JEE Main 2023 10th April Evening Shift Question Paper with Solutions

To find the maximum force exerted by the spring on the block, we can use Hooke's law and the properties of simple harmonic motion.

First, let's find the angular frequency $\omega$:

$\omega = \frac{2\pi}{T}$

where $T = 3.14\,\mathrm{s}$ is the time period.

$\omega = \frac{2\pi}{3.14} \approx 2\,\mathrm{rad/s}$

Now, let's find the maximum velocity $v_{max}$ of the block:

$v_{max} = \omega A$

where $A = 1\,\mathrm{m}$ is the amplitude.

$v_{max} = 2\,\mathrm{rad/s} \times 1\,\mathrm{m} = 2\,\mathrm{m/s}$

Next, we can find the spring constant $k$ using the mass of the block $m = 5\,\mathrm{kg}$ and the angular frequency $\omega$:

$\omega^2 = \frac{k}{m} \Rightarrow k = m\omega^2$

$k = 5\,\mathrm{kg} \times (2\,\mathrm{rad/s})^2 = 20\,\mathrm{N/m}$

Finally, we can find the maximum force exerted by the spring on the block. At maximum displacement, the force is given by Hooke's law:

$F_{max} = kA$

$F_{max} = 20\,\mathrm{N/m} \times 1\,\mathrm{m} = 20\,\mathrm{N}$

The maximum force exerted by the spring on the block is $20\,\mathrm{N}$.

Let's use the formula for the remaining mass of a radioactive substance after a certain time:

$N(t) = N_0(1/2)^{t/T}$

where N(t) is the mass at time t, N₀ is the initial mass, T is the half-life, and t is the time elapsed.

We are given that $\frac{7}{8}$th of the original mass has disintegrated. Therefore, the remaining mass is $\frac{1}{8}$th of the original mass:

$\frac{N(t)}{N_0} = \frac{1}{8}$

Using the formula, we have:

$\frac{1}{8} = (1/2)^{t/T}$

Taking the logarithm of both sides:

$\log_{1/2}\frac{1}{8} = \frac{t}{T}$

$3 = \frac{t}{T}$

Now, solving for t:

$t = 3T$

So, the time taken for disintegrating $\frac{7}{8}$th part of the original mass is 3T.

Let the initial speed of both projectiles be v. The maximum height attained by a projectile can be calculated using the formula:

$H = \frac{v^2 \sin^2 \theta}{2g}$

where H is the maximum height, v is the initial speed, θ is the angle of projection, and g is the acceleration due to gravity.

For the projectile projected at 30°, the maximum height is:

$H_1 = \frac{v^2 \sin^2 30^{\circ}}{2g} = \frac{v^2 \times \frac{1}{4}}{2g} = \frac{v^2}{8g}$

For the projectile projected at 60°, the maximum height is:

$H_2 = \frac{v^2 \sin^2 60^{\circ}}{2g} = \frac{v^2 \times \frac{3}{4}}{2g} = \frac{3v^2}{8g}$

Now, let's find the ratio of the maximum heights:

$\frac{H_1}{H_2} = \frac{\frac{v^2}{8g}}{\frac{3v^2}{8g}} = \frac{v^2}{3v^2}$

The v² terms cancel out, and we get:

$\frac{H_1}{H_2} = \frac{1}{3}$

Therefore, the ratio of the maximum heights attained by the two projectiles is 1 : 3

In this experiment, the vernier callipers have a least count of 0.1 mm, which is equal to 0.01 cm.

First, let's calculate the zero error since the zero of the vernier scale does not coincide with the zero of the main scale when the two jaws are joined together. The zero error can be found using the formula:

Zero Error = (Number of divisions coinciding) × Least Count

Zero Error = 6 × 0.01 cm = 0.06 cm

Since the zero of the vernier scale lies to the right of the main scale zero, the zero error is positive.

Now, let's find the main scale reading (MSR) when measuring the diameter of the spherical bob. The MSR is the value just before the zero of the vernier scale, which is 3.2 cm in this case.

