JEE Main 2024 1st February Morning Shift Question Paper Question Paper with Solutions | JEE Main Mock Test | JeeMate

Q 1PhysicsSection A • MCQalternating current

+4 / -1

A parallel plate capacitor has a capacitance

\mathrm{C}=200~ \mathrm{pF}

. It is connected to

230 \mathrm{~V}

ac supply with an angular frequency

300~ \mathrm{rad} / \mathrm{s}

. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :

A

14.3 ~\mu \mathrm{A}

and

143 ~\mu \mathrm{A}

B

13.8 ~\mu \mathrm{A}

and

13.8 ~\mu \mathrm{A}

C

13.8 ~\mu \mathrm{A}

and

138 ~\mu \mathrm{A}

D

1.38 ~\mu \mathrm{A}

and

1.38 ~\mu \mathrm{A}

Official Correct Answer:Option (B)

Step-by-Step Solution:

To solve this problem, we need to understand how to calculate the rms current in an AC circuit containing a capacitor, as well as the concept of displacement current.

First, the rms (root mean square) value of the current ( $I_{\text{rms}}$ ) in a capacitor when connected to an AC supply is given by:

$I_{\text{rms}} = V_{\text{rms}} \cdot \omega C$

where $V_{\text{rms}}$ is the rms voltage, $\omega$ is the angular frequency, and $C$ is the capacitance.

For an AC supply, the rms voltage is related to the peak voltage ( $V_0$ ) by:

$V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$

However, we've been given the rms voltage directly, which is $230\, \text{V}$ . Therefore, we can use this value directly in our calculations.

Given:

$C = 200\, \text{pF} = 200 \times 10^{-12}\, \text{F}$

$\omega = 300\, \text{rad/s}$

$V_{\text{rms}} = 230\, \text{V}$

Substitute these values into the equation for rms current:

$I_{\text{rms}} = V_{\text{rms}} \cdot \omega C = 230 \times 300 \times 200 \times 10^{-12} = 13.8 \times 10^{-6}\, \text{A}$

Therefore, the rms value of the conduction current is $13.8\, \mu \text{A}$ .

Now, for the displacement current (which is essentially the current that would "flow" through the dielectric of the capacitor as a result of the changing electric field). In an AC circuit, the displacement current and the conduction current are the same. That's because the displacement current is necessary to sustain the changing electric field between the plates of the capacitor, and this changing field is what causes the conduction current in the leads and the rest of the circuit.

Hence, the rms value of the displacement current is the same as the conduction current, which is $13.8\, \mu \text{A}$ .

So, the correct option is:

Option B

$13.8\, \mu \text{A}$ and $13.8\, \mu \text{A}$

Q 2PhysicsSection A • MCQheat and thermodynamics

+4 / -1

The pressure and volume of an ideal gas are related as

\mathrm{PV}^{\frac{3}{2}}=\mathrm{K}

(Constant). The work done when the gas is taken from state

A\left(P_1, V_1, T_1\right)

to state

B\left(P_2, V_2, T_2\right)

is :

A

2\left(\mathrm{P}_2 \sqrt{\mathrm{V}_2}-\mathrm{P}_1 \sqrt{\mathrm{V}_1}\right)

B

2\left(\sqrt{\mathrm{P}_1} \mathrm{~V}_1-\sqrt{\mathrm{P}_2} \mathrm{~V}_2\right)

C

2\left(\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1\right)

D

2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right)

Official Correct Answer:Option (D)

Step-by-Step Solution:

To find the work done by the gas when it goes from state A to state B, we can look at the definition of work done on or by a gas in a thermodynamic process. For a quasi-static process, the work done $W$ is given by the integral of the pressure $P$ with respect to the volume $V$ :

$W = \int_{V_1}^{V_2} P dV$

Given the relationship $PV^{\frac{3}{2}} = K$ , we can solve for $P$ :

$P = \frac{K}{V^{\frac{3}{2}}}$

Now, substitute $P$ into the work integral and evaluate it:

$W = \int_{V_1}^{V_2} \frac{K}{V^{\frac{3}{2}}} dV = K \int_{V_1}^{V_2} V^{-\frac{3}{2}} dV$

To integrate this, we'll use the power rule for integration. The integral of $V^{-\frac{3}{2}}$ is $-2V^{-\frac{1}{2}}$ , so the work done is:

$W = K \left[-2V^{-\frac{1}{2}}\right]_{V_1}^{V_2} = K \left(-2V_2^{-\frac{1}{2}} + 2V_1^{-\frac{1}{2}} \right)$

We can rewrite $V^{-\frac{1}{2}}$ as $\frac{1}{\sqrt{V}}$ :

$W = K \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} \right)$

Since $P_2V_2^{\frac{3}{2}} = K$ and $P_1V_1^{\frac{3}{2}} = K$ , we can express $K$ in terms of $P_1$ and $V_1$ (or $P_2$ and $V_2$ , but we’ll use $P_1$ and $V_1$ for now):

$W = P_1V_1^{\frac{3}{2}} \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} \right)$

$W = -2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_2}} + 2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_1}}$

$W = -2P_1V_1\frac{\sqrt{V_1}}{\sqrt{V_2}} + 2P_1V_1$

Since $\frac{\sqrt{V_1}}{\sqrt{V_2}} = \sqrt{\frac{V_1}{V_2}}$ , and recalling $P_2V_2^{\frac{3}{2}} = K$ once more:

$W = -2P_1V_1\sqrt{\frac{V_1}{V_2}} + 2P_1V_1$

$W = -2 \frac{P_1V_1^{\frac{3}{2}}}{\sqrt{V_2}} + 2P_1V_1$

Putting $P_1V_1^{\frac{3}{2}}$ back as $K$ :

$W = -2\frac{K}{\sqrt{V_2}} + 2P_1V_1$

$W = -2P_2V_2 \sqrt{V_2}\frac{1}{\sqrt{V_2}} + 2P_1V_1$

$W = -2P_2V_2 + 2P_1V_1$

So the work done is:

$W = 2P_1V_1 - 2P_2V_2$

Therefore, the correct answer matching the given options is:

Option D: $2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right)$

Shortcut :

For

\mathrm{PV}^{\mathrm{x}}=

constant

If work done by gas is asked then

\begin{aligned} \mathrm{W} & =\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\mathrm{x}} \\ & \text { Here } \mathrm{x}=\frac{3}{2} \\ \therefore \mathrm{W} & =\frac{\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1}{-\frac{1}{2}} \\ = & 2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right) \end{aligned}

Q 3PhysicsSection B • Numericalcenter of mass

+4 / -1

The identical spheres each of mass

2 \mathrm{M}

are placed at the corners of a right angled triangle with mutually perpendicular sides equal to

4 \mathrm{~m}

each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is

\frac{4 \sqrt{2}}{x}

, where the value of

x

is ___________ .

Answer Format:Enter numerical value rounded to the nearest integer

Q 4PhysicsSection A • MCQcircular motion

+4 / -1

A ball of mass

0.5 \mathrm{~kg}

is attached to a string of length

50 \mathrm{~cm}

. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is

400 \mathrm{~N}

. The maximum possible value of angular velocity of the ball in

\mathrm{rad} / \mathrm{s}

is, :

A

1600

B

20

C

40

D

1000

Q 5PhysicsSection A • MCQproperties of matter

+4 / -1

With rise in temperature, the Young's modulus of elasticity :

A

changes erratically

B

increases

C

decreases

D

remains unchanged

Q 6PhysicsSection B • Numericalelectrostatics

+4 / -1

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle

\theta

with each other. When suspended in water the angle remains the same. If density of the material of the sphere is

1.5 \mathrm{~g} / \mathrm{cc}

, the dielectric constant of water will be __________.

(Take density of water

=1 \mathrm{~g} / \mathrm{cc}

)

Answer Format:Enter numerical value rounded to the nearest integer

Q 7PhysicsSection A • MCQunits and measurements

+4 / -1

The radius

(\mathrm{r})

, length

(l)

and resistance

(\mathrm{R})

of a metal wire was measured in the laboratory as

\begin{aligned} & \mathrm{r}=(0.35 \pm 0.05) ~\mathrm{cm} \\ & \mathrm{R}=(100 \pm 10) ~\mathrm{ohm} \\ & l=(15 \pm 0.2)~ \mathrm{cm} \end{aligned}

The percentage error in resistivity of the material of the wire is :

A

37.3 \%

B

25.6 \%

C

35.6 \%

D

39.9 \%

Official Correct Answer:Option (D)

Step-by-Step Solution:

To calculate the percentage error in the resistivity of the material of the wire, we need to understand the formula for resistivity. The resistivity $\rho$ of a wire is given by:

$\rho = \frac{RA}{l}$

where:

$R$ is the resistance

$A$ is the cross-sectional area of the wire

$l$ is the length of the wire

The cross-sectional area $A$ of the wire with radius $r$ is:

$A = \pi r^2$

We can plug this into the equation for resistivity to get:

$\rho = \frac{R \pi r^2}{l}$

Now, to find the percentage error in resistivity, we need to find the percentage errors in $R$ , $r$ , and $l$ and then use the following rule for combining errors:

For a given function, $f = f(x,y,z,...)$ , where $x, y, z,...$ are the measured quantities with possible errors, the percentage error in $f$ , denoted as $(\delta f)_{\%}$ , can be approximated by adding the relative percentage errors of the input quantities. If $f$ has the form of a product and quotient of the measured quantities as in our case ( $\rho = \frac{R \pi r^2}{l}$ ), the percentage error in $f$ is given by:

$(\delta f)_{\%} = (\delta x)_{\%} + (\delta y)_{\%} + (\delta z)_{\%} + ...$

Where $(\delta x)_{\%}$ , $(\delta y)_{\%}$ , and $(\delta z)_{\%}$ are the percentage errors in each measured quantity $x, y, z, ...$ respectively.

For our case:

The percentage error in radius $(\delta r)_{\%}$ is given by the error in $r$ divided by the average radius and then multiplied by 100:

$(\delta r)_{\%} = \left(\frac{0.05}{0.35}\right) \times 100$

The percentage error in resistance $(\delta R)_{\%}$ is:

$(\delta R)_{\%} = \left(\frac{10}{100}\right) \times 100$

The percentage error in length $(\delta l)_{\%}$ is:

$(\delta l)_{\%} = \left(\frac{0.2}{15}\right) \times 100$

Now let's calculate each:

$(\delta r)_{\%} = \left(\frac{0.05}{0.35}\right) \times 100 \approx 14.29\%$

$(\delta R)_{\%} = \left(\frac{10}{100}\right) \times 100 = 10\%$

$(\delta l)_{\%} = \left(\frac{0.2}{15}\right) \times 100 \approx 1.33\%$

However, since the area $A$ is proportional to $r^2$ , the percentage error in $A$ will be twice the percentage error in $r$ . Thus:

$(\delta A)_{\%} = 2 \times (\delta r)_{\%} = 2 \times 14.29\% \approx 28.58\%$

Finally, we add the percentage errors to find the percentage error in resistivity:

$(\delta \rho)_{\%} = (\delta R)_{\%} + (\delta A)_{\%} + (\delta l)_{\%}$

$(\delta \rho)_{\%} = 10\% + 28.58\% + 1.33\% \approx 39.91\%$

This calculation gives us a value close to 39.91%, which means the correct option is closest to this value. Thus, the best answer is:

Option D $39.9\%$

Q 8PhysicsSection A • MCQatoms and nuclei

+4 / -1

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly :

A

13.6 \mathrm{eV}

B

1.5 \mathrm{eV}

C

12.1 \mathrm{eV}

D

1.9 \mathrm{eV}

Official Correct Answer:Option (C)

Step-by-Step Solution:

The energy $E_n$ of an electron in the $n$ th energy level of a hydrogen atom is given by the formula:

$E_n = -\frac{13.6 \, \text{eV}}{n^2}$

where:

$E_n$ is the energy in electron volts (eV),

$13.6 \, \text{eV}$ is the ionization energy of hydrogen (the energy required to remove the electron from the ground state, $n=1$ ),

$n$ is the principal quantum number (an integer starting from 1).

The Balmer Series

The Balmer series is a specific set of spectral lines of hydrogen that are visible to the naked eye. It results from the electron making transitions between energy levels, specifically from levels with $n \geq 3$ (i.e., from higher energy states) down to $n = 2$ . These transitions release energy in the form of electromagnetic radiation, which we observe as the Balmer series.

Ground State Energy ( $E_1$ ): The energy of the electron in the ground state ( $n=1$ ) of hydrogen is:

$E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV}$

This represents the electron's energy level when it's closest to the nucleus.

Energy for the $n=3$ Level ( $E_3$ ): For an electron in the $n=3$ level, its energy is:

$E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -1.51 \, \text{eV}$

Minimum Energy for Balmer Series Transition: The minimum energy transition within the Balmer series is from $n=3$ to $n=2$ , but to understand the total energy required for an electron to emit radiation in the Balmer series, starting from the ground state, we consider the energy needed to first excite the electron from $n=1$ to $n=3$ .

Energy Difference ( $\Delta E$ ): The energy emitted as radiation for the electron to transition from the ground state to a state where it can participate in the Balmer series is the difference between the ground state energy and the energy of the $n=3$ state:

$\Delta E = E_3 - E_1 = -1.51 \, \text{eV} - (-13.6 \, \text{eV}) = 12.09 \, \text{eV}$

Note

This calculation shows that the minimum energy required by a hydrogen atom in the ground state to emit radiation in the Balmer series is approximately 12.1 eV. The energy difference essentially represents the energy that must be supplied to an electron to excite it from the ground state ( $n=1$ ) to an excited state ( $n=3$ ) from which it can then transition to $n=2$ , emitting radiation observable as part of the Balmer series.

