JEE Main 2024 8th April Evening Shift Question Paper with Solutions

$\begin{aligned} & \mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{Fr}}+\mathrm{W}_{\mathrm{s}}=\Delta \mathrm{KE} \\ & 5 \times 10 \times 5-0.5 \times 5 \times 10 \times \mathrm{x}-\frac{1}{2} \mathrm{Kx}^2=0-0 \\ & 250=25 \mathrm{x}+50 \mathrm{x}^2 \\ & 2 \mathrm{x}^2+\mathrm{x}-10=0 \\ & \mathrm{x}=2\end{aligned}$

$$\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ & \frac{1}{v}-\frac{1}{-30}=\frac{1}{10} \\ & \frac{1}{v}=\frac{1}{10}-\frac{1}{30}=\frac{3-1}{30}=\frac{1}{15} \\ & v=15 \end{aligned}$$

$\Rightarrow 30 \mathrm{~cm} \text { right of third }$

To determine the coefficient of kinetic friction, let's analyze the time taken by the object to slide down each plane and use the equations of motion for both scenarios.

First, consider the perfectly smooth $45^{\circ}$ inclined plane (no friction). The acceleration of the object on this plane can be calculated using the component of gravitational force parallel to the incline. Since there is no friction, the only force acting down the plane is the component of the gravitational force:

$a_{\text{smooth}} = g \sin 45^{\circ} = \frac{g}{\sqrt{2}}$

Let the time taken to slide down this smooth plane be $t_{\text{smooth}}$. The distance $d$ covered by the object can be expressed using the equation of motion:

$d = \frac{1}{2} a_{\text{smooth}} t_{\text{smooth}}^2 = \frac{1}{2} \frac{g}{\sqrt{2}} t_{\text{smooth}}^2$

Now, consider the rough $45^{\circ}$ inclined plane. The acceleration down the rough plane can be found by considering both the component of gravitational force and the kinetic friction force. Here, the friction force is $f_k = \mu_k N$, where $N = mg \cos 45^\circ = \frac{mg}{\sqrt{2}}$. Thus the frictional force is:

$f_k = \mu_k \cdot \frac{mg}{\sqrt{2}}$

The net force acting on the object down the plane would be:

$F_{\text{net}} = mg \sin 45^{\circ} - f_k = \frac{mg}{\sqrt{2}} - \mu_k \cdot \frac{mg}{\sqrt{2}}$

So the net acceleration $a_{\text{rough}}$ is:

$a_{\text{rough}} = \frac{F_{\text{net}}}{m} = \frac{g}{\sqrt{2}} - \mu_k \cdot \frac{g}{\sqrt{2}} = \frac{g}{\sqrt{2}} (1 - \mu_k)$

The time taken to slide down the rough plane is given as $nt_{\text{smooth}}$. So, the distance $d$ can be written as:

$d = \frac{1}{2} a_{\text{rough}} (nt_{\text{smooth}})^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}} (1 - \mu_k)\right) (nt_{\text{smooth}})^2$

Setting the distances equal for both cases, we get:

$\frac{1}{2} \frac{g}{\sqrt{2}} t_{\text{smooth}}^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}} (1 - \mu_k)\right) (nt_{\text{smooth}})^2$

Simplifying, we find:

$t_{\text{smooth}}^2 = (1 - \mu_k) n^2 t_{\text{smooth}}^2$

Therefore:

$1 = (1 - \mu_k) n^2$

Solving for $\mu_k$, we get:

$\mu_k = 1 - \frac{1}{n^2}$

Thus, the coefficient of kinetic friction between the object and the surface of inclined plane is:

Option A: $1 - \frac{1}{\mathrm{n}^2}$

When an electric kettle is used to heat water, the time taken to boil the water depends on the power of the heating element. The power supplied to the heating element is inversely proportional to the heating time. If we want to reduce the boiling time, the power needs to be increased. The power of the heating element is given by:

$P = \frac{V^2}{R}$

where $P$ is the power, $V$ is the voltage, and $R$ is the resistance of the heating element. The resistance $R$ of the heating element is proportional to its length $L$ while the material and cross-sectional area remain constant.