Next, let's find the vernier scale reading (VSR) when measuring the diameter. The VSR is the product of the coinciding division number and the least count:

VSR = (Number of divisions coinciding) × Least Count

VSR = 4 × 0.01 cm = 0.04 cm

Now, we can find the total reading, which is the sum of the MSR and VSR:

Total Reading = MSR + VSR = 3.2 cm + 0.04 cm = 3.24 cm

Since there is a positive zero error, we need to subtract it from the total reading to get the corrected diameter:

Corrected Diameter = Total Reading - Zero Error = 3.24 cm - 0.06 cm = 3.18 cm

So, the diameter of the bob is measured as 3.18 cm.

Electron drifts from lower potential to higher potential on curved path.

In a Young's double slit experiment, the intensity at a point on the screen is given by the formula :

$I = 4I_0\cos^2\left(\frac{\pi d\sin\theta}{\lambda}\right)$

where $I_0$ is the intensity at the center of the pattern, $d$ is the distance between the slits, $\theta$ is the angle between the line joining the point to the center of the pattern and the line passing through the center of the pattern and the slits, and $\lambda$ is the wavelength of the light used.

The phase difference between the waves from the two slits at a point on the screen is given by :

$\Delta \phi = \frac{2\pi d\sin\theta}{\lambda}$

For a phase difference of $\pi/3$ between the waves from the two slits at point P, we have :

$\frac{\Delta \phi}{\pi} = \frac{2d\sin\theta}{\lambda} = \frac{1}{3}$

For a phase difference of $\pi/2$ between the waves from the two slits at point Q, we have :

$\frac{\Delta \phi}{\pi} = \frac{2d\sin\theta}{\lambda} = \frac{1}{2}$

Solving for $\sin\theta$ in both cases, we get:

$\sin\theta_P = \frac{\lambda}{6d},\quad \sin\theta_Q = \frac{\lambda}{4d}$

Substituting these values in the formula for intensity, we get :

$\frac{I_P}{I_Q} = \frac{4\cos^2\left(\frac{\pi}{6}\right)}{4\cos^2\left(\frac{\pi}{4}\right)} = \frac{3}{2}$

Therefore, the ratio of intensities at points P and Q is 3 : 2

$$\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\frac{150 \times 150}{300}+25=100 \Omega \\ & \mathrm{I}_1=\frac{5}{10}=0.05 \mathrm{~A} \\ & \mathrm{I}_2=\mathrm{I}_4=\frac{0.05}{2}=0.025 \mathrm{~A} \\ & \frac{\mathrm{I}_1}{\mathrm{I}_2}=2 \end{aligned}$$

The energy difference formula for transitions between energy levels in a hydrogen atom, which is given by

$ \Delta E = -13.6 \, \text{eV} \times \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $

where ($n_1$) and ($n_2$) are the initial and final energy levels, respectively. For the Lyman series, the electron transitions to the ground state (($n_1$ = 1)), and for the Balmer series, the electron transitions to the first excited state (($n_1$ = 2)).

From the given information, we have the lowest wavelength in the Lyman series, ($\lambda_1 = 917 \, \text{Å}$). Therefore, the energy difference ($\Delta E$) for the Lyman series is

$ \Delta E = \frac{hc}{\lambda_1} $

where (h) is Planck's constant and (c) is the speed of light.

Similarly, for the Balmer series, the energy difference ($\Delta E$) is

$ \Delta E = -13.6 \, \text{eV} \times \left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) = -13.6 \, \text{eV} \times \frac{1}{4} $

The corresponding wavelength ($\lambda_2$) is

$ \lambda_2 = \frac{hc}{\Delta E} $

By comparing ($\Delta E$) for the Lyman and Balmer series, you found that

$ \frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} = \frac{1}{4} $

Therefore, the lowest wavelength in the Balmer series is

$ \lambda_2 = 4 \lambda_1 = 4 \times 917 \, \text{Å} = 3668 \, \text{Å} $

Both Statement I and Statement II are true.

Statement I: For diamagnetic substances, the magnetic susceptibility (χ) lies between -1 and 0. This is because diamagnetic substances have a negative magnetic susceptibility, which means they have a tendency to oppose the applied magnetic field.