Q 9PhysicsSection A • MCQalternating current

+4 / -1

In series LCR circuit, the capacitance is changed from

C

to

4 C

. To keep the resonance frequency unchanged, the new inductance should be:

A

increased by

2 \mathrm{~L}

B

reduced by

\frac{1}{4} \mathrm{~L}

C

reduced by

\frac{3}{4} \mathrm{~L}

D

increased to

4 \mathrm{~L}

Q 10PhysicsSection A • MCQlaws of motion

+4 / -1

Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04 , the acceleration of the system in

\mathrm{ms}^{-2}

is :

(Consider that the string is massless and unstretchable and the pulley is also massless and frictionless) :

qImage-motion-of-connected-bodies-laws-of-motion-mechanics-Jeemate

qImage-motion-of-connected-bodies-laws-of-motion-mechanics-Jeemate

A

1.2

B

4

C

3

D

2

Q 11PhysicsSection A • MCQsimple harmonic motion

+4 / -1

A simple pendulum of length

1 \mathrm{~m}

has a wooden bob of mass

1 \mathrm{~kg}

. It is struck by a bullet of mass

10^{-2} \mathrm{~kg}

moving with a speed of

2 \times 10^2 \mathrm{~ms}^{-1}

. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use

\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2

)

A

0.20 \mathrm{~m}

B

0.40 \mathrm{~m}

C

0.30 \mathrm{~m}

D

0.35 \mathrm{~m}

Official Correct Answer:Option (A)

Step-by-Step Solution:

The initial momentum of the system (bullet + bob) is the momentum of the bullet because the bob is initially at rest. The momentum of the bullet is given by its mass times its velocity :

$p_{\text{initial}} = m_{\text{bullet}} \times v_{\text{bullet}}$

After the collision, the bullet and the bob move together with a common velocity. Let's denote this common velocity as $v'$ . The final momentum $p_{\text{final}}$ is the combined mass of the bullet and bob times the common velocity :

$p_{\text{final}} = (m_{\text{bullet}} + m_{\text{bob}}) \times v'$

According to the principle of conservation of linear momentum,

$p_{\text{initial}} = p_{\text{final}}$

$m_{\text{bullet}} \times v_{\text{bullet}} = (m_{\text{bullet}} + m_{\text{bob}}) \times v'$

Plugging in the values :

$(10^{-2} \text{ kg}) \times (2 \times 10^2 \text{ m/s}) = (10^{-2} \text{ kg} + 1 \text{ kg}) \times v'$

Solving for $v'$ :

$v' = \frac{10^{-2} \times 2 \times 10^2}{10^{-2} + 1} = \frac{2}{1.01} \approx 1.98 \text{ m/s}$

After the collision, the system has some kinetic energy which will be completely converted to potential energy at the maximum height $h$ that the bob reaches. Using the principle of conservation of energy :

$\text{Kinetic Energy (KE)}_{\text{initial}} = \text{Potential Energy (PE)}_{\text{final}}$

$\frac{1}{2}(m_{\text{bullet}} + m_{\text{bob}})v'^2 = (m_{\text{bullet}} + m_{\text{bob}})gh$

Isolating $h$ , we get :

$h = \frac{\frac{1}{2}(m_{\text{bullet}} + m_{\text{bob}})v'^2}{(m_{\text{bullet}} + m_{\text{bob}})g} = \frac{v'^2}{2g}$

Substituting the values for $v'$ and $g$ :

$h = \frac{(1.98)^2}{2 \times 10} = \frac{3.9204}{20} = 0.19602 \text{ m}$

Looking at the given options, the result most closely matches Option A :

$\boxed{0.20 \text{ m}}$

Therefore, the height to which the bob rises before swinging back is approximately $0.20 \text{ m}$ .

Q 12PhysicsSection A • MCQcapacitor

+4 / -1

Two identical capacitors have same capacitance

C

. One of them is charged to the potential

V

and other to the potential

2 \mathrm{~V}

. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :

A

\frac{1}{4} \mathrm{CV}^2

B

\frac{3}{4} \mathrm{CV}^2

C

\frac{1}{2} \mathrm{CV}^2

D

2 \mathrm{CV}^2

Official Correct Answer:Option (A)

Step-by-Step Solution:

To solve the problem, let's start by considering the energy stored in each capacitor before they are connected together.

The energy stored in a capacitor is given by the formula:

$E = \frac{1}{2} C V^2$

Where $E$ is the energy, $C$ is the capacitance, and $V$ is the potential.

For the first capacitor charged to the potential $V$ , the energy stored is:

$E_1 = \frac{1}{2} C V^2$

For the second capacitor charged to the potential $2V$ , the energy stored is:

$E_2 = \frac{1}{2} C (2V)^2 = \frac{1}{2} C \cdot 4V^2 = 2 CV^2$

The total initial energy stored in the system is the sum of $E_1$ and $E_2$ :

$E_{\text{initial}} = E_1 + E_2 = \frac{1}{2} CV^2 + 2 CV^2 = \frac{5}{2} CV^2$

When the two capacitors are connected together, their potentials will become equal because they are identical capacitors. Let's denote this final potential as $V_f$ . The total charge before and after the connection remains constant because charge is conserved. Therefore, we can write:

$C \cdot V + C \cdot 2V = 2C \cdot V_f$

Simplifying this equation gives us the final potential:

$V + 2V = 2 V_f$

$3V = 2 V_f$

$V_f = \frac{3}{2} V$

The final energy stored in the system when the capacitors are connected is now the total energy stored across both capacitors at the final potential $V_f$ :

$E_{\text{final}} = 2 \cdot \frac{1}{2} C V_f^2 = 2 \cdot \frac{1}{2} C \left(\frac{3}{2} V\right)^2 = C \cdot \frac{9}{4} V^2$

The decrease in energy $\Delta E$ of the combined system is the initial energy minus the final energy:

$\Delta E = E_{\text{initial}} - E_{\text{final}}$

$\Delta E = \frac{5}{2} CV^2 - \frac{9}{4} CV^2$

$\Delta E = \frac{10}{4} CV^2 - \frac{9}{4} CV^2$

$\Delta E = \frac{1}{4} CV^2$

The correct answer is:

Option A: $\frac{1}{4} CV^2$

Q 13PhysicsSection A • MCQunits and measurements

+4 / -1

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is :

A

\frac{5}{11}

B

\frac{10}{11}

C

\frac{50}{11}

D

\frac{1}{2}

Official Correct Answer:Option (A)

Step-by-Step Solution:

The least count of a Vernier caliper is the smallest measurement that can be obtained with it and is determined by the difference in the measurement of one main scale division and one Vernier scale division.

In this case, 10 divisions on the main scale coincide with 11 divisions on the Vernier scale. This means that 11 divisions on the Vernier scale are equal in length to 10 divisions on the main scale. Since each division on the main scale is 5 units, this is equal to:

$10 \text{ divisions on main scale} \times 5 \text{ units per division} = 50 \text{ units}$

The length of 10 divisions on the main scale (or 50 units) is therefore equal to the length of 11 divisions on the Vernier scale. This means that:

$1 \text{ division on the Vernier scale} = \frac{50 \text{ units}}{11}$

Therefore, the least count of the Vernier caliper, which is the difference between one division on the main scale and one division on the Vernier scale, can be calculated as follows:

$\text{Least count} = \text{value of one main scale division} - \text{value of one Vernier scale division}$

$\text{Least count} = 5 \text{ units} - \frac{50 \text{ units}}{11}$

$\text{Least count} = \frac{55 \text{ units}}{11} - \frac{50 \text{ units}}{11}$

$\text{Least count} = \frac{5 \text{ units}}{11}$

Thus, the least count of the Vernier caliper is

$\frac{5 \text{ units}}{11}$

Therefore, the correct answer is:

Option A: $\frac{5}{11}$

Q 14PhysicsSection A • MCQgravitation

+4 / -1

If

\mathrm{R}

is the radius of the earth and the acceleration due to gravity on the surface of earth is

g=\pi^2 \mathrm{~m} / \mathrm{s}^2

, then the length of the second's pendulum at a height

\mathrm{h}=2 R

from the surface of earth will be, :

A

\frac{1}{9} \mathrm{~m}

B

\frac{8}{9} \mathrm{~m}

C

\frac{2}{9} \mathrm{~m}

D

\frac{4}{9} \mathrm{~m}

Q 15PhysicsSection A • MCQdual nature of radiation

+4 / -1

The de Broglie wavelengths of a proton and an

\alpha

particle are

\lambda

and

2 \lambda

respectively. The ratio of the velocities of proton and

\alpha

particle will be :

A

8: 1

B

1: 2

C

1: 8

D

4: 1

Official Correct Answer:Option (A)

Step-by-Step Solution:

To find the ratio of velocities of two particles based on their de Broglie wavelengths, we can use the de Broglie wavelength formula, which relates the momentum of a particle to its wavelength. The de Broglie's wavelength formula is given by:

$\lambda = \frac{h}{p}$

where:

$\lambda$ is the de Broglie wavelength,

$h$ is the Planck constant, and

$p$ is the momentum of the particle.

The momentum $p$ of a particle can also be expressed as the product of its mass $m$ and velocity $v$ :

$p = mv$

So, we can rewrite the de Broglie wavelength equation in terms of mass and velocity:

$\lambda = \frac{h}{mv}$

For the proton (let's use subscript $p$ for proton), the wavelength is $\lambda$ :

$\lambda_p = \frac{h}{m_p v_p}$

For the $\alpha$ particle (let's use subscript $\alpha$ for alpha), the wavelength is $2\lambda$ :

$2\lambda = \frac{h}{m_\alpha v_\alpha}$

We are interested in finding the ratio of the velocities $\frac{v_p}{v_\alpha}$ . Using the given data about wavelengths:

$\lambda_p = \lambda$

$\lambda_\alpha = 2\lambda$

Using the de Broglie equation for both particles:

$\frac{h}{m_p v_p} = \lambda$

$\frac{h}{m_\alpha v_\alpha} = 2\lambda$

Dividing the second equation by the first equation gives us:

$\frac{\frac{h}{m_\alpha v_\alpha}}{\frac{h}{m_p v_p}} = \frac{2\lambda}{\lambda}$

$\frac{m_p v_p}{m_\alpha v_\alpha} = \frac{2}{1}$

$\frac{v_p}{v_\alpha} = \frac{2m_\alpha}{m_p}$

Since we know an $\alpha$ particle consists of 2 protons and 2 neutrons (essentially four nucleons), the mass of an $\alpha$ particle is roughly four times the mass of a proton ( $m_\alpha \approx 4m_p$ ).

Substituting $m_\alpha$ with $4m_p$ in the equation:

$\frac{v_p}{v_\alpha} = \frac{2(4m_p)}{m_p}$

$\frac{v_p}{v_\alpha} = 2 \cdot 4$

$\frac{v_p}{v_\alpha} = 8$

Therefore, the ratio of the velocities of proton to $\alpha$ particle is $8:1$ , which corresponds to Option A.

Q 16PhysicsSection A • MCQcurrent electricity

+4 / -1

A galvanometer has a resistance of

50 ~\Omega

and it allows maximum current of

5 \mathrm{~mA}

. It can be converted into voltmeter to measure upto

100 \mathrm{~V}

by connecting in series a resistor of resistance :

A

19500 \Omega

B

5975 \Omega

C

20050 \Omega

D

19950 \Omega

Q 17PhysicsSection A • MCQunits and measurements

+4 / -1

The dimensional formula of angular impulse is :

A

\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}\right]

B

\left[\mathrm{M} \mathrm{L}^{-2} \mathrm{~T}^{-1}\right]

C

\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1}\right]

D

\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-1}\right]

Q 18PhysicsSection B • Numericalmotion

+4 / -1

A particle is moving in one dimension (along

x

axis) under the action of a variable force. It's initial position was

16 \mathrm{~m}

right of origin. The variation of its position

(x)

with time

(t)

is given as

x=-3 t^3+18 t^2+16 t

, where

x

is in

\mathrm{m}

and

\mathrm{t}

is in

\mathrm{s}

.

The velocity of the particle when its acceleration becomes zero is _________

\mathrm{m} / \mathrm{s}

.

Answer Format:Enter numerical value rounded to the nearest integer

Q 19PhysicsSection B • Numericalelectromagnetic induction

+4 / -1

A rectangular loop of sides

12 \mathrm{~cm}

and

5 \mathrm{~cm}

, with its sides parallel to the

x

-axis and

y

-axis respectively, moves with a velocity of

5 \mathrm{~cm} / \mathrm{s}

in the positive

x

axis direction, in a space containing a variable magnetic field in the positive

z

direction. The field has a gradient of

10^{-3} \mathrm{~T} / \mathrm{cm}

along the negative

x

direction and it is decreasing with time at the rate of

10^{-3} \mathrm{~T} / \mathrm{s}

. If the resistance of the loop is

6 \mathrm{~m} \Omega

, the power dissipated by the loop as heat is __________

\times 10^{-9} \mathrm{~W}

.

Answer Format:Enter numerical value rounded to the nearest integer

Q 20PhysicsSection B • Numericalmagnetics

+4 / -1

A regular polygon of 6 sides is formed by bending a wire of length

4 \pi

meter.

If an electric current of

4 \pi \sqrt{3}

A is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be

x \times 10^{-7} \mathrm{~T}

.

The value of

x

is _________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 21PhysicsSection A • MCQelectronic devices

+4 / -1

In the given circuit if the power rating of Zener diode is

10 \mathrm{~mW}

, the value of series resistance

R_s

to regulate the input unregulated supply is :

qImage-semiconductor-and-p-n-junction-diode-electronic-devices-modern-physics-Jeemate

A

10 \mathrm{k} \Omega

B

10 \Omega

C

1 \mathrm{k} \Omega

D

\frac{3}{7} \mathrm{k} \Omega<\mathrm{R}_{\mathrm{s}}<\frac{3}{5} \mathrm{k} \Omega

Q 22PhysicsSection B • Numericalatoms and nuclei

+4 / -1

The radius of a nucleus of mass number 64 is 4.8 fermi. Then the mass number of another nucleus having radius of 4 fermi is

\frac{1000}{x}

, where

x

is _______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:27

Step-by-Step Solution:

According to the empirical formula relating the radius of a nucleus ( $R$ ) with its mass number ( $A$ ), we know that the radius of a nucleus is proportional to the cube root of its mass number. This relationship is given as:

$R = R_0 A^{1/3}$

where $R_0$ is a constant with an approximate value of 1.2 fermis.

Given that for a nucleus with a mass number 64 has a radius of 4.8 fermis, we can write:

$4.8 \text{ fermi} = R_0 \times 64^{1/3}$

Now another nucleus has a radius of 4 fermis:

$4 \text{ fermi} = R_0 \times A'^{1/3}$

Where $A'$ is the mass number of the other nucleus.

Let's solve for $R_0$ from the first equation:

$R_0 = \frac{4.8 \text{ fermi}}{64^{1/3}}$

Now we're going to find the mass number $A'$ using the second equation and substituting $R_0$ from the above:

$4 \text{ fermi} = \left( \frac{4.8 \text{ fermi}}{64^{1/3}} \right) \times A'^{1/3}$

Now, we want to find $A'$ in terms of $x$ as given by the equation in the question:

$A' = \frac{1000}{x}$

Substitute $A'$ in the equation above, we get:

$4 = \left( \frac{4.8}{64^{1/3}} \right) \times \left( \frac{1000}{x} \right)^{1/3}$

Let's solve for $x$ :

$(4)^3 = \left( \frac{4.8}{64^{1/3}} \right)^3 \times \frac{1000}{x}$

$4^3 \times x = \left( \frac{4.8}{64^{1/3}} \right)^3 \times 1000$

$x = \left( \frac{\left( \frac{4.8}{64^{1/3}} \right)^3 \times 1000}{4^3} \right)$

Now calculate the values:

$x = \left( \frac{\left( \frac{4.8}{4} \right)^3 \times 1000}{64} \right)$

$x = \left( \frac{1.2^3 \times 1000}{64} \right)$

$x = \left( \frac{1.728 \times 1000}{64} \right)$

$x = \left( \frac{1728}{64} \right)$

$x = 27$

Therefore, the value of $x$ is 27.