So, we can write:

$R \propto L$

To achieve boiling in 15 minutes instead of 20 minutes, the power needs to increase, which implies the resistance must decrease. Let the initial length of the heating element be $L_0$, and let the new length needed be $L$. The time taken to heat is inversely proportional to the power:

$\frac{T_1}{T_2} = \frac{P_2}{P_1}$

Given that:

$T_1 = 20 \text{ minutes}$

$T_2 = 15 \text{ minutes}$

We need to find the ratio:

$\frac{20}{15} = \frac{P_2}{P_1}$

Simplifying:

$\frac{4}{3} = \frac{P_2}{P_1}$

The power is inversely proportional to the resistance:

$\frac{P_2}{P_1} = \frac{R_1}{R_2}$

Thus:

$\frac{4}{3} = \frac{R_1}{R_2}$

Since resistance is proportional to length:

$\frac{R_1}{R_2} = \frac{L_1}{L_2}$

Hence:

$\frac{4}{3} = \frac{L_1}{L_2}$

Simplifying, we find:

$L_2 = \frac{3}{4} L_1$

This means the length of the heating element should be decreased to $\frac{3}{4}$ of its initial length.

Thus, the correct option is:

Option C: decreased, $\frac{3}{4}$

To find the new terminal velocity of the larger drop formed by the coalescence of 8 smaller droplets, we need to understand the relationship between the radius of the droplets and their terminal velocity.

The terminal velocity for a small spherical droplet falling through the air is given by the Stokes' law:

$v_t = \frac{2}{9} \frac{r^2 (\rho - \rho_{\text{air}}) g}{\eta}$

where:

• $v_t$ is the terminal velocity.
• $r$ is the radius of the droplet.
• $\rho$ is the density of the droplet.
• $\rho_{\text{air}}$ is the density of the air.
• $g$ is the acceleration due to gravity.
• $\eta$ is the viscosity of the air.

Given that the radius of the small droplets is $0.01 \ \text{mm}$ and their terminal velocity is $10 \ \text{cm/s}$, we now need to determine the radius of the larger drop formed by the coalescence of 8 smaller droplets.

When droplets coalesce, the volume of the larger drop is equal to the sum of the volumes of the smaller droplets. The volume of a sphere is given by:

$V = \frac{4}{3} \pi r^3$

Therefore, the volume of the large drop (V_large) can be calculated by:

$V_{\text{large}} = 8 \times V_{\text{small}} = 8 \times \left( \frac{4}{3} \pi r_{\text{small}}^3 \right)$

Let the radius of the larger drop be $R$. Then:

$\frac{4}{3} \pi R^3 = 8 \times \left( \frac{4}{3} \pi r_{\text{small}}^3 \right)$

Simplifying, we get:

$R^3 = 8 r_{\text{small}}^3$

Taking the cube root on both sides:

$R = 2 r_{\text{small}}$

Therefore, the radius of the larger drop is twice the radius of the smaller droplet:

$R = 2 \times 0.01 \ \text{mm} = 0.02 \ \text{mm}$

The terminal velocity of a droplet is proportional to the square of its radius. Therefore:

$v_{t_{\text{large}}} \propto R^2$

Given that the terminal velocity of the smaller droplets is 10 cm/s, the terminal velocity of the larger drop (formed by coalescing 8 smaller droplets) is:

$v_{t_{\text{large}}} = 10 \ \text{cm/s} \times \left( \frac{R}{r_{\text{small}}} \right)^2$

$v_{t_{\text{large}}} = 10 \ \text{cm/s} \times \left( \frac{0.02 \ \text{mm}}{0.01 \ \text{mm}} \right)^2$

$v_{t_{\text{large}}} = 10 \ \text{cm/s} \times \left( 2 \right)^2$

$v_{t_{\text{large}}} = 10 \ \text{cm/s} \times 4$

$v_{t_{\text{large}}} = 40 \ \text{cm/s}$

Therefore, the new terminal velocity of the larger drop will be 40 cm/s.