Statement II: Diamagnetic substances, when placed in an external magnetic field, experience a force that tends to move them from stronger to weaker parts of the field. This is due to the fact that diamagnetic substances oppose the applied magnetic field and try to minimize their exposure to it.

We can write the given equation as:

$y = \sqrt{(\sin\omega t)^2 + (\cos\omega t)^2} \cos\left(\omega t - \arctan\frac{\sin\omega t}{\cos\omega t}\right)$

Using the identity $\sin^2\theta + \cos^2\theta = 1$, we get:

$y = \sqrt{1 + \sin 2\omega t} \cos\left(\omega t - \frac{\pi}{4}\right)$

The amplitude of the motion is the maximum value of $|y|$, which occurs when $\sin 2\omega t = 1$, i.e., at $t = \frac{\pi}{4\omega} + \frac{n\pi}{\omega}$, where $n$ is an integer. Substituting this value of $t$ in the above equation, we get:

$|y_{\text{max}}| = \sqrt{1 + \sin \frac{\pi}{2}} = \sqrt{2}$

Therefore, the amplitude of the motion is $\sqrt{2}$

The bandwidth of an amplitude-modulated wave is given by $2B$, where $B$ is the bandwidth of the modulating signal.

In this case, the modulating signal is a message signal of frequency $3~\mathrm{kHz}$, so the bandwidth of the amplitude modulated wave is $2(3~\mathrm{kHz})=6~\mathrm{kHz}$.

Therefore, the correct answer is $6~\mathrm{kHz}$.

When a metal sheet of thickness ($\frac{2d}{3}$) is introduced between the plates of a capacitor, it divides the capacitor into two separate capacitors. The metal sheet acts as a new plate in each capacitor, and because the metal is a conductor, it is at the same potential as the plates on either side.

The capacitance of the original capacitor with air between the plates is given by:

$\mathrm{C}_1 = \frac{\epsilon_0 \mathrm{A}}{\mathrm{d}}$

where ($\epsilon_0$) is the permittivity of free space, ($\mathrm{A}$) is the area of one of the plates, and ($\mathrm{d}$) is the distance between the plates.

When the metal sheet of thickness ($\frac{2d}{3}$) is introduced, the distance between the plates is reduced by ($\frac{2d}{3}$), so the capacitance of each of the new capacitors is:

$\mathrm{C}' = \frac{\epsilon_0 \mathrm{A}}{\mathrm{d}-\frac{2d}{3}}$

Simplifying, we get:

$\mathrm{C}' = \frac{3\epsilon_0 \mathrm{A}}{\mathrm{d}}$

Since the two capacitors are in parallel with each other, the total capacitance is:

$\mathrm{C}_2 = 2\mathrm{C}' = 2 \times \frac{3\epsilon_0 \mathrm{A}}{\mathrm{d}} = 3\mathrm{C}_1$

So the ratio of the capacitances is:

$\frac{\mathrm{C}_2}{\mathrm{C}_1} = 3:1$

In this problem, a square loop is inside a long solenoid, and there's a varying current flowing through the solenoid. Because the current is changing, it induces a changing magnetic field inside the solenoid.

According to Faraday's law of electromagnetic induction, a changing magnetic field will induce an electromotive force (emf) in a loop placed in that field. In this case, the loop is the square loop inside the solenoid.

The formula used here is based on Faraday's law, which states that the induced emf in a loop is equal to the rate of change of magnetic flux through the loop. This is given by:

$\text{emf} = -\frac{d \Phi}{dt}$

where $\Phi$ is the magnetic flux.

The magnetic field inside a solenoid is given by $B = \mu_0 n I$, where $\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length in the solenoid, and $I$ is the current through the solenoid.

The magnetic flux through the square loop is then given by $\Phi = B \cdot A = \mu_0 n I A$, where $A$ is the area of the loop.

When the current is sinusoidal, i.e., $I(t) = I_0 \sin(\omega t)$, its derivative with respect to time is $dI/dt = I_0 \omega \cos(\omega t)$, where $\omega$ is the angular frequency.