Q 23PhysicsSection B • Numericalcurrent electricity

+4 / -1

The current in a conductor is expressed as

I=3 t^2+4 t^3

, where

I

is in Ampere and

t

is in second. The amount of electric charge that flows through a section of the conductor during

t=1 \mathrm{~s}

to

t=2 \mathrm{~s}

is __________ C.

Answer Format:Enter numerical value rounded to the nearest integer

Q 24PhysicsSection B • Numericalproperties of matter

+4 / -1

A plane is in level flight at constant speed and each of its two wings has an area of

40 \mathrm{~m}^2

. If the speed of the air is

180 \mathrm{~km} / \mathrm{h}

over the lower wing surface and

252 \mathrm{~km} / \mathrm{h}

over the upper wing surface, the mass of the plane is ___________ kg.

(Take air density to be

1 \mathrm{~kg} \mathrm{~m}^{-3}

and

\mathrm{g}=10 \mathrm{~ms}^{-2}

)

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:9600

Step-by-Step Solution:

To solve this problem, we need to employ Bernoulli's equation, which is applied to describe the behavior of fluid flow. For a fluid in steady flow, the principle states that the sum of the pressure potential energy density, kinetic energy density, and the gravitational potential energy density has the same value at all points along a streamline. Since the plane is in level flight, we can ignore changes in gravitational potential energy. Bernoulli's equation can be written as:

$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2,$

where,

$P_1$ and $P_2$ are the static pressures on the lower and upper surfaces of the wing respectively,
$v_1$ is the airspeed over the lower surface of the wing,
$v_2$ is the airspeed over the upper surface of the wing,
$\rho$ is the density of air,
$P_2 > P_1$ because $v_2 > v_1$ .

First, let's convert the airspeeds to $\mathrm{m/s}$ :

$v_1 = 180 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{3600 \mathrm{s}} = 50 \mathrm{m/s},$

$v_2 = 252 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{3600 \mathrm{s}} = 70 \mathrm{m/s}.$

The pressure difference between the lower and upper wing surfaces can thus be calculated using Bernoulli's equation:

$\Delta P = P_1 - P_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2.$

Substitute the values (with $\rho = 1 \mathrm{~kg/m}^3$ ):

$\Delta P = \frac{1}{2} (1 \mathrm{~kg/m}^3) (70 \mathrm{m/s})^2 - \frac{1}{2} (1 \mathrm{~kg/m}^3) (50 \mathrm{m/s})^2,$

$\Delta P = \frac{1}{2} (4900 \mathrm{~kg/m \cdot s}^2) - \frac{1}{2} (2500 \mathrm{~kg/m \cdot s}^2),$

$\Delta P = \frac{1}{2} (2400 \mathrm{~kg/m \cdot s}^2),$

$\Delta P = 1200 \mathrm{~N/m}^2.$

The lift force generated by the pressure difference over one wing is $\Delta P \times \text{wing area}$ , and since there are two wings, we must double the lift force generated by one wing to find the total lift force sustaining the plane. The weight of the plane is effectively the lift force when in level flight at constant speed. So:

$\text{Lift} = 2 \times \Delta P \times \text{wing area},$

$\text{Lift} = 2 \times 1200 \mathrm{~N/m}^2 \times 40 \mathrm{~m}^2,$

$\text{Lift} = 2 \times 48000 \mathrm{N},$

$\text{Lift} = 96000 \mathrm{N}.$

To find the mass $m$ of the plane, we use Newton's second law, where lift force is equal to the weight ( $mg$ , where $g$ is the acceleration due to gravity):

$m g = \text{Lift},$

$m = \frac{\text{Lift}}{g},$

$m = \frac{96000 \mathrm{N}}{10 \mathrm{m/s}^2},$

$m = 9600 \mathrm{kg}.$

Therefore, the mass of the plane is 9600 kg.

Q 25PhysicsSection A • MCQwave optics

+4 / -1

A monochromatic light of wavelength

6000 ~\mathring{A}

is incident on the single slit of width

0.01 \mathrm{~mm}

. If the diffraction pattern is formed at the focus of the convex lens of focal length

20 \mathrm{~cm}

, the linear width of the central maximum is :

A

12 \mathrm{~mm}

B

24 \mathrm{~mm}

C

60 \mathrm{~mm}

D

120 \mathrm{~mm}

Official Correct Answer:Option (B)

Step-by-Step Solution:

To find the linear width of the central maximum in a single-slit diffraction pattern, we can use the formula that relates the position of the first minima on either side of the central maximum. The angle, $\theta$ , at which the first minimum occurs is given by:

$a \sin(\theta) = m\lambda$

Where:

$a$ is the width of the slit

$\lambda$ is the wavelength of monochromatic light

$m$ is the order number of the minimum (for the first minimum, $m = \pm 1$ )

In our case, we want to find the position of the first minima to determine the width of the central maximum on the screen. So we will use $m = 1$ and $m = -1$ which correspond to the first minima on either side of the central peak.

Given the width of the slit $a = 0.01$ mm, which we need to convert to meters for consistency with the wavelength:

$a = 0.01 \times 10^{-3} \text{m}$

And the wavelength $\lambda = 6000 \mathring{A}$ , also converting to meters:

$\lambda = 6000 \times 10^{-10} \text{m}$

We're using a convex lens of focal length $f = 20$ cm to project the pattern onto a screen. We'll need to convert the focal length to meters as well:

$f = 20 \times 10^{-2} \text{m}$

We can calculate the angle $\theta$ needed for the first minima using the approximation of small angles, where $\sin(\theta) \approx \theta$ :

$a \theta = m\lambda$

Now, solve for $\theta$ for the first minimum (m = 1):

$\theta = \frac{\lambda}{a}$

Plugging in the values:

$\theta = \frac{6000 \times 10^{-10}}{0.01 \times 10^{-3}}$

$\theta = \frac{6000 \times 10^{-10}}{10^{-5}}$

$\theta = 6000 \times 10^{-5}$

Now, linear width of the central maximum on the screen (from -m to m, or -1 to +1) will be twice the distance from the center to the first minimum, which can be found using the focal length of the lens $f$ and the angle $\theta$ :

$y = 2f\theta$

Plugging the focal length and calculated theta:

$y = 2 \times 20 \times 10^{-2} \times 6000 \times 10^{-5}$

$y = 40 \times 10^{-2} \times 6000 \times 10^{-5}$

$y = 240 \times 10^{-3} \text{m}$

$y = 24 \times 10^{-2} \text{m}$

$y = 24 \text{mm}$

This means the linear width of the central maximum is 24 mm. Hence, the correct answer is Option B.

Q 26PhysicsSection B • Numericalwaves

+4 / -1

A tuning fork resonates with a sonometer wire of length

1 \mathrm{~m}

stretched with a tension of

6 \mathrm{~N}

. When the tension in the wire is changed to

54 \mathrm{~N}

, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is ________________

\mathrm{Hz}

.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:6

Step-by-Step Solution:

To solve this problem, we'll have to use the relationship between the frequency of a vibrating string and the tension applied to it. When a tuning fork resonates with a sonometer wire, their frequencies are equal.

The frequency of a vibrating string is given by the formula:

$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

Where:

$f$ is the frequency of the vibration

$L$ is the length of the wire

$T$ is the tension in the wire

$\mu$ is the linear mass density of the wire

The linear mass density ( $\mu$ ) of the wire remains constant.

Initially, when the string resonates with the tuning fork, the frequency of both is given by:

$f_1 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}}$

Here $T_1 = 6 \mathrm{~N}$ and $L = 1 \mathrm{~m}$ , so we have:

$f_1 = \frac{1}{2 \cdot 1} \sqrt{\frac{6}{\mu}} = \frac{1}{2} \sqrt{\frac{6}{\mu}}$

Now, when the tension is changed to $T_2 = 54 \mathrm{~N}$ , the frequency of the wire changes to $f_2$ and it is given by:

$f_2 = \frac{1}{2L} \sqrt{\frac{T_2}{\mu}}$

Since $L$ and $\mu$ remain the same, substituting $T_2$ :

$f_2 = \frac{1}{2 \cdot 1} \sqrt{\frac{54}{\mu}} = \frac{1}{2} \sqrt{\frac{54}{\mu}}$

Notice that $54 = 6 \times 9$ , therefore:

$f_2 = \frac{1}{2} \sqrt{\frac{6 \times 9}{\mu}} = \frac{1}{2} \sqrt{9} \sqrt{\frac{6}{\mu}} = \frac{3}{2} \sqrt{\frac{6}{\mu}} = 3 f_1$

When the tension was increased, the frequency became thrice the original frequency.

Since the second instance of the string produces 12 beats per second this means that the frequency of the tuning fork (and original string) and the new frequency (of the string with higher tension) differ by 12 Hz. If $f_F$ is the frequency of the tuning fork, then:

Either $f_2 = f_F + 12$ Hz or $f_2 = f_F - 12$ Hz.

Since we have determined $f_2 = 3 f_1$ and $f_1 = f_F$ , we can state:

Either $3f_F = f_F + 12$ or $3f_F = f_F - 12$ .

If $3f_F = f_F - 12$ , then :

$3f_F - f_F = -12$

$2f_F = -12$

This result is not possible since frequency cannot be negative.

So, we must consider the correct equation which is:

$f_2 = f_F + 12$

Now, substituting $f_2 = 3f_F$ :

$3f_F = f_F + 12$

$2f_F = 12$

$f_F = 6 \text{ Hz}$

Thus, the frequency of the tuning fork is 6 Hz.

Q 27PhysicsSection B • Numericalray optics

+4 / -1

The distance between object and its 3 times magnified virtual image as produced by a convex lens is

20 \mathrm{~cm}

. The focal length of the lens used is __________

\mathrm{cm}

.

Answer Format:Enter numerical value rounded to the nearest integer

Q 28PhysicsSection A • MCQheat and thermodynamics

+4 / -1

Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :

A

\frac{3}{2} \mathrm{R}

B

\frac{7}{4} \mathrm{R}

C

\frac{5}{2} \mathrm{R}

D

\frac{9}{4} \mathrm{R}

Q 29PhysicsSection A • MCQcircular motion

+4 / -1

A particle moving in a circle of radius

\mathrm{R}

with uniform speed takes time

\mathrm{T}

to complete one revolution.

If this particle is projected with the same speed at an angle

\theta

to the horizontal, the maximum height attained by it is equal to

4 R

. The angle of projection

\theta

is then given by :

A

\sin ^{-1}\left[\frac{2 \mathrm{gT}^2}{\pi^2 \mathrm{R}}\right]^{\frac{1}{2}}

B

\sin ^{-1}\left[\frac{\pi^2 \mathrm{R}}{2 \mathrm{gT}^2}\right]^{\frac{1}{2}}

C

\cos ^{-1}\left[\frac{\pi \mathrm{R}}{2 \mathrm{gT}^2}\right]^{\frac{1}{2}}

D

\cos ^{-1}\left[\frac{2 \mathrm{gT}^2}{\pi^2 \mathrm{R}}\right]^{\frac{1}{2}}

Q 30PhysicsSection A • MCQcurrent electricity

+4 / -1

The reading in the ideal voltmeter

(\mathrm{V})

shown in the given circuit diagram is :

qImage-kirchhoffs-circuit-laws-current-electricity-electricity-Jeemate

A

3 \mathrm{~V}

B

10 \mathrm{~V}

C

5 \mathrm{~V}

D

0 \mathrm{~V}

Q 31ChemistrySection A • MCQcoordination compounds

+4 / -1

Given below are two statements :

Statement (I) : A solution of

\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}

is green in colour.

Statement (II) : A solution of

\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}

is colourless.

In the light of the above statements, choose the most appropriate answer from the options given below :

A

Statement I is incorrect but Statement II is correct

B

Both Statement I and Statement II are correct

C

Both Statement I and Statement II are incorrect

D

Statement I is correct but Statement II is incorrect

Q 32ChemistrySection A • MCQcoordination compounds

+4 / -1

Which of the following complex is homoleptic?

A

\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]

B

\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}

C

\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}

D

\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}

Q 33ChemistrySection A • MCQchemical bonding and molecular structure

+4 / -1

Arrange the bonds in order of increasing ionic character in the molecules.

\mathrm{LiF}

,

\mathrm{K}_2 \mathrm{O}, \mathrm{N}_2, \mathrm{SO}_2

and

\mathrm{ClF}_3

:

A

\mathrm{N}_2<\mathrm{SO}_2<\mathrm{ClF}_3<\mathrm{K}_2 \mathrm{O}<\mathrm{LiF}

B

\mathrm{ClF}_3<\mathrm{N}_2<\mathrm{SO}_2<\mathrm{K}_2 \mathrm{O}<\mathrm{LiF}

C

\mathrm{LiF}<\mathrm{K}_2 \mathrm{O}<\mathrm{ClF}_3<\mathrm{SO}_2<\mathrm{N}_2

D

\mathrm{N}_2<\mathrm{ClF}_3<\mathrm{SO}_2<\mathrm{K}_2 \mathrm{O}<\mathrm{LiF}

Official Correct Answer:Option (A)

Step-by-Step Solution:

To determine the order of increasing ionic character in the bonds of the given molecules, we should consider the concept of electronegativity, which is a measure of the tendency of an atom to attract a bonding pair of electrons. The larger the difference in electronegativity between two atoms, the more ionic the bond is. Covalent bonds, on the other hand, have smaller differences in electronegativity.

We can qualitatively assess the ionic character of the bonds in each molecule as follows :

-

\mathrm{N_2}

: Both nitrogen atoms have the same electronegativity since it is a diatomic molecule of the same element, therefore the bond is purely covalent with no ionic character.

-

\mathrm{SO_2}

: Sulfur and oxygen have different electronegativities, with oxygen being more electronegative than sulfur, but not as drastically different as in ionic compounds. Therefore, the bond has some ionic character but is primarily covalent.