To solve this question, we will use the fact that for an object in a circular orbit, the centripetal force required to keep the object in orbit is provided by the gravitational force between the object and the planet it is orbiting. This principle gives us the relationship between the speed of the satellite, its orbital radius, and the mass of the planet.

The formula for the gravitational force is given by:

$F = \frac{G M m}{r^2}$

where:

• $F$ is the gravitational force between the two masses,
• $G$ is the gravitational constant,
• $M$ is the mass of the planet,
• $m$ is the mass of the satellite, and
• $r$ is the distance between the center of the planet and the satellite (radius of the orbit).

The centripetal force required to keep the satellite in orbit is given by:

$F_c = \frac{m v^2}{r}$

where:

• $F_c$ is the centripetal force,
• $v$ is the orbital speed of the satellite, and
• $r$ is the radius of the orbit.

Since the gravitational force is providing the centripetal force, we can set the two forces equal to each other to find the relationship between speed and radius:

$\frac{G M m}{r^2} = \frac{m v^2}{r}$

By simplifying this, we find the equation for the orbital speed of a satellite:

$v = \sqrt{\frac{G M}{r}}$

Now, we can compare the speeds of satellites $A$ and $B$ based on their radii. Satellite $A$ orbits at a radius of $4R$ with a speed of $3v$, and we need to find the speed of satellite $B$ which orbits at a radius of $R$.

Substituting the respective radii into the speed equation:

• For satellite $A$, $v_A = 3v = \sqrt{\frac{G M}{4R}}$
• For satellite $B$, we want to find $v_B$ given that its radius is $R$: $v_B = \sqrt{\frac{G M}{R}}$

To find the ratio of $v_B$ to $3v$ (or the speed of $A$), we can write:

$v_B = \sqrt{\frac{G M}{R}} = \sqrt{4} \sqrt{\frac{G M}{4R}} = 2 \cdot \sqrt{\frac{G M}{4R}}$

Given that $\sqrt{\frac{G M}{4R}}$ is the speed of satellite $A$ divided by 3 ($3v$), we find that:

$v_B = 2 \cdot 3v = 6v$

Therefore, the correct answer is Option A: $6 v$.

$\begin{gathered} R_H=\frac{100 \times 100}{1000}=10 \Omega \\ i_H=\sqrt{\frac{62.5}{10}}=2.5 \mathrm{~A} \\ V_H=25 \mathrm{~V} \\ \Rightarrow \quad i_b=\frac{75}{10}=7.5 \mathrm{~A} \\ \Rightarrow \quad i_R=5 \mathrm{~A} \\ R=\frac{25}{5}=5 \Omega \end{gathered}$

$$\begin{aligned} & v_\omega(1)+v_k(0.8)=\left(v_\omega+v_k\right)(0.9) \\ & \Rightarrow \frac{v_\omega}{v_k}=1 \end{aligned}$$

To find the ratio of the magnetic field at $\frac{a}{2}$ and $2a$ distances from the axis of a long straight wire, we use Ampère's Law, which relates the magnetic field around a current-carrying conductor to the current enclosed by it.

Ampère’s Law is given by:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$

where:

$\vec{B}$ is the magnetic field,

$\mu_0$ is the permeability of free space,

$I_{\text{enc}}$ is the enclosed current.

For a point inside the wire (at radius $r=a/2$):

The current enclosed by a radius $r$ is proportional to the area of the cross-section at radius $r$.

The area of the cross-section at radius $r$ is given by:

$\pi \left( \frac{a}{2} \right)^2 = \frac{\pi a^2}{4}$

The total current $I$ is uniformly distributed, thus the current enclosed $I_{\text{enc}}$ at radius $r = \frac{a}{2}$ is:

$I_{\text{enc}} = I \times \frac{\text{Area enclosed}}{\text{Total area}} = I \times \frac{\frac{\pi a^2}{4}}{\pi a^2} = \frac{I}{4}$

Applying Ampère’s Law inside the conductor, we get:

$B \cdot 2 \pi \left( \frac{a}{2} \right) = \mu_0 \left( \frac{I}{4} \right)$

So,

$B \cdot \pi a = \frac{\mu_0 I}{4}$

Therefore, the magnetic field inside the wire at $r = \frac{a}{2}$ is:

$B_{\frac{a}{2}} = \frac{\mu_0 I}{4 \pi a}$

For a point outside the wire (at radius $r = 2a$):

The total current enclosed by a radius $r = 2a$ is the entire current $I$.