Hence, the rate of change of flux becomes:

$\frac{d \Phi}{dt} = \mu_0 n A \frac{dI}{dt} = \mu_0 n A I_0 \omega \cos(\omega t)$

The emf, which is equal to the negative of the rate of change of flux, will have a maximum value (the amplitude) when $\cos(\omega t) = 1$, giving:

$ \text{Emf amplitude} = \mu_0 n A I_0 \omega$<br><br>$ = 4\pi \times 10^{-7} \, \text{T m/A} \times \left(\frac{50}{10^{-2}}\right) \, \text{turns/m} \times (2 \times 10^{-2} \, \text{m})^2 \times 2.5 \, \text{A} \times 700 \, \text{rad/s} $

which simplifies to:

$ \text{Emf amplitude} = 44 \times 10^{-4} \, \text{V} $

So, the value of $x$ in the question is $44$

$$\begin{aligned} & e E=\frac{\mathrm{mV}^2}{r} \\ & e \cdot \frac{2 \mathrm{~K} \lambda}{r}=\frac{\mathrm{mV}^2}{r} \\ & V=\sqrt{\frac{e \cdot 2 \mathrm{k} \lambda}{\mathrm{m}}} \\ & =\sqrt{\frac{1.6 \times 10^{-19} \times 2 \times 9 \times 10^9 \times 2 \times 10^{-8}}{9 \times 10^{-31}}} \\ & =8 \times 10^6 \mathrm{~m} / \mathrm{s} \end{aligned}$$

In an electromagnetic wave, the maximum value of the electric field E is related to the maximum value of the magnetic field B by the equation

$E = cB$

where c is the speed of light in a vacuum, which is approximately $3 × 10^8 m/s$.

Given that the amplitude of the magnetic field B is $6.0 × 10^{-7} T$, we can substitute these values into the equation to find the maximum value of the electric field:

$E = (3 × 10^8 m/s) \times (6.0 × 10^{-7} T) = 180 V/m$

Therefore, 180 V/m is the correct answer.

First, let's find the total weight of the elevator and passengers:

Total weight = (mass of passengers + mass of elevator) × g

Total weight = (600 kg + 800 kg) × 10 m/s²

Total weight = 1400 kg × 10 m/s² = 14,000 N

Now, we need to calculate the total force acting on the elevator as it moves upwards. Since the elevator is moving at a constant speed, the net force acting on it is zero. Therefore, the tension in the cable must balance the total weight and frictional force:

Tension = Total weight + Frictional force Tension = 14,000 N + 2,000 N = 16,000 N

The power used by the motor can be calculated using the formula:

Power = Force × Velocity

Here, the force is the tension in the cable, and the velocity is the speed of the elevator:

Power = 16,000 N × 3 m/s = 48,000 W

To convert the power to kilowatts, divide by 1,000:

Power = 48,000 W / 1,000 = 48 kW

So, the maximum power used by the motor is 48 kW.

By equation of continuity

$$\begin{aligned} & \mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2 \\ & \mathrm{~V}_2=\frac{2 \times 4}{10 \times 10^{-2}}=80 \mathrm{~cm} / \mathrm{s} \end{aligned}$$

$$\begin{aligned} & e V_0=\frac{h c}{\lambda}-\phi \\ & V_0=\frac{h c}{e \lambda}-\left(\frac{\phi}{e}\right)=\frac{h v}{e}-\left(\frac{\phi}{e}\right) \end{aligned}$$

When $V_0=0$,

$$\begin{aligned} \frac{\phi}{e} & =\left(\frac{h v}{e}\right) \\ \phi & =h v=\frac{6.626 \times 10^{-34} \times 5 \times 10^{14}}{1.6 \times 10^{-19}} \\ & =\frac{6.626 \times 5}{1.6} \times 10^{-1} \\ & =\frac{6.626 \times 5}{16} \approx 2.07 \mathrm{eV} \end{aligned}$$

From I^st lens :

$\frac{1}{\mathrm{v}}+\frac{1}{6}=\frac{1}{24}$

$$\begin{aligned} \frac{1}{\mathrm{v}} & =\frac{1}{24}-\frac{1}{6}=-\frac{1}{8} \\ \mathrm{v} & =-8 \mathrm{~cm} \end{aligned}$$

From II^nd lens :

$\frac{1}{\mathrm{v}}+\frac{1}{18}=\frac{1}{9}$

$$\begin{aligned} \frac{1}{\mathrm{v}} & =\frac{1}{18} \\ \mathrm{v} & =18 \end{aligned}$$



So distance between object and its image

= 6 + 10 + 18 = 34 cm

Given the formula for elongation in a material due to a force:

$\Delta L = \frac{FL}{AY}$

where:

• F is the force applied,
• L is the original length,
• A is the cross-sectional area of the material, and
• Y is Young's modulus of the material.