-

\mathrm{ClF_3}

: Chlorine and fluorine have different electronegativities, with fluorine being more electronegative. However, the difference is not as large as between metals and nonmetals, so while this bond has ionic character, it is not as ionic as a bond between a metal and a nonmetal.

-

\mathrm{K_2O}

: Potassium is a metal with low electronegativity, and oxygen is a nonmetal with high electronegativity. Therefore, bonds between potassium and oxygen will have a high degree of ionic character.

-

\mathrm{LiF}

: Lithium is a metal with low electronegativity, and fluorine is a nonmetal with very high electronegativity. The difference in electronegativity is very large, indicating a very high ionic character, even more so than in

\mathrm{K_2O}

because fluorine is more electronegative than oxygen.

Given these considerations, the correct order from least to most ionic character is as follows :

\mathrm{N_2}<\mathrm{SO_2}<\mathrm{ClF_3}<\mathrm{K_2O}<\mathrm{LiF}

Therefore, the correct option is :

Option A : $\mathrm{N_2}<\mathrm{SO_2}<\mathrm{ClF_3}<\mathrm{K_2O}<\mathrm{LiF}$

Q 34ChemistrySection A • MCQatomic structure

+4 / -1

In case of isoelectronic species the size of

\mathrm{F}^{-}, \mathrm{Ne}

and

\mathrm{Na}^{+}

is affected by :

A

Nuclear charge

(\mathrm{z})

B

None of the factors because their size is the same

C

Electron-electron interaction in the outer orbitals

D

Principal quantum number (n)

Official Correct Answer:Option (A)

Step-by-Step Solution:

Option A: Nuclear charge $(\mathrm{z})$ is the correct answer.

Isoelectronic species are atoms and ions that have the same number of electrons. When comparing the sizes of isoelectronic species, the major factor that affects their size is the nuclear charge $(\mathrm{z})$ , which is the charge of the nucleus or the number of protons in the nucleus. Here’s why:

The more protons there are in the nucleus (higher nuclear charge), the more positively charged the nucleus is, and thus the stronger it can attract the electrons towards it. This results in a smaller radius for the species. Conversely, if there are fewer protons in the nucleus, the attraction to the electrons is weaker, leading to a larger radius. Let's compare the species given:

$\mathrm{F}^{-}$ (fluoride ion) has 9 protons in the nucleus and 10 electrons.
$\mathrm{Ne}$ (neon atom) has 10 protons in the nucleus and 10 electrons.
$\mathrm{Na}^{+}$ (sodium ion) has 11 protons in the nucleus and 10 electrons.

Despite having the same number of electrons, the $\mathrm{Na}^{+}$ ion will be the smallest among the three due to the higher nuclear charge of 11 protons, which attracts the same number of electrons more strongly than $\mathrm{Ne}$ with 10 protons and $\mathrm{F}^{-}$ with 9 protons. Hence, $\mathrm{Na}^{+}$ has a lower atomic radius compared to $\mathrm{Ne}$ , which in turn is smaller than $\mathrm{F}^{-}$ .

Considering Option B (size being the same), Option C (electron-electron interaction), and Option D (principal quantum number), none of these factors are as significant as the difference in nuclear charge for isoelectronic species.

The principal quantum number (n) is the same for all these isoelectronic species since they all have their outer electrons in the second energy level (n=2) in the ground state. Electron-electron interactions do contribute to energy levels and could affect the size slightly, but the primary and determining factor for isoelectronic species remains the nuclear charge, as it directly influences the effective nuclear charge felt by the electrons. Therefore, Option A is the best answer.

Q 35ChemistrySection B • Numericalchemical bonding and molecular structure

+4 / -1

The lowest oxidation number of an atom in a compound

\mathrm{A}_2 \mathrm{B}

is -2 . The number of electrons in its valence shell is _______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 36ChemistrySection B • Numericalionic equilibrium

+4 / -1

\mathrm{K}_{\mathrm{a}}

for

\mathrm{CH}_3 \mathrm{COOH}

is

1.8 \times 10^{-5}

and

\mathrm{K}_{\mathrm{b}}

for

\mathrm{NH}_4 \mathrm{OH}

is

1.8 \times 10^{-5}

. The

\mathrm{pH}

of ammonium acetate solution will be _________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 37ChemistrySection A • MCQcompounds containing nitrogen

+4 / -1

Given below are two statements :

Statement (I) : Aminobenzene and aniline are same organic compounds.

Statement (II) : Aminobenzene and aniline are different organic compounds.

In the light of the above statements, choose the most appropriate answer from the options given below :

A

Both Statement I and Statement II are correct

B

Statement I is incorrect but Statement II is correct

C

Statement I is correct but Statement II is incorrect

D

Both Statement I and Statement II are incorrect

Q 38ChemistrySection A • MCQredox reactions

+4 / -1

In acidic medium,

\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7

shows oxidising action as represented in the half reaction:

\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}+\mathrm{XH}^{+}+\mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A}+\mathrm{ZH}_2 \mathrm{O}

\mathrm{X}, \mathrm{Y}, \mathrm{Z}

and

\mathrm{A}

are respectively are :

A

14,7,6

and

\mathrm{Cr}^{3+}

B

14,6,7

and

\mathrm{Cr}^{3+}

C

8,4,6

and

\mathrm{Cr}_2 \mathrm{O}_3

D

8,6,4

and

\mathrm{Cr}_2 \mathrm{O}_3

Q 39ChemistrySection A • MCQbasics of organic chemistry

+4 / -1

Ionic reactions with organic compounds proceed through :

(A) homolytic bond cleavage

(B) heterolytic bond cleavage

(C) free radical formation

(D) primary free radical

(E) secondary free radical

Choose the correct answer from the options given below :

A

(A) only

B

(B) only

C

(C) only

D

(D) and (E) only

Q 40ChemistrySection B • Numericalp block elements

+4 / -1

The number of white coloured salts, among the following is

(a)

\mathrm{SrSO}_4

(b)

\mathrm{Mg}\left(\mathrm{NH}_4\right) \mathrm{PO}_4

(c)

\mathrm{BaCrO}_4

(d)

\mathrm{Mn}(\mathrm{OH})_2

(e)

\mathrm{PbSO}_4

(f)

\mathrm{PbCrO}_4

(g)

\mathrm{AgBr}

(h)

\mathrm{PbI}_2

(i)

\mathrm{CaC}_2 \mathrm{O}_4

(j)

\left[\mathrm{Fe}(\mathrm{OH})_2\left(\mathrm{CH}_3 \mathrm{COO}\right)\right]

Answer Format:Enter numerical value rounded to the nearest integer

Q 41ChemistrySection B • Numericalbasics of organic chemistry

+4 / -1

Total number of deactivating groups in aromatic electrophilic substitution reaction among the following is _______ .

qImage-electron-displacement-effect-basics-of-organic-chemistry-organic-chemistry-Jeemate

qImage-electron-displacement-effect-basics-of-organic-chemistry-organic-chemistry-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Q 42ChemistrySection A • MCQbiomolecules

+4 / -1

If one strand of a DNA has the sequence ATGCTTCA, sequence of the bases in complementary strand is :

A

CATTAGCT

B

TACGAAGT

C

ATGCGACT

D

GTACTTAC

Q 43ChemistrySection A • MCQchemical bonding and molecular structure

+4 / -1

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A):

\mathrm{PH}_3

has lower boiling point than

\mathrm{NH}_3

.

Reason (R) : In liquid state

\mathrm{NH}_3

molecules are associated through vander Waal's forces, but

\mathrm{PH}_3

molecules are associated through hydrogen bonding.

In the light of the above statements, choose the most appropriate answer from the options given below :

A

Both (A) and (R) are correct and (R) is the correct explanation of (A)

B

(A) is not correct but (R) is correct

C

(A) is correct but (R) is not correct

D

Both

(\mathbf{A})

and

(\mathbf{R})

are correct but

(\mathbf{R})

is not the correct explanation of

(\mathbf{A})

Official Correct Answer:Option (C)

Step-by-Step Solution:

The correct answer is Option C: (A) is correct but (R) is not correct.

Assertion (A): $\mathrm{PH}_3$ has lower boiling point than $\mathrm{NH}_3$ .

This assertion is true. The boiling point of ammonia ( $\mathrm{NH}_3$ ) is higher than that of phosphine ( $\mathrm{PH}_3$ ). Ammonia has a boiling point of about -33.34°C, whereas phosphine has a boiling point of about -87.7°C. The reason for the higher boiling point of ammonia as compared to phosphine lies in the strength and type of intermolecular forces present in the substances.

Reason (R) : In liquid state $\mathrm{NH}_3$ molecules are associated through Vander Waals' forces, but $\mathrm{PH}_3$ molecules are associated through hydrogen bonding.

This reason is incorrect. Ammonia ( $\mathrm{NH}_3$ ) can form hydrogen bonds due to the presence of a highly electronegative nitrogen atom bonded to hydrogen atoms. This allows $\mathrm{NH}_3$ molecules to strongly associate with each other through hydrogen bonding, which is a much stronger intermolecular force than Van der Waals forces. This hydrogen bonding is responsible for the relatively high boiling point of ammonia.

On the other hand, phosphine ( $\mathrm{PH}_3$ ) does not form hydrogen bonds because the electronegativity difference between phosphorus and hydrogen is not significant enough to enable the formation of hydrogen bonds. Instead, phosphine molecules are associated mainly through weaker Van der Waals forces, leading to a lower boiling point when compared to ammonia.

In conclusion, the assertion that $\mathrm{PH}_3$ has a lower boiling point than $\mathrm{NH}_3$ is correct, due to the hydrogen bonding present in $\mathrm{NH}_3$ and absent in $\mathrm{PH}_3$ . However, the reason provided is incorrect because it incorrectly states that $\mathrm{PH}_3$ forms hydrogen bonds and that $\mathrm{NH}_3$ is associated through Van der Waals forces. It is actually the other way around, so the correct answer is that (A) is true but (R) is false.

Q 44ChemistrySection A • MCQhaloalkanes and haloarenes

+4 / -1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Haloalkanes react with KCN to form alkyl cyanides as a main product while with

\mathrm{AgCN}

form isocyanide as the main product.

Reason (R) :

\mathrm{KCN}

and

\mathrm{AgCN}

both are highly ionic compounds.

In the light of the above statements, choose the most appropriate answer from the options given below :

A

(A) is correct but

(\mathbf{R})

is not correct

B

Both (A) and (R) are correct and (R) is the correct explanation of (A)

C

(A) is not correct but (R) is correct

D

Both (A) and (R) are correct but (R) is not the correct explanation of (A)

Q 45ChemistrySection A • MCQatomic structure

+4 / -1

According to the wave-particle duality of matter by de-Broglie, which of the following graph plot presents most appropriate relationship between wavelength of electron

(\lambda)

and momentum of electron

(p)

?

A

B

C

D

Q 46ChemistrySection B • Numericalp block elements

+4 / -1

Among the following oxides of p-block elements, number of oxides having amphoteric nature is ________.

\mathrm{Cl}_2 \mathrm{O}_7, \mathrm{CO}, \mathrm{PbO}_2, \mathrm{~N}_2 \mathrm{O}, \mathrm{NO}, \mathrm{Al}_2 \mathrm{O}_3, \mathrm{SiO}_2, \mathrm{~N}_2 \mathrm{O}_5, \mathrm{SnO}_2

Answer Format:Enter numerical value rounded to the nearest integer

Q 47ChemistrySection A • MCQsolutions

+4 / -1

We have three aqueous solutions of

\mathrm{NaCl}

labelled as '

\mathrm{A}

', '

\mathrm{B}

' and '

\mathrm{C}

' with concentration

0.1 \mathrm{M}

,

0.01 \mathrm{M}

and

0.001 \mathrm{M}

, respectively. The value of van 't Hoff factor(i) for these solutions will be in the order :

A

\mathrm{i}_{\mathrm{A}}<\mathrm{i}_{\mathrm{C}}<\mathrm{i}_{\mathrm{B}}

B

\mathrm{i}_{\mathrm{A}}<\mathrm{i}_{\mathrm{B}}<\mathrm{i}_{\mathrm{C}}

C

\mathrm{i}_{\mathrm{A}}>\mathrm{i}_{\mathrm{B}}>\mathrm{i}_{\mathrm{C}}

D

\mathrm{i}_{\mathrm{A}}=\mathrm{i}_{\mathrm{B}}=\mathrm{i}_{\mathrm{C}}

Official Correct Answer:Option (B)

Step-by-Step Solution:

Salt	Values of i (for different conc. of a Salt)
NaCl	0.1 M	0.01 M	0.001 M
	1.87	1.94	1.94

The van 't Hoff factor (i) is used to describe the number of particles a solute formula unit produces in a solution. For an electrolyte like $\mathrm{NaCl}$ , which dissociates completely in very dilute solutions, the theoretical value of $i$ is approximately 2, since $\mathrm{NaCl}$ dissociates into $\mathrm{Na}^+$ and $\mathrm{Cl}^-$ ions.

In real scenarios, as the concentration of the solution decreases (making the solution more dilute), the interaction between the ions decreases, allowing more complete dissociation. Therefore, for practical purposes, the van 't Hoff factor $i$ approaches its theoretical maximum value as concentration decreases. Thus, for $\mathrm{NaCl}$ solutions of concentrations $0.1 \mathrm{M}$ , $0.01 \mathrm{M}$ , and $0.001 \mathrm{M}$ , the fact that $\mathrm{NaCl}$ dissociates more completely in more dilute solutions implies that $i$ increases with decreasing concentration.

Hence, the order of $i$ based on the concentration would be $\mathrm{i}_{\mathrm{A}} < \mathrm{i}_{\mathrm{B}} < \mathrm{i}_{\mathrm{C}}$ , as concentration $0.1 \mathrm{M} > 0.01 \mathrm{M} > 0.001 \mathrm{M}$ , respectively. Thus, option B correctly describes the order of the van 't Hoff factors for these solutions.

Q 48ChemistrySection A • MCQbasics of organic chemistry

+4 / -1

Which of the following compound will most easily be attacked by an electrophile?

A

B

C

D

Q 49ChemistrySection A • MCQpractical organic chemistry

+4 / -1

Given below are two statements :

Statement (I) : Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution.

Statement (II) : In this titration phenolphthalein can be used as indicator.

In the light of the above statements, choose the most appropriate answer from the options given below :

A

Both Statement I and Statement II are correct

B

Statement I is incorrect but Statement II is correct

C

Both Statement I and Statement II are incorrect

D

Statement I is correct but Statement II is incorrect

Q 50ChemistrySection A • MCQredox reactions

+4 / -1

Which of the following reactions are disproportionation reactions?

