Option (C)

Step-by-Step Solution:

The energy of a single photon is calculated using the formula $E = h\nu$ , where $h$ is Planck's constant and $\nu$ is the frequency of the light.

Given:

Frequency, $\nu = 2.47 \times 10^{15}$ Hz
Planck's constant, $h = 6.63 \times 10^{-34}$ J s

Substitute the values into the formula: $' in math mode at position 1:$ ̲ E = (6.63 \tim…" style="color:#cc0000">$ E = (6.63 \times 10^{-34}) \times (2.47 \times 10^{15})$

First, multiply the coefficients: $' in math mode at position 1:$ ̲ 6.63 \times 2.…" style="color:#cc0000">$ 6.63 \times 2.47$

Break it down step by step: $' in math mode at position 1:$ ̲ 6.63 \times 2.…" style="color:#cc0000">$ 6.63 \times 2.47 = 6.63 \times (2 + 0.4 + 0.07)$ $' in math mode at position 1:$ ̲ = (6.63 \times…" style="color:#cc0000">$ = (6.63 \times 2) + (6.63 \times 0.4) + (6.63 \times 0.07)$ $' in math mode at position 1:$ ̲ = 13.26 + 2.65…" style="color:#cc0000">$ = 13.26 + 2.652 + 0.4641$ $' in math mode at position 1:$ ̲ = 16.3761" style="color:#cc0000">$ = 16.3761$

Now, multiply the powers of 10: $' in math mode at position 1:$ ̲ 10^{-34} \time…" style="color:#cc0000">$ 10^{-34} \times 10^{15} = 10^{-34 + 15} = 10^{-19}$

Combine the results: $' in math mode at position 1:$ ̲ E = 16.3761 \t…" style="color:#cc0000">$ E = 16.3761 \times 10^{-19} \text{ J}$

Convert to standard scientific notation by adjusting the decimal: $' in math mode at position 1:$ ̲ 16.3761 \times…" style="color:#cc0000">$ 16.3761 \times 10^{-19} = 1.63761 \times 10^1 \times 10^{-19} = 1.63761 \times 10^{-18} \text{ J}$

Consider significant figures. Both given values ( $6.63$ and $2.47$ ) have three significant figures. Round $1.63761 \times 10^{-18}$ to three significant figures:

The third digit is $3$ (in $1.63$ ), and the next digit $7 \geq 5$ , so round up: $1.64 \times 10^{-18}$ J.

Compare with the options:

A: $8.041 \times 10^{-40}$ J
B: $2.680 \times 10^{-19}$ J
C: $1.640 \times 10^{-18}$ J
D: $6.111 \times 10^{-17}$ J

The value

1.64 \times 10^{-18}

J matches option C (

1.640 \times 10^{-18}

J).

Hence, the correct answer is Option C.

Q 35ChemistrySection A • MCQperiodic classification of elements

+4 / -1

Similarity in chemical properties of the atoms of elements in a group of the Periodic table is most closely related to:

A

atomic numbers

B

atomic masses

C

number of principal energy levels

D

number of valence electrons

Q 36ChemistrySection A • MCQperiodic classification of elements

+4 / -1

Which of the following arrangements represents the increasing order (smallest to largest) of ionic radii of the given species O $^{2-}$ , S $^{2-}$ , N $^{3-}$ , P $^{3-}$ ?

A

O $^{2-}$ < N $^{3-}$ < S $^{2-}$ < P $^{3-}$

B

O $^{2-}$ < P $^{3-}$ < N $^{3-}$ < S $^{2-}$

C

N $^3$ < O $^{2-}$ < P $^{3-}$ < S $^{2-}$

D

N $^{3-}$ < S $^{2-}$ < O $^{2-}$ < P $^{3-}$

Official Correct Answer:Option (A)

Step-by-Step Solution:

To determine the increasing order of ionic radii for the species O2-, S2-, N3-, and P3-, we need to analyze their positions in the periodic table and their electron configurations. Ionic radius depends on the number of electrons and the effective nuclear charge.

First, let's find the electron count for each ion:

Oxygen (atomic number 8) in O2- has 8 + 2 = 10 electrons.
Sulfur (atomic number 16) in S2- has 16 + 2 = 18 electrons.
Nitrogen (atomic number 7) in N3- has 7 + 3 = 10 electrons.
Phosphorus (atomic number 15) in P3- has 15 + 3 = 18 electrons.

This shows that O2- and N3- are isoelectronic (both have 10 electrons, like neon), while S2- and P3- are isoelectronic (both have 18 electrons, like argon).

For isoelectronic species, ionic radius decreases as nuclear charge (atomic number) increases because a higher nuclear charge pulls the electron cloud closer. Comparing O2- (atomic number 8) and N3- (atomic number 7):

N3- has lower nuclear charge (7) than O2- (8), so N3- has a larger ionic radius than O2-.
Thus, O2- < N3-.

Similarly, comparing S2- (atomic number 16) and P3- (atomic number 15):

P3- has lower nuclear charge (15) than S2- (16), so P3- has a larger ionic radius than S2-.
Thus, S2- < P3-.

Now, we compare across the two groups. O2- and N3- are in period 2 (principal quantum number n=2), while S2- and P3- are in period 3 (n=3). Ions in higher periods have larger radii due to more electron shells. Therefore, both S2- and P3- are larger than O2- and N3-.

Combining the comparisons:

O2- is the smallest (period 2, highest nuclear charge in its group).
N3- is next (period 2, larger than O2- due to lower nuclear charge).
S2- is next (period 3, smaller than P3- due to higher nuclear charge).
P3- is the largest (period 3, lowest nuclear charge in its group).

So the order is: O2- < N3- < S2- < P3-.

Comparing with the options:

A: O2- < N3- < S2- < P3-
B: O2- < P3- < N3- < S2-
C: N3- < O2- < P3- < S2-
D: N3- < S2- < O2- < P3-

Option A matches our derived order.

Hence, the correct answer is Option A.

Q 37ChemistrySection A • MCQchemical bonding and molecular structure

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Which of the following molecules has two sigma ( $\sigma$ ) and two pi ( $\pi$ ) bonds?

A

C $_2$ H $_4$

B

N $_2$ F $_2$

C

C $_2$ H $_2$ Cl $_2$

D

HCN

Official Correct Answer:Option (D)

Step-by-Step Solution:

To determine which molecule has two sigma (σ) and two pi (π) bonds, we need to analyze the bonding in each option. A sigma bond is formed by the head-on overlap of atomic orbitals and is present in all single bonds. A pi bond is formed by the sideway overlap of atomic orbitals and is found in double bonds (one π bond) or triple bonds (two π bonds).

Starting with option A: C₂H₄ (ethylene). The structure is H₂C=CH₂. The carbon-carbon double bond consists of one σ bond and one π bond. Each carbon is bonded to two hydrogen atoms via single bonds, which are σ bonds. So, there are four C-H σ bonds and one C-C σ bond, totaling five σ bonds. The double bond contributes one π bond. Therefore, C₂H₄ has five σ bonds and one π bond, which does not match the requirement of two σ and two π bonds.

Next, option B: N₂F₂ (dinitrogen difluoride). The common structure is F-N=N-F, with a double bond between the nitrogen atoms. The double bond has one σ and one π bond. Each nitrogen-fluorine bond is a single bond, which is a σ bond. So, there are two N-F σ bonds and one N-N σ bond, totaling three σ bonds. The double bond contributes one π bond. Therefore, N₂F₂ has three σ bonds and one π bond, which does not match.

Moving to option C: C₂H₂Cl₂ (dichloroethene). This has isomers, but both isomers have a carbon-carbon double bond. For example, in 1,2-dichloroethene (ClHC=CHCl), the C=C bond has one σ and one π bond. There are two C-Cl single bonds (σ bonds) and two C-H single bonds (σ bonds). So, total σ bonds are one (from C=C) + two (from C-Cl) + two (from C-H) = five σ bonds, and one π bond. The same count applies to the 1,1-dichloroethene isomer (H₂C=CCl₂). Therefore, C₂H₂Cl₂ has five σ bonds and one π bond, which does not match.

Finally, option D: HCN (hydrogen cyanide). The structure is H-C≡N. The carbon-nitrogen triple bond consists of one σ bond and two π bonds. The hydrogen-carbon bond is a single bond, which is a σ bond. So, total σ bonds are one (from H-C) + one (from C≡N) = two σ bonds. The triple bond contributes two π bonds. Therefore, HCN has two σ bonds and two π bonds, matching the requirement.

Hence, the correct answer is Option D.

Q 38ChemistrySection A • MCQthermodynamics

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The (S°) of the following substances are: CH $_4$ (g) 186.2 JK $^{-1}$ mol $^{-1}$ , O $_2$ (g) 205.2 JK $^{-1}$ mol $^{-1}$ , CO $_2$ (g) 213.6 JK $^{-1}$ mol $^{-1}$ , H $_2$ O(g) 69.9 JK $^{-1}$ mol $^{-1}$ . The entropy change ( $\Delta S^\circ$ ) for the reaction CH $_4$ (g) + 2O $_2$ (g) → CO $_2$ (g) + 2H $_2$ O(l) is:

A

$-312.5$ J K $^{-1}$ mol $^{-1}$

B

$-242.8$ JK $^{-1}$ mol $^{-1}$

C

$-108.1$ JK $^{-1}$ mol $^{-1}$

D

$-37.6$ J K $^{-1}$ mol $^{-1}$

Official Correct Answer:Option (B)

Step-by-Step Solution:

The standard molar entropies (S°) given are:

$S^\circ(\text{CH}_4\text{(g)}) = 186.2 \text{ J K}^{-1} \text{ mol}^{-1}$

$S^\circ(\text{O}_2\text{(g)}) = 205.2 \text{ J K}^{-1} \text{ mol}^{-1}$

$S^\circ(\text{CO}_2\text{(g)}) = 213.6 \text{ J K}^{-1} \text{ mol}^{-1}$

$S^\circ(\text{H}_2\text{O(g)}) = 69.9 \text{ J K}^{-1} \text{ mol}^{-1}$

However, the reaction is: $\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$ which has water in liquid form as a product. The given entropy is labeled for H₂O(g), but the value 69.9 J K⁻¹ mol⁻¹ is typical for liquid water (H₂O(l)). Therefore, we will use this value for H₂O(l), assuming a labeling error. So, $S^\circ(\text{H}_2\text{O(l)}) = 69.9 \text{ J K}^{-1} \text{ mol}^{-1}$ .

The standard entropy change for the reaction, ΔS°, is calculated as the difference between the sum of the standard entropies of the products and the sum of the standard entropies of the reactants:

$\Delta S^\circ = \left[ \sum S^\circ_{\text{products}} \right] - \left[ \sum S^\circ_{\text{reactants}} \right]$

First, identify the products and reactants:

Products: 1 mol of CO₂(g) and 2 mol of H₂O(l).

Reactants: 1 mol of CH₄(g) and 2 mol of O₂(g).

Sum of standard entropies of products:

$\sum S^\circ_{\text{products}} = S^\circ(\text{CO}_2) + 2 \times S^\circ(\text{H}_2\text{O(l)})$

Substitute the values:

$= 213.6 + 2 \times 69.9$

Multiply 2 by 69.9:

$2 \times 69.9 = 139.8$

Now add to 213.6:

$213.6 + 139.8 = 353.4$

So, $\sum S^\circ_{\text{products}} = 353.4 \text{ J K}^{-1} \text{ mol}^{-1}$

Sum of standard entropies of reactants:

$\sum S^\circ_{\text{reactants}} = S^\circ(\text{CH}_4) + 2 \times S^\circ(\text{O}_2)$

Substitute the values:

$= 186.2 + 2 \times 205.2$

Multiply 2 by 205.2:

$2 \times 205.2 = 410.4$

Now add to 186.2:

$186.2 + 410.4 = 596.6$

So, $\sum S^\circ_{\text{reactants}} = 596.6 \text{ J K}^{-1} \text{ mol}^{-1}$

Now, calculate ΔS°:

$\Delta S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}} = 353.4 - 596.6$

Subtract 596.6 from 353.4:

$353.4 - 596.6 = -243.2$

So, $\Delta S^\circ = -243.2 \text{ J K}^{-1} \text{ mol}^{-1}$

Now, compare with the given options:

A. $-312.5 \text{ J K}^{-1} \text{ mol}^{-1}$

B. $-242.8 \text{ J K}^{-1} \text{ mol}^{-1}$

C. $-108.1 \text{ J K}^{-1} \text{ mol}^{-1}$

D. $-37.6 \text{ J K}^{-1} \text{ mol}^{-1}$

The calculated value is -243.2 J K⁻¹ mol⁻¹, which is closest to option B (-242.8 J K⁻¹ mol⁻¹). Considering possible rounding in the options or given values, and as the correct answer is indicated as option B, we select this option.

Hence, the correct answer is Option B.

Q 39ChemistrySection A • MCQthermodynamics

+4 / -1

The standard enthalpy of formation ( $\Delta_fH^\circ_{298}$ ) for methane, CH $_4$ is $-74.9$ kJ mol $^{-1}$ . In order to calculate the average energy given out in the formation of a C-H bond from this it is necessary to know which one of the following?

A

The dissociation energy of the hydrogen molecule, H $_2$ .

B

The first four ionisation energies of carbon.

C

The dissociation energy of H $_2$ and enthalpy and sublimation of carbon (graphite).

D

The first four ionisation energies of carbon and electron affinity of hydrogen.

Q 40ChemistrySection A • MCQchemical equilibrium

+4 / -1

What happens when an inert gas is added to an equilibrium keeping volume unchanged?

A

More product will form

B

Less product will form

C

More reactant will form

D

Equilibrium will remain unchanged

Official Correct Answer:Option (D)

Step-by-Step Solution:

When an inert gas is added to a system at equilibrium with the volume kept unchanged, we need to analyze the effect on the equilibrium. An inert gas does not react with any of the reactants or products. Therefore, it does not participate in the chemical reaction.

The equilibrium constant for a reaction, denoted as $K_c$ , is expressed in terms of concentrations. For a general reaction:

aA + bB \rightleftharpoons cC + dD

the equilibrium constant is:

K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}

where $[A]$ , $[B]$ , $[C]$ , and $[D]$ are the molar concentrations of the species. Concentration is defined as the number of moles per unit volume, so $[A] = \frac{n_A}{V}$ , and similarly for others.

When an inert gas is added at constant volume:

The volume $V$ remains the same.
The number of moles of the reactants and products (e.g., $n_A$ , $n_B$ , $n_C$ , $n_D$ ) does not change because the inert gas does not react with them.

Therefore, the concentrations of the reactants and products remain unchanged. For example:

[A]_{\text{new}} = \frac{n_A}{V} = [A]_{\text{initial}}

and similarly for $B$ , $C$ , and $D$ .

Since the concentrations are unchanged, the reaction quotient $Q_c$ (which has the same expression as $K_c$ ) remains equal to $K_c$ :

Q_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} = K_c

Because $Q_c = K_c$ , the system remains at equilibrium. There is no shift in the equilibrium position towards reactants or products.

Although the total pressure increases due to the addition of the inert gas, pressure changes only affect the equilibrium if the volume changes (which alters concentrations) or if the reaction involves a change in the number of moles of gas. Here, volume is constant, and the inert gas does not alter the mole numbers of the reacting species, so the equilibrium is unaffected.

Hence, the correct answer is Option D.

Q 41ChemistrySection A • MCQionic equilibrium

+4 / -1

The conjugate base of hydrazoic acid is:

A

N $^{-3}$

B

N $_3^-$

C

N $_2^-$

D

HN $_3^-$

Q 42ChemistrySection A • MCQhydrogen

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Hydrogen peroxide acts both as an oxidising and as a reducing agent depending upon the nature of the reacting species. In which of the following cases H $_2$ O $_2$ acts as a reducing agent in acid medium?

A

MnO $_4^-$

B

Cr $_2$ O $_7^{2-}$

C

SO $_3^{2-}$

D

KI

Q 43ChemistrySection A • MCQbasics of organic chemistry

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Which one of the following acids does not exhibit optical isomerism?

A

Lactic acid

B

Tartaric acid

C

Maleic acid

D

$\alpha$ -amino acids

Official Correct Answer:Option (C)

Step-by-Step Solution:

To determine which acid does not exhibit optical isomerism, we must recall that optical isomerism arises when a molecule has a chiral center, typically a carbon atom with four different substituents, and the molecule lacks a plane of symmetry. A chiral molecule is non-superimposable on its mirror image. Let's examine each option step by step.

Starting with option A: Lactic acid. Its molecular formula is $CH_{3}CH(OH)COOH$ . The central carbon atom (the one attached to the OH group) is bonded to four different groups: a hydrogen atom (H), a hydroxyl group (OH), a methyl group ( $CH_{3}$ ), and a carboxyl group ( $COOH$ ). Since all four groups are distinct, this carbon is a chiral center. Therefore, lactic acid exhibits optical isomerism and exists as two enantiomers.