The ratio of the elongations in the two wires A and B is given by:

$\frac{\Delta L_1}{\Delta L_2} = \frac{F_1}{F_2} \times \frac{A_2}{A_1} \times \frac{Y_2}{Y_1}$

Since the same force is applied on both wires (i.e., ($F_1/F_2$ = 1)), the areas are in the ratio 1:3 (i.e., ($A_2/A_1$ = 3)), and the Young's moduli are in the ratio 1:4 (i.e., ($Y_2/Y_1$ = 4)), substituting these values into the equation gives:

$\frac{\Delta L_1}{\Delta L_2} = 1 \times 3 \times 4 = 12$

So, the ratio of the elongations is 12:1, which indicates that wire A will elongate 12 times more than wire B when the same force is applied.

When a bar magnet is released from rest along the axis of a very long vertical copper tube, it will move down with an almost constant speed after some time.

As the magnet falls, it moves through the copper tube, inducing eddy currents in the tube. These eddy currents, in turn, create an opposing magnetic field that opposes the motion of the magnet. The opposing force generated by the eddy currents acts as a damping force, which slows down the magnet's acceleration. Eventually, the magnet reaches a terminal velocity, at which point the gravitational force pulling the magnet downward is balanced by the opposing force from the eddy currents.

At this terminal velocity, the magnet moves down with an almost constant speed.

The internal energy (U) of a gas depends on its degrees of freedom (f).

For a monatomic gas like neon, the degrees of freedom are f = 3 (translational). For a diatomic gas like oxygen, the degrees of freedom are f = 5 (3 translational + 2 rotational).

The internal energy for each component of the gas mixture can be calculated using the formula:

$U = \frac{f}{2}nRT$

where n is the number of moles, R is the universal gas constant, and T is the temperature.

For the 4 moles of neon (monatomic):

$U_{Ne} = \frac{3}{2} \cdot 4RT = 6RT$

For the 2 moles of oxygen (diatomic):

$U_{O_2} = \frac{5}{2} \cdot 2RT = 5RT$

Now, to find the total internal energy, we sum the internal energies of the individual components:

$U_{total} = U_{Ne} + U_{O_2} = 6RT + 5RT = 11RT$

Thus, the total internal energy of the system is 11RT.

The average velocity is defined as the total displacement divided by the total time. Here, the person travels the same distance $x$ twice, once with velocity $v_1$ and once with velocity $v_2$.

The time to travel distance $x$ with velocity $v_1$ is $t_1 = \frac{x}{v_1}$, and the time to travel distance $x$ with velocity $v_2$ is $t_2 = \frac{x}{v_2}$.

The total time is then

$t = t_1 + t_2 = \frac{x}{v_1} + \frac{x}{v_2}$.

The total displacement is $2x$. So, the average velocity $v$ is given by

$v = \frac{\text{total displacement}}{\text{total time}} = \frac{2x}{\frac{x}{v_1} + \frac{x}{v_2}} = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}}$

Multiplying both sides by $2$, we get

$\frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2}$

An adiabatic process is one in which there is no heat exchange between a system (in this case, the gas) and its surroundings. This happens because the system is perfectly insulated or the process occurs very quickly, not allowing for heat exchange.

Given the options:

Option A: There is no heat supplied to the system.

• This statement is TRUE. In an adiabatic process, there is no heat exchange between the system and its surroundings, as mentioned above.

Option B: The temperature of the gas increases.

• This statement is TRUE. When a gas is compressed adiabatically, the work done on the gas increases its internal energy, which in turn increases the temperature of the gas.