(A)

\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}

(B)

3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 <br><br>\mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}

(C)

2 \mathrm{KMnO}_4 \longrightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2

(D)

2 \mathrm{MnO}_4^{-}+3 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 5 \mathrm{MnO}_2+4 \mathrm{H}^{+}

Choose the correct answer from the options given below :

A

(A), (B)

B

(A), (D)

C

(B), (C), (D)

D

(A), (B), (C)

Official Correct Answer:Option (A)

Step-by-Step Solution:

A disproportionation reaction is a chemical reaction in which a single substance is simultaneously reduced and oxidized, forming two different products. To identify disproportionation reactions, one should look for a specified element in the reactant that gains electrons (reduction) and for the same element that lose electrons (oxidation).

Here's an analysis of each reaction :

(A) $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$

Here, copper (Cu) is in the +1 oxidation state and forms two products, where it is oxidized to the +2 state in $\mathrm{Cu}^{2+}$ and reduced to the 0 state in Cu. This is a classic example of disproportionation. So, (A) is true for disproportionation.

(B) $3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$

In this reaction, $\mathrm{MnO}_4^{2-}$ (where Mn is in +6 oxidation state) is converted into $\mathrm{MnO}_4^{-}$ (where Mn is in +7 oxidation state, oxidation has occurred) and $\mathrm{MnO}_2$ (where Mn is in +4 oxidation state, reduction has occurred). Here, again, we have the same element, manganese (Mn), undergoing both reduction and oxidation. Hence, (B) is also a disproportionation reaction.

(C) $2 \mathrm{KMnO}_4 \longrightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2$

For the reaction with potassium permanganate, $\mathrm{KMnO}_4$ , manganese starts in the +7 oxidation state. It forms $\mathrm{K}_2 \mathrm{MnO}_4$ where Mn is in +6 state (reduction) and $\mathrm{MnO}_2$ where Mn is in +4 state (further reduction), but no increase in oxidation state of manganese is observed. Instead, oxygen is released, so the manganese is not being oxidized in any of its products; it is only reduced twice. This is not a disproportionation reaction because the same element should be undergoing both oxidation and reduction, which is not the case here. Hence, (C) is not a disproportionation reaction.

(D) $2 \mathrm{MnO}_4^{-}+3 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 5 \mathrm{MnO}_2+4 \mathrm{H}^{+}$

In this reaction, the $\mathrm{MnO}_4^{-}$ ion (where Mn is at +7 oxidation state) and $\mathrm{Mn}^{2+}$ ion (where Mn is at +2 oxidation state) react to form $\mathrm{MnO}_2$ (where Mn is at +4 oxidation state). This reaction is not a case of disproportionation; it is a comproportionation or synproportionation reaction, where two species with the same element in different oxidation states produce a product with the element at an intermediate oxidation state.

Therefore, the disproportionation reactions among the given choices are:

(A) $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$

(B) $3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$

Option A would be the correct choice: (A), (B).

Q 51ChemistrySection B • Numericalbasics of organic chemistry

+4 / -1

Number of optical isomers possible for 2-chlorobutane ________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 52ChemistrySection B • Numericalsome basic concepts of chemistry

+4 / -1

Consider the following reaction :

3 \mathrm{PbCl}_2+2\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \rightarrow \mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2+6 \mathrm{NH}_4 \mathrm{Cl}

If

72 ~\mathrm{mmol}

of

\mathrm{PbCl}_2

is mixed with

50 ~\mathrm{mmol}

of

\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4

, then the amount of

\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2

formed is ________ mmol (nearest integer).

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:24

Step-by-Step Solution:

To solve this problem, we will use the stoichiometry of the balanced chemical reaction. The balanced equation shows that 3 moles of $\mathrm{PbCl}_2$ react with 2 moles of $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$ to produce 1 mole of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ .

The reaction is:

3 \mathrm{PbCl}_2 + 2\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \rightarrow \mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2 + 6 \mathrm{NH}_4 \mathrm{Cl}

The molar ratio of $\mathrm{PbCl}_2$ to $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ is 3:1, and the molar ratio of $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$ to $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ is 2:1. We need to determine which reactant is the limiting reagent because it will dictate the amount of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ produced.

The stoichiometric calculations are as follows:

For $\mathrm{PbCl}_2$ :

\text{Moles of }\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2\text{ formed from } \mathrm{PbCl}_2 = \frac{72 \text{ mmol of } \mathrm{PbCl}_2}{3 \text{ mmol of } \mathrm{PbCl}_2\text{/mmol of } \mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2} = 24 \text{ mmol}

For $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$ :

\text{Moles of }\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2\text{ formed from } \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 = \frac{50 \text{ mmol of } \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4}{2 \text{ mmol of } \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4\text{/mmol of } \mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2} = 25 \text{ mmol}

Now we can identify the limiting reagent by comparing the two amounts of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ that could be produced. The smaller quantity will be the actual amount produced since the limiting reagent restricts the reaction.

Since the $\mathrm{PbCl}_2$ can produce only 24 mmol of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ versus the 25 mmol that could be produced by $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$ , $\mathrm{PbCl}_2$ is the limiting reagent.

Therefore, the amount of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ formed is 24 mmol (as a nearest integer).

Q 53ChemistrySection A • MCQcompounds containing nitrogen

+4 / -1

Given below are two statements :

Statement (I) : The

\mathrm{NH}_2

group in Aniline is ortho and para directing and a powerful activating group.

Statement (II) : Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation).

In the light of the above statements, choose the most appropriate answer from the options given below :

A

Both Statement I and Statement II are correct

B

Both Statement I and Statement II are incorrect

C

Statement I is correct but Statement II is incorrect

D

Statement I is incorrect but Statement II is correct

Q 54ChemistrySection A • MCQaldehydes ketones and carboxylic acids

+4 / -1

Identify

A

and

B

in the following sequence of reaction

qImage-preparation-methods-for-both-aldehydes-and-ketones-aldehydes-ketones-and-carboxylic-acids-organic-chemistry-Jeemate

qImage-preparation-methods-for-both-aldehydes-and-ketones-aldehydes-ketones-and-carboxylic-acids-organic-chemistry-Jeemate

A

B

C

D

Q 55ChemistrySection A • MCQpractical organic chemistry

+4 / -1

In Kjeldahl's method for estimation of nitrogen,

\mathrm{CuSO}_4

acts as :

A

catalytic agent

B

hydrolysis agent

C

reducing agent

D

oxidising agent

Q 56ChemistrySection B • Numericalelectrochemistry

+4 / -1

The potential for the given half cell at

298 \mathrm{~K}

is (-) __________

\times 10^{-2} \mathrm{~V}

\begin{aligned} & 2 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2(\mathrm{~g}) \\ & {\left[\mathrm{H}^{+}\right]=1 \mathrm{M}, \mathrm{P}_{\mathrm{H}_2}=2 \mathrm{~atm}} \end{aligned}

(Given :

2.303 \mathrm{RT} / \mathrm{F}=0.06 \mathrm{~V}, \log 2=0.3

)

Answer Format:Enter numerical value rounded to the nearest integer

Q 57ChemistrySection A • MCQthermodynamics

+4 / -1

Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following :

A

\mathrm{q}=0, \Delta \mathrm{T}=0, \mathrm{w}=0

B

\mathrm{q}=0, \Delta \mathrm{T} \neq 0, \mathrm{w}=0

C

\mathrm{q} \neq 0, \Delta \mathrm{T}=0, \mathrm{w}=0

D

\mathrm{q}=0, \Delta \mathrm{T}<0, \mathrm{w} \neq 0

Q 58ChemistrySection B • Numericalchemical kinetics and nuclear chemistry

+4 / -1

The ratio of

\frac{{ }^{14} \mathrm{C}}{{ }^{12} \mathrm{C}}

in a piece of wood is

\frac{1}{8}

part that of atmosphere. If half life of

{ }^{14} \mathrm{C}

is 5730 years, the age of wood sample is ________ years.

Answer Format:Enter numerical value rounded to the nearest integer

Q 59ChemistrySection B • Numericalchemical bonding and molecular structure

+4 / -1

The number of molecules/ion/s having trigonal bipyramidal shape is _______.

\mathrm{PF}_5, \mathrm{BrF}_5, \mathrm{PCl}_5,\left[\mathrm{Pt} \mathrm{Cl}_4\right]^{2-}, \mathrm{BF}_3, \mathrm{Fe}(\mathrm{CO})_5

Answer Format:Enter numerical value rounded to the nearest integer

Q 60ChemistrySection A • MCQaldehydes ketones and carboxylic acids

+4 / -1

Match List - I with List - II.

List I (Reactions)	List II (Reagents)
(A)	(I) CH₃MgBr, H₂O
(B) C₆H₅COC₆H₅ ⟶ C₆H₅CH=C₆H₅	(II) Zn(Hg) and conc. HCl
(C) C₆H₅CHO ⟶ C₆H₅CH(OH)CH₃	(III) NaBH₄, H⁺
(D)	(IV) DIBAL-H, H₂O

Choose the correct answer from the options given below :

A

(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

B

(A)-(IV), (B)-(II), (C)-(III), (D)-(I)

C

(A)-(IV), (B)-(II), (C)-(I), (D)-(III)

D

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Q 61MathematicsSection A • MCQmatrices and determinants

+4 / -1

If the system of equations

\begin{aligned} & 2 x+3 y-z=5 \\ & x+\alpha y+3 z=-4 \\ & 3 x-y+\beta z=7 \end{aligned}

has infinitely many solutions, then

13 \alpha \beta

is equal to :

A

1110

B

1120

C

1210

D

1220

Q 62MathematicsSection A • MCQtrigonometric ratios and identities

+4 / -1

If

\tan \mathrm{A}=\frac{1}{\sqrt{x\left(x^2+x+1\right)}}, \tan \mathrm{B}=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}

and

\tan \mathrm{C}=\left(x^{-3}+x^{-2}+x^{-1}\right)^{1 / 2}, 0<\mathrm{A}, \mathrm{B}, \mathrm{C}<\frac{\pi}{2}

, then

\mathrm{A}+\mathrm{B}

is equal to :

A

\mathrm{C}

B

\pi-C

C

2 \pi-C

D

\frac{\pi}{2}-\mathrm{C}

Q 63MathematicsSection A • MCQarea under the curves

+4 / -1

The area enclosed by the curves

x y+4 y=16

and

x+y=6

is equal to :

A

28-30 \log _{\mathrm{e}} 2

B

30-28 \log _{\mathrm{e}} 2

C

30-32 \log _{\mathrm{e}} 2

D

32-30 \log _{\mathrm{e}} 2

Official Correct Answer:Option (C)

Step-by-Step Solution:

To find the enclosed area between the two curves $x y+4 y=16$ and $x+y=6$ , we need to determine the region of intersection and integrate the difference of the functions over the interval where they intersect.

First, let's solve the equations simultaneously to find the points of intersection.

The second curve $x+y=6$ can be rewritten as $y=6-x$ .

Substitute $y=6-x$ into the first equation:

$x(6-x) + 4(6-x) = 16$

$6x - x^2 + 24 - 4x = 16$

$-x^2 + 2x + 8 = 0$

This can be simplified to:

$x^2 - 2x - 8 = 0$

Using the quadratic formula, we can find the roots of this equation:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}$

$x = \frac{2 \pm \sqrt{4 + 32}}{2}$

$x = \frac{2 \pm \sqrt{36}}{2}$

$x = \frac{2 \pm 6}{2}$

So we have two solutions:

$x_1 = \frac{2 + 6}{2} = 4$

$x_2 = \frac{2 - 6}{2} = -2$

For $x_1 = 4$ , substitute this back into the line equation $x+y=6$ :

$4+y=6$

$y=2$

For $x_2 = -2$ , do the same:

$-2+y=6$

$y=8$

So we have two points of intersection: $(4,2)$ and $(-2,8)$ .

qImage-area-bounded-between-the-curves-area-under-the-curves-calculus-Jeemate

Now, to find the area between $x y + 4y = 16$ and $x + y = 6$ , we'll integrate the top function minus the bottom function from $x = -2$ to $x = 4$ . First, we need to express $y$ from both equations in terms of $x$ :

For $x y + 4 y = 16$ , express $y$ in terms of $x$ :

$y(x+4) = 16$

$y = \frac{16}{x+4}$

For $x + y = 6$ , we have:

$y = 6 - x$

The area $A$ is therefore given by:

$A = \int\limits_{-2}^{4}\left( 6 - x - \frac{16}{x+4} \right) dx$

Now, we integrate:

$A = \int\limits_{-2}^{4} (6 - x) dx - \int_{-2}^{4} \frac{16}{x + 4} dx$

$A = \left[6x - \frac{x^2}{2}\right]_{-2}^{4} - \left[16 \ln|x + 4|\right]_{-2}^{4}$

$A = \left( 6 \times 4 - \frac{1}{2} \times 4^2 \right) - \left( 6 \times -2 - \frac{1}{2} \times -2^2 \right) - 16 \left( \ln|4+4| - \ln|-2+4| \right)$

$A = \left( 24 - 8 \right) - \left( -12 - 2 \right) - 16 \left( \ln 8 - \ln 2 \right)$

$A = 16 + 14 - 16 \left( \ln (2^3) - \ln 2 \right)$

$A = 30 - 16 \left( 3 \ln 2 - \ln 2 \right)$

$A = 30 - 16 (2 \ln 2)$

$A = 30- 32 \ln 2$

Q 64MathematicsSection A • MCQlimits continuity and differentiability

+4 / -1

Let

f: \mathbf{R} \rightarrow \mathbf{R}

be defined as :

f(x)= \begin{cases}\frac{a-b \cos 2 x}{x^2} ; & x<0 \\ x^2+c x+2 ; & 0 \leq x \leq 1 \\ 2 x+1 ; & x>1\end{cases}

If

f

is continuous everywhere in

\mathbf{R}

and

m

is the number of points where

f

is NOT differential then

\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}

equals :

A

1

B

4

C

3

D

2

Official Correct Answer:Option (D)

Step-by-Step Solution:

At

\mathrm{x}=1, \mathrm{f}(\mathrm{x})

is continuous therefore,

\mathrm{f}\left(1^{-}\right)=\mathrm{f}(1)=\mathrm{f}\left(1^{+}\right)

f(1)=3+c

.........(1)

\begin{aligned} & \mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 2(1+\mathrm{h})+1 \\ & \mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 3+2 \mathrm{~h}=3 .........(2) \end{aligned}

from (1) and (2)

\mathrm{c}=0

at

\mathrm{x}=0, \mathrm{f}(\mathrm{x})

is continuous therefore,

\begin{aligned} & \mathrm{f}\left(0^{-}\right)=\mathrm{f}(0)=\mathrm{f}\left(0^{+}\right) ........(3) \\ & \mathrm{f}(0)=\mathrm{f}\left(0^{+}\right)=2 ...........(4) \end{aligned}