Next, option B: Tartaric acid. Its molecular formula is $HOOCCH(OH)CH(OH)COOH$ . This molecule has two carbon atoms, each bonded to an OH group. Each of these carbons is attached to four different groups: H, OH, COOH, and the other $CH(OH)COOH$ group. Thus, both carbons are chiral centers. Tartaric acid exists in three stereoisomeric forms: the d-form, the l-form (which are enantiomers and optically active), and the meso-form (which has a plane of symmetry and is optically inactive due to internal compensation). However, since tartaric acid can exist in optically active forms (the d- and l-enantiomers), it does exhibit optical isomerism.

Now, option C: Maleic acid. Maleic acid is the cis-isomer of butenedioic acid, with the formula $HOOCCH=CHCOOH$ . The structure can be represented as:

$\begin{array}{c} \text{H} & & \text{COOH} \\ & \backslash & \\ & C = C \\ & / & \\ \text{COOH} & & \text{H} \end{array}$

In maleic acid, the double bond between the two carbon atoms restricts rotation. Each carbon in the double bond is sp² hybridized and bonded to three atoms: one hydrogen (H), one carboxyl group (COOH), and the other carbon via the double bond. Since neither carbon has four substituents, there are no chiral centers. Additionally, the molecule is planar and has a plane of symmetry that bisects the double bond and aligns the two identical halves, making it achiral. Therefore, maleic acid does not exhibit optical isomerism.

Finally, option D: α-amino acids. These have the general formula $R-CH(NH_{2})COOH$ , where the α-carbon is bonded to four groups: a hydrogen atom (H), an amino group ( $NH_{2}$ ), a carboxyl group ( $COOH$ ), and a side chain (R). For most α-amino acids, R is different from H, $NH_{2}$ , and $COOH$ , making the α-carbon a chiral center. For example, alanine (R = $CH_{3}$ ) has a chiral center and exhibits optical isomerism. However, glycine is an exception where R = H, so the α-carbon has two identical H atoms and is not chiral. Despite glycine not exhibiting optical isomerism, the class of α-amino acids as a whole includes compounds that do exhibit optical isomerism (like alanine). Therefore, this option does not universally lack optical isomerism.

Comparing all options, lactic acid (A) and tartaric acid (B) exhibit optical isomerism due to chiral centers. α-amino acids (D) include some that exhibit optical isomerism. Maleic acid (C) has no chiral center and is achiral, so it does not exhibit optical isomerism.

Hence, the correct answer is Option C.

Q 44ChemistrySection A • MCQhydrocarbons

+4 / -1

qImage----Jeemate

on mercuration-demercuration produces the major product:

A

qImage----Jeemate

B

qImage----Jeemate

C

qImage----Jeemate

D

qImage----Jeemate

Q 45ChemistrySection A • MCQhydrocarbons

+4 / -1

In the presence of peroxide, HCl and HI do not give anti-Markownikoff's addition of alkenes because:

A

One of the steps is endothermic in HCl and HI

B

Both HCl and HI are strong acids

C

HCl is oxidizing and the HI is reducing

D

All the steps are exothermic is HCl and HI

Q 46ChemistrySection A • MCQenvironmental chemistry

+4 / -1

Global warming is due to increase of:

A

methane and nitrous oxide in atmosphere

B

methane and CO $_2$ in atmosphere

C

methane and O $_3$ in atmosphere

D

methane and CO in atmosphere

Official Correct Answer:Option (B)

Step-by-Step Solution:

Global warming refers to the long-term rise in Earth's average surface temperature, primarily caused by the accumulation of greenhouse gases in the atmosphere. These gases trap heat from the sun, preventing it from escaping back into space, which leads to a warming effect known as the greenhouse effect.

Now, let us examine the options one by one:

Option A mentions methane and nitrous oxide. Methane (CH $_4$ ) is a potent greenhouse gas, and nitrous oxide (N $_2$ O) also contributes to global warming. However, carbon dioxide (CO $_2$ ) is the most significant greenhouse gas in terms of total emissions and long-term impact, so this pair is incomplete.

Option B mentions methane and CO $_2$ . Methane is a major greenhouse gas with high heat-trapping ability, though it has a shorter atmospheric lifetime than CO $_2$ . Carbon dioxide is the primary driver of global warming, accounting for the largest share of human-induced greenhouse gas emissions from sources like fossil fuel combustion and deforestation. Together, methane and CO $_2$ represent two of the most critical gases responsible for global warming.

Option C mentions methane and ozone (O $_3$ ). While ozone in the lower atmosphere (troposphere) acts as a greenhouse gas, its contribution is secondary compared to CO $_2$ . Ozone is also not directly emitted in large quantities but forms from other pollutants. This pair does not emphasize the dominant role of CO $_2$ .

Option D mentions methane and carbon monoxide (CO). Carbon monoxide is not a direct greenhouse gas. It can indirectly contribute to warming by reacting to form CO $_2$ or ozone, but it is not a primary cause. Methane is relevant, but CO $_2$ 's absence makes this pair incorrect.

Scientific consensus identifies carbon dioxide as the principal greenhouse gas, with methane being the second most significant. Human activities have substantially increased atmospheric concentrations of both, making them key contributors to global warming. Hence, Option B accurately identifies the increase of methane and CO $_2$ as the cause.

So, the answer is Option B.

Q 47ChemistrySection A • MCQsolid state

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In a monoclinic unit cell, the relation of sides and angles are respectively:

A

a = b $\neq$ c and $\alpha = \beta = \gamma = 90$ °

B

a $\neq$ b $\neq$ c and $\alpha = \beta = \gamma = 90$ °

C

a $\neq$ b $\neq$ c and $\beta = \gamma = 90$ ° $\neq \alpha$

D

a $\neq$ b $\neq$ c and $\alpha \neq \beta \neq \gamma \neq 90$ °

Official Correct Answer:Option (C)

Step-by-Step Solution:

In crystallography, the monoclinic crystal system is one of the seven crystal systems. To solve this, we need to recall the defining characteristics of a monoclinic unit cell regarding the lengths of its edges (denoted as a, b, c) and the angles between them (denoted as α, β, γ).

First, for the sides (edge lengths) in a monoclinic unit cell, all three edges are of different lengths. This means that a is not equal to b, b is not equal to c, and a is not equal to c. So, we have $a \neq b \neq c$ .

Second, for the angles, in a monoclinic system, two of the angles are exactly 90 degrees, and one angle is not equal to 90 degrees. Specifically, in the standard convention, the angle β (between the a-axis and c-axis) is not 90 degrees, while the angles α (between the b-axis and c-axis) and γ (between the a-axis and b-axis) are both 90 degrees. However, the labeling of axes can vary, and the monoclinic system can be described equivalently by having any one angle not equal to 90 degrees while the other two are 90 degrees, as long as the side lengths are unequal.

Now, let's examine the given options:

Option A states: a = b $\neq$ c and $\alpha = \beta = \gamma = 90$ °. This describes a tetragonal system, where two sides are equal and all angles are 90 degrees, which is not monoclinic.

Option B states: a $\neq$ b $\neq$ c and $\alpha = \beta = \gamma = 90$ °. This describes an orthorhombic system, where all sides are unequal but all angles are 90 degrees, which is not monoclinic.

Option C states: a $\neq$ b $\neq$ c and $\beta = \gamma = 90$ ° $\neq \alpha$ . This means the sides are all unequal, and angles β and γ are 90 degrees, while α is not 90 degrees. Although the standard monoclinic system has β not equal to 90 degrees (with α and γ at 90 degrees), this option describes a valid monoclinic system because the geometry is equivalent under a relabeling of axes. For example, if we swap the b and c axes, the non-right angle becomes α, matching this description. Thus, this option correctly represents the monoclinic system.

Option D states: a $\neq$ b $\neq$ c and $\alpha \neq \beta \neq \gamma \neq 90$ °. This describes a triclinic system, where all angles are different and not 90 degrees, which is not monoclinic.

Therefore, after comparing all options, Option C matches the monoclinic unit cell characteristics when considering the flexibility in axis labeling.

Hence, the correct answer is Option C.

Q 48ChemistrySection A • MCQelectrochemistry

+4 / -1

How many electrons would be required to deposit 6.35 g of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate? (Atomic mass of copper = 63.5u, N $_A$ = Avogadro's constant):

A

$\frac{N_A}{20}$

B

$\frac{N_A}{10}$

C

$\frac{N_A}{5}$

D

$\frac{N_A}{2}$

Q 49ChemistrySection A • MCQchemical kinetics

+4 / -1

The rate coefficient (k) for a particular reactions is $1.3 \times 10^{-4}$ M $^{-1}$ s $^{-1}$ at 100°C, and $1.3 \times 10^{-3}$ M $^{-1}$ s $^{-1}$ at 150°C. What is the energy of activation (E $_A$ ) (in kJ) for this reaction? (R = molar gas constant = 8.314 JK $^{-1}$ mol $^{-1}$ )

A

16

B

60

C

99

D

132

Official Correct Answer:Option (B)

Step-by-Step Solution:

The activation energy ( $E_A$ ) can be found using the Arrhenius equation, which relates the rate constant ( $k$ ) to temperature ( $T$ ) and activation energy. The formula for two different temperatures is:

$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_A}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

where:

$k_1$ and $k_2$ are the rate constants at temperatures $T_1$ and $T_2$ respectively,
$R$ is the gas constant (8.314 J K⁻¹ mol⁻¹),
$E_A$ is the activation energy in J/mol.

First, convert the temperatures from Celsius to Kelvin:

$T_1 = 100^\circ \text{C} + 273 = 373 \text{K}$

$T_2 = 150^\circ \text{C} + 273 = 423 \text{K}$

Given:

$k_1 = 1.3 \times 10^{-4} \text{M}^{-1} \text{s}^{-1}$

$k_2 = 1.3 \times 10^{-3} \text{M}^{-1} \text{s}^{-1}$

Calculate the ratio $\frac{k_2}{k_1}$ :

$\frac{k_2}{k_1} = \frac{1.3 \times 10^{-3}}{1.3 \times 10^{-4}} = 10$

Take the natural logarithm of the ratio:

$\ln\left(\frac{k_2}{k_1}\right) = \ln(10) \approx 2.302585$

Now, compute the difference in the reciprocal temperatures:

$\frac{1}{T_1} = \frac{1}{373} \approx 0.002681$

$\frac{1}{T_2} = \frac{1}{423} \approx 0.002364$

$\frac{1}{T_1} - \frac{1}{T_2} = 0.002681 - 0.002364 = 0.000317$

Alternatively, use the exact fraction:

$\frac{1}{T_1} - \frac{1}{T_2} = \frac{T_2 - T_1}{T_1 T_2} = \frac{423 - 373}{373 \times 423} = \frac{50}{373 \times 423}$

Calculate $373 \times 423$ :

$373 \times 400 = 149200$

$373 \times 23 = 373 \times 20 + 373 \times 3 = 7460 + 1119 = 8579$

$373 \times 423 = 149200 + 8579 = 157779$

So,

$\frac{1}{T_1} - \frac{1}{T_2} = \frac{50}{157779} \approx 0.000317$

Now rearrange the Arrhenius equation to solve for $E_A$ :

$E_A = \frac{R \cdot \ln\left(\frac{k_2}{k_1}\right)}{\frac{1}{T_1} - \frac{1}{T_2}}$

Substitute the values:

$E_A = \frac{8.314 \times 2.302585}{0.000317}$

First, compute the numerator:

$8.314 \times 2.302585 = 19.14369169$

Now divide by the denominator:

$E_A = \frac{19.14369169}{0.000317} \approx 60402.77 \text{J/mol}$

Alternatively, using the exact fraction:

$E_A = \frac{8.314 \times 2.302585 \times 157779}{50}$

First, compute $8.314 \times 157779 = 1311774.606$

Then, $1311774.606 \times 2.302585 \approx 3020890.06$

Then, divide by 50:

$\frac{3020890.06}{50} = 60417.8012 \text{J/mol}$

Both methods give approximately 60400 J/mol. Convert to kJ/mol by dividing by 1000:

$E_A \approx 60.4 \text{kJ/mol}$

Rounding to the nearest whole number, $E_A \approx 60 \text{kJ/mol}$ .

Comparing with the options:

A. 16
B. 60
C. 99
D. 132

Hence, the correct answer is Option B.

Q 50ChemistrySection A • MCQp block elements

+4 / -1

Which of the following xenon-oxo compounds may not be obtained by hydrolysis of xenon fluorides?

A

XeO $_2$ F $_2$

B

XeOF $_4$

C

XeO $_3$

D

XeO $_4$

Official Correct Answer:Option (D)

Step-by-Step Solution:

To solve this question, we need to determine which xenon-oxo compound cannot be obtained through the hydrolysis of xenon fluorides. Xenon fluorides include XeF2, XeF4, and XeF6. Hydrolysis involves reacting these fluorides with water to form various xenon-oxygen compounds. Let's examine the hydrolysis products of each xenon fluoride step by step.

First, consider XeF2. When hydrolyzed, it decomposes to xenon gas, hydrogen fluoride, and oxygen or hydrogen peroxide. The reactions are:

$2XeF_2 + 2H_2O \rightarrow 2Xe + 4HF + O_2$

or alternatively,

$XeF_2 + 2H_2O \rightarrow Xe + 2HF + H_2O_2$

In both cases, no stable xenon-oxo compound is formed. Therefore, hydrolysis of XeF2 does not yield any of the listed oxo compounds.

Next, consider XeF4. It undergoes slow hydrolysis and disproportionation. The reaction is:

$6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2$

Here, xenon trioxide (XeO3) is produced as a stable product. Thus, hydrolysis of XeF4 yields XeO3.

Now, consider XeF6, which undergoes hydrolysis in stages:

Partial hydrolysis produces xenon oxytetrafluoride (XeOF4):

$XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$

Further hydrolysis of XeOF4 produces xenon dioxide difluoride (XeO2F2):

$XeOF_4 + H_2O \rightarrow XeO_2F_2 + 2HF$

Complete hydrolysis produces xenon trioxide (XeO3):

$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$

Therefore, hydrolysis of XeF6 yields XeOF4, XeO2F2, and XeO3.

Summarizing the compounds obtained by hydrolysis:

XeO2F2 is obtained from hydrolysis of XeOF4 (which comes from XeF6).
XeOF4 is obtained from partial hydrolysis of XeF6.
XeO3 is obtained from hydrolysis of XeF4 or complete hydrolysis of XeF6.

Now, consider xenon tetroxide (XeO4). It is not prepared by hydrolysis of any xenon fluoride. Instead, it is synthesized by treating sodium perxenate (Na4XeO6) with concentrated sulfuric acid:

$Na_4XeO_6 + 2H_2SO_4 \rightarrow XeO_4 + 2Na_2SO_4 + 2H_2O$

No hydrolysis reaction of xenon fluorides produces XeO4.

Therefore, among the options, XeO4 may not be obtained by hydrolysis of xenon fluorides.

Hence, the correct answer is Option D.

Q 51ChemistrySection A • MCQd and f block elements

+4 / -1

Which one of the following exhibits the large number of oxidation states?

A

Ti(22)

B

V(23)

C

Cr(24)

D

Mn(25)

Official Correct Answer:Option (D)

Step-by-Step Solution:

The question asks which element exhibits the largest number of oxidation states among titanium (Ti, atomic number 22), vanadium (V, 23), chromium (Cr, 24), and manganese (Mn, 25). Oxidation states represent the hypothetical charges an atom would have if all bonds were ionic, and transition metals show variable oxidation states due to the involvement of both ns and (n-1)d electrons in bonding.

First, recall the electron configurations:

Titanium (Ti, 22): [Ar] 4s² 3d²
Vanadium (V, 23): [Ar] 4s² 3d³
Chromium (Cr, 24): [Ar] 4s¹ 3d⁵ (exception for half-filled stability)
Manganese (Mn, 25): [Ar] 4s² 3d⁵

Now, list the oxidation states each element can exhibit:

Titanium (Ti): The common oxidation states are +2 (losing 4s electrons), +3 (losing 4s and one 3d electron), and +4 (losing 4s and both 3d electrons). Less common states include 0 and +1, but typically, Ti shows 3 main oxidation states: +2, +3, +4.

Vanadium (V): The oxidation states include +2 (losing 4s electrons), +3 (losing 4s and one 3d electron), +4 (losing 4s and two 3d electrons), and +5 (losing 4s and all three 3d electrons). Less common states like +1 exist, but V primarily exhibits 4 oxidation states: +2, +3, +4, +5.

Chromium (Cr): Oxidation states range from +1 to +6. Specifically:

+1: losing the 4s electron (rare)
+2: losing the 4s electron
+3: losing 4s and one 3d electron
+4: losing 4s and two 3d electrons
+5: losing 4s and three 3d electrons
+6: losing 4s and all five 3d electrons (as in CrO₃)

Thus, chromium exhibits 6 oxidation states: +1, +2, +3, +4, +5, +6.

Manganese (Mn): Oxidation states range from +1 to +7. Specifically:

+1: losing one 4s electron (in some complexes)
+2: losing both 4s electrons (e.g., MnCl₂)
+3: losing both 4s and one 3d electron (e.g., Mn₂O₃)
+4: losing both 4s and two 3d electrons (e.g., MnO₂)
+5: losing both 4s and three 3d electrons (e.g., K₃MnO₄)
+6: losing both 4s and four 3d electrons (e.g., K₂MnO₄)
+7: losing both 4s and all five 3d electrons (e.g., KMnO₄)

Manganese exhibits 7 oxidation states: +1, +2, +3, +4, +5, +6, +7. Additionally, manganese can show states like -3, -2, -1, and 0 in specific compounds, but the positive states from +1 to +7 are well-documented.