Option C: There is no change in the internal energy.

• This statement is NOT TRUE. In an adiabatic process, the change in internal energy of the system is equal to the work done on the system. When the gas is compressed, work is done on the gas, which increases its internal energy.

Option D: The change in the internal energy is equal to the work done on the gas.

• This statement is TRUE. As mentioned above, in an adiabatic process, the change in internal energy of the system is equal to the work done on the system.

So, Option C is the statement that is not true for an adiabatic process.

For a satellite orbiting the Earth at a height equal to Earth's radius, its distance from the center of the Earth will be 2R, where R is the radius of the Earth.

Using the formula for the gravitational force:

$F = G\frac{Mm}{r^2}$

where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance from the center of the Earth.

The centripetal force acting on the satellite is given by:

$F_c = \frac{mv^2}{r}$

Equating the gravitational force and the centripetal force, we get:

$G\frac{Mm}{(2R)^2} = \frac{mv^2}{2R}$

Solving for the orbital speed v, we get:

$v^2 = \frac{GM}{2R}$

The circumference of the satellite's orbit is given by:

$C = 2\pi(2R) = 4\pi R$

The time period T of the satellite's orbit can be calculated as the ratio of the circumference to the orbital speed:

$T = \frac{C}{v} = \frac{4\pi R}{\sqrt{\frac{GM}{2R}}}$

Given that the acceleration due to gravity at the Earth's surface is $g = \pi^2 \,\mathrm{m/s^2}$, we can express the gravitational constant G in terms of the Earth's radius R and mass M:

$g = \frac{GM}{R^2} \Rightarrow GM = gR^2 = \pi^2 R^2$

Substituting the expression for GM into the equation for the time period T, we get:

$T = \frac{4\pi R}{\sqrt{\frac{\pi^2 R^2}{2R}}} = \frac{4\pi R}{\sqrt{\frac{\pi^2 R}{2}}}$

$T = \frac{4\pi R}{\sqrt{\pi^2}\sqrt{\frac{R}{2}}} = \frac{4\pi R}{\pi\sqrt{\frac{R}{2}}} = \frac{4R}{\sqrt{\frac{R}{2}}}$

Multiplying the numerator and denominator by $\sqrt{2}$, we get:

$T = \frac{4R\sqrt{2}}{\sqrt{R}} = 4\sqrt{2R} = \sqrt{32R}$

Let the point where the force acts be A, the origin of the coordinate system (0, 0, 0), and let the point about which the torque is calculated be B (2, -3, 0). The force vector is given by $\vec{F} = -P\hat{k}$.

To find the torque, we first find the position vector of point A with respect to point B:

$\vec{r}_{AB} = \vec{r}_A - \vec{r}_B = (0 - 2)\hat{i} + (0 - (-3))\hat{j} + (0 - 0)\hat{k} = -2\hat{i} + 3\hat{j}$

To calculate the cross product, we can use the determinant method with a 3x3 matrix:

$$\vec{\tau} = \vec{r}_{AB} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 0 \\ 0 & 0 & -P \\ \end{vmatrix}$$

Now, we will calculate the cross product components by expanding the determinant along the first row:

1. $$\tau_i = \hat{i} \begin{vmatrix} 3 & 0 \\ 0 & -P \\ \end{vmatrix} = \hat{i}((3)(-P) - (0)(0)) = -3P\hat{i}$$

2. $$\tau_j = -\hat{j} \begin{vmatrix} -2 & 0 \\ 0 & -P \\ \end{vmatrix} = -\hat{j}((-2)(-P) - (0)(0)) = -2P\hat{j}$$

(Notice the negative sign in front of the $\hat{j}$ term, as it comes from the expansion of the determinant.)

3. $$\tau_k = \hat{k} \begin{vmatrix} -2 & 3 \\ 0 & 0 \\ \end{vmatrix} = \hat{k}((-2)(0) - (3)(0)) = 0\hat{k}$$

Now, combine the components to get the torque vector:

$\vec{\tau} = -3P\hat{i} - 2P\hat{j} + 0\hat{k} = -3P\hat{i} - 2P\hat{j}$

Comparing this to the given torque vector $\vec{\tau} = P(a\hat{i} + b\hat{j})$, we find that:

$a = -3$

$b = -2$

Thus, the ratio $\frac{a}{b} = \frac{-3}{-2} = \frac{x}{2}$.