Applying Ampère’s Law outside the conductor, we get:

$B \cdot 2 \pi (2a) = \mu_0 I$

So,

$B \cdot 4 \pi a = \mu_0 I$

Therefore, the magnetic field outside the wire at $r = 2a$ is:

$B_{2a} = \frac{\mu_0 I}{4 \pi a}$

Hence, the ratio of the magnetic field at $\frac{a}{2}$ and $2a$ is:

$\frac{B_{\frac{a}{2}}}{B_{2a}} = \frac{\frac{\mu_0 I}{4 \pi a}}{\frac{\mu_0 I}{4 \pi a}} = 1:1$

So, the correct option is:

Option C: $1:1$

1. Pitch of the screw gauge: 1 mm


2. Number of divisions on the circular scale: 100 divisions


3. Zero error: The zero of the circular scale lies 5 divisions below the reference line, indicating a positive zero error.


4. Measurement data:

• 4 linear scale divisions are visible.


• 60 divisions on the circular scale coincide with the reference line.

Step-by-Step Calculation

1. Least Count of the screw gauge:

$\text{Least Count} = \frac{\text{Pitch}}{\text{Number of Divisions on Circular Scale}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$

1. Main Scale Reading (MSR):

The linear scale shows 4 divisions, so the main scale reading is:

$\text{MSR} = 4 \text{ mm}$

1. Circular Scale Reading (CSR):

60 divisions coincide with the reference line, so the circular scale reading is:

$\text{CSR} = 60 \times \text{Least Count} = 60 \times 0.01 \text{ mm} = 0.60 \text{ mm}$

1. Zero Error:

The zero error is 5 divisions below the reference line, indicating a positive zero error:

$\text{Zero Error} = +5 \times \text{Least Count} = +5 \times 0.01 \text{ mm} = +0.05 \text{ mm}$

1. Total Reading without considering zero error:

$\text{Total Reading (without zero error)} = \text{MSR} + \text{CSR} = 4 \text{ mm} + 0.60 \text{ mm} = 4.60 \text{ mm}$

1. Corrected Reading considering zero error:

Since the zero error is positive, we subtract it from the total reading:

$\text{Corrected Reading} = \text{Total Reading (without zero error)} - \text{Zero Error} = 4.60 \text{ mm} - 0.05 \text{ mm} = 4.55 \text{ mm}$

Conclusion

The diameter of the wire is:

Option C: 4.55 mm

To find the angle of projection for a projectile to have the same horizontal range and maximum height, we need to express both the range and the maximum height in terms of the projectile's initial velocity and the angle of projection, and then set them equal to each other.

The formula for the horizontal range $R$ of a projectile is given by:

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

where $v$ is the initial velocity of the projectile, $g$ is the acceleration due to gravity, and $\theta$ is the angle of projection.

The formula for the maximum height $H$ reached by the projectile is:

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

To have the same numerical value for $R$ and $H$, we set them equal to each other:

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

Simplifying this equation, we get:

$2 \sin 2\theta = \sin^2\theta.$

Using the double-angle formula, $\sin 2\theta = 2 \sin\theta \cos\theta$, we can rewrite the equation as:

$4 \sin\theta \cos\theta = \sin^2\theta.$

This can be simplified further to:

$4 \sin\theta \cos\theta = (\sin\theta)^2.$

Dividing both sides by $\sin\theta$ (assuming $\sin\theta \neq 0$), we get:

$4 \cos\theta = \sin\theta.$

Now, dividing both sides by $\cos\theta$, we have:

$4 = \tan\theta.$

So, the angle of projection $\theta$ is:

$\theta = \tan^{-1}(4).$

Therefore, the correct answer is:

Option D $\tan ^{-1}(4)$.