Comparing the number of oxidation states:

Ti: 3 main states (+2, +3, +4)
V: 4 main states (+2, +3, +4, +5)
Cr: 6 states (+1 to +6)
Mn: 7 states (+1 to +7)

Manganese exhibits the largest number of oxidation states (7) among the given options.

The reason for manganese having the highest number is its electron configuration ([Ar] 4s² 3d⁵). It can lose up to 7 electrons (both 4s and all five 3d electrons) to achieve the +7 state, and it also forms stable compounds in lower oxidation states due to the half-filled d-subshell stability.

Hence, the correct answer is Option D.

Q 52ChemistrySection A • MCQd and f block elements

+4 / -1

Copper becomes green when exposed to moist air for a long period. This is due to:

A

the formation of a layer of cupric oxide on the surface of copper.

B

the formation of a layer of basic carbonate of copper on the surface of copper.

C

the formation of a layer of cupric hydroxide on the surface of copper.

D

the formation of basic copper sulphate layer on the surface of the metal.

Q 53ChemistrySection A • MCQcoordination compounds

+4 / -1

Among the following species the one which causes the highest CFSE, $\Delta_0$ as a ligand is:

A

CN $^-$

B

NH $_3$

C

F $^-$

D

CO

Q 54ChemistrySection A • MCQcoordination compounds

+4 / -1

Which one of the following complexes will most likely absorb visible light? (At. nos. Sc = 21, Ti = 22, V = 23, Zn = 30)

A

[Sc(H $_2$ O) $_6$ ] $^{3+}$

B

[Ti(NH $_3$ ) $_6$ ] $^{4+}$

C

[V(NH $_3$ ) $_6$ ] $^{3+}$

D

[Zn(NH $_3$ ) $_6$ ] $^{2+}$

Official Correct Answer:Option (C)

Step-by-Step Solution:

To determine which complex is most likely to absorb visible light, we need to understand that absorption of visible light in transition metal complexes occurs due to d-d transitions. These transitions happen when an electron moves from a lower energy d orbital to a higher energy d orbital within the same subshell. For this to occur, the central metal ion must have partially filled d orbitals. If the d orbitals are completely empty (d⁰) or completely filled (d¹⁰), no d-d transitions are possible, and the complex will not absorb visible light in the typical range for such transitions.

Let's examine the electron configurations of the metal ions in each complex:

Scandium (Sc, atomic number 21): The neutral atom has configuration [Ar] 4s² 3d¹. For Sc³⁺, it loses three electrons (two from 4s and one from 3d), resulting in [Ar] 3d⁰.
Titanium (Ti, atomic number 22): Neutral atom is [Ar] 4s² 3d². For Ti⁴⁺, it loses four electrons (two from 4s and two from 3d), giving [Ar] 3d⁰.
Vanadium (V, atomic number 23): Neutral atom is [Ar] 4s² 3d³. For V³⁺, it loses three electrons (two from 4s and one from 3d), so the configuration is [Ar] 3d².
Zinc (Zn, atomic number 30): Neutral atom is [Ar] 4s² 3d¹⁰. For Zn²⁺, it loses two electrons from the 4s orbital, resulting in [Ar] 3d¹⁰.

Now, analyze each complex:

Option A: [Sc(H₂O)₆]³⁺
The central ion is Sc³⁺, which has a 3d⁰ configuration. Since there are no d electrons, d-d transitions cannot occur. Water (H₂O) is a weak field ligand, but even with splitting, no electrons are present to transition. Thus, this complex does not absorb visible light via d-d transitions.

Option B: [Ti(NH₃)₆]⁴⁺
The central ion is Ti⁴⁺, with a 3d⁰ configuration. Again, no d electrons are present, so d-d transitions are impossible. Ammonia (NH₃) is a strong field ligand, but without electrons, splitting is irrelevant. This complex will not absorb visible light due to d-d transitions.

Option C: [V(NH₃)₆]³⁺
The central ion is V³⁺, with a 3d² configuration. Partially filled d orbitals allow d-d transitions. Ammonia (NH₃) is a strong field ligand, creating a large splitting energy (Δ₀) in the octahedral complex. For d², the electrons can occupy the t₂g orbitals, and transitions to the e_g level are possible. This complex can absorb visible light.

Option D: [Zn(NH₃)₆]²⁺
The central ion is Zn²⁺, with a 3d¹⁰ configuration. The d orbitals are fully filled, so no empty orbitals are available for d-d transitions. Ammonia (NH₃) causes splitting, but electrons cannot move within the d subshell as all orbitals are occupied. Thus, this complex does not absorb visible light via d-d transitions.

Comparing all options, only [V(NH₃)₆]³⁺ (option C) has a metal ion with partially filled d orbitals (d²), enabling d-d transitions and absorption of visible light.

Hence, the correct answer is Option C.

Q 55ChemistrySection A • MCQhaloalkanes and haloarenes

+4 / -1

Conversion of benzene diazonium chloride to chlorobenzene is an example of which of the following reactions?

A

Claisen

B

Friedel-craft

C

Sandmeyer

D

Wurtz

Q 56ChemistrySection A • MCQhydrocarbons

+4 / -1

The major product obtained in the photo catalysed bromination of 2-methylbutane is:

A

1-bromo-2-methylbutane

B

1-bromo-3-methylbutane

C

2-bromo-3-methylbutane

D

2-bromo-2-methylbutane

Official Correct Answer:Option (D)

Step-by-Step Solution:

The photo-catalysed bromination of 2-methylbutane is a free radical substitution reaction. Under UV light, bromine molecules ( $Br_2$ ) undergo homolytic cleavage to form bromine radicals ( $Br^\bullet$ ). These radicals abstract hydrogen atoms from the alkane, leading to alkyl radicals. The stability of the alkyl radical determines the major product, as bromination is selective for the most stable radical.

The structure of 2-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$ . We identify the types of hydrogen atoms:

The hydrogen on the carbon at position 2 (the carbon with the methyl group) is tertiary (3°), as this carbon is bonded to three other carbons. There is one such hydrogen.
The hydrogens on the carbon at position 3 ( $CH_2$ ) are secondary (2°). There are two such hydrogens.
The hydrogens on the terminal methyl groups (positions 1, 4, and the methyl branch on carbon 2) are primary (1°). Specifically:
- Position 1 ( $CH_3$ - attached to carbon 2): 3 hydrogens.
- The methyl branch on carbon 2: 3 hydrogens.
- Position 4 ( $CH_3$ ): 3 hydrogens.
Total primary hydrogens: 9.

The stability order of alkyl radicals is tertiary > secondary > primary. Bromine radicals preferentially abstract the tertiary hydrogen due to its higher stability, even though there is only one tertiary hydrogen compared to multiple primary and secondary hydrogens.

Abstraction of the tertiary hydrogen at carbon 2:

CH_3-\underset{\underset{CH_3}{|}}{CH}-CH_2-CH_3 + Br^\bullet \rightarrow CH_3-\underset{\underset{CH_3}{|}}{\overset{\bullet}{C}}-CH_2-CH_3 + HBr

This forms a tertiary alkyl radical at carbon 2. This radical then reacts with a bromine molecule:

CH_3-\underset{\underset{CH_3}{|}}{\overset{\bullet}{C}}-CH_2-CH_3 + Br_2 \rightarrow CH_3-\underset{\underset{CH_3}{|}}{C}-CH_2-CH_3 + Br^\bullet

Br

The product is 2-bromo-2-methylbutane ( $CH_3-CBr(CH_3)-CH_2-CH_3$ ).

Comparing with the options:

Option A: 1-bromo-2-methylbutane ( $BrCH_2-CH(CH_3)-CH_2-CH_3$ ) is a primary bromide.
Option B: 1-bromo-3-methylbutane likely refers to $CH_3CH(CH_3)CH_2CH_2Br$ , which is also a primary bromide.
Option C: 2-bromo-3-methylbutane ( $CH_3CHBrCH(CH_3)CH_3$ ) is a secondary bromide.
Option D: 2-bromo-2-methylbutane ( $CH_3CBr(CH_3)CH_2CH_3$ ) is a tertiary bromide.

Due to the high selectivity of bromine for tertiary hydrogens (relative rate approximately 1600 times that of primary), the major product is the tertiary bromide, 2-bromo-2-methylbutane.

Hence, the correct answer is Option D.

Q 57ChemistrySection A • MCQalcohols phenols and ethers

+4 / -1

In the Victor-Meyer's test, the colour given by 1°, 2° and 3° alcohols are respectively:

A

Red, colourless, blue

B

Red, blue, colourless

C

Colourless, red, blue

D

Red, blue, violet

Q 58ChemistrySection A • MCQaldehydes ketones and carboxylic acids

+4 / -1

Phthalic acid reacts with resorcinol in the presence of concentrated H $_2$ SO $_4$ to give:

A

Phenolphthalein

B

Alizarin

C

Coumarin

D

Fluorescein

Q 59ChemistrySection A • MCQchemistry in everyday life

+4 / -1

Aminoglycosides are usually used as:

A

antibiotic

B

analgesic

C

hypnotic

D

antifertility

Q 60ChemistrySection A • MCQbiomolecules

+4 / -1

Which of the following will not show mutarotation?

A

Maltose

B

Lactose

C

Glucose

D

Sucrose

Q 61MathematicsSection A • MCQquadratic equations

+4 / -1

The sum of the roots of the equation, $x^2 + |2x - 3| - 4 = 0$ , is:

A

2

B

$-2$

C

$\sqrt{2}$

D

$-\sqrt{2}$

Official Correct Answer:Option (C)

Step-by-Step Solution:

We are given the equation $x^2 + |2x - 3| - 4 = 0$ . The absolute value $|2x - 3|$ changes behavior at the point where $2x - 3 = 0$ , which is $x = \frac{3}{2}$ . Therefore, we need to solve the equation in two cases: when $x < \frac{3}{2}$ and when $x \geq \frac{3}{2}$ .

Case 1: $x < \frac{3}{2}$

For $x < \frac{3}{2}$ , $2x - 3 < 0$ , so $|2x - 3| = -(2x - 3) = -2x + 3$ . Substituting this into the equation:

$x^2 + (-2x + 3) - 4 = 0$

Simplify:

$x^2 - 2x + 3 - 4 = 0$

$x^2 - 2x - 1 = 0$

Solve this quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = -2$ , and $c = -1$ .

Discriminant $D = b^2 - 4ac = (-2)^2 - 4(1)(-1) = 4 + 4 = 8$ .

So, $x = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$ .

The roots are $x = 1 + \sqrt{2}$ and $x = 1 - \sqrt{2}$ .

Since this case is for $x < \frac{3}{2} = 1.5$ , we check:

$1 + \sqrt{2} \approx 1 + 1.414 = 2.414 > 1.5$ → not valid.

$1 - \sqrt{2} \approx 1 - 1.414 = -0.414 < 1.5$ → valid.

Thus, only $x = 1 - \sqrt{2}$ is valid in this case.

Case 2: $x \geq \frac{3}{2}$

For $x \geq \frac{3}{2}$ , $2x - 3 \geq 0$ , so $|2x - 3| = 2x - 3$ . Substituting:

$x^2 + (2x - 3) - 4 = 0$

Simplify:

$x^2 + 2x - 3 - 4 = 0$

$x^2 + 2x - 7 = 0$

Solve using quadratic formula: $a = 1$ , $b = 2$ , $c = -7$ .

Discriminant $D = 2^2 - 4(1)(-7) = 4 + 28 = 32$ .

So, $x = \frac{-2 \pm \sqrt{32}}{2} = \frac{-2 \pm 4\sqrt{2}}{2} = -1 \pm 2\sqrt{2}$ .

The roots are $x = -1 + 2\sqrt{2}$ and $x = -1 - 2\sqrt{2}$ .

Check $x \geq 1.5$ :

$-1 + 2\sqrt{2} \approx -1 + 2 \times 1.414 = -1 + 2.828 = 1.828 > 1.5$ → valid.

$-1 - 2\sqrt{2} \approx -1 - 2.828 = -3.828 < 1.5$ → not valid.

Thus, only $x = -1 + 2\sqrt{2}$ is valid in this case.

The roots of the equation are $x = 1 - \sqrt{2}$ and $x = -1 + 2\sqrt{2}$ . The sum of the roots is:

$(1 - \sqrt{2}) + (-1 + 2\sqrt{2}) = 1 - \sqrt{2} - 1 + 2\sqrt{2} = (1 - 1) + (-\sqrt{2} + 2\sqrt{2}) = 0 + \sqrt{2} = \sqrt{2}$

Comparing with the options:

A. 2

B. -2

C. $\sqrt{2}$

D. $-\sqrt{2}$

Hence, the correct answer is Option C.

Q 62MathematicsSection A • MCQcomplex numbers

+4 / -1

Let $z \neq -i$ be any complex number such that $\frac{z-1}{z+1}$ is a purely imaginary number. Then $z + \frac{1}{z}$ is:

A

0

B

any non-zero real number other than 1

C

any non-zero real number.

D

a purely imaginary number.

Q 63MathematicsSection A • MCQpermutations and combinations

+4 / -1

8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such numbers in which the odd digits do no occupy odd places, is:

A

160

B

120

C

60

D

48

Official Correct Answer:Option (B)

Step-by-Step Solution:

To solve this problem, we need to form 8-digit numbers using the digits 1, 1, 2, 2, 2, 3, 4, 4, such that no odd digit occupies an odd place. The odd places in an 8-digit number are positions 1, 3, 5, and 7, while the even places are positions 2, 4, 6, and 8. The odd digits are 1 and 3 (since 1 and 3 are odd), and there are three odd digits in total: two 1's and one 3. The even digits are 2 and 4, with five even digits in total: three 2's and two 4's.

The condition requires that odd digits do not occupy odd places. Therefore, all odd digits must be placed in even places. Since there are four even places but only three odd digits, one of the even places must be occupied by an even digit. Additionally, all four odd places must be occupied by even digits because no odd digit can be there.

We can break down the arrangement into steps:

First, we need to choose which even position (out of positions 2, 4, 6, and 8) will be occupied by an even digit. The remaining three even positions will be occupied by the three odd digits. The number of ways to choose which even position gets the even digit is given by the combination formula. Since there are four even positions and we choose one, the number of ways is $\binom{4}{1} = 4$ .

Next, we arrange the three odd digits in the remaining three even positions. The odd digits are two identical 1's and one 3. The number of distinct arrangements for these three digits in three positions is calculated by considering the repetitions. We choose a position for the digit 3 (since the two 1's are identical), which can be done in $\binom{3}{1} = 3$ ways. The other two positions automatically get the 1's. Thus, there are 3 ways to arrange the odd digits.

Now, we have five positions that must be filled with even digits: the four odd places (positions 1, 3, 5, 7) and the one chosen even position (from step 1). The even digits available are three identical 2's and two identical 4's. We need to arrange these five digits in the five positions. The number of distinct arrangements is given by the multinomial coefficient, which accounts for identical items. Specifically, we choose positions for the three 2's (or equivalently for the two 4's). The number of ways is $\binom{5}{3} = 10$ (since choosing three positions for the 2's leaves the remaining two for the 4's).

To find the total number of such 8-digit numbers, we multiply the number of choices from each step: the number of ways to choose the even position for the even digit, the number of ways to arrange the odd digits in the remaining even positions, and the number of ways to arrange the even digits in the five designated positions. Therefore, the total is $4 \times 3 \times 10 = 120$ .

This satisfies the condition that no odd digit occupies an odd place, as all odd digits are in even positions, and all odd positions have even digits.

Hence, the correct answer is Option B.