Therefore, $x = 3$.

Let's analyze both statements:

Statement I: Rotation of the earth shows an effect on the value of acceleration due to gravity (g).

This is true. The value of $g$ is affected by the rotation of the Earth. The centripetal force due to the Earth's rotation causes a reduction in the perceived gravitational acceleration for objects on the surface. This effect is zero at the poles and increases towards the equator. The formula for the effective acceleration due to gravity at a latitude $\theta$ is:

$g_{\text{effective}} = g - R\omega^2\cos^2(\theta)$

where:

• $g$ is the theoretical acceleration due to gravity without Earth's rotation,
• $R$ is the radius of the Earth,
• $\omega$ is the angular velocity of the Earth's rotation, and
• $\theta$ is the latitude.

Statement II: The effect of rotation of the earth on the value of 'g' at the equator is minimum and that at the pole is maximum.

This is false. For clarification, the effect of the Earth's rotation on the value of $g$ is maximum at the equator and minimum at the poles. At the equator, the centrifugal force is highest because of the maximum velocity due to Earth's rotation, which decreases the effect of gravity more than at any other latitude. At the poles, the centrifugal force is zero since there is no rotational velocity contributing to a centripetal effect; therefore, the effect of rotation on $g$ is minimum (zero).

With these considerations, the correct option is:

Option B: Statement I is true but Statement II is false.

The resistance of a material can be calculated using the formula:

$R = \rho \frac{L}{A}$

where

• $R$ is the resistance,
• $\rho$ (rho) is the resistivity or specific resistance of the material,
• $L$ is the length (or distance over which the resistance is being measured), and
• $A$ is the cross-sectional area through which the current flows.

In the context of a rectangular parallelepiped, the "length" and "cross-sectional area" can vary depending on which faces of the shape are considered.

In this particular calculation, the resistance is being measured between the two smaller faces of the parallelepiped, which are squares of side length 1 cm:

The cross-sectional area $A$ should be:

$A = 100 \, \text{cm} \times 1 \, \text{cm} = 100 \, \text{cm}^2$

Converting this to meters gives:

$A = 100 \, \text{cm}^2 = 1 \, \text{m} \times 0.01 \, \text{m} = 0.01 \, \text{m}^2$

So, if we use these values for the length $L$ and cross-sectional area $A$ in the resistance formula, we get:

$R = \rho \frac{L}{A} = 3 \times 10^{-7} \Omega \, m \times \frac{0.01 \, m}{0.01 \, \text{m}^2} = 3 \times 10^{-7} \Omega$

Therefore, the resistance between the two smaller faces of the rectangular parallelepiped is $3 \times 10^{-7} \Omega$.

The correct answer is Both A and R are correct and R is the correct explanation of A.

Explanation:

Assertion A: An electric fan continues to rotate for some time after the current is switched off. This is a correct statement. When you switch off the fan, it doesn't stop immediately but continues to rotate for some time.

Reason R: Fan continues to rotate due to inertia of motion. This is also a correct statement. Inertia is the resistance of any physical object to any change in its state of motion. This includes changes to the object's speed or direction of motion. An object will stay in its state of motion unless a force acts on it. In the case of the fan, after the current is switched off, the fan blades have inertia and continue to move due to this inertia until the frictional forces (like air resistance and friction in the fan's bearings) cause it to stop. Therefore, inertia of motion is the correct reason for the fan's continued motion after the current is switched off.

A. 16 g of $\mathrm{CH_4~(g)}$

The molar mass of $\mathrm{CH_4}$ (Methane) is 16 g/mol. Therefore, 16 g of $\mathrm{CH_4}$ is equivalent to 1 mole of $\mathrm{CH_4}$. Furthermore, each molecule of $\mathrm{CH_4}$ has 10 electrons (6 from Carbon and 4 from Hydrogen). As a result, 1 mole of $\mathrm{CH_4}$ (or $6.022 \times 10^{23}$ molecules of $\mathrm{CH_4}$) would have $10 \times 6.022 \times 10^{23} = 60.2 \times 10^{23}$ electrons. This matches with (II).