To find the inductance of the coil, we need to analyze the given circuit and use the information provided. The circuit consists of a resistor and an inductor in series, connected to an AC supply. Here are the given values:

1. Resistance, $R = 90 \Omega$

2. Supply voltage, $V_{\text{total}} = 120 \mathrm{~V}$

3. Frequency, $f = 60 \mathrm{~Hz}$

4. Voltage across the resistor, $V_R = 36 \mathrm{~V}$

First, we calculate the current through the resistor (which is the same as the current through the inductor, since they are in series) using Ohm's law:

$I = \frac{V_R}{R} = \frac{36}{90} = 0.4 \mathrm{~A}$

Next, we find the total impedance $Z$ of the series combination from the total supply voltage:

$V_{\text{total}} = I \cdot Z$

$Z = \frac{V_{\text{total}}}{I} = \frac{120}{0.4} = 300 \Omega$

We know that the total impedance in a series circuit consisting of a resistor and an inductor is given by:

$Z = \sqrt{R^2 + (X_L)^2}$

where $X_L$ is the inductive reactance. Rearrange this to solve for $X_L$:

$X_L = \sqrt{Z^2 - R^2} = \sqrt{(300)^2 - (90)^2} = \sqrt{90000 - 8100} = \sqrt{81900} \approx 286 \Omega$

Now, we use the inductive reactance formula to find the inductance $L$:

$X_L = 2\pi f L$

$L = \frac{X_L}{2\pi f} = \frac{286}{2 \pi \cdot 60} \approx \frac{286}{376.99} \approx 0.76 \mathrm{~H}$

Thus, the inductance of the coil is:

Option B: 0.76 H

To find the frequency of the plane progressive wave given by the equation $y = 2 \cos 2 \pi(330 \mathrm{t} - x) \mathrm{m}$, we start by analyzing the general form of a wave equation.

The general form of a wave equation is:

where:

• $A$ is the amplitude of the wave.
• $f$ is the frequency of the wave.
• $t$ is the time variable.
• $k$ is the wave number, defined as $\frac{2 \pi}{\lambda}$ where $\lambda$ is the wavelength.
• $x$ is the spatial variable.
• $\phi$ is the phase constant.

By comparing the given wave equation with the general form, we have:

We observe that the term $2 \pi(330t - x)$ corresponds to $2 \pi ft - kx$ in the general form.

From this, it is clear that:

• $2 \pi ft \rightarrow 2 \pi \cdot 330 t$
• Therefore, $f = 330$ Hz

So, the frequency of the wave is 330 Hz.

Thus, the correct answer is:

Option C: 330 Hz

To determine the dimensions of $\epsilon_{\mathrm{o}} \mathrm{E}^2$, we need to first understand the dimensional formulas of each component in the expression.

1. Permittivity of free space, $\epsilon_{\mathrm{o}}$:

The permittivity of free space has the dimensions: $\left[\epsilon_{\mathrm{o}}\right] = [\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2]$.

2. Electric field, $\mathrm{E}$:

The electric field $\mathrm{E}$ has the dimensions: $\left[\mathrm{E}\right] = [\mathrm{M} \mathrm{L} \mathrm{T}^{-3} \mathrm{A}^{-1}]$.