Q 64MathematicsSection A • MCQsequences and series

+4 / -1

Let G be the geometric mean of two positive numbers a and b, and M be the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ . If $\frac{1}{M}$ : G is 4 : 5, then a : b can be:

A

1 : 4

B

1 : 2

C

2 : 3

D

3 : 4

Official Correct Answer:Option (A)

Step-by-Step Solution:

We are given two positive numbers $a$ and $b$ . The geometric mean $G$ of $a$ and $b$ is defined as $G = \sqrt{ab}$ . The arithmetic mean $M$ of $\frac{1}{a}$ and $\frac{1}{b}$ is given by $M = \frac{\frac{1}{a} + \frac{1}{b}}{2}$ . Simplifying $M$ , we get:

$M = \frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{\frac{b + a}{ab}}{2} = \frac{a + b}{2ab}.$

The problem states that $\frac{1}{M} : G = 4 : 5$ , which means the ratio of $\frac{1}{M}$ to $G$ is 4 to 5. This can be written as:

$\frac{\frac{1}{M}}{G} = \frac{4}{5} \quad \Rightarrow \quad \frac{1}{M \cdot G} = \frac{4}{5} \quad \Rightarrow \quad M \cdot G = \frac{5}{4}.$

Substituting the expressions for $M$ and $G$ , we have:

$\left( \frac{a + b}{2ab} \right) \cdot \sqrt{ab} = \frac{5}{4}.$

Simplify the left side:

$\frac{a + b}{2ab} \cdot \sqrt{ab} = \frac{a + b}{2 \sqrt{ab} \cdot \sqrt{ab}} \cdot \sqrt{ab} = \frac{a + b}{2 \sqrt{ab}}.$

Note that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ , so:

$\frac{a + b}{2 \sqrt{ab}} = \frac{a + b}{2 \sqrt{a} \sqrt{b}}.$

Set $k = \sqrt{\frac{a}{b}}$ . This implies $k^2 = \frac{a}{b}$ , so $a = k^2 b$ . Substitute this into the equation:

$\frac{k^2 b + b}{2 \sqrt{(k^2 b) \cdot b}} = \frac{b(k^2 + 1)}{2 \sqrt{k^2 b^2}} = \frac{b(k^2 + 1)}{2 \cdot k b} = \frac{k^2 + 1}{2k}.$

We now have:

$\frac{k^2 + 1}{2k} = \frac{5}{4}.$

Solve for $k$ :

Multiply both sides by $2k$ :

$k^2 + 1 = \frac{5}{4} \cdot 2k = \frac{10k}{4} = \frac{5k}{2}.$

Multiply both sides by 2 to eliminate the denominator:

$2(k^2 + 1) = 5k \quad \Rightarrow \quad 2k^2 + 2 = 5k \quad \Rightarrow \quad 2k^2 - 5k + 2 = 0.$

Solve this quadratic equation using the quadratic formula:

$k = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4}.$

This gives two solutions:

$k = \frac{5 + 3}{4} = \frac{8}{4} = 2 \quad \text{and} \quad k = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2}.$

Recall that $k = \sqrt{\frac{a}{b}}$ , so:

If $k = 2$ , then $\sqrt{\frac{a}{b}} = 2$ , so $\frac{a}{b} = 4$ , giving $a : b = 4 : 1$ .
If $k = \frac{1}{2}$ , then $\sqrt{\frac{a}{b}} = \frac{1}{2}$ , so $\frac{a}{b} = \frac{1}{4}$ , giving $a : b = 1 : 4$ .

Now, check the options:

A. $1 : 4$
B. $1 : 2$
C. $2 : 3$
D. $3 : 4$

The ratio $a : b = 1 : 4$ (from $k = \frac{1}{2}$ ) corresponds to option A. The ratio $4 : 1$ is not listed in the options.

Verify $a : b = 1 : 4$ :

Let $a = 1$ , $b = 4$ . Then:

$G = \sqrt{1 \cdot 4} = \sqrt{4} = 2,$

$M = \frac{\frac{1}{1} + \frac{1}{4}}{2} = \frac{1 + 0.25}{2} = \frac{1.25}{2} = 0.625,$

$\frac{1}{M} = \frac{1}{0.625} = 1.6,$

Ratio $\frac{1}{M} : G = 1.6 : 2 = 1.6 \div 2 = 0.8 = \frac{4}{5}$ , which matches $4 : 5$ .

Other options do not satisfy:

Option B ( $1:2$ ): $G = \sqrt{2} \approx 1.414$ , $M = \frac{1 + 0.5}{2} = 0.75$ , $\frac{1}{M} \approx 1.333$ , ratio $\approx 1.333 : 1.414 \approx 0.943 \neq 0.8$ .
Option C ( $2:3$ ): $G = \sqrt{6} \approx 2.45$ , $M = \frac{0.5 + \frac{1}{3}}{2} = \frac{5/6}{2} \approx 0.4167$ , $\frac{1}{M} \approx 2.4$ , ratio $\approx 2.4 : 2.45 \approx 0.98 \neq 0.8$ .
Option D ( $3:4$ ): $G = \sqrt{12} \approx 3.464$ , $M = \frac{\frac{1}{3} + \frac{1}{4}}{2} = \frac{7/12}{2} \approx 0.2917$ , $\frac{1}{M} \approx 3.4286$ , ratio $\approx 3.4286 : 3.464 \approx 0.99 \neq 0.8$ .

Thus, $a : b = 1 : 4$ is the only ratio from the options that satisfies the condition.

Hence, the correct answer is Option A.

Q 65MathematicsSection A • MCQsequences and series

+4 / -1

The least positive integer n such that $1 - \frac{2}{3} - \frac{2}{3^2} - \ldots - \frac{2}{3^{n-1}} \lt \frac{1}{100}$ , is:

A

4

B

5

C

6

D

7

Q 66MathematicsSection A • MCQbinomial theorem

+4 / -1

If $1 + x^4 + x^5 = \sum_{i=0}^{5} a_i(1+x)^i$ , for all $x$ in R, then $a_2$ is:

A

$-4$

B

6

C

$-8$

D

10

Q 67MathematicsSection A • MCQbinomial theorem

+4 / -1

If $\left(2 + \frac{x}{3}\right)^{55}$ is expanded in the ascending powers of x and the coefficients of powers of x in two consecutive terms of the expansion are equal, then these terms are:

A

7 $^{th}$ and 8 $^{th}$

B

8 $^{th}$ and 9 $^{th}$

C

28 $^{th}$ and 29 $^{th}$

D

27 $^{th}$ and 28 $^{th}$

Q 68MathematicsSection A • MCQstraight lines

+4 / -1

If a line intercepted between the coordinate axes is trisected at a point A(4, 3), which is nearer to x-axis, then its equation is:

A

$4x - 3y = 7$

B

$3x + 2y = 18$

C

$3x + 8y = 36$

D

$x + 3y = 13$

Official Correct Answer:Option (B)

Step-by-Step Solution:

We are given that a line segment intercepted between the coordinate axes is trisected at the point A(4, 3), and A is nearer to the x-axis. We need to find the equation of this line.

Let the line intersect the x-axis at point P(a, 0) and the y-axis at point Q(0, b). The segment PQ is trisected, meaning it is divided into three equal parts by two points. Since A(4, 3) is nearer to the x-axis, it must be the trisection point closer to P (the x-intercept). Therefore, A divides PQ in the ratio 1:2, where PA:AQ = 1:2 (the part from P to A is one-third of PQ).

Using the section formula, if a point divides the line segment joining (x₁, y₁) and (x₂, y₂) in the ratio m:n, its coordinates are:

$\left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)$

Here, P(a, 0) and Q(0, b), and A(4, 3) divides PQ in the ratio m:n = 1:2. Substituting:

For the x-coordinate:

$4 = \frac{1 \cdot 0 + 2 \cdot a}{1 + 2} = \frac{2a}{3}$

Solving for a:

$4 = \frac{2a}{3}$

Multiply both sides by 3:

$12 = 2a$

Divide both sides by 2:

$a = 6$

For the y-coordinate:

$3 = \frac{1 \cdot b + 2 \cdot 0}{1 + 2} = \frac{b}{3}$

Solving for b:

$3 = \frac{b}{3}$

Multiply both sides by 3:

$b = 9$

So, the x-intercept is 6 and the y-intercept is 9. The equation of a line in intercept form is:

$\frac{x}{a} + \frac{y}{b} = 1$

Substituting a = 6 and b = 9:

$\frac{x}{6} + \frac{y}{9} = 1$

To eliminate denominators, multiply both sides by the least common multiple of 6 and 9, which is 18:

$18 \cdot \frac{x}{6} + 18 \cdot \frac{y}{9} = 18 \cdot 1$

Simplify each term:

$3x + 2y = 18$

Thus, the equation is 3x + 2y = 18.

Now, comparing with the options:

A. 4x - 3y = 7

B. 3x + 2y = 18

C. 3x + 8y = 36

D. x + 3y = 13

Option B matches our equation. To verify, check if A(4, 3) lies on this line and if it trisects PQ with intercepts (6, 0) and (0, 9).

Substitute x = 4, y = 3 into 3x + 2y:

$3(4) + 2(3) = 12 + 6 = 18$

This equals the right-hand side, so A lies on the line.

For trisection: the point dividing PQ from P(6, 0) to Q(0, 9) in the ratio 1:2 is:

$\left( \frac{1 \cdot 0 + 2 \cdot 6}{1 + 2}, \frac{1 \cdot 9 + 2 \cdot 0}{1 + 2} \right) = \left( \frac{12}{3}, \frac{9}{3} \right) = (4, 3)$

This confirms A(4, 3) is the trisection point nearer to the x-axis (since the other trisection point at ratio 2:1 is (2, 6), which has a larger y-coordinate and is farther from the x-axis).

Hence, the correct answer is Option B.

Q 69MathematicsSection A • MCQstraight lines

+4 / -1

If the three distinct lines $x + 2ay + a = 0$ , $x + 3by + b = 0$ and $x + 4ay + a = 0$ are concurrent, then the point $(a, b)$ lies on a:

A

circle

B

hyperbola

C

straight line

D

parabola

Official Correct Answer:Option (C)

Step-by-Step Solution:

The three given lines are:

1. $x + 2ay + a = 0$

2. $x + 3by + b = 0$

3. $x + 4ay + a = 0$

Since the lines are concurrent, they all pass through a common point. For three lines to be concurrent, the determinant of their coefficients must be zero. The general form of a line is $px + qy + r = 0$ . For the given lines:

- Line 1: coefficients are $p_1 = 1$ , $q_1 = 2a$ , $r_1 = a$

- Line 2: coefficients are $p_2 = 1$ , $q_2 = 3b$ , $r_2 = b$

- Line 3: coefficients are $p_3 = 1$ , $q_3 = 4a$ , $r_3 = a$

The determinant condition is:

$\begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4a & a \\ \end{vmatrix} = 0$

Expanding the determinant:

$= 1 \cdot \left( (3b)(a) - (b)(4a) \right) - 2a \cdot \left( (1)(a) - (b)(1) \right) + a \cdot \left( (1)(4a) - (3b)(1) \right)$

Compute each part:

First part: $(3b)(a) - (b)(4a) = 3ab - 4ab = -ab$

Second part: $-2a \cdot (a - b) = -2a(a - b) = -2a^2 + 2ab$

Third part: $a \cdot (4a - 3b) = 4a^2 - 3ab$

Summing them up:

$-ab + (-2a^2 + 2ab) + (4a^2 - 3ab) = -ab - 2a^2 + 2ab + 4a^2 - 3ab$

Combine like terms:

For $a^2$ terms: $-2a^2 + 4a^2 = 2a^2$

For $ab$ terms: $-ab + 2ab - 3ab = (-1 + 2 - 3)ab = -2ab$

So, the expression becomes:

$2a^2 - 2ab = 0$

Factor out 2:

$2(a^2 - ab) = 0$

Divide both sides by 2:

$a^2 - ab = 0$

Factor:

$a(a - b) = 0$

So, either $a = 0$ or $a - b = 0$ (i.e., $a = b$ ).

Now, check for distinct lines. If $a = 0$ :

- Line 1: $x + 2(0)y + 0 = 0$ → $x = 0$

- Line 2: $x + 3by + b = 0$

- Line 3: $x + 4(0)y + 0 = 0$ → $x = 0$

Lines 1 and 3 are identical (both are $x = 0$ ), violating distinctness. Thus, $a = 0$ is invalid.

Therefore, $a = b$ . The point $(a, b)$ satisfies $b = a$ , meaning it lies on the straight line $y = x$ .

Hence, the correct answer is Option C.

Q 70MathematicsSection A • MCQcircles

+4 / -1

For the two circles $x^2 + y^2 = 16$ and $x^2 + y^2 - 2y = 0$ , there is/are:

A

one pair of common tangents

B

two pair of common tangents

C

three pair of common tangents

D

no common tangent

Official Correct Answer:Option (D)

Step-by-Step Solution:

We are given two circles:

First circle: $x^2 + y^2 = 16$

Second circle: $x^2 + y^2 - 2y = 0$

To find the number of common tangents, we first rewrite both equations in standard form to identify their centers and radii.

For the first circle: $x^2 + y^2 = 16$ . This is already in standard form, so it can be written as $(x - 0)^2 + (y - 0)^2 = 4^2$ . Therefore, the center is $C_1 = (0, 0)$ and the radius is $r_1 = 4$ .

For the second circle: $x^2 + y^2 - 2y = 0$ . We complete the square for the $y$ -terms. Rewrite as $x^2 + (y^2 - 2y) = 0$ . To complete the square, add and subtract 1: $x^2 + (y^2 - 2y + 1) = 1$ , which simplifies to $x^2 + (y - 1)^2 = 1$ . Therefore, the center is $C_2 = (0, 1)$ and the radius is $r_2 = 1$ .

Now, we have:

Circle 1: Center $(0, 0)$ , Radius $4$
Circle 2: Center $(0, 1)$ , Radius $1$

Next, we calculate the distance $d$ between the centers $C_1(0, 0)$ and $C_2(0, 1)$ :

$d = \sqrt{(0 - 0)^2 + (0 - 1)^2} = \sqrt{0 + 1} = \sqrt{1} = 1$

We compare this distance with the sum and difference of the radii:

Sum of radii: $r_1 + r_2 = 4 + 1 = 5$
Difference of radii: $|r_1 - r_2| = |4 - 1| = 3$

The relative positions of the circles determine the number of common tangents:

If $d > r_1 + r_2$ , the circles are separate and have 4 common tangents (2 direct, 2 transverse).
If $d = r_1 + r_2$ , the circles touch externally and have 3 common tangents.
If $|r_1 - r_2| < d < r_1 + r_2$ , the circles intersect at two points and have 2 common tangents.
If $d = |r_1 - r_2|$ , the circles touch internally and have 1 common tangent.
If $d < |r_1 - r_2|$ , one circle lies entirely inside the other without touching, and there are no common tangents.

Here, $d = 1$ and $|r_1 - r_2| = 3$ . Since $1 < 3$ , we have $d < |r_1 - r_2|$ . This means circle 2 lies entirely inside circle 1 without touching it.

To confirm, we check for points of intersection by solving the equations simultaneously:

Equation of circle 1: $x^2 + y^2 = 16$

Equation of circle 2: $x^2 + y^2 - 2y = 0$

Subtract circle 2 from circle 1:

$(x^2 + y^2) - (x^2 + y^2 - 2y) = 16 - 0$

$x^2 + y^2 - x^2 - y^2 + 2y = 16$

$2y = 16$

$y = 8$

Substitute $y = 8$ into circle 1:

$x^2 + (8)^2 = 16$

$x^2 + 64 = 16$

$x^2 = 16 - 64 = -48$

This gives $x^2 = -48$ , which has no real solutions. Therefore, the circles do not intersect and do not touch.

Since circle 2 is entirely inside circle 1 without touching, there are no common tangents. Any tangent to circle 2 (the inner circle) will intersect circle 1 (the outer circle) at two points, as verified by testing tangents at various points on circle 2. For example:

At point $(0, 0)$ on circle 2, the tangent is $y = 0$ (x-axis). Substituting into circle 1: $x^2 + 0^2 = 16$ gives $x = \pm 4$ , so it intersects circle 1 at two points.
At point $(0, 2)$ on circle 2, the tangent is $y = 2$ . Substituting into circle 1: $x^2 + 2^2 = 16$ gives $x^2 = 12$ , so $x = \pm 2\sqrt{3}$ , again two intersection points.
At point $(1, 1)$ on circle 2, the tangent is $x = 1$ . Substituting into circle 1: $1^2 + y^2 = 16$ gives $y^2 = 15$ , so $y = \pm \sqrt{15}$ , two intersection points.

Thus, no line is tangent to both circles.

Hence, the correct answer is Option D.

Q 71MathematicsSection A • MCQparabola

+4 / -1

Two tangents are drawn from a point $(-2, -1)$ to the curve, $y^2 = 4x$ . If $\alpha$ is the angle between them, then $|\tan\alpha|$ is equal to:

A

$\frac{1}{3}$

B

$\frac{1}{\sqrt{3}}$

C

$\sqrt{3}$

D

3

Official Correct Answer:Option (D)

Step-by-Step Solution:

We are given a point $(-2, -1)$ and the parabola $y^2 = 4x$ . We need to find $|\tan \alpha|$ , where $\alpha$ is the angle between the two tangents drawn from this point to the parabola.

First, recall that for a parabola $y^2 = 4ax$ , the equation of a tangent with slope $m$ is $y = mx + \frac{a}{m}$ . Comparing $y^2 = 4x$ with $y^2 = 4ax$ , we get $4a = 4$ , so $a = 1$ . Therefore, the equation of a tangent to the parabola $y^2 = 4x$ with slope $m$ is:

$y = mx + \frac{1}{m}$

Since the tangent passes through the point $(-2, -1)$ , we substitute $x = -2$ and $y = -1$ into the tangent equation:

$-1 = m \cdot (-2) + \frac{1}{m}$

Simplify the equation:

$-1 = -2m + \frac{1}{m}$

To eliminate the denominator, multiply both sides by $m$ :

$-1 \cdot m = \left(-2m + \frac{1}{m}\right) \cdot m$

This gives:

$-m = -2m^2 + 1$

Bring all terms to one side to form a quadratic equation:

$2m^2 - m - 1 = 0$

Solve for $m$ using the quadratic formula. Here, $a = 2$ , $b = -1$ , and $c = -1$ . The discriminant $D$ is:

$D = b^2 - 4ac = (-1)^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9$

So, the solutions are:

$m = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-1) \pm \sqrt{9}}{2 \cdot 2} = \frac{1 \pm 3}{4}$

Thus, the two slopes are:

$m_1 = \frac{1 + 3}{4} = \frac{4}{4} = 1$

$m_2 = \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2}$

The angle $\alpha$ between two lines with slopes $m_1$ and $m_2$ is given by:

$\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Substitute $m_1 = 1$ and $m_2 = -\frac{1}{2}$ :

$m_1 - m_2 = 1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2}$

$1 + m_1 m_2 = 1 + (1) \cdot \left(-\frac{1}{2}\right) = 1 - \frac{1}{2} = \frac{1}{2}$

Therefore,

$\tan \alpha = \left| \frac{\frac{3}{2}}{\frac{1}{2}} \right| = \left| \frac{3}{2} \times \frac{2}{1} \right| = |3| = 3$

Hence, $|\tan \alpha| = 3$ .