B. 1 g of $\mathrm{H_2~(g)}$

The molar mass of $\mathrm{H_2}$ (Hydrogen) is 2 g/mol. Therefore, 1 g of $\mathrm{H_2}$ is equivalent to 0.5 moles of $\mathrm{H_2}$. The volume that a given quantity of gas occupies is proportional to the number of moles of gas. At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. Therefore, 0.5 moles of gas would occupy $0.5 \times 22.4 = 11.2$ liters. The closest match is (IV) with 11.4 liters (the small discrepancy may be due to rounding or slightly different conditions than STP).

C. 1 mole of $\mathrm{N_2~(g)}$

The molar mass of $\mathrm{N_2}$ (Nitrogen) is 28 g/mol. Therefore, 1 mole of $\mathrm{N_2}$ weighs 28 g. This matches with (I).

D. 0.5 mol of $\mathrm{SO_2~(g)}$

The molar mass of $\mathrm{SO_2}$ (Sulfur Dioxide) is 64 g/mol. Therefore, 0.5 moles of $\mathrm{SO_2}$ weigh $0.5 \times 64 = 32$ g. This matches with (III).

Thus, the correct matches are:

A - II

B - IV

C - I

D - III

$\mathrm{X} \stackrel{\mathrm{Na}_2 \mathrm{O}_2}{\longrightarrow} \mathrm{Y} \stackrel{\mathrm{BaCl}_2}{\longrightarrow} \underset{\left[\mathrm{BaSO}_4\right]}{\mathrm{Z}}$

Methionine: $\mathrm{C}_5 \mathrm{H}_{11} \mathrm{NO}_2 \mathrm{S}$

• Washing soda is a hydrated form of sodium carbonate, with the chemical formula Na₂CO₃·10H₂O. This means that each molecule of washing soda contains 10 water molecules.


• Soda ash is the anhydrous form of sodium carbonate, with the chemical formula Na₂CO₃. This means that it does not contain any water molecules.

Soda ash is typically produced by heating washing soda to drive off the water molecules. Washing soda can be produced by dissolving soda ash in water and then recrystallizing it.

The correct answer is Option D: 10 and 0.

$\mathrm{OCH}_3$ is ortho/para directing, major product will be obtained from para attack.

The energy levels of a hydrogen-like atom (such as Li²⁺ in this case) can be calculated using the formula:

$E = -R_H \cdot \frac{Z^2}{n^2}$

where:

• $E$ is the energy of the electron,
• $R_H$ is the Rydberg constant (in joules),
• $Z$ is the atomic number, and
• $n$ is the principal quantum number.

The change in energy (∆E) between two energy levels is given by the difference in the energies of the two levels. In this case, the electron is excited from the nth level to the (n+1)th level, and the energy of the radiation is given as $1.47 \times 10^{-17}$ J. Therefore:

$\Delta E = E_{n+1} - E_n = 1.47 \times 10^{-17} \mathrm{~J}$

Substituting the formula for $E_{n}$ and $E_{n+1}$:

$\Delta E = -R_H \cdot \frac{Z^2}{(n+1)^2} - \left(-R_H \cdot \frac{Z^2}{n^2}\right)$

Simplifying this, we get:

$\Delta E = R_H \cdot Z^2 \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$

We know that $Z=3$ for $\mathrm{Li}^{2+}$, $R_H = 2.18 \times 10^{-18} \mathrm{~J}$ and $\Delta E = 1.47 \times 10^{-17} \mathrm{~J}$.

We substitute the given values into the formula

$1.47 \times 10^{-17} \mathrm{~J} = 2.18 \times 10^{-18} \mathrm{~J} \times 3^2 \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$

Simplify to

$\frac{1.47}{1.96} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$

Solving this quadratic equation for n, we get two possible solutions. However, we disregard the negative solution since n (the principal quantum number) can't be negative. So, the only valid solution is n=1.

Thus, the value of n is 1.