Now, let's calculate the dimensions of $\epsilon_{\mathrm{o}} \mathrm{E}^2$:

\begin{equation} \left[\epsilon_{\mathrm{o}} \mathrm{E}^2 \right] = \left[\epsilon_{\mathrm{o}}\right] \left[\mathrm{E}\right]^2 = \left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2\right] \left([\mathrm{M} \mathrm{L} \mathrm{T}^{-3} \mathrm{A}^{-1}]^2\right] = \left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2\right] \left([\mathrm{M}^2 \mathrm{L}^2 \mathrm{T}^{-6} \mathrm{A}^{-2}]\right] \end{equation}

Combining the dimensions:

\begin{equation} \left[\epsilon_{\mathrm{o}} \mathrm{E}^2 \right] = \left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2\right] \left[\mathrm{M}^2 \mathrm{L}^2 \mathrm{T}^{-6} \mathrm{A}^{-2}\right] = \left[\mathrm{M}^{-1 + 2} \mathrm{L}^{-3 + 2} \mathrm{T}^{4 - 6} \mathrm{A}^{2 - 2}\right] = \left[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}\right] \end{equation}

Therefore, the dimensions of $\epsilon_{\mathrm{o}} \mathrm{E}^2$ are $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$, which corresponds to Option A.

Hence, the correct answer is Option A: $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$.

For minimum length of $P Q$,

$$\begin{aligned} & \mathrm{V}_{\text {diodes }}=1.8+3.2=5 \mathrm{~V} \\ & \therefore \quad I_{P Q}=\frac{20}{20} \times 5 \\ & =5 \mathrm{~cm} \\ & \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad & E_2 \cos \theta=E_3 \cos \phi \\ & \frac{q_2}{\left(2^2+4^2\right)} \frac{4}{\sqrt{2^2+4^2}}=\frac{q_3}{4^2+3^2} \frac{4}{\sqrt{4^2+3^2}} \\ & \frac{q_2}{q_3}=\frac{20}{25} \frac{\sqrt{20}}{5}=\frac{4}{5} \times \frac{2 \sqrt{5}}{5} \\ \Rightarrow \quad & n=5 \end{aligned}$$

To determine the nuclear binding energy of the isotope $_5^{12}B$, we need to consider the mass defect concept. The mass defect is the difference between the sum of the individual masses of nucleons (protons and neutrons) and the actual mass of the nucleus.

Let's calculate the mass defect first. The isotope $_5^{12}B$ has 5 protons and 7 neutrons (since the total number of nucleons is 12). Therefore, the mass defect can be written as:

$\Delta M = (5 M_p + 7 M_n) - M_0$

Once we have the mass defect, the binding energy can be found using Einstein's mass-energy equivalence principle, which is given by:

$E = \Delta M \cdot c^2$

Substituting the mass defect into this equation, we get:

$E = ((5 M_p + 7 M_n) - M_0) \cdot c^2$

Therefore, the correct answer is Option A:

$(5 M_p + 7 M_n - M_0) \cdot c^2$

$$\begin{aligned} & m v_r \frac{d v_r}{d r}=m \omega^2 r \\ & \frac{v_v^2}{2}=\frac{\omega^2\left(r^2-1\right)}{2} \\ & v_r=\omega \sqrt{8}=2 \omega \sqrt{2} \end{aligned}$$

To determine the correct relation between the kinetic energies of a proton ($\mathrm{K}_{\mathrm{p}}$) and an electron ($\mathrm{K}_{\mathrm{e}}$) when they have the same de Broglie wavelength, we need to use the de Broglie wavelength formula:

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

where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, and $p$ is the momentum of the particle.

The momentum $p$ of a particle is given by:

$p = \sqrt{2mK}$

where $m$ is the mass of the particle and $K$ is its kinetic energy. For a proton and an electron with the same de Broglie wavelength:

$\lambda_{\mathrm{p}} = \lambda_{\mathrm{e}}$

This implies the momenta should be the same:

$p_{\mathrm{p}} = p_{\mathrm{e}}$

Thus, we can write:

$\sqrt{2 m_{\mathrm{p}} K_{\mathrm{p}}} = \sqrt{2 m_{\mathrm{e}} K_{\mathrm{e}}}$

Squaring both sides to eliminate the square roots:

$2 m_{\mathrm{p}} K_{\mathrm{p}} = 2 m_{\mathrm{e}} K_{\mathrm{e}}$

$m_{\mathrm{p}} K_{\mathrm{p}} = m_{\mathrm{e}} K_{\mathrm{e}}$

Rearranging to solve for $K_{\mathrm{p}}$ in terms of $K_{\mathrm{e}}$:

$K_{\mathrm{p}} = \frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} K_{\mathrm{e}}$

Since the mass of a proton $m_{\mathrm{p}}$ is much greater than the mass of an electron $m_{\mathrm{e}}$:

$m_{\mathrm{p}} \gg m_{\mathrm{e}}$

This means:

$\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}} \ll 1$

Therefore, it implies:

$K_{\mathrm{p}} < K_{\mathrm{e}}$

So, the correct option is:

Option C: $\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}$

To solve for the amplitude of oscillation, we start by using the properties of simple harmonic motion (SHM). In SHM, the total energy of the system is conserved and is given by the sum of kinetic energy (KE) and potential energy (PE).

Given:

• Mass $m = 0.2 \ \mathrm{kg}$
• Frequency $f = \left(\frac{25}{\pi}\right) \ \mathrm{Hz}$
• Position $x = 0.04 \ \mathrm{m}$
• KE at $x = 0.04 \ \mathrm{m}$ is $0.5 \ \mathrm{J}$
• PE at $x = 0.04 \ \mathrm{m}$ is $0.4 \ \mathrm{J}$

The total mechanical energy (E) of the SHM system can be found by summing the given kinetic and potential energies:

$E = KE + PE = 0.5 \ \mathrm{J} + 0.4 \ \mathrm{J} = 0.9 \ \mathrm{J}$

For simple harmonic motion, the total energy (E) is also related to the amplitude (A) by the following formula:

$E = \frac{1}{2} k A^2$

where $k$ is the spring constant. First, we need to find the angular frequency $\omega$:

$\omega = 2 \pi f = 2 \pi \left(\frac{25}{\pi}\right) \ \mathrm{Hz} = 50 \ \mathrm{rad/s}$

The spring constant $k$ can be calculated using the relationship between $m$, $\omega$, and $k$:

$\omega = \sqrt{\frac{k}{m}} \Rightarrow k = m \omega^2 = 0.2 \times (50)^2 = 500 \ \mathrm{N/m}$

Now, substituting $k$ back into the energy equation, we solve for the amplitude $A$:

$0.9 = \frac{1}{2} \times 500 \times A^2 \Rightarrow A^2 = \frac{0.9 \times 2}{500} \Rightarrow A^2 = \frac{1.8}{500} \Rightarrow A^2 = 0.0036 \Rightarrow A = \sqrt{0.0036} = 0.06 \ \mathrm{m}$

Converting $A$ from meters to centimeters:

$A = 0.06 \ \mathrm{m} \times 100 = 6 \ \mathrm{cm}$

Thus, the amplitude of oscillation is $6 \ \mathrm{cm}$.

To find the heat given to a diatomic gas during an isobaric (constant pressure) expansion, we can use the formula that relates the work done by the gas, the heat added to the system, and the change in the internal energy of the system. The first law of thermodynamics states that:

$\Delta Q = \Delta U + W$

where:

• $\Delta Q$ is the heat added to the system,
• $\Delta U$ is the change in internal energy of the system, and
• $W$ is the work done by the gas.

For an isobaric process, the work done $W$ is given by:

$W = P \Delta V$

We're given that $W = 100 \ \mathrm{J}$ for this process, so:

$W = 100 \ \mathrm{J}$

The change in internal energy $\Delta U$ for an ideal gas can also be related to the temperature change and the specific heat capacity at constant volume $C_v$. Using the equation:

$\Delta U = nC_v\Delta T$

However, without direct values for $n$, $\Delta T$, or $C_v$, we need to rely on the relation between the provided work and the heat capacity ratio $\gamma$ to find the heat added. For a diatomic gas, $\gamma = C_p/C_v$. The heat added at constant pressure can also be described as:

$\Delta Q = nC_p\Delta T$

Since we know that for an ideal gas, the work done on the gas during an isobaric process is related to the heat added by the ratio of the specific heats ($\gamma$), we can use the fact that $C_p = \gamma C_v$, and relating that to the work done, we get:

$\Delta Q = \frac{\gamma}{\gamma - 1} W$

Substituting the known values, with $\gamma = 1.4$, and $W = 100 \ \mathrm{J}$, we find:

$\Delta Q = \frac{1.4}{1.4 - 1} \times 100 \ \mathrm{J} = \frac{1.4}{0.4} \times 100 \ \mathrm{J} = 3.5 \times 100 \ \mathrm{J} = 350 \ \mathrm{J}$

Therefore, the heat given to the gas is $\Delta Q = 350 \ \mathrm{J}$, so the correct option is:

Option C

350 J

In the given hypothetical fission reaction:

$^{236}_{92} X \rightarrow \, ^{141}_{56} Y + \, ^{92}_{36} Z + 3 R$

We need to determine the identity of particles denoted by $R$. Let's use the conservation of charge and mass number (nucleon number) to identify $R$.

First, for the conservation of nucleon number (mass number), we have:

$236 = 141 + 92 + 3 \times A_R$

Where $A_R$ is the mass number of $R$. This simplifies to:

$236 = 233 + 3A_R$

$3A_R = 236 - 233$

$3A_R = 3$

$A_R = 1$

Next, we use the conservation of charge (atomic number), we have:

$92 = 56 + 36 + 3Z_R$

Where $Z_R$ is the atomic number of $R$. This simplifies to:

$92 = 92 + 3Z_R$

$3Z_R = 92 - 92$

$3Z_R = 0$

$Z_R = 0$

Since the particle $R$ has a mass number of 1 and atomic number of 0, it must be a neutron.

So, the identity of the emitted particles $R$ is:

Option B: Neutron

$$\begin{aligned} & \text { Width central maxima }=10 \text { fringe width }=\frac{2 \lambda \theta}{a}=10 \beta \\ & \frac{2 \lambda \theta}{a}=10 \times \frac{\lambda D}{d} \\ & a=\frac{d}{5}=\frac{10^{-3}}{5}=2 \times 10^{-4} \end{aligned}$$

To determine the new angular velocity of the system, we use the principle of conservation of angular momentum. When no external torque acts on a system, its angular momentum remains constant. Let's denote the initial angular momentum and the final angular momentum, respectively, as $L_{\text{initial}}$ and $L_{\text{final}}$.

The initial angular momentum of the system is given by:

$L_{\text{initial}} = I_{\text{initial}} \cdot \omega$

Here, $I_{\text{initial}}$ is the moment of inertia of the first disc. For a thin circular disc, the moment of inertia about its center is:

$I_{\text{initial}} = \frac{1}{2} M R^2$

Thus,

$L_{\text{initial}} = \left( \frac{1}{2} M R^2 \right) \omega$

When the second disc is placed gently on the first disc, the two discs rotate together with a common angular velocity $\omega'$. The moment of inertia of the second disc is:

$I_{\text{second}} = \frac{1}{2} \left( \frac{M}{2} \right) R^2 = \frac{1}{4} M R^2$

The combined moment of inertia of the system after placing the second disc is:

$I_{\text{final}} = I_{\text{initial}} + I_{\text{second}} = \frac{1}{2} M R^2 + \frac{1}{4} M R^2 = \frac{3}{4} M R^2$

Thus, the final angular momentum of the system is:

$L_{\text{final}} = I_{\text{final}} \cdot \omega' = \left( \frac{3}{4} M R^2 \right) \omega'$

By the conservation of angular momentum:

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

This simplifies to:

$\left( \frac{1}{2} M R^2 \right) \omega = \left( \frac{3}{4} M R^2 \right) \omega'$

Solving for $\omega'$:

$\omega' = \frac{\left( \frac{1}{2} M R^2 \right) \omega}{\left( \frac{3}{4} M R^2 \right)} = \frac{\omega}{\frac{3}{2}} = \frac{2}{3} \omega$

So, the new angular velocity of the system is:

$\omega' = \frac{2}{3} \omega$

Thus, the correct answer is:

Option D: $\frac{2}{3} \omega$