Comparing with the options, we see that 3 corresponds to option D.

Hence, the correct answer is Option D.

Q 72MathematicsSection A • MCQellipse

+4 / -1

The minimum area of a triangle formed by any tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{81} = 1$ and the co-ordinate axes is:

A

12

B

18

C

26

D

36

Official Correct Answer:Option (D)

Step-by-Step Solution:

We are given the ellipse equation: $\frac{x^2}{16} + \frac{y^2}{81} = 1$ . Here, $a^2 = 16$ so $a = 4$ , and $b^2 = 81$ so $b = 9$ . We need to find the minimum area of the triangle formed by any tangent to this ellipse and the coordinate axes.

First, recall the equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ in terms of slope $m$ . It is given by: $y = mx \pm \sqrt{a^2 m^2 + b^2}$ Substituting $a^2 = 16$ and $b^2 = 81$ , we get: $y = mx \pm \sqrt{16m^2 + 81}$

This tangent intersects the coordinate axes. To find the x-intercept, set $y = 0$ : $0 = mx \pm \sqrt{16m^2 + 81}$ Solving for $x$ : $mx = \mp \sqrt{16m^2 + 81}$ $x = \frac{\mp \sqrt{16m^2 + 81}}{m}$ The absolute value of the x-intercept is: $\left| x \right| = \frac{\sqrt{16m^2 + 81}}{\left| m \right|}$

To find the y-intercept, set $x = 0$ : $y = m(0) \pm \sqrt{16m^2 + 81} = \pm \sqrt{16m^2 + 81}$ The absolute value of the y-intercept is: $\left| y \right| = \sqrt{16m^2 + 81}$

The area $A$ of the triangle formed by the tangent and the axes is half the product of the intercepts: $A = \frac{1}{2} \times \left| x \right| \times \left| y \right| = \frac{1}{2} \times \frac{\sqrt{16m^2 + 81}}{\left| m \right|} \times \sqrt{16m^2 + 81}$ Simplifying: $A = \frac{1}{2} \times \frac{16m^2 + 81}{\left| m \right|}$ Since the area is positive and the expression depends on $m^2$ , we can assume $m > 0$ without loss of generality. Thus: $A = \frac{1}{2} \times \frac{16m^2 + 81}{m} = \frac{1}{2} \left( 16m + \frac{81}{m} \right)$

To minimize $A$ , we take the derivative with respect to $m$ and set it to zero. Let: $A(m) = \frac{1}{2} \left( 16m + \frac{81}{m} \right)$ The derivative is: $A'(m) = \frac{1}{2} \left( 16 - \frac{81}{m^2} \right)$ Set $A'(m) = 0$ : $\frac{1}{2} \left( 16 - \frac{81}{m^2} \right) = 0$ $16 - \frac{81}{m^2} = 0$ $16 = \frac{81}{m^2}$ $m^2 = \frac{81}{16}$ $m = \frac{9}{4} \quad \text{(since$ m > 0 $)}$

To confirm this is a minimum, check the second derivative: $A''(m) = \frac{1}{2} \left( \frac{162}{m^3} \right) = \frac{81}{m^3}$ For $m = \frac{9}{4} > 0$ , $A''(m) > 0$ , so it is a minimum.

Now substitute $m = \frac{9}{4}$ into the area formula: $A = \frac{1}{2} \left( 16 \times \frac{9}{4} + \frac{81}{\frac{9}{4}} \right)$ Simplify inside the parentheses: $16 \times \frac{9}{4} = 36$ $\frac{81}{\frac{9}{4}} = 81 \times \frac{4}{9} = 36$ So: $A = \frac{1}{2} (36 + 36) = \frac{1}{2} \times 72 = 36$

Therefore, the minimum area is 36. Comparing with the options: A. 12, B. 18, C. 26, D. 36. Hence, the correct answer is Option D.

Q 73MathematicsSection A • MCQmathematical reasoning

+4 / -1

Let p, q, r denote arbitrary statements. Then the logically equivalent of the statement $p \Rightarrow (q \vee r)$ is:

A

$(p \vee q) \Rightarrow r$

B

$(p \Rightarrow q) \vee (p \Rightarrow r)$

C

$(p \Rightarrow \sim q) \wedge (p \Rightarrow r)$

D

$(p \Rightarrow q) \wedge (p \Rightarrow \sim r)$

Official Correct Answer:Option (B)

Step-by-Step Solution:

We are given the statement $p \Rightarrow (q \vee r)$ and need to find which option is logically equivalent to it. Remember that $a \Rightarrow b$ is equivalent to $\sim a \vee b$ . So, let's rewrite the given statement:

$p \Rightarrow (q \vee r) = \sim p \vee (q \vee r)$

Since disjunction (OR) is associative, we can write this as:

$\sim p \vee q \vee r$

Now, we'll check each option by converting them into equivalent forms using the same implication rule.

Option A: $(p \vee q) \Rightarrow r$

Rewrite the implication:

$(p \vee q) \Rightarrow r = \sim (p \vee q) \vee r$

Apply De Morgan's law to $\sim (p \vee q)$ :

$(\sim p \wedge \sim q) \vee r$

This is not the same as $\sim p \vee q \vee r$ because it has a conjunction ( $\wedge$ ) and different terms. For example, if $p$ is false, $q$ is false, and $r$ is false, the original statement is true (since $\sim p$ is true), but this expression becomes $(\text{true} \wedge \text{true}) \vee \text{false} = \text{true} \vee \text{false} = \text{true}$ . However, if $p$ is true, $q$ is false, and $r$ is false, the original is false (true implies false), but this expression is $(\text{false} \wedge \text{true}) \vee \text{false} = \text{false} \vee \text{false} = \text{false}$ , which matches. But let's test another case: $p$ false, $q$ true, $r$ false. Original: false implies (true or false) = false implies true = true. Option A: $(\text{false} \vee \text{true}) \Rightarrow \text{false} = \text{true} \Rightarrow \text{false} = \text{false}$ . Not the same. So, not equivalent.

Option B: $(p \Rightarrow q) \vee (p \Rightarrow r)$

Rewrite each implication:

$p \Rightarrow q = \sim p \vee q$

$p \Rightarrow r = \sim p \vee r$

So the expression becomes:

$(\sim p \vee q) \vee (\sim p \vee r)$

Since disjunction is associative and commutative, rearrange:

$\sim p \vee \sim p \vee q \vee r$

Simplify ( $\sim p \vee \sim p = \sim p$ ):

$\sim p \vee q \vee r$

This matches the original expression exactly. So, option B is equivalent.

Option C: $(p \Rightarrow \sim q) \wedge (p \Rightarrow r)$

Rewrite each implication:

$p \Rightarrow \sim q = \sim p \vee \sim q$

$p \Rightarrow r = \sim p \vee r$

So the expression is:

$(\sim p \vee \sim q) \wedge (\sim p \vee r)$

Factor out $\sim p$ using distribution:

$\sim p \vee (\sim q \wedge r)$

This is not the same as $\sim p \vee q \vee r$ . For example, if $p$ is true, $q$ is true, and $r$ is false, the original statement is true (true implies (true or false) = true implies true = true), but this expression is $\text{false} \vee (\text{false} \wedge \text{false}) = \text{false} \vee \text{false} = \text{false}$ . Not equivalent.

Option D: $(p \Rightarrow q) \wedge (p \Rightarrow \sim r)$

Rewrite each implication:

$p \Rightarrow q = \sim p \vee q$

$p \Rightarrow \sim r = \sim p \vee \sim r$

So the expression is:

$(\sim p \vee q) \wedge (\sim p \vee \sim r)$

Factor out $\sim p$ :

$\sim p \vee (q \wedge \sim r)$

This is not the same as $\sim p \vee q \vee r$ . For example, if $p$ is true, $q$ is false, and $r$ is true, the original statement is true (true implies (false or true) = true implies true = true), but this expression is $\text{false} \vee (\text{false} \wedge \text{false}) = \text{false} \vee \text{false} = \text{false}$ . Not equivalent.

Therefore, only option B is logically equivalent to the given statement. Hence, the correct answer is Option B.

Q 74MathematicsSection A • MCQstatistics

+4 / -1

Let $\bar{X}$ and M.D. be the mean and the mean deviation about $\bar{X}$ of n observations $x_i$ , i = 1, 2, n. If each of the observations is increased by 5, then the new mean and the mean deviation about the new mean, respectively, are:

A

$\bar{X}$ , M.D.

B

$\bar{X} + 5$ , M.D.

C

$\bar{X}$ , M.D. +5

D

$\bar{X} + 5$ , M.D. +5

Official Correct Answer:Option (B)

Step-by-Step Solution:

We are given n observations $x_i$ for $i = 1, 2, \ldots, n$ . The mean of these observations is denoted by $\bar{X}$ , and the mean deviation about $\bar{X}$ is denoted by M.D. The mean deviation is defined as:

$\text{M.D.} = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{X}|$

Each observation is increased by 5, so the new observations are $y_i = x_i + 5$ for all $i$ . We need to find the new mean and the new mean deviation about this new mean.

First, let's find the new mean, which we'll call $\bar{Y}$ . The new mean is the average of the new observations:

$\bar{Y} = \frac{1}{n} \sum_{i=1}^{n} y_i = \frac{1}{n} \sum_{i=1}^{n} (x_i + 5)$

We can split this summation into two parts:

$\bar{Y} = \frac{1}{n} \left( \sum_{i=1}^{n} x_i + \sum_{i=1}^{n} 5 \right)$

The first sum is the sum of the original observations, which is $n \bar{X}$ because $\bar{X} = \frac{1}{n} \sum_{i=1}^{n} x_i$ , so $\sum_{i=1}^{n} x_i = n \bar{X}$ . The second sum is the sum of $n$ fives, which is $5n$ . Substituting these:

$\bar{Y} = \frac{1}{n} \left( n \bar{X} + 5n \right) = \frac{1}{n} \times n (\bar{X} + 5) = \bar{X} + 5$

So the new mean is $\bar{X} + 5$ .

Next, we find the new mean deviation about the new mean $\bar{Y}$ . The new mean deviation is:

$\text{New M.D.} = \frac{1}{n} \sum_{i=1}^{n} |y_i - \bar{Y}|$

Substitute $y_i = x_i + 5$ and $\bar{Y} = \bar{X} + 5$ :

$y_i - \bar{Y} = (x_i + 5) - (\bar{X} + 5) = x_i + 5 - \bar{X} - 5 = x_i - \bar{X}$

Therefore, the absolute difference is:

$|y_i - \bar{Y}| = |x_i - \bar{X}|$

Now plug this into the formula for the new mean deviation:

$\text{New M.D.} = \frac{1}{n} \sum_{i=1}^{n} |y_i - \bar{Y}| = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{X}|$

But $\frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{X}|$ is exactly the original mean deviation M.D. So:

$\text{New M.D.} = \text{M.D.}$

Thus, the new mean is $\bar{X} + 5$ and the new mean deviation is M.D. (unchanged).

Comparing with the options:

A. $\bar{X}$ , M.D.

B. $\bar{X} + 5$ , M.D.

C. $\bar{X}$ , M.D. +5

D. $\bar{X} + 5$ , M.D. +5

Option B matches our result.

Hence, the correct answer is Option B.

Q 75MathematicsSection A • MCQsets and relations

+4 / -1

A relation on the set A = {x : |x| < 3, x $\in$ Z}, where Z is the set of integers is defined by R = {(x, y) : y = |x|, x $\neq$ $-1$ }. Then the number of elements in the power set of R is:

A

32

B

16

C

8

D

64

Official Correct Answer:Option (B)

Step-by-Step Solution:

First, we need to understand the set A. The set A is defined as A = {x : |x| < 3, x ∈ Z}, where Z is the set of integers. This means A consists of all integers x such that the absolute value of x is less than 3. The condition |x| < 3 translates to -3 < x < 3. Since x must be an integer, we list all integers satisfying this inequality: x = -2, because | -2 | = 2 < 3; x = -1, because | -1 | = 1 < 3; x = 0, because |0| = 0 < 3; x = 1, because |1| = 1 < 3; and x = 2, because |2| = 2 < 3. Note that x = -3 is not included because | -3 | = 3 is not less than 3, and similarly x = 3 is not included. Therefore, A = {-2, -1, 0, 1, 2}.

Next, we define the relation R on the set A. The relation is given as R = {(x, y) : y = |x|, x ≠ -1}. This means R consists of ordered pairs (x, y) where y is the absolute value of x, and x is not equal to -1. Since R is a relation on A, both x and y must be elements of A.

We now list all possible pairs (x, y) that satisfy the condition. The elements of A are x = -2, -1, 0, 1, 2. However, the condition x ≠ -1 excludes x = -1. So we consider the remaining x values: -2, 0, 1, 2.

For each x, we compute y = |x|:

When x = -2, y = | -2 | = 2, so the pair is (-2, 2).
When x = 0, y = |0| = 0, so the pair is (0, 0).
When x = 1, y = |1| = 1, so the pair is (1, 1).
When x = 2, y = |2| = 2, so the pair is (2, 2).

We must ensure that each y is in A. Checking: y = 2 is in A, y = 0 is in A, y = 1 is in A, and y = 2 is in A. Also, there is no pair for x = -1 because it is excluded. Therefore, the relation R consists of the ordered pairs: (-2, 2), (0, 0), (1, 1), and (2, 2). Thus, R has 4 elements.

Now, we need to find the power set of R, denoted P(R). The power set of a set is the set of all its subsets. If a set has n elements, the number of elements in its power set is $2^n$ . Here, R has 4 elements, so the number of elements in P(R) is $2^4$ .

Calculating: $2^4 = 2 \times 2 \times 2 \times 2 = 16$ . Therefore, the power set of R has 16 elements.

Looking at the options: A. 32, B. 16, C. 8, D. 64. The value 16 corresponds to option B. Hence, the correct answer is Option B.

Q 76MathematicsSection A • MCQmatrices and determinants

+4 / -1

If $A = \begin{bmatrix} 1 & 2 & x \\ 3 & -1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} y \\ x \\ 1 \end{bmatrix}$ be such that AB = $\begin{bmatrix} 6 \\ 8 \end{bmatrix}$ , then:

A

$y = 2x$

B

$y = -2x$

C

$y = x$

D

$y = -x$

Official Correct Answer:Option (A)

Step-by-Step Solution:

We are given two matrices: matrix A and matrix B. Matrix A is a 2x3 matrix defined as $A = \begin{bmatrix} 1 & 2 & x \\ 3 & -1 & 2 \end{bmatrix}$ and matrix B is a 3x1 column vector defined as $B = \begin{bmatrix} y \\ x \\ 1 \end{bmatrix}$ . The product AB is given to be $\begin{bmatrix} 6 \\ 8 \end{bmatrix}$ . We need to find the relationship between y and x from the options.

To compute the product AB, we multiply matrix A by matrix B. Since A is 2x3 and B is 3x1, the result will be a 2x1 matrix. Each element of the product is obtained by taking the dot product of each row of A with the column vector B.

For the first element (top entry) of AB, we take the first row of A, which is [1, 2, x], and dot it with B, which is [y, x, 1]. This gives: $(1 \times y) + (2 \times x) + (x \times 1) = y + 2x + x = y + 3x$ .

For the second element (bottom entry) of AB, we take the second row of A, which is [3, -1, 2], and dot it with B: $(3 \times y) + (-1 \times x) + (2 \times 1) = 3y - x + 2$ .

Therefore, the product AB is: $\begin{bmatrix} y + 3x \\ 3y - x + 2 \end{bmatrix}$ .

We are told that AB equals $\begin{bmatrix} 6 \\ 8 \end{bmatrix}$ . So we set up the following equations:

Equation 1: $y + 3x = 6$

Equation 2: $3y - x + 2 = 8$

Simplify Equation 2 by subtracting 2 from both sides: $3y - x = 6$ .

Now we have the system:

$y + 3x = 6 \quad \text{(Equation 1)}$

$3y - x = 6 \quad \text{(Equation 2)}$

We can solve this system using elimination. Multiply Equation 1 by 3 to align coefficients:

$3(y + 3x) = 3 \times 6$

$3y + 9x = 18 \quad \text{(Equation 3)}$

Subtract Equation 2 from Equation 3:

$(3y + 9x) - (3y - x) = 18 - 6$

$3y + 9x - 3y + x = 12$

$10x = 12$

Divide both sides by 10: $x = \frac{12}{10} = \frac{6}{5}$ .

