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

In expression $a \times 10^b$, If $a \leq 5 ; a \approx 1$ by round off

$\Rightarrow$ Order $B$

The critical angle is the angle of incidence at which light is refracted along the boundary, meaning the angle of refraction is $90^{\circ}$. The relationship between the critical angle and the refractive indices of two media can be understood using Snell's Law, given as:

$n_1 \sin(\theta_c) = n_2 \sin(90^{\circ})$

Here, $n_1$ is the refractive index of the first medium, $n_2$ is the refractive index of the second medium, and $\theta_c$ is the critical angle. Since $\sin(90^{\circ}) = 1$, the equation simplifies to:

$n_1 \sin(\theta_c) = n_2$

Given that the critical angle $\theta_c$ is $45^{\circ}$, we have:

$\sin(45^{\circ}) = \frac{\sqrt{2}}{2}$

Substituting this value in the simplified Snell's Law equation, we get:

$n_1 \left( \frac{\sqrt{2}}{2} \right) = n_2$

Therefore,

$n_1 = n_2 \cdot \frac{2}{\sqrt{2}}$

$n_1 = n_2 \cdot \sqrt{2}$

Hence, the ratio of the refractive indices of the first and second media is:

$n_1 : n_2 = \sqrt{2} : 1$

The correct answer is therefore:

Option D

$\sqrt{2}: 1$

Bernoulli's equation relates the pressure, velocity, and height in a flowing fluid and is derived from the principle of conservation of energy. The correct form of Bernoulli's equation is:

$P + \rho g h + \frac{1}{2} \rho v^2 = \text{constant}$

where:

• P is the pressure within the fluid.
• $\rho$ is the density of the fluid.
• g is the acceleration due to gravity.
• h is the height above a reference point.
• v is the velocity of the fluid.

Looking at the options given:

1. Option A: $P + \frac{1}{2} \rho g h + \frac{1}{2} \rho v^2 = \text{constant}$
2. Option B: $P + m g h + \frac{1}{2} m v^2 = \text{constant}$
3. Option C: $P + \rho g h + \rho v^2 = \text{constant}$
4. Option D: $P + \rho g h + \frac{1}{2} \rho v^2 = \text{constant}$

Option D correctly represents Bernoulli's equation in its proper form. Therefore, the correct answer is:

Option D

Paramagnetic substances exhibit specific characteristics in the presence of an external magnetic field. Understanding these characteristics will help us choose the most appropriate answer. Let's discuss each statement individually:

A. align themselves along the directions of external magnetic field.
This is true. Paramagnetic substances have unpaired electrons, and under the influence of an external magnetic field, the atomic dipoles (due to those unpaired electrons) tend to align themselves in the direction of the magnetic field.

B. attract strongly towards external magnetic field.
This statement is the cause of some confusion. Paramagnetic substances are attracted towards an external magnetic field, but the keyword here is strongly. Compared to ferromagnetic substances, the attraction is relatively weak. Given the context of the options provided, saying they are "strongly" attracted could be misleading, but it is undeniable they are attracted nevertheless.

C. has susceptibility little more than zero.
This is correct. The magnetic susceptibility of paramagnetic materials is positive, meaning they are attracted to magnetic fields, but it is small in magnitude, typically greater than zero but far less than the susceptibility of ferromagnetic materials.

D. move from a region of strong magnetic field to weak magnetic field.
This statement is incorrect for paramagnetic substances. They are attracted to magnetic