\mathrm{f}\left(0^{-}\right)

has to be equal to 2

\lim\limits_{h \rightarrow 0} \frac{a-b \cos (2 h)}{h^2}

=

\lim\limits_{h \rightarrow 0} \frac{a-b\left\{1-\frac{4 h^2}{2 !}+\frac{16 h^4}{4 !}+\ldots\right\}}{h^2}

=

\lim\limits_{h \rightarrow 0} \frac{a-b+b\left\{2 h^2-\frac{2}{3} h^4 \ldots\right\}}{h^2}

for limit to exist

a-b=0

and limit is

2 b

from (3), (4) and (5)

\mathrm{a}=\mathrm{b}=1

checking differentiability at

\mathrm{x}=0

\begin{aligned} & \text { LHD : } \lim _{h \rightarrow 0} \frac{\frac{1-\cos 2 h}{h^2}-2}{-h} \\ & \lim _{h \rightarrow 0} \frac{1-\left(1-\frac{4 h^2}{2 !}+\frac{16 h^4}{4 !} \ldots\right)-2 h^2}{-h^3}=0 \\ & \text { RHD : } \lim _{h \rightarrow 0} \frac{(0+h)^2+2-2}{h}=0 \end{aligned}

Function is differentiable at every point in its domain

\therefore \mathrm{m}=0

m + a + b + c = 0 + 1 + 1 + 0 = 2

Q 65MathematicsSection A • MCQhyperbola

+4 / -1

For

0<\theta<\pi / 2

, if the eccentricity of the hyperbola

x^2-y^2 \operatorname{cosec}^2 \theta=5

is

\sqrt{7}

times eccentricity of the

ellipse

x^2 \operatorname{cosec}^2 \theta+y^2=5

, then the value of

\theta

is :

A

\frac{\pi}{6}

B

\frac{5 \pi}{12}

C

\frac{\pi}{3}

D

\frac{\pi}{4}

Official Correct Answer:Option (C)

Step-by-Step Solution:

To find the value of $\theta$ , we need to determine the relationship between the eccentricities of the given hyperbola and ellipse. Let's start by writing down the standard forms of ellipse and hyperbola and then relate them to the given equations.

The standard form of an ellipse is:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

The eccentricity $e$ of an ellipse is given by:

$e = \sqrt{1 - \frac{b^2}{a^2}}\ \text{for}\ a > b$

For the given ellipse:

$x^2 \operatorname{cosec}^2 \theta + y^2 = 5$

We can compare it with the standard form by writing it as:

$\frac{x^2}{\frac{5}{\operatorname{cosec}^2 \theta}} + \frac{y^2}{5} = 1$

From this we have $a^2 = \frac{5}{\operatorname{cosec}^2 \theta}$ and $b^2 = 5$ . The eccentricity of the ellipse (let's call it $e_1$ ) is:

$e_1 = \sqrt{1 - \frac{5}{a^2}} = \sqrt{1 - \frac{5}{\frac{5}{\operatorname{cosec}^2 \theta}}} = \sqrt{1 - \operatorname{sin}^2 \theta} = \operatorname{cos} \theta$

The standard form of a hyperbola is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

The eccentricity $e$ of a hyperbola is given by:

$e = \sqrt{1 + \frac{b^2}{a^2}}$

For the given hyperbola:

$x^2 - y^2 \operatorname{cosec}^2 \theta = 5$

We can rewrite it as:

$\frac{x^2}{5} - \frac{y^2}{\frac{5}{\operatorname{cosec}^2 \theta}} = 1$

From this we have $a^2 = 5$ and $b^2 = \frac{5}{\operatorname{cosec}^2 \theta}$ . The eccentricity of the hyperbola (let's call it $e_2$ ) is:

$e_2 = \sqrt{1 + \frac{\frac{5}{\operatorname{cosec}^2 \theta}}{5}} = \sqrt{1 + \operatorname{sin}^2 \theta}$

According to the question the eccentricity of the hyperbola is $\sqrt{7}$ times the eccentricity of the ellipse, so we can write:

$e_2 = \sqrt{7} \cdot e_1$

Substitute $e_1$ and $e_2$ from the above:

$\sqrt{1 + \operatorname{sin}^2 \theta} = \sqrt{7} \cdot \operatorname{cos} \theta$

Square both sides to eliminate the square root:

$1 + \operatorname{sin}^2 \theta = 7 \cdot \operatorname{cos}^2 \theta$

$1 + \operatorname{sin}^2 \theta = 7 \cdot (1 - \operatorname{sin}^2 \theta)$

$1 + \operatorname{sin}^2 \theta = 7 - 7 \cdot \operatorname{sin}^2 \theta$

$8 \cdot \operatorname{sin}^2 \theta = 6$

$\operatorname{sin}^2 \theta = \frac{6}{8} = \frac{3}{4}$

$\operatorname{sin} \theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

Looking at the options provided and knowing the sine values for common angles, we can see that $\operatorname{sin} \theta = \frac{\sqrt{3}}{2}$ corresponds to $\theta = \frac{\pi}{3}$ .

The correct answer is:

Option C.

$\frac{\pi}{3}$

Q 66MathematicsSection A • MCQ3d geometry

+4 / -1

If the shortest distance between the lines

\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}

and

\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}

is 1 , then the sum of all possible values of

\lambda

is :

A

0

B

2 \sqrt{3}

C

3 \sqrt{3}

D

-2 \sqrt{3}

Official Correct Answer:Option (B)

Step-by-Step Solution:

Given the two lines:

$L_1: \frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1}$

$L_2: \frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1}$

We observe that these lines are not parallel as their directional vectors are not proportional. The directional vector for $L_1$ is $(-2,1,1)$ and for $L_2$ is $(1,-2,1)$ . The shortest distance between two skew (non-intersecting and non-parallel) lines in the three-dimensional space is along the line that is perpendicular to both lines. This implies we can find a vector that is perpendicular to both directional vectors by taking their cross product.

The directional vector for $L_1$ is $d_1 = \langle -2, 1, 1 \rangle$ , and for $L_2$ is $d_2 = \langle 1, -2, 1 \rangle$ . The cross product of $d_1$ and $d_2$ , which will be perpendicular to both lines, is given by:

$d = d_1 \times d_2 = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \\ \end{matrix} \right|$

Expanding the determinate gives:

$d = \mathbf{i}((1)(1) - (1)(-2)) - \mathbf{j}((-2)(1) - (1)(1)) + \mathbf{k}((-2)(-2) - (1)(1))$

$d = \mathbf{i}(1 + 2) - \mathbf{j}(-2 - 1) + \mathbf{k}(4 - 1)$

$d = 3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}$

$d = \langle 3, 3, 3 \rangle$

The shortest distance $D$ between the two lines can then be given by the formula:

$D = \frac{\left| (\mathbf{a_2} - \mathbf{a_1}) \cdot d \right|}{\|d\|}$

Where $\mathbf{a_1}$ and $\mathbf{a_2}$ are position vectors to any points on line $L_1$ and line $L_2$ , respectively, and ' $\cdot$ ' denotes the dot product.

From the lines' equations, we can choose a point on each line (when the parameter is zero). Thus, for $L_1$ , let's choose the point $A(\lambda, 2, 1)$ , and for $L_2$ , let's choose the point $B(\sqrt{3}, 1, 2)$ . These points correspond to the vectors $\mathbf{a_1} = \langle \lambda, 2, 1 \rangle$ and $\mathbf{a_2} = \langle \sqrt{3}, 1, 2 \rangle$ , respectively.

The vector $\mathbf{a_2} - \mathbf{a_1}$ is:

$\mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3}, 1, 2 \rangle - \langle \lambda, 2, 1 \rangle$

$\mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3} - \lambda, 1 - 2, 2 - 1 \rangle$

$\mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3} - \lambda, -1, 1 \rangle$

We can now substitute this, along with $d$ , into the distance formula:

$D = \frac{\left| (\langle \sqrt{3} - \lambda, -1, 1 \rangle) \cdot \langle 3, 3, 3 \rangle \right|}{\|\langle 3, 3, 3 \rangle\|}$

$D = \frac{\left| 3(\sqrt{3} - \lambda) + 3(-1) + 3(1) \right|}{\sqrt{3^2 + 3^2 + 3^2}}$

$D = \frac{\left| 3\sqrt{3} - 3\lambda - 3 + 3 \right|}{\sqrt{27}}$

$D = \frac{\left| 3\sqrt{3} - 3\lambda \right|}{3\sqrt{3}}$

$D = \frac{\left| \sqrt{3} - \lambda \right|}{\sqrt{3}}$

Given that the shortest distance $D$ between the lines is 1, we can equate the above result to 1, and solve for $\lambda$ :

$\frac{\left| \sqrt{3} - \lambda \right|}{\sqrt{3}} = 1$

$\left| \sqrt{3} - \lambda \right| = \sqrt{3}$

This absolute value equation gives us two possible cases:

Case 1: $\sqrt{3} - \lambda = \sqrt{3}$ , which gives $\lambda = 0$ .

Case 2: $\sqrt{3} - \lambda = -\sqrt{3}$ , which gives $\lambda = 2\sqrt{3}$ .

Therefore, the sum of all possible values of $\lambda$ is:

$\lambda_{sum} = \lambda_1 + \lambda_2 = 0 + 2\sqrt{3} = 2\sqrt{3}$

Hence, option B ( $2 \sqrt{3}$ ) is the correct answer.

Q 67MathematicsSection A • MCQquadratic equation and inequalities

+4 / -1

Let

\mathbf{S}=\left\{x \in \mathbf{R}:(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=10\right\}

. Then the number of elements in

\mathrm{S}

is :

A

4

B

0

C

2

D

1

Q 68MathematicsSection A • MCQdefinite integration

+4 / -1

The value of the integral

\int\limits_0^{\pi / 4} \frac{x \mathrm{~d} x}{\sin ^4(2 x)+\cos ^4(2 x)}

equals :

A

\frac{\sqrt{2} \pi^2}{8}

B

\frac{\sqrt{2} \pi^2}{16}

C

\frac{\sqrt{2} \pi^2}{32}

D

\frac{\sqrt{2} \pi^2}{64}

Official Correct Answer:Option (C)

Step-by-Step Solution:

Take

I=\int\limits_0^{\pi / 4} \frac{x d x}{\sin ^4(2 x)+\cos ^4(2 x)}

Let

2 x=t

2 d x=d t

d x=\frac{d t}{2}

\begin{aligned} & I=\int\limits_0^{\pi / 2} \frac{t / 2 \cdot 1 / 2 d t}{\sin ^4 t+\cos ^4 t} \\ & I=\frac{1}{4} \int\limits_0^{\pi / 2} \frac{t d t}{\sin ^4 t+\cos ^4 t} \\ & =\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-t\right) d t}{\sin ^4(\pi / 2-t)+\cos ^4(\pi / 2-t)} \\ & =\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-t\right)}{\sin ^4 t+\cos ^4 t}\end{aligned}

\begin{aligned} & 2 I=\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\frac{\pi}{2}}{\sin ^4 t+\cos ^4 t} d t \\ & 2 I=\frac{\pi}{8} \int\limits_0^{\pi / 2} \frac{d t}{\sin ^4 t+\cos ^4 t} \\ & 2 I=\frac{\pi}{8} \int\limits_0^{\pi / 2} \frac{\sec ^4 t}{1+\tan ^4 t} d t\end{aligned}

\begin{aligned} & \text { Put tant }=y \\ & \sec ^2 t d t=d y \\ & 2 I=\frac{\pi}{8} \int\limits_0^{\infty} \frac{\left(1+y^2\right) d y}{1+y^4} \\ & =\frac{\pi}{8} \int\limits_0^{\infty} \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}-2+2} d y \\ & = \frac{\pi}{8} \int\limits_0^{\infty} \frac{\left(1+\frac{1}{y^2}\right) d y}{2+\left(y-\frac{1}{y}\right)^2}\end{aligned}

\begin{aligned} & \text { Put, } y-\frac{1}{y}=4 \\ & 2 I=\frac{\pi}{8} \int\limits_{-\infty}^{\infty} \frac{d u}{2+u^2} \\ & =\frac{\pi}{8 \sqrt{2}}\left[\tan ^{-1} \frac{y}{\sqrt{2}}\right]_{-\infty}^{\infty} \\ & =\frac{\sqrt{2} \pi^2}{32}\end{aligned}

Q 69MathematicsSection B • Numericaldifferential equations

+4 / -1

If

x=x(t)

is the solution of the differential equation

(t+1) \mathrm{d} x=\left(2 x+(t+1)^4\right) \mathrm{dt}, x(0)=2

, then,

x(1)

equals _________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 70MathematicsSection A • MCQcircle

+4 / -1

Let

C: x^2+y^2=4

and

C^{\prime}: x^2+y^2-4 \lambda x+9=0

be two circles. If the set of all values of

\lambda

so that the circles

\mathrm{C}

and

\mathrm{C}

intersect at two distinct points, is

\mathrm{R}-[\mathrm{a}, \mathrm{b}]

, then the point

(8 \mathrm{a}+12,16 \mathrm{~b}-20)

lies on the curve :

A

x^2+2 y^2-5 x+6 y=3

B

5 x^2-y=-11

C

x^2-4 y^2=7

D

6 x^2+y^2=42

Official Correct Answer:Option (D)

Step-by-Step Solution:

\begin{aligned} & C: x^2+y^2=4 \Rightarrow C(0,0), r_1=2 \\ & C^{\prime}: x^2+y^2-4 \lambda x+9=0 \Rightarrow C^{\prime}(2 \lambda, 0), r_2=\sqrt{4 \lambda^2-9} \\ & \left|r_1-r_2\right| < C C^{\prime} < \left|r_1+r_2\right| \\ & \left|2-\sqrt{4 \lambda^2-9}\right|<|2 \lambda|<2+\sqrt{4 \lambda^2-9} \\ & |2 \lambda|>\left|2-\sqrt{4 \lambda^2-9}\right| \\ & \Rightarrow 4 \lambda^2 > 4+4 \lambda^2-9-4 \sqrt{4 \lambda^2-9} \\ & 4 \sqrt{4 \lambda^2-9}+5>0 \Rightarrow \lambda \in R\end{aligned}

\begin{aligned} & |2 \lambda|<2+\sqrt{4 \lambda^2-9} \\ & \Rightarrow 4 \lambda^2<4+4 \lambda^2-9+4 \sqrt{\left(4 \lambda^2\right)-9} \\ & 5<4 \sqrt{4 \lambda^2-9} \text { and } \lambda^2 \geq \frac{9}{4}\end{aligned}

\begin{aligned} & \frac{25}{16} < 4 \lambda^2-9 \\ & \Rightarrow \lambda^2>\frac{169}{64} \\ & \lambda \in\left(-\infty, \frac{-13}{8}\right) \cup\left(\frac{13}{8}, \infty\right) \\ & \lambda \in R-\left[\frac{-13}{8}, \frac{13}{8}\right] \\ & a=\frac{-13}{8}, b=\frac{13}{8} \\ & \Rightarrow(8 a+12,16 b-20)=(-1,6) \\ & \Rightarrow 6(-1)^2+(6)^2=42\end{aligned}