Substitute $x = \frac{6}{5}$ into Equation 1 to find y:

$y + 3 \times \frac{6}{5} = 6$

$y + \frac{18}{5} = 6$

Convert 6 to a fraction with denominator 5: $6 = \frac{30}{5}$

$y = \frac{30}{5} - \frac{18}{5} = \frac{12}{5}$

So we have $x = \frac{6}{5}$ and $y = \frac{12}{5}$ .

Now, check the relationship: $\frac{y}{x} = \frac{12/5}{6/5} = \frac{12}{5} \times \frac{5}{6} = \frac{12}{6} = 2$ , which means $y = 2x$ .

Verify by substituting into the original equations:

Equation 1: $y + 3x = \frac{12}{5} + 3 \times \frac{6}{5} = \frac{12}{5} + \frac{18}{5} = \frac{30}{5} = 6$ , correct.

Equation 2: $3y - x + 2 = 3 \times \frac{12}{5} - \frac{6}{5} + 2 = \frac{36}{5} - \frac{6}{5} + \frac{10}{5} = \frac{40}{5} = 8$ , correct.

Comparing with the options:

A. $y = 2x$

B. $y = -2x$

C. $y = x$

D. $y = -x$

Hence, the correct answer is Option A.

Q 77MathematicsSection A • MCQmatrices and determinants

+4 / -1

If $\begin{vmatrix} a^2 & b^2 & c^2\\ (a+\lambda)^2 & (b+\lambda)^2 & (c+\lambda)^2 \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2 \end{vmatrix}$ = $k\lambda \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix}$ , $\lambda \neq 0$ , then k is equal to:

A

4 $\lambda$ abc

B

$-4\lambda$ abc

C

$4\lambda^2$

D

$-4\lambda^2$

Official Correct Answer:Option (C)

Step-by-Step Solution:

$\Delta_1 = \begin{vmatrix} a^2 & b^2 & c^2 \\ (a + \lambda)^2 & (b + \lambda)^2 & (c + \lambda)^2 \\ (a - \lambda)^2 & (b - \lambda)^2 & (c - \lambda)^2 \end{vmatrix}$

$\Delta_1 = \begin{vmatrix} a^2 & b^2 & c^2 \\ a^2 + 2a\lambda + \lambda^2 & b^2 + 2b\lambda + \lambda^2 & c^2 + 2c\lambda + \lambda^2 \\ a^2 - 2a\lambda + \lambda^2 & b^2 - 2b\lambda + \lambda^2 & c^2 - 2c\lambda + \lambda^2 \end{vmatrix}$

Operation 1: $R_2 \to R_2 - R_3$

The elements of $R_2$ become $(a^2 + 2a\lambda + \lambda^2) - (a^2 - 2a\lambda + \lambda^2) = 4a\lambda$ .

$\Delta_1 = \begin{vmatrix} a^2 & b^2 & c^2 \\ 4a\lambda & 4b\lambda & 4c\lambda \\ a^2 - 2a\lambda + \lambda^2 & b^2 - 2b\lambda + \lambda^2 & c^2 - 2c\lambda + \lambda^2 \end{vmatrix}$

Operation 2: $R_3 \to R_3 + \frac{1}{2} R_2 - R_1$

For the first column: $(a^2 - 2a\lambda + \lambda^2) + \frac{1}{2}(4a\lambda) - a^2 = \lambda^2$ .

$\Delta_1 = \begin{vmatrix} a^2 & b^2 & c^2 \\ 4a\lambda & 4b\lambda & 4c\lambda \\ \lambda^2 & \lambda^2 & \lambda^2 \end{vmatrix}$

$\Delta_1 = (4\lambda)(\lambda^2) \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} = 4\lambda^3 \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix}$

$4\lambda^3 \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} = k\lambda \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix}$

$k = 4\lambda^2$

Q 78MathematicsSection A • MCQmatrices and determinants

+4 / -1

If $f(\theta) = \begin{vmatrix} 1 & \cos\theta & 1 \\ -\sin\theta & 1 & -\cos\theta \\ -1 & \sin\theta & 1 \end{vmatrix}$ and A and B are respectively the maximum and the minimum values of $f(\theta)$ , then (A, B) is equal to:

A

(3, $-1$ )

B

$(4, 2-\sqrt{2})$

C

$(2+\sqrt{2}, 2-\sqrt{2})$

D

$(2+\sqrt{2}, -1)$

Q 79MathematicsSection A • MCQinverse trigonometric functions

+4 / -1

Statement I: The equation $(\sin^{-1}x)^3 + (\cos^{-1}x)^3 - a\pi^3 = 0$ has a solution for all $a \geq \frac{1}{32}$ .

Statement II: For any x $\in$ R, $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ and $0 \leq \left(\sin^{-1}x - \frac{\pi}{4}\right)^2 \leq \frac{9\pi^2}{16}$ .

A

Both statements I and II are true.

B

Both statements I and II are false.

C

Statement I is true and statement II is false.

D

Statement I is false and statement II is true.

Official Correct Answer:Option (D)

Step-by-Step Solution:

Let $f(x) = (\sin^{-1}x)^3 + (\cos^{-1}x)^3$ . Using the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$ , let $\sin^{-1}x = t$ , where $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ .

$f(x) = t^3 + \left(\frac{\pi}{2} - t\right)^3$

$A^3 + B^3 = (A+B)(A^2 - AB + B^2)$ :

$f(x) = \left(t + \frac{\pi}{2} - t\right) \left[t^2 - t\left(\frac{\pi}{2} - t\right) + \left(\frac{\pi}{2} - t\right)^2\right]$

$f(x) = \frac{\pi}{2} \left[t^2 - \frac{\pi t}{2} + t^2 + \frac{\pi^2}{4} - \pi t + t^2\right]$

$f(x) = \frac{\pi}{2} \left[3t^2 - \frac{3\pi t}{2} + \frac{\pi^2}{4}\right]$

$f(x) = \frac{3\pi}{2} \left[t^2 - \frac{\pi t}{2} + \frac{\pi^2}{12}\right]$

$f(x) = \frac{3\pi}{2} \left[\left(t - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{48}\right]$

For $x \in [-1, 1]$ , the range of $t = \sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ .

The range of $\left(t - \frac{\pi}{4}\right)$ is $[-\frac{3\pi}{4}, \frac{\pi}{4}]$ .

Squaring this interval: $0 \le \left(\sin^{-1}x - \frac{\pi}{4}\right)^2 \le \left(-\frac{3\pi}{4}\right)^2 = \frac{9\pi^2}{16}$

Thus, Statement II is true.

Using the range of $\left(t - \frac{\pi}{4}\right)^2$ in the expression for $f(x)$ :

Minimum value: $f(x)_{min} = \frac{3\pi}{2} \left[0 + \frac{\pi^2}{48}\right] = \frac{\pi^3}{32}$

Maximum value: $f(x)_{max} = \frac{3\pi}{2} \left[\frac{9\pi^2}{16} + \frac{\pi^2}{48}\right] = \frac{3\pi}{2} \left[\frac{28\pi^2}{48}\right] = \frac{7\pi^3}{8}$

The given equation $(\sin^{-1}x)^3 + (\cos^{-1}x)^3 = a\pi^3$ has a solution if $a\pi^3 \in [\frac{\pi^3}{32}, \frac{7\pi^3}{8}]$ .

This implies $\frac{1}{32} \le a \le \frac{7}{8}$

In the context of the given statements, the condition $a \ge \frac{1}{32}$ identifies the lower bound necessary for a solution to exist. Hence, Statement I is false.

Q 80MathematicsSection A • MCQapplication of derivatives

+4 / -1

If $f(x) = x^2 - x + 5$ , $x > \frac{1}{2}$ , and g(x) is its inverse function, then g'(7) equals:

A

$-\frac{1}{3}$

B

$\frac{1}{13}$

C

$\frac{1}{3}$

D

$-\frac{1}{13}$

Official Correct Answer:Option (C)

Step-by-Step Solution:

We are given the function $f(x) = x^2 - x + 5$ for $x > \frac{1}{2}$ , and we know that $g(x)$ is its inverse function. We need to find $g'(7)$ .

To find the derivative of the inverse function at a point, we use the formula:

$g'(y) = \frac{1}{f'(g(y))}$

So, for $y = 7$ , we have:

$g'(7) = \frac{1}{f'(g(7))}$

This means we need to find two things: first, the value of $g(7)$ , which is the $x$ such that $f(x) = 7$ , and second, the derivative $f'(x)$ evaluated at that $x$ .

Start by solving $f(x) = 7$ :

$x^2 - x + 5 = 7$

Subtract 7 from both sides:

$x^2 - x + 5 - 7 = 0$

$x^2 - x - 2 = 0$

Now, solve this quadratic equation. We factor it:

$x^2 - x - 2 = (x - 2)(x + 1) = 0$

So, the solutions are $x = 2$ and $x = -1$ .

But the domain is restricted to $x > \frac{1}{2}$ . Since $-1 < \frac{1}{2}$ , we discard $x = -1$ . Thus, $x = 2$ is the solution.

Therefore, $g(7) = 2$ .

Next, find the derivative of $f(x)$ . The function is $f(x) = x^2 - x + 5$ , so:

$f'(x) = \frac{d}{dx}(x^2 - x + 5) = 2x - 1$

Now, evaluate $f'(x)$ at $x = g(7) = 2$ :

$f'(2) = 2(2) - 1 = 4 - 1 = 3$

Finally, substitute back into the inverse derivative formula:

$g'(7) = \frac{1}{f'(g(7))} = \frac{1}{f'(2)} = \frac{1}{3}$

So, $g'(7) = \frac{1}{3}$ .

Looking at the options:

A. $-\frac{1}{3}$

B. $\frac{1}{13}$

C. $\frac{1}{3}$

D. $-\frac{1}{13}$

Hence, the correct answer is Option C.

Q 81MathematicsSection A • MCQlimits continuity and differentiability

+4 / -1

Let f, g : R → R be two functions defined by $f(x) = \begin{cases} x\sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$ and $g(x) = xf(x)$ . Statement I: f is a continuous function at x = 0. Statement II: g is a differentiable function at x = 0.

A

Both statement I and II are false.

B

Both statement I and II are true.

C

Statement I is true, statement II is false.

D

Statement I is false, statement II is true.

Official Correct Answer:Option (B)

Step-by-Step Solution:

We are given two functions, $f$ and $g$ , both mapping real numbers to real numbers. The function $f(x)$ is defined as:

$f(x) = \begin{cases} x \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$

and $g(x)$ is defined as $g(x) = x \cdot f(x)$ . We need to verify two statements: Statement I claims that $f$ is continuous at $x = 0$ , and Statement II claims that $g$ is differentiable at $x = 0$ .

First, let's check Statement I: continuity of $f$ at $x = 0$ . For $f$ to be continuous at $x = 0$ , the limit of $f(x)$ as $x$ approaches 0 must equal $f(0)$ . We know $f(0) = 0$ . So, we need to compute:

$\lim_{x \to 0} f(x) = \lim_{x \to 0} x \sin\left(\frac{1}{x}\right)$

Since $|\sin(\theta)| \leq 1$ for any real $\theta$ , we have:

$\left| x \sin\left(\frac{1}{x}\right) \right| \leq |x| \cdot 1 = |x|$

This implies:

$-|x| \leq x \sin\left(\frac{1}{x}\right) \leq |x|$

As $x$ approaches 0, both $-|x|$ and $|x|$ approach 0. By the Squeeze Theorem,

$\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$

Thus, $\lim_{x \to 0} f(x) = 0 = f(0)$ , so $f$ is continuous at $x = 0$ . Statement I is true.

Next, check Statement II: differentiability of $g$ at $x = 0$ . First, express $g(x)$ explicitly. Since $g(x) = x \cdot f(x)$ , and using the definition of $f(x)$ :

For $x \neq 0$ , $f(x) = x \sin\left(\frac{1}{x}\right)$ , so:

$g(x) = x \cdot \left( x \sin\left(\frac{1}{x}\right) \right) = x^2 \sin\left(\frac{1}{x}\right)$

At $x = 0$ , $g(0) = 0 \cdot f(0) = 0 \cdot 0 = 0$ . Thus,

$g(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$

To check differentiability at $x = 0$ , we compute the derivative using the limit definition:

$g'(0) = \lim_{h \to 0} \frac{g(0 + h) - g(0)}{h} = \lim_{h \to 0} \frac{g(h) - 0}{h} = \lim_{h \to 0} \frac{g(h)}{h}$

For $h \neq 0$ , $g(h) = h^2 \sin\left(\frac{1}{h}\right)$ , so:

$\frac{g(h)}{h} = \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} = h \sin\left(\frac{1}{h}\right)$

Thus,

$g'(0) = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right)$

Again, $|\sin(\theta)| \leq 1$ , so:

$\left| h \sin\left(\frac{1}{h}\right) \right| \leq |h| \cdot 1 = |h|$

This implies:

$-|h| \leq h \sin\left(\frac{1}{h}\right) \leq |h|$

As $h$ approaches 0, both $-|h|$ and $|h|$ approach 0. By the Squeeze Theorem,

$\lim_{h \to 0} h \sin\left(\frac{1}{h}\right) = 0$

Therefore, $g'(0)$ exists and equals 0, so $g$ is differentiable at $x = 0$ . Statement II is true.

Both statements are true. Now, reviewing the options:

A. Both statement I and II are false.

B. Both statement I and II are true.

C. Statement I is true, statement II is false.

D. Statement I is false, statement II is true.

Hence, the correct answer is Option B.

Q 82MathematicsSection A • MCQapplication of derivatives

+4 / -1

Let $f$ and $g$ be two differentiable functions on R such that $f'(x) > 0$ and $g'(x) < 0$ for all $x \in R$ . Then for all x:

A

$f(g(x)) > f(g(x-1))$

B

$f(g(x)) > f(g(x+1))$

C

$g(f(x)) > g(f(x-1))$

D

$g(f(x)) < g(f(x+1))$

Official Correct Answer:Option (B)

Step-by-Step Solution:

We are given that $f$ and $g$ are differentiable functions on $\mathbb{R}$ with $f'(x) > 0$ and $g'(x) < 0$ for all $x \in \mathbb{R}$ . This means $f$ is strictly increasing and $g$ is strictly decreasing. A strictly increasing function preserves inequalities: if $a > b$ , then $f(a) > f(b)$ . A strictly decreasing function reverses inequalities: if $a > b$ , then $g(a) < g(b)$ .

Now, let's evaluate each option step by step.

Starting with Option A: $f(g(x)) > f(g(x-1))$ .

Compare $x$ and $x-1$ . Since $x > x-1$ (as $x - (x-1) = 1 > 0$ ), and $g$ is strictly decreasing, we have $g(x) < g(x-1)$ . Now, because $f$ is strictly increasing, $g(x) < g(x-1)$ implies $f(g(x)) < f(g(x-1))$ . This contradicts the option's claim $f(g(x)) > f(g(x-1))$ . Thus, Option A is false.

Next, Option B: $f(g(x)) > f(g(x+1))$ .

Compare $x$ and $x+1$ . Since $x < x+1$ (as $x+1 - x = 1 > 0$ ), and $g$ is strictly decreasing, we have $g(x) > g(x+1)$ . Now, because $f$ is strictly increasing, $g(x) > g(x+1)$ implies $f(g(x)) > f(g(x+1))$ . This matches the option's statement. Thus, Option B appears true, but we'll verify the others.

Option C: $g(f(x)) > g(f(x-1))$ .

Compare $x$ and $x-1$ . Since $x > x-1$ , and $f$ is strictly increasing, we have $f(x) > f(x-1)$ . Now, because $g$ is strictly decreasing, $f(x) > f(x-1)$ implies $g(f(x)) < g(f(x-1))$ . This contradicts the option's claim $g(f(x)) > g(f(x-1))$ . Thus, Option C is false.

Option D: $g(f(x)) < g(f(x+1))$ .

Compare $x$ and $x+1$ . Since $x < x+1$ , and $f$ is strictly increasing, we have $f(x) < f(x+1)$ . Now, because $g$ is strictly decreasing, $f(x) < f(x+1)$ implies $g(f(x)) > g(f(x+1))$ . This contradicts the option's claim $g(f(x)) < g(f(x+1))$ . Thus, Option D is false.

Therefore, only Option B is correct. Hence, the correct answer is Option B.

Q 83MathematicsSection A • MCQindefinite integration

+4 / -1

The integral $\int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$ is equal to:

A

$\frac{1}{(1+\cot^3 x)} + c$

B

$-\frac{1}{3(1+\tan^3 x)} + c$

C

$\frac{\sin^3 x}{(1+\cos^3 x)} + c$

D

$-\frac{\cos^3 x}{3(1+\sin^3 x)} + c$

Official Correct Answer:Option (B)

Step-by-Step Solution:

We start with the integral: $\int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$

First, recall the algebraic identity for the sum of cubes: $\sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)$ Since $\sin^2 x + \cos^2 x = 1$ , this simplifies to: $\sin^3 x + \cos^3 x = (\sin x + \cos x)(1 - \sin x \cos x)$

Therefore, the denominator becomes: $(\sin^3 x + \cos^3 x)^2 = [(\sin x + \cos x)(1 - \sin x \cos x)]^2 = (\sin x + \cos x)^2 (1 - \sin x \cos x)^2$

The integral now is: $\int \frac{\sin^2 x \cos^2 x}{(\sin x + \cos x)^2 (1 - \sin x \cos x)^2} dx$

Notice that the numerator $\sin^2 x \cos^2 x = (\sin x \cos x)^2$ , so we write: $\int \frac{(\sin x \cos x)^2}{(\sin x + \cos x)^2 (1 - \sin x \cos x)^2} dx$

To simplify, we express everything in terms of $\tan x$ . Divide both numerator and denominator by $\cos^4 x$ :

Numerator: $\frac{\sin^2 x \cos^2 x}{\cos^4 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$

Denominator: $\frac{(\sin^3 x + \cos^3 x)^2}{\cos^4 x} = \left( \frac{\sin^3 x}{\cos^3 x} + \frac{\cos^3 x}{\cos^3 x} \right)^2 \cdot \frac{\cos^6 x}{\cos^4 x} = (\tan^3 x + 1)^2 \cos^2 x$

The integral becomes: $\int \frac{\tan^2 x}{(\tan^3 x + 1)^2 \cos^2 x} dx$

We know that $\frac{1}{\cos^2 x} dx = \sec^2 x dx = d(\tan x)$ . Let $t = \tan x$ , so $dt = \sec^2 x dx = \frac{1}{\cos^2 x} dx$ .

Substituting $t$ and $dt$ , the integral simplifies to: $\int \frac{t^2}{(t^3 + 1)^2} dt$

Now, solve this integral using substitution. Let $u = t^3 + 1$ , then $du = 3t^2 dt$ , which gives $t^2 dt = \frac{du}{3}$ .

Substituting: $\int \frac{t^2}{(t^3 + 1)^2} dt = \int \frac{1}{u^2} \cdot \frac{du}{3} = \frac{1}{3} \int u^{-2} du = \frac{1}{3} \cdot \frac{u^{-1}}{-1} + c = -\frac{1}{3u} + c$

Substitute back $u = t^3 + 1$ : $-\frac{1}{3(t^3 + 1)} + c$

Since $t = \tan x$ , we have: $-\frac{1}{3(\tan^3 x + 1)} + c = -\frac{1}{3(1 + \tan^3 x)} + c$

Comparing with the options:

Option A: $\frac{1}{(1+\cot^3 x)} + c$
Option B: $-\frac{1}{3(1+\tan^3 x)} + c$
Option C: $\frac{\sin^3 x}{(1+\cos^3 x)} + c$
Option D: $-\frac{\cos^3 x}{3(1+\sin^3 x)} + c$

Our result matches option B.

Hence, the correct answer is Option B.

Q 84MathematicsSection A • MCQdefinite integration

+4 / -1

If [ ] denotes the greatest integer function, then the integral $\int_0^{\pi} [\cos x] dx$ is equal to:

A

$\frac{\pi}{2}$

B

0

C

$-1$

D

$-\frac{\pi}{2}$

Official Correct Answer:Option (D)

Step-by-Step Solution:

The integral to evaluate is $\int_0^{\pi} [\cos x] dx$ , where $[ \cdot ]$ denotes the greatest integer function. This function returns the greatest integer less than or equal to its argument. The cosine function, $\cos x$ , decreases from 1 to -1 as $x$ goes from 0 to $\pi$ . The greatest integer function $[ \cos x ]$ will change values at points where $\cos x$ is an integer, specifically at $x = 0$ , $x = \pi/2$ , and $x = \pi$ , because $\cos x$ takes integer values 1, 0, and -1 at these points, respectively. To compute the integral, split it at the discontinuity points, specifically at $x = \pi/2$ : $' in math mode at position 1:$ ̲ \int_0^{\pi} […" style="color:#cc0000">$ \int_0^{\pi} [\cos x] dx = \int_0^{\pi/2} [\cos x] dx + \int_{\pi/2}^{\pi} [\cos x] dx$ Now, evaluate each integral separately. First, consider $\int_0^{\pi/2} [\cos x] dx$ . - At $x = 0$ , $\cos 0 = 1$ , so $[ \cos 0 ] = [1] = 1$ . - For $x \in (0, \pi/2]$ , $\cos x$ is in the interval $(0, 1]$ . Since $\cos x \gt 0$ and $\cos x \lt 1$ for $x \gt 0$ , the greatest integer less than or equal to $\cos x$ is 0. At $x = \pi/2$ , $\cos(\pi/2) = 0$ , so $[0] = 0$ . Thus, in the interval $[0, \pi/2]$ , $[ \cos x ] = 1$ only at the single point $x = 0$ , and $[ \cos x ] = 0$ for all $x \in (0, \pi/2]$ . In the Riemann integral, the value at a single point does not affect the result because it has measure zero. Therefore, the function $[ \cos x ]$ is effectively 0 almost everywhere in $[0, \pi/2]$ , and the integral is: $' in math mode at position 1:$ ̲ \int_0^{\pi/2}…" style="color:#cc0000">$ \int_0^{\pi/2} [\cos x] dx = \int_0^{\pi/2} 0 dx = 0$ To be rigorous, consider the limit as $\epsilon \to 0^+$ : $' in math mode at position 1:$ ̲ \int_0^{\pi/2}…" style="color:#cc0000">$ \int_0^{\pi/2} [\cos x] dx = \lim_{\epsilon \to 0^+} \int_{\epsilon}^{\pi/2} [\cos x] dx$ For $\epsilon \gt 0$ and small, $x \in [\epsilon, \pi/2]$ , $\cos x \in (0, 1)$ , so $[ \cos x ] = 0$ . Thus: $' in math mode at position 1:$ ̲ \int_{\epsilon…" style="color:#cc0000">$ \int_{\epsilon}^{\pi/2} 0 dx = 0$ As $\epsilon \to 0^+$ , the integral remains 0. Next, consider $\int_{\pi/2}^{\pi} [\cos x] dx$ . - At $x = \pi/2$ , $\cos(\pi/2) = 0$ , so $[0] = 0$ . - For $x \in (\pi/2, \pi]$ , $\cos x$ is in the interval $(-1, 0)$ . For example, at $x = \pi$ , $\cos \pi = -1$ , so $[-1] = -1$ . For any $y \in (-1, 0)$ , the greatest integer less than or equal to $y$ is $-1$ (e.g., $[-0.5] = -1$ ). Thus, in the interval $[\pi/2, \pi]$ , $[ \cos x ] = 0$ only at the single point $x = \pi/2$ , and $[ \cos x ] = -1$ for all $x \in (\pi/2, \pi]$ . Again, the value at a single point does not affect the integral, so $[ \cos x ]$ is effectively $-1$ almost everywhere in $[\pi/2, \pi]$ . Therefore: $' in math mode at position 1:$ ̲ \int_{\pi/2}^{…" style="color:#cc0000">$ \int_{\pi/2}^{\pi} [\cos x] dx = \int_{\pi/2}^{\pi} (-1) dx = -1 \cdot \left( \pi - \frac{\pi}{2} \right) = -1 \cdot \frac{\pi}{2} = -\frac{\pi}{2}$ To be rigorous, consider the limit as $\delta \to 0^+$ : $' in math mode at position 1:$ ̲ \int_{\pi/2}^{…" style="color:#cc0000">$ \int_{\pi/2}^{\pi} [\cos x] dx = \lim_{\delta \to 0^+} \left( \int_{\pi/2 + \delta}^{\pi} [\cos x] dx + \int_{\pi/2}^{\pi/2 + \delta} [\cos x] dx \right)$ For $x \in [\pi/2 + \delta, \pi]$ , $\cos x \in [-1, -\sin \delta) \subset (-1, 0)$ , so $[ \cos x ] = -1$ . Thus: $' in math mode at position 1:$ ̲ \int_{\pi/2 + …" style="color:#cc0000">$ \int_{\pi/2 + \delta}^{\pi} (-1) dx = -1 \cdot \left( \pi - \left( \frac{\pi}{2} + \delta \right) \right) = -\left( \frac{\pi}{2} - \delta \right)$ For $x \in [\pi/2, \pi/2 + \delta]$ , $[ \cos x ]$ is bounded between $-1$ and $0$ , so the integral is bounded in absolute value by $\delta \cdot 1 = \delta$ . As $\delta \to 0^+$ , this part approaches 0. Thus: $' in math mode at position 1:$ ̲ \lim_{\delta \…" style="color:#cc0000">$ \lim_{\delta \to 0^+} \left( -\left( \frac{\pi}{2} - \delta \right) + \text{bounded term} \right) = -\frac{\pi}{2}$ Now, sum the two integrals: $' in math mode at position 1:$ ̲ \int_0^{\pi} […" style="color:#cc0000">$ \int_0^{\pi} [\cos x] dx = \int_0^{\pi/2} [\cos x] dx + \int_{\pi/2}^{\pi} [\cos x] dx = 0 + \left( -\frac{\pi}{2} \right) = -\frac{\pi}{2}$ Therefore, the integral evaluates to $-\pi/2$ . Hence, the correct answer is Option D.

So, the answer is $-\frac{\pi}{2}$ .

Q 85MathematicsSection A • MCQdefinite integration

+4 / -1

If for a continuous function f(x), $\int_{-\pi}^{t} (f(x) + x) dx = \pi^2 - t^2$ , for all $t \geq -\pi$ , then $f\left(-\frac{\pi}{3}\right)$ is equal to:

A

$\pi$

B

$\frac{\pi}{2}$

C

$\frac{\pi}{3}$

D

$\frac{\pi}{6}$

Official Correct Answer:Option (A)

Step-by-Step Solution:

We are given that for a continuous function $f(x)$ , the equation $\int_{-\pi}^{t} (f(x) + x) dx = \pi^2 - t^2$ holds for all $t \geq -\pi$ . We need to find the value of $f\left(-\frac{\pi}{3}\right)$ .

Since the equation is true for all $t \geq -\pi$ , we can differentiate both sides with respect to $t$ to find an expression for $f(t)$ . According to the Fundamental Theorem of Calculus, if $F(t) = \int_{a}^{t} g(x) dx$ , then $F'(t) = g(t)$ . Here, the left side is $\int_{-\pi}^{t} (f(x) + x) dx$ , so we let $g(x) = f(x) + x$ .

Differentiating both sides of the given equation with respect to $t$ :

Left side derivative: $\frac{d}{dt} \left[ \int_{-\pi}^{t} (f(x) + x) dx \right] = f(t) + t$ .

Right side derivative: $\frac{d}{dt} \left[ \pi^2 - t^2 \right] = -2t$ .

So, we have:

$f(t) + t = -2t$

Now, solve for $f(t)$ :

Subtract $t$ from both sides:

$f(t) = -2t - t$

$f(t) = -3t$

Therefore, the function is $f(t) = -3t$ .

Now, substitute $t = -\frac{\pi}{3}$ to find $f\left(-\frac{\pi}{3}\right)$ :

$f\left(-\frac{\pi}{3}\right) = -3 \times \left(-\frac{\pi}{3}\right)$

Simplify:

$f\left(-\frac{\pi}{3}\right) = -3 \times -\frac{\pi}{3} = 3 \times \frac{\pi}{3} = \pi$

To verify, substitute $f(x) = -3x$ into the original integral:

$\int_{-\pi}^{t} (f(x) + x) dx = \int_{-\pi}^{t} (-3x + x) dx = \int_{-\pi}^{t} (-2x) dx$

Integrate $-2x$ :

$\int -2x dx = -2 \times \frac{x^2}{2} = -x^2$

Evaluate from $-\pi$ to $t$ :

$\left[ -x^2 \right]_{-\pi}^{t} = -t^2 - \left( -(-\pi)^2 \right) = -t^2 - (-\pi^2) = -t^2 + \pi^2 = \pi^2 - t^2$

This matches the right side of the given equation, confirming that $f(x) = -3x$ is correct.

Hence, $f\left(-\frac{\pi}{3}\right) = \pi$ , which corresponds to option A.

So, the answer is Option A.

Q 86MathematicsSection A • MCQdifferential equations

+4 / -1

The general solution of the differential equation, $\sin 2x\left(\frac{dy}{dx} - \sqrt{\tan x}\right) - y = 0$ , is:

A

$y\sqrt{\tan x} = x + c$

B

$y\sqrt{\cot x} = \tan x + c$

C

$y\sqrt{\tan x} = \cot x + c$

D

$y\sqrt{\cot x} = x + c$

Official Correct Answer:Option (D)

Step-by-Step Solution:

The given differential equation is:

$\sin 2x\left(\frac{dy}{dx} - \sqrt{\tan x}\right) - y = 0$

First, expand the equation:

$\sin 2x \cdot \frac{dy}{dx} - \sin 2x \cdot \sqrt{\tan x} - y = 0$

Recall that $\sin 2x = 2 \sin x \cos x$ , so substitute:

$2 \sin x \cos x \cdot \frac{dy}{dx} - 2 \sin x \cos x \cdot \sqrt{\tan x} - y = 0$

Rearrange the terms to isolate the derivative:

$2 \sin x \cos x \cdot \frac{dy}{dx} - y = 2 \sin x \cos x \cdot \sqrt{\tan x}$

Divide both sides by $\sin x \cos x$ :

$\frac{2 \sin x \cos x \cdot \frac{dy}{dx}}{\sin x \cos x} - \frac{y}{\sin x \cos x} = \frac{2 \sin x \cos x \cdot \sqrt{\tan x}}{\sin x \cos x}$

Simplify:

$2 \frac{dy}{dx} - \frac{y}{\sin x \cos x} = 2 \sqrt{\tan x}$

Note that $\frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$ , so:

$2 \frac{dy}{dx} - 2 \frac{y}{\sin 2x} = 2 \sqrt{\tan x}$

Divide both sides by 2:

$\frac{dy}{dx} - \frac{y}{\sin 2x} = \sqrt{\tan x}$

This is a linear differential equation of the form $\frac{dy}{dx} + P(x) y = Q(x)$ , where $P(x) = -\frac{1}{\sin 2x}$ and $Q(x) = \sqrt{\tan x}$ .

The integrating factor (I.F.) is $e^{\int P(x) dx} = e^{\int -\frac{1}{\sin 2x} dx}$ .

Compute $\int -\frac{1}{\sin 2x} dx$ . Since $\frac{1}{\sin 2x} = \csc 2x$ , we have $\int -\csc 2x dx$ .

Let $\theta = 2x$ , so $d\theta = 2 dx$ and $dx = \frac{d\theta}{2}$ :

$\int -\csc \theta \cdot \frac{d\theta}{2} = -\frac{1}{2} \int \csc \theta d\theta$

The integral of $\csc \theta$ is $\ln |\csc \theta - \cot \theta| + C$ , so:

$-\frac{1}{2} \int \csc \theta d\theta = -\frac{1}{2} \ln |\csc \theta - \cot \theta| + C$

Substitute back $\theta = 2x$ :

$-\frac{1}{2} \ln |\csc 2x - \cot 2x| + C$

Simplify $\csc 2x - \cot 2x$ :

$\csc 2x - \cot 2x = \frac{1}{\sin 2x} - \frac{\cos 2x}{\sin 2x} = \frac{1 - \cos 2x}{\sin 2x}$

Use identities $1 - \cos 2x = 2 \sin^2 x$ and $\sin 2x = 2 \sin x \cos x$ :

$\frac{2 \sin^2 x}{2 \sin x \cos x} = \frac{\sin x}{\cos x} = \tan x$

Thus:

$\int -\frac{1}{\sin 2x} dx = -\frac{1}{2} \ln |\tan x| + C$

The integrating factor is:

$I.F. = e^{-\frac{1}{2} \ln |\tan x|} = e^{\ln |\tan x|^{-1/2}} = |\tan x|^{-1/2}$

Since $\sqrt{\tan x}$ is in the equation, we consider $\tan x \gt 0$ , so drop the absolute value:

$I.F. = (\tan x)^{-1/2} = \frac{1}{\sqrt{\tan x}} = \sqrt{\cot x}$

The solution of the linear differential equation is:

$y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$

$y \sqrt{\cot x} = \int \sqrt{\tan x} \cdot \sqrt{\cot x} dx + C$

Simplify the integrand:

$\sqrt{\tan x} \cdot \sqrt{\cot x} = \sqrt{\tan x \cdot \cot x} = \sqrt{1} = 1$

Thus:

$y \sqrt{\cot x} = \int 1 dx + C = x + C$

Comparing with the options:

A. $y\sqrt{\tan x} = x + c$

B. $y\sqrt{\cot x} = \tan x + c$

C. $y\sqrt{\tan x} = \cot x + c$

D. $y\sqrt{\cot x} = x + c$

Hence, the correct answer is Option D.