Q 71MathematicsSection A • MCQcomplex numbers

+4 / -1

Let

\mathrm{S}=|\mathrm{z} \in \mathrm{C}:| z-1 \mid=1

and

(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2} \mid

. Let

z_1, z_2 \in \mathrm{S}

be such that

\left|z_1\right|=\max\limits_{z \in s}|z|

and

\left|z_2\right|=\min\limits _{z \in S}|z|

. Then

\left|\sqrt{2} z_1-z_2\right|^2

equals :

A

1

B

4

C

3

D

2

Official Correct Answer:Option (D)

Step-by-Step Solution:

Let

z=x+i y

\begin{aligned} & |z-1|=1 \Rightarrow|x+i y-1|=1 \\ & (x-1)^2+y^2=1 .......(1) \end{aligned}

\begin{aligned} & (\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2} \text { (Given) } \\ & (\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2} \\ & (\sqrt{2}-1) x+y=\sqrt{2} .........(2)\end{aligned}

Solving (1) and (2), we get

\begin{aligned} & (x-1)^2+(\sqrt{2}-(\sqrt{2}-1) x)^2=1 \\ & \left(x^2-2 x+1\right)+2+(\sqrt{2}-1)^2 x^2-2 \sqrt{2}(\sqrt{2}-1) x=0 \\ & x^2\left(1+(\sqrt{2}-1)^2\right)+x(-2-2 \sqrt{2}(\sqrt{2}-1))+2=0 \\ & x^2(4-2 \sqrt{2})+x(2 \sqrt{2}-6)+2=0 \\ & x^2(2-\sqrt{2})+x(\sqrt{2}-3)+1=0 \end{aligned}

\begin{aligned} & \Rightarrow x=1 \text { and } x=\frac{1}{2-\sqrt{2}} ..........(3) \\ & \text { When } x=1, y=1 \Rightarrow z_2=1+i \\ & \text { When } x=\frac{1}{2-\sqrt{2}}, y=\sqrt{2}-\frac{1}{\sqrt{2}} \\ & \Rightarrow z_1=\left(1+\frac{1}{\sqrt{2}}\right)+\frac{i}{\sqrt{2}}\end{aligned}

Now,

\begin{aligned} & \left|\sqrt{2} z_1-z_2\right|^2 \\ & =\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^2 \\ & =(\sqrt{2})^2 \\ & =2 \end{aligned}

Q 72MathematicsSection A • MCQvector algebra

+4 / -1

Let

\overrightarrow{\mathrm{a}}=-5 \hat{i}+\hat{j}-3 \hat{k}, \overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}-4 \hat{k}

and

\overrightarrow{\mathrm{c}}=(((\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \hat{i}) \times \hat{i}) \times \hat{i}

. Then

\vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})

is equal to :

A

-12

B

-10

C

-13

D

-15

Q 73MathematicsSection A • MCQfunctions

+4 / -1

Let

f: \mathbf{R} \rightarrow \mathbf{R}

and

g: \mathbf{R} \rightarrow \mathbf{R}

be defined as

f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, & x>0 \\ \mathrm{e}^{-x}, & x \leq 0\end{array}\right.

and

g(x)=\left\{\begin{array}{ll}x, & x \geqslant 0 \\ \mathrm{e}^x, & x<0\end{array}\right.

. Then, gof :

\mathbf{R} \rightarrow \mathbf{R}

is :

A

one-one but not onto

B

neither one-one nor onto

C

onto but not one-one

D

both one-one and onto

Q 74MathematicsSection B • Numericalsequences and series

+4 / -1

Let

3,7,11,15, \ldots, 403

and

2,5,8,11, \ldots, 404

be two arithmetic progressions. Then the sum, of the common terms in them, is equal to ___________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:6699

Step-by-Step Solution:

To find the common terms in the two given arithmetic progressions (AP), we need to first identify the common difference for each sequence and then find the sequence that represents their overlap by employing the concept of least common multiple (LCM).

The first AP is:

$3, 7, 11, 15, \ldots, 403$

The common difference ( $d_1$ ) for the first AP can be calculated by subtracting the first term from the second term:

$d_1 = 7 - 3 = 4$

The second AP is:

$2, 5, 8, 11, \ldots, 404$

The common difference ( $d_2$ ) for the second AP is:

$d_2 = 5 - 2 = 3$

To find the terms common to both sequences, we need to find a term that appears in both sequences. Any common term must be of the form $3 + 4k$ and $2 + 3l$ for some integers $k$ and $l$ . We want to find when these two forms will give us the same number, so we set them equal to each other:

$3 + 4k = 2 + 3l$

Rearranging the terms gives us:

$4k - 3l = 2 - 3$

This simplifies to:

$4k - 3l = -1 ....... (1)$

The solutions to equation $(1)$ will give us the common terms. Notice this is a Diophantine equation (A Diophantine equation is a polynomial equation, usually with two or more variables,) and has an infinite number of solutions. Let's find one such solution. We can see that:

$k = 1 \quad \text{yields} \quad 4(1) - 3l = -1 \implies 4 - 3l = -1 \implies 3l = 5 \implies l = 1\frac{2}{3}$

This is not an integer solution for $l$ , so $k = 1$ does not work. Trying $k = 2$ gives:

$4(2) - 3l = -1 \implies 8 - 3l = -1 \implies 3l = 9 \implies l = 3$

Now we've found integers $k = 2$ and $l = 3$ that satisfy the equation. The corresponding term in both sequences would be:

$3 + 4(2) = 3 + 8 = 11 \quad \text{and} \quad 2 + 3(3) = 2 + 9 = 11$

Since $11$ is a common term, we can assert that every common term in both APs will be of the form $11 + m(4 \times 3)$ , where $m$ is a non-negative integer, and $4 \times 3 = 12$ is the LCM of the common differences of the two APs. Thus, the general form for the common terms would be:

$11 + 12m$

Now we are to find all terms that are common up to $403$ in the first sequence and up to $404$ in the second sequence. Because the first sequence doesn't exceed $403$ , we'll use this as our limit:

$11 + 12m \leq 403$

To find the largest possible integer value for $m$ , we solve the inequality:

$12m \leq 403 - 11$

$12m \leq 392$

$m \leq 32\frac{2}{3}$

Since $m$ has to be an integer, the largest possible value for $m$ is $32$ . Therefore, the common terms are generated by $m = 0, 1, 2, \ldots, 32$ . There are $32 + 1 = 33$ terms in total.

We will now sum these up. The sum of an AP is given by the formula:

$S = \frac{n}{2}(a_1 + a_n)$

Where $S$ is the sum, $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the last term. Using the formula:

$S = \frac{33}{2}(11 + (11 + 12 \times 32))$

$S = \frac{33}{2}(11 + 11 + 384)$

$S = \frac{33}{2}(406)$

$S = 33 \times 203$

$S = 6699$

Therefore, the sum of the common terms in the two arithmetic progressions is 6699.

Q 75MathematicsSection A • MCQpermutations and combinations

+4 / -1

If

\mathrm{n}

is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then

\mathrm{n}

is equal to :

A

47

B

53

C

51

D

43

Q 76MathematicsSection A • MCQdifferential equations

+4 / -1

Let

y=y(x)

be the solution of the differential equation

\frac{\mathrm{d} y}{\mathrm{~d} x}=2 x(x+y)^3-x(x+y)-1, y(0)=1

.

Then,

\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2

equals :

A

\frac{4}{4+\sqrt{\mathrm{e}}}

B

\frac{3}{3-\sqrt{\mathrm{e}}}

C

\frac{2}{1+\sqrt{\mathrm{e}}}

D

\frac{1}{2-\sqrt{\mathrm{e}}}

Q 77MathematicsSection B • Numericalpermutations and combinations

+4 / -1

The number of elements in the set

\mathrm{S}=\{(x, y, z): x, y, z \in \mathbf{Z}, x+2 y+3 z=42, x, y, z \geqslant 0\}

equals __________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 78MathematicsSection B • Numericalbinomial theorem

+4 / -1

If the Coefficient of

x^{30}

in the expansion of

\left(1+\frac{1}{x}\right)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 ; x \neq 0

is

\alpha

, then

|\alpha|

equals ___________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:678

Step-by-Step Solution:

\begin{aligned} & \text { Coefficient of } x^{30} \text { in } \frac{(1+x)^6\left(1+x^2\right)^7\left(1-x^3\right)^8}{x^6} \\ & \Rightarrow \text { Coefficient of } x^{36} \text { in }(1+x)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 \\ & \Rightarrow \text { General term }={ }^6 C_{r_1}{ }^7 C_{r_2}{ }^8 C_{r_3}(-1)^{r_3} x^{r_1+2 r_2+3 r_3} \\ & \Rightarrow r_1+2 r_2+3 r_3=36\end{aligned}

\text { Case-I : } \begin{array}{|c|c|c|} \hline \mathrm{r}_1 & \mathrm{r}_2 & \mathrm{r}_3 \\ \hline 0 & 6 & 8 \\ \hline 2 & 5 & 8 \\ \hline 4 & 4 & 8 \\ \hline 6 & 3 & 8 \\ \hline \end{array}

r_1+2 r_2=12 \quad\left(\right.

Taking

\left.r_3=8\right)

\begin{aligned} &\text { Case-II :}\\ &\begin{array}{|l|l|l|} \hline r_1 & r_2 & r_3 \\ \hline 1 & 7 & 7 \\ \hline 3 & 6 & 7 \\ \hline 5 & 5 & 7 \\ \hline \end{array} \end{aligned}

r_1+2 r_2=15\left(\right.

Taking

\left.r_3=7\right)

\begin{aligned} &\text { Case-III : }\\ &\begin{array}{|l|l|l|} \hline r_1 & r_2 & r_3 \\ \hline 4 & 7 & 6 \\ \hline 6 & 6 & 6 \\ \hline \end{array} \end{aligned}

\begin{aligned} & \text { Coefficient}=7+(15 \times 21)+(15 \times 35)+(35) \\ & -(6 \times 8)-(20 \times 7 \times 8)-(6 \times 21 \times 8)+(15 \times 28) \\ & +(7 \times 28)=-678=\alpha \\ & |\alpha|=678\end{aligned}

Q 79MathematicsSection A • MCQstatistics

+4 / -1

Let the median and the mean deviation about the median of 7 observation

170,125,230,190,210

, a, b be 170 and

\frac{205}{7}

respectively. Then the mean deviation about the mean of these 7 observations is :

A

31

B

28

C

30

D

32

Q 80MathematicsSection A • MCQsequences and series

+4 / -1

Let

3, a, b, c

be in A.P. and

3, a-1, b+1, c+9

be in G.P. Then, the arithmetic mean of

a, b

and

c

is :

A

-4

B

-1

C

13

D

11

Official Correct Answer:Option (D)

Step-by-Step Solution:

Since $3, a, b, c$ are in arithmetic progression (A.P.), the common difference can be calculated using the term $a$ (the second term) as follows:

$d = a - 3$

The nth term of an A.P. is given by the formula:

$T_n = a + (n-1)d$

So, using this formula, we can express $b$ and $c$ in terms of $a$ and $d$ :

$b = a + d$

$c = a + 2d$

Substituting $d = a - 3$ into these expressions:

$b = a + (a - 3)$

$c = a + 2(a - 3)$

Therefore:

$b = 2a - 3$

$c = 3a - 6$

Now, let's consider that $3, a-1, b+1, c+9$ are in geometric progression (G.P.). For terms in a G.P., the ratio (common ratio, r) between consecutive terms is constant. So:

$\frac{a - 1}{3} = \frac{b + 1}{a - 1} = \frac{c + 9}{b + 1}$

Now, we will establish the relation between the terms using the property of G.P.:

$\frac{a - 1}{3} = \frac{b + 1}{a - 1}$

$(a - 1)^2 = 3(b + 1)$

$a^2 - 2a + 1 = 3b + 3$

Substituting $b = 2a - 3$ , we get:

$a^2 - 2a + 1 = 3(2a - 3) + 3$

$a^2 - 2a + 1 = 6a - 9 + 3$

$a^2 - 8a + 7 = 0$

Solving this quadratic equation:

$(a - 7)(a - 1) = 0$

Hence, $a = 7$ or $a = 1$ . However, if $a = 1$ , the terms $3, a-1, b+1, c+9$ cannot form a G.P. as it would involve division by zero. Therefore, $a = 7$ . We use this value to find $b$ and $c$ :

$b = 2a - 3 = 2(7) - 3 = 14 - 3 = 11$

$c = 3a - 6 = 3(7) - 6 = 21 - 6 = 15$

Now we can find the arithmetic mean ( $A$ ) of $a$ , $b$ , and $c$ :

$A = \frac{a + b + c}{3}$

$A = \frac{7 + 11 + 15}{3}$

$A = \frac{33}{3}$

$A = 11$

Hence, the arithmetic mean of $a$ , $b$ , and $c$ is $11$ , which corresponds to Option D.