Q 87MathematicsSection A • MCQvector algebra

+4 / -1

If $\hat{x}$ , $\hat{y}$ and $\hat{z}$ are three unit vectors in threedimensional space, then the minimum value of $|\hat{x} + \hat{y}|^2 + |\hat{y} + \hat{z}|^2 + |\hat{z} + \hat{x}|^2$

A

$\frac{3}{2}$

B

3

C

$3\sqrt{3}$

D

6

Official Correct Answer:Option (B)

Step-by-Step Solution:

We are given three unit vectors $\hat{x}$ , $\hat{y}$ , and $\hat{z}$ in three-dimensional space and need to find the minimum value of the expression $|\hat{x} + \hat{y}|^2 + |\hat{y} + \hat{z}|^2 + |\hat{z} + \hat{x}|^2$ .

First, recall that for any vector $\vec{a}$ , $|\vec{a}|^2 = \vec{a} \cdot \vec{a}$ . Using this, expand each term in the expression.

Start with $|\hat{x} + \hat{y}|^2$ :

$|\hat{x} + \hat{y}|^2 = (\hat{x} + \hat{y}) \cdot (\hat{x} + \hat{y}) = \hat{x} \cdot \hat{x} + 2 \hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{y}$

Since $\hat{x}$ and $\hat{y}$ are unit vectors, $\hat{x} \cdot \hat{x} = 1$ and $\hat{y} \cdot \hat{y} = 1$ , so:

$|\hat{x} + \hat{y}|^2 = 1 + 2 \hat{x} \cdot \hat{y} + 1 = 2 + 2 \hat{x} \cdot \hat{y}$

Similarly, for $|\hat{y} + \hat{z}|^2$ :

$|\hat{y} + \hat{z}|^2 = (\hat{y} + \hat{z}) \cdot (\hat{y} \cdot \hat{z}) = \hat{y} \cdot \hat{y} + 2 \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{z} = 1 + 2 \hat{y} \cdot \hat{z} + 1 = 2 + 2 \hat{y} \cdot \hat{z}$

And for $|\hat{z} + \hat{x}|^2$ :

$|\hat{z} + \hat{x}|^2 = (\hat{z} + \hat{x}) \cdot (\hat{z} + \hat{x}) = \hat{z} \cdot \hat{z} + 2 \hat{z} \cdot \hat{x} + \hat{x} \cdot \hat{x} = 1 + 2 \hat{z} \cdot \hat{x} + 1 = 2 + 2 \hat{z} \cdot \hat{x}$

Now, sum these expressions:

$|\hat{x} + \hat{y}|^2 + |\hat{y} + \hat{z}|^2 + |\hat{z} + \hat{x}|^2 = (2 + 2 \hat{x} \cdot \hat{y}) + (2 + 2 \hat{y} \cdot \hat{z}) + (2 + 2 \hat{z} \cdot \hat{x}) = 6 + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x})$

Let $s = \hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}$ . The expression becomes:

$6 + 2s$

To minimize the entire expression, minimize $s$ . Since $\hat{x}$ , $\hat{y}$ , and $\hat{z}$ are unit vectors, the dot products equal the cosines of the angles between them:

$\hat{x} \cdot \hat{y} = \cos \theta_{xy}, \quad \hat{y} \cdot \hat{z} = \cos \theta_{yz}, \quad \hat{z} \cdot \hat{x} = \cos \theta_{zx}$

So, $s = \cos \theta_{xy} + \cos \theta_{yz} + \cos \theta_{zx}$ .

Consider the magnitude of the sum $\hat{x} + \hat{y} + \hat{z}$ :

$|\hat{x} + \hat{y} + \hat{z}|^2 = (\hat{x} + \hat{y} + \hat{z}) \cdot (\hat{x} + \hat{y} + \hat{z}) = \hat{x} \cdot \hat{x} + \hat{y} \cdot \hat{y} + \hat{z} \cdot \hat{z} + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) = 3 + 2s$

Since $|\hat{x} + \hat{y} + \hat{z}|^2 \geq 0$ , we have:

$3 + 2s \geq 0 \quad \Rightarrow \quad s \geq -\frac{3}{2}$

The minimum value of $s$ is $-\frac{3}{2}$ , achieved when $|\hat{x} + \hat{y} + \hat{z}|^2 = 0$ , meaning $\hat{x} + \hat{y} + \hat{z} = \vec{0}$ . This occurs when the vectors are coplanar and equally spaced at 120 degrees to each other.

For example, take:

$\hat{x} = (1, 0, 0), \quad \hat{y} = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right), \quad \hat{z} = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}, 0\right)$

Verify these are unit vectors:

Magnitude of $\hat{y}$ : $\sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$ , similarly for $\hat{z}$ .

Compute dot products:

$\hat{x} \cdot \hat{y} = (1)\left(-\frac{1}{2}\right) + (0)\left(\frac{\sqrt{3}}{2}\right) = -\frac{1}{2}$

$\hat{y} \cdot \hat{z} = \left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) = \frac{1}{4} - \frac{3}{4} = -\frac{1}{2}$

$\hat{z} \cdot \hat{x} = \left(-\frac{1}{2}\right)(1) + \left(-\frac{\sqrt{3}}{2}\right)(0) = -\frac{1}{2}$

So, $s = -\frac{1}{2} + (-\frac{1}{2}) + (-\frac{1}{2}) = -\frac{3}{2}$ .

Substitute $s = -\frac{3}{2}$ into the expression:

$6 + 2s = 6 + 2\left(-\frac{3}{2}\right) = 6 - 3 = 3$

Verify with the example vectors:

$\hat{x} + \hat{y} = \left(1 - \frac{1}{2}, 0 + \frac{\sqrt{3}}{2}, 0\right) = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right), \quad |\hat{x} + \hat{y}|^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$

$\hat{y} + \hat{z} = \left(-\frac{1}{2} - \frac{1}{2}, \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}, 0\right) = (-1, 0, 0), \quad |\hat{y} + \hat{z}|^2 = (-1)^2 = 1$

$\hat{z} + \hat{x} = \left(-\frac{1}{2} + 1, -\frac{\sqrt{3}}{2} + 0, 0\right) = \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}, 0\right), \quad |\hat{z} + \hat{x}|^2 = \left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$

Sum: $1 + 1 + 1 = 3$ .

Since $s \geq -\frac{3}{2}$ , the expression $6 + 2s \geq 6 + 2(-\frac{3}{2}) = 3$ , and we achieved 3, it is the minimum.

Hence, the correct answer is Option B.

Q 88MathematicsSection A • MCQ3d geometry

+4 / -1

A symmetrical form of the line of intersection of the planes $x = ay + b$ and $z = cy + d$ is:

A

$\frac{x-b}{a} = \frac{y-1}{1} = \frac{z-d}{c}$

B

$\frac{x-b-a}{a} = \frac{y-1}{1} = \frac{z-d-c}{c}$

C

$\frac{x-a}{b} = \frac{y-0}{1} = \frac{z-c}{d}$

D

$\frac{x-b-a}{b} = \frac{y-1}{0} = \frac{z-d-c}{d}$

Official Correct Answer:Option (B)

Step-by-Step Solution:

We are given two planes: $x = ay + b$ and $z = cy + d$ . To find the symmetrical form of their line of intersection, we first rewrite the plane equations in standard form. The first plane is $x - ay = b$ , and the second plane is $-cy + z = d$ .

The normal vector to the first plane is $\langle 1, -a, 0 \rangle$ and to the second plane is $\langle 0, -c, 1 \rangle$ . The direction vector of the line of intersection is perpendicular to both normals, so we compute their cross product:

$' in math mode at position 1:$ ̲ \vec{d} = \beg…" style="color:#cc0000">$ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -a & 0 \\ 0 & -c & 1 \\ \end{vmatrix} = \hat{i}[(-a)(1) - (0)(-c)] - \hat{j}[(1)(1) - (0)(0)] + \hat{k}[(1)(-c) - (-a)(0)]$

Simplifying each component:

$' in math mode at position 1:$ ̲ \hat{i}[-a - 0…" style="color:#cc0000">$ \hat{i}[-a - 0] - \hat{j}[1 - 0] + \hat{k}[-c - 0] = -a\hat{i} - \hat{j} - c\hat{k}$

So, the direction vector is $\langle -a, -1, -c \rangle$ , which can be simplified to $\langle a, 1, c \rangle$ by multiplying by $-1$ . Thus, the direction ratios are $a$ , $1$ , and $c$ .

Next, we find a point on the line by choosing a value for $y$ . Let $y = 1$ :

From the first plane: $x = a(1) + b = a + b$
From the second plane: $z = c(1) + d = c + d$

So, the point is $(a + b, 1, c + d)$ .

The symmetrical form of a line is $\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$ , where $(x_0, y_0, z_0)$ is a point on the line and $\langle l, m, n \rangle$ is the direction vector. Substituting the point $(a + b, 1, c + d)$ and direction ratios $\langle a, 1, c \rangle$ :

$' in math mode at position 1:$ ̲ \frac{x - (a +…" style="color:#cc0000">$ \frac{x - (a + b)}{a} = \frac{y - 1}{1} = \frac{z - (c + d)}{c}$

Rewriting the numerators:

$' in math mode at position 1:$ ̲ \frac{x - a - …" style="color:#cc0000">$ \frac{x - a - b}{a} = \frac{y - 1}{1} = \frac{z - c - d}{c}$

Now, comparing with the options:

Option A: $\frac{x - b}{a} = \frac{y - 1}{1} = \frac{z - d}{c}$ corresponds to the point $(b, 1, d)$ , which does not satisfy the first plane unless $a = 0$ (since $b = a(1) + b$ implies $a = 0$ ), so it is incorrect.
Option B: $\frac{x - b - a}{a} = \frac{y - 1}{1} = \frac{z - d - c}{c}$ matches exactly with our derived form.
Option C: $\frac{x - a}{b} = \frac{y - 0}{1} = \frac{z - c}{d}$ has denominators $b$ and $d$ instead of $a$ and $c$ , and the point $(a, 0, c)$ may not lie on the planes (e.g., $a = a(0) + b$ implies $a = b$ , which is not necessarily true).
Option D: $\frac{x - b - a}{b} = \frac{y - 1}{0} = \frac{z - d - c}{d}$ has a denominator of $0$ for $y$ , which is invalid since the direction vector has a non-zero $y$ -component ( $m = 1$ ).

Therefore, only option B is correct.

Hence, the correct answer is Option B.

Q 89MathematicsSection A • MCQ3d geometry

+4 / -1

If the distance between planes, $4x - 2y - 4z + 1 = 0$ and $4x - 2y - 4z + d = 0$ is 7, then d is:

A

41 or $-42$

B

42 or $-43$

C

$-41$ or 43

D

$-42$ or 44

Official Correct Answer:Option (C)

Step-by-Step Solution:

The given planes are $4x - 2y - 4z + 1 = 0$ and $4x - 2y - 4z + d = 0$ . These planes are parallel because they have the same normal vector $\langle 4, -2, -4 \rangle$ . The distance between two parallel planes $ax + by + cz + d_1 = 0$ and $ax + by + cz + d_2 = 0$ is given by the formula:

$\text{Distance} = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$

For the first plane, $4x - 2y - 4z + 1 = 0$ , we have $d_1 = 1$ . For the second plane, $4x - 2y - 4z + d = 0$ , we have $d_2 = d$ . The coefficients are $a = 4$ , $b = -2$ , and $c = -4$ .

First, compute the denominator $\sqrt{a^2 + b^2 + c^2}$ :

$a^2 = 4^2 = 16$

$b^2 = (-2)^2 = 4$

$c^2 = (-4)^2 = 16$

Adding these: $16 + 4 + 16 = 36$

So, $\sqrt{36} = 6$

The distance is given as 7, so:

$7 = \frac{|1 - d|}{6}$

Multiply both sides by 6 to isolate the absolute value:

$7 \times 6 = |1 - d|$

$42 = |1 - d|$

The absolute value equation $|1 - d| = 42$ means that $1 - d$ could be 42 or $1 - d$ could be -42. We solve both cases.

Case 1: $1 - d = 42$

Subtract 1 from both sides:

$-d = 42 - 1$

$-d = 41$

Multiply both sides by -1:

$d = -41$

Case 2: $1 - d = -42$

Subtract 1 from both sides:

$-d = -42 - 1$

$-d = -43$

Multiply both sides by -1:

$d = 43$

Therefore, the possible values of $d$ are $-41$ and $43$ .

Comparing with the options:

A. 41 or $-42$

B. 42 or $-43$

C. $-41$ or 43

D. $-42$ or 44

The values $-41$ and $43$ match option C.

Hence, the correct answer is Option C.

Q 90MathematicsSection A • MCQprobability

+4 / -1

A number x is chosen at random from the set {1, 2, 3, 4, ..., 100}. Define the event: A = the chosen number x satisfies $\frac{(x-10)(x-50)}{(x-30)} \geq 0$ . Then P(A) is:

A

0.71

B

0.70

C

0.51

D

0.20

Official Correct Answer:Option (A)

Step-by-Step Solution:

We are given a set of numbers from 1 to 100, and we choose a number $x$ at random. The event $A$ is defined by the inequality $\frac{(x-10)(x-50)}{(x-30)} \geq 0$ . We need to find the probability $P(A)$ , which is the number of integers $x$ in $\{1, 2, \ldots, 100\}$ that satisfy this inequality divided by 100.

First, note that the expression is undefined when the denominator is zero, i.e., at $x = 30$ . Therefore, $x = 30$ cannot satisfy the inequality and must be excluded from the favorable outcomes.

The critical points where the numerator or denominator is zero are $x = 10$ , $x = 50$ , and $x = 30$ . These points divide the real number line into intervals. We will test the sign of the expression in each interval to determine where the inequality holds. The intervals are:

$x < 10$
$10 < x < 30$
$30 < x < 50$
$x > 50$

We analyze each interval by picking a test point and evaluating the sign of each factor:

For $x < 10$ $x < 10$ (e.g., $x = 5$ $x = 5$ ):
- $(x-10) = 5-10 = -5 < 0$
- $(x-50) = 5-50 = -45 < 0$
- $(x-30) = 5-30 = -25 < 0$
- Numerator: $(-) \times (-) = +$
- Denominator: $-$
- Fraction: $\frac{+}{-} = -$ (negative)
Since the expression is negative, it does not satisfy $\geq 0$ $\geq 0$ .
For $10 < x < 30$ $10 < x < 30$ (e.g., $x = 20$ $x = 20$ ):
- $(x-10) = 20-10 = 10 > 0$
- $(x-50) = 20-50 = -30 < 0$
- $(x-30) = 20-30 = -10 < 0$
- Numerator: $(+) \times (-) = -$
- Denominator: $-$
- Fraction: $\frac{-}{-} = +$ (positive)
Since the expression is positive, it satisfies $\geq 0$ $\geq 0$ .
For $30 < x < 50$ $30 < x < 50$ (e.g., $x = 40$ $x = 40$ ):
- $(x-10) = 40-10 = 30 > 0$
- $(x-50) = 40-50 = -10 < 0$
- $(x-30) = 40-30 = 10 > 0$
- Numerator: $(+) \times (-) = -$
- Denominator: $+$
- Fraction: $\frac{-}{+} = -$ (negative)
Since the expression is negative, it does not satisfy $\geq 0$ $\geq 0$ .
For $x > 50$ $x > 50$ (e.g., $x = 60$ $x = 60$ ):
- $(x-10) = 60-10 = 50 > 0$
- $(x-50) = 60-50 = 10 > 0$
- $(x-30) = 60-30 = 30 > 0$
- Numerator: $(+) \times (+) = +$
- Denominator: $+$
- Fraction: $\frac{+}{+} = +$ (positive)
Since the expression is positive, it satisfies $\geq 0$ $\geq 0$ .

Next, we check the critical points:

At $x = 10$ : $\frac{(10-10)(10-50)}{(10-30)} = \frac{0 \times (-40)}{-20} = \frac{0}{-20} = 0 \geq 0$ , so it satisfies.
At $x = 50$ : $\frac{(50-10)(50-50)}{(50-30)} = \frac{40 \times 0}{20} = \frac{0}{20} = 0 \geq 0$ , so it satisfies.
At $x = 30$ : The expression is undefined, so it does not satisfy.

Combining the results, the inequality holds for:

$x = 10$
$10 < x < 30$
$x = 50$
$x > 50$

In interval notation, this is $x \in [10, 30) \cup [50, \infty)$ . Since our set is $\{1, 2, \ldots, 100\}$ , we consider integers from 1 to 100.

The favorable integers are:

From 10 to 29 inclusive (since 30 is excluded): $x = 10, 11, \ldots, 29$ .
From 50 to 100 inclusive: $x = 50, 51, \ldots, 100$ .

Now, we count these integers:

Number of integers from 10 to 29: First term is 10, last term is 29, so the count is $29 - 10 + 1 = 20$ .
Number of integers from 50 to 100: First term is 50, last term is 100, so the count is $100 - 50 + 1 = 51$ .

Total favorable outcomes = $20 + 51 = 71$ .

Total possible outcomes = 100 (since the set has 100 integers).

Therefore, $P(A) = \frac{71}{100} = 0.71$ .

Comparing with the options:

A. 0.71
B. 0.70
C. 0.51
D. 0.20

Hence, the correct answer is Option A.

JEE Main 2014 12th April Morning Shift Question Paper with Solutions