Q 81MathematicsSection A • MCQmatrices and determinants

+4 / -1

If

\mathrm{A}=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right], \mathrm{C}=\mathrm{ABA}^{\mathrm{T}}

and

\mathrm{X}=\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{~A}

, then

\operatorname{det} \mathrm{X}

is equal to :

A

243

B

729

C

27

D

891

Official Correct Answer:Option (B)

Step-by-Step Solution:

The solution involves understanding matrix operations and properties such as multiplication, transpose, and determinant. Given $\mathrm{A}$ , $\mathrm{B}$ , and that $\mathrm{C} = \mathrm{ABA}^{\mathrm{T}}$ , and $\mathrm{X} = \mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}$ , we find $\operatorname{det} \mathrm{X}$ as follows:

First, we express $|\mathrm{C}|$ , the determinant of $\mathrm{C}$ , in terms of the determinants of $\mathrm{A}$ and $\mathrm{B}$ , using the property that $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$ :

$\begin{aligned} |\mathrm{C}| &= |\mathrm{ABA}^{\mathrm{T}}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{A}^{\mathrm{T}}| \\ &= |\mathrm{A}|^2 \cdot |\mathrm{B}| \quad \text{since } |\mathrm{A}^{\mathrm{T}}| = |\mathrm{A}|. \end{aligned}$

Next, we find $|\mathrm{X}|$ , using the property $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$ and substituting $|\mathrm{C}|$ :

$\begin{aligned} |\mathrm{X}| &= |\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}| = |\mathrm{A}^{\mathrm{T}}| \cdot |\mathrm{C}|^2 \cdot |\mathrm{A}| \\ &= |\mathrm{A}|^2 \cdot |\mathrm{C}|^2. \end{aligned}$

Substituting the expression for $|\mathrm{C}|$ obtained earlier:

$\begin{aligned} |\mathrm{X}| &= |\mathrm{A}|^2 \cdot (|\mathrm{A}|^2 \cdot |\mathrm{B}|)^2 \\ &= (|\mathrm{A}|^3 \cdot |\mathrm{B}|)^2. \end{aligned}$

The determinants of $\mathrm{A}$ and $\mathrm{B}$ are calculated as:

|A|=\left|\begin{array}{cc} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{array}\right|=2+1=3

|B|=\left|\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right|=1

Finally, substituting these values into our expression for $|\mathrm{X}|$ :

$\begin{aligned} |\mathrm{X}| &= (3^3 \cdot 1)^2 \\ &= 729. \end{aligned}$

Therefore, $\operatorname{det} \mathrm{X} = 729$ .

Q 82MathematicsSection B • Numericallimits continuity and differentiability

+4 / -1

Let

\{x\}

denote the fractional part of

x

and

f(x)=\frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, x \neq 0

. If

\mathrm{L}

and

\mathrm{R}

respectively denotes the left hand limit and the right hand limit of

f(x)

at

x=0

, then

\frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right)

is equal to ___________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:18

Step-by-Step Solution:

Finding right hand limit

R =

\begin{gathered} \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) =\lim _{h \rightarrow 0} f(h) \end{gathered}

\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right) \sin ^{-1}(1-h)}{h\left(1-h^2\right)} \\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{\cos ^{-1}\left(1-\mathrm{h}^2\right)}{\mathrm{h}}\left(\frac{\sin ^{-1} 1}{1}\right)\end{aligned}

Let

\cos ^{-1}\left(1-h^2\right)=\theta \Rightarrow \cos \theta=1-h^2

\begin{aligned} & =\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{\theta}{\sqrt{1-\cos \theta}} \\ & =\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{1}{\sqrt{\frac{1-\cos \theta}{\theta^2}}} \\ & =\frac{\pi}{2} \frac{1}{\sqrt{1 / 2}} \\ & \therefore \mathrm{R}=\frac{\pi}{\sqrt{2}} \end{aligned}

Now finding left hand limit

\begin{aligned} & L=\lim _{x \rightarrow 0^{-}} f(x) =\lim _{h \rightarrow 0} f(-h) \end{aligned}

\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-\{-h\}^2\right) \sin ^{-1}(1-\{-h\})}{\{-h\}-\{-h\}^3} \\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-(-h+1)^2\right) \sin ^{-1}(1-(-h+1))}{(-h+1)-(-h+1)^3} \\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(-h^2+2 h\right) \sin ^{-1} h}{(1-h)\left(1-(1-h)^2\right)}\end{aligned}

\begin{aligned} & =\lim _{h \rightarrow 0}\left(\frac{\pi}{2}\right) \frac{\sin ^{-1} h}{\left(1-(1-h)^2\right)} \\ & =\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{-h^2+2 h}\right) \\ & =\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{h}\right)\left(\frac{1}{-h+2}\right) \\ & \quad L=\frac{\pi}{4}\end{aligned}

\begin{aligned} & \frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right)=\frac{32}{\pi^2}\left(\frac{\pi^2}{2}+\frac{\pi^2}{16}\right) \\ & =16+2 \\ & =18\end{aligned}

Q 83MathematicsSection B • Numericalsets and relations

+4 / -1

Let

A=\{1,2,3, \ldots, 20\}

. Let

R_1

and

R_2

two relation on

A

such that

R_1=\{(a, b): b

is divisible by

a\}

R_2=\{(a, b): a

is an integral multiple of

b\}

.

Then, number of elements in

R_1-R_2

is equal to _____________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:46

Step-by-Step Solution:

To determine the number of elements in $R_1 - R_2$ , let's first articulate the meaning of both relations on set $A = \{1, 2, 3, \ldots, 20\}$ :

$R_1$ includes pairs $(a, b)$ where $b$ is divisible by $a$ . This includes pairs like $(1,1), (1,2), \ldots, (1,20)$ for $1$ ; similar series for $2$ up to $(2,20)$ (excluding odd numbers); for $3$ up to $(3,18)$ ; and so on, reflecting the divisibility condition.

\begin{aligned} & R_1:\left\{\begin{array}{l}(1,1),(1,2) \ldots,(1,20), \\ (2,2),(2,4) \ldots,(2,20), \\ (3,3),(3,6) \ldots,(3,18), \\ (4,4),(4,8) \ldots,(4,20), \\ (5,5)(5,10) \ldots,(5,20), \\ (6,6),(6,12),(6,18),(7,7),(7,14), \\ (8,8),(8,16),(9,9),(9,18)(10,10), \\ (10,20),(11,11),(12,12) \ldots,(20,20)\end{array}\right\} \\ & n\left(R_1\right)=66\end{aligned}

$R_2$ consists of pairs $(a, b)$ where $a$ is an integral multiple of $b$ . Essentially, this relationship is the reverse of $R_1$ . However, for the essence of $R_1 - R_2$ , the key overlap comes with pairs where $a = b$ , since those are the pairs that clearly reflect both conditions identically (i.e., a number is always divisible by itself, and it is always a multiple of itself). There are 20 such pairs corresponding to each integer in $A$ from 1 to 20, resulting in the common elements between $R_1$ and $R_2$ (the intersection $R_1 \cap R_2$ ) being 20 pairs.

\begin{aligned} & \mathrm{R}_1 \cap \mathrm{R}_2=\{(1,1),(2,2), \ldots(20,20)\} \\ & \mathrm{n}\left(\mathrm{R}_1 \cap \mathrm{R}_2\right)=20 \end{aligned}

The difference $R_1 - R_2$ seeks elements present in $R_1$ but not in $R_2$ . Given that $R_1$ and $R_2$ share 20 elements that are identical, to find $R_1 - R_2$ , we subtract these 20 common elements from the total in $R_1$ , resulting in $66 - 20 = 46$ pairs.

\begin{aligned} & \mathrm{n}\left(\mathrm{R}_1-\mathrm{R}_2\right)=\mathrm{n}\left(\mathrm{R}_1\right)-\mathrm{n}\left(\mathrm{R}_1 \cap \mathrm{R}_2\right) \\ & =\mathrm{n}\left(\mathrm{R}_1\right)-20 \\ & =66-20 \\ & \mathrm{R}_1-\mathrm{R}_2=46 \text { Pair }\end{aligned}

Q 84MathematicsSection A • MCQapplication of derivatives

+4 / -1

If

5 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2, \forall x \neq 0

and

y=9 x^2 f(x)

, then

y

is strictly increasing in :

A

\left(0, \frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)

B

\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)

C

\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)

D

\left(-\infty, \frac{1}{\sqrt{5}}\right) \cup\left(0, \frac{1}{\sqrt{5}}\right)

Q 85MathematicsSection B • Numericalparabola

+4 / -1

Let the line

\mathrm{L}: \sqrt{2} x+y=\alpha

pass through the point of the intersection

\mathrm{P}

(in the first quadrant) of the circle

x^2+y^2=3

and the parabola

x^2=2 y

. Let the line

\mathrm{L}

touch two circles

\mathrm{C}_1

and

\mathrm{C}_2

of equal radius

2 \sqrt{3}

. If the centres

Q_1

and

Q_2

of the circles

C_1

and

C_2

lie on the

y

-axis, then the square of the area of the triangle

\mathrm{PQ}_1 \mathrm{Q}_2

is equal to ___________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 86MathematicsSection B • Numericaldefinite integration

+4 / -1

If

\int\limits_{-\pi / 2}^{\pi / 2} \frac{8 \sqrt{2} \cos x \mathrm{~d} x}{\left(1+\mathrm{e}^{\sin x}\right)\left(1+\sin ^4 x\right)}=\alpha \pi+\beta \log _{\mathrm{e}}(3+2 \sqrt{2})

, where

\alpha, \beta

are integers, then

\alpha^2+\beta^2

equals :

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:8

Step-by-Step Solution:

I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{\left(1+e^{\sin x}\right)\left(1+\sin ^4 x\right)} d x

............(1)

Apply king's rule

I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x\left(e^{\sin x}\right)}{\left(1+e^{\sin x}\right)\left(1+\sin ^4 x\right)} d x

..........(2)

Adding (1) and (2), we get

\begin{aligned} & 2 I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} d x \\ & I=\int\limits_0^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} d x, \end{aligned}

Let

\sin x=t

\begin{aligned} & I=8 \sqrt{2} \int\limits_0^1 \frac{d t}{1+t^4} \\ & =4 \sqrt{2} \int\limits_0^1 \frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}} d t \end{aligned}

\begin{aligned} & =4 \sqrt{2} \int\limits_0^1 \frac{\left(1+\frac{1}{t^2}\right) d t}{\left(t-\frac{1}{t^2}\right)^2+2}-4 \sqrt{2} \int\limits_0^1 \frac{\left(1-\frac{1}{t^2}\right) d t}{\left(t+\frac{1}{t^2}\right)^2-2} \\ & =4 \sqrt{2} \cdot \frac{1}{\sqrt{2}}\left(\left.\tan ^{-1} \frac{t-\frac{1}{t}}{\sqrt{2}}\right|_0 ^1-4 \sqrt{2} \cdot \frac{1}{2 \sqrt{2}}\left[\log \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|\right]_0^1\right. \\ & =2 \pi-2 \log \left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right|\end{aligned}

\begin{aligned} & =2 \pi+2 \log (3+2 \sqrt{2})=\alpha \pi+\beta \log _e(3+2 \sqrt{2}) \\ & \Rightarrow \alpha=2, \beta=2 \\ & \Rightarrow \alpha^2+\beta^2=8\end{aligned}

Q 87MathematicsSection B • Numericalcomplex numbers

+4 / -1

Let

\mathrm{P}=\{\mathrm{z} \in \mathbb{C}:|z+2-3 i| \leq 1\}

and

\mathrm{Q}=\{\mathrm{z} \in \mathbb{C}: z(1+i)+\bar{z}(1-i) \leq-8\}

. Let in

\mathrm{P} \cap \mathrm{Q}

,

|z-3+2 i|

be maximum and minimum at

z_1

and

z_2

respectively. If

\left|z_1\right|^2+2\left|z_2\right|^2=\alpha+\beta \sqrt{2}

, where

\alpha, \beta

are integers, then

\alpha+\beta

equals _____________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 88MathematicsSection A • MCQellipse

+4 / -1

Let

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \mathrm{a}>\mathrm{b}

be an ellipse, whose eccentricity is

\frac{1}{\sqrt{2}}

and the length of the latusrectum is

\sqrt{14}

. Then the square of the eccentricity of

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

is :

A

3

B

{7 \over 2}

C

{3 \over 2}

D

{5 \over 2}

Official Correct Answer:Option (C)

Step-by-Step Solution:

Given the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a > b$ , the eccentricity $e$ is given by the formula:

$e = \sqrt{1 - \left(\frac{b}{a}\right)^2}$

It is provided that the eccentricity $e$ is $\frac{1}{\sqrt{2}}$ (given), so we can equate the two expressions for eccentricity:

$\frac{1}{\sqrt{2}} = \sqrt{1 - \left(\frac{b}{a}\right)^2}$

Squaring both sides to eliminate the square root gives:

$\frac{1}{2} = 1 - \left(\frac{b}{a}\right)^2$

$\left(\frac{b}{a}\right)^2 = 1 - \frac{1}{2}$

$\left(\frac{b}{a}\right)^2 = \frac{1}{2}$

Taking the square root on both sides:

$\frac{b}{a} = \frac{1}{\sqrt{2}}$

$a = b\sqrt{2}$

Now, for the ellipse, the length of the latus rectum is given by the formula:

$\text{Length of Latus Rectum (L)} = \frac{2b^2}{a}$

It's provided that the length of the latus rectum $L$ is $\sqrt{14}$ , so substitute the known values to find $b$ :

$\sqrt{14} = \frac{2b^2}{b\sqrt{2}} = \frac{2b}{\sqrt{2}}$

$b\sqrt{2} = \sqrt{14}$

$b^2 = \frac{14}{2}$

$b^2 = 7$

And since $a = b\sqrt{2}$ , we can find $a^2$ :

$a^2 = (b\sqrt{2})^2$

$a^2 = 7 \cdot 2$

$a^2 = 14$

Now we have an ellipse with $a^2 = 14$ and $b^2 = 7$ . The equation of a hyperbola similar to the given ellipse but with the terms subtracted is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

For the hyperbola, the square of the eccentricity $e'$ is given by:

$(e')^2 = 1 + \frac{b^2}{a^2}$

Substitute the values we've found for $a^2$ and $b^2$ into the formula for the square of the hyperbola's eccentricity:

$(e')^2 = 1 + \frac{b^2}{a^2}$

$(e')^2 = 1 + \frac{7}{14}$

$(e')^2 = 1 + \frac{1}{2}$

$(e')^2 = \frac{3}{2}$

Therefore, the square of the eccentricity of the hyperbola is $\frac{3}{2}$ , which corresponds to option C.

Q 89MathematicsSection A • MCQprobability

+4 / -1

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is :

A

\frac{2}{5}

B

\frac{2}{7}

C

\frac{1}{7}

D

\frac{1}{5}

Q 90MathematicsSection B • Numerical3d geometry

+4 / -1

Let the line of the shortest distance between the lines

\begin{aligned} & \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\ & \mathrm{L}_2: \overrightarrow{\mathrm{r}}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k}) \end{aligned}

intersect

\mathrm{L}_1

and

\mathrm{L}_2

at

\mathrm{P}

and

\mathrm{Q}

respectively. If

(\alpha, \beta, \gamma)

is the mid point of the line segment

\mathrm{PQ}

, then

2(\alpha+\beta+\gamma)

is equal to ____________.

Answer Format:Enter numerical value rounded to the nearest integer

JEE Main 2024 1st February Morning Shift Question Paper with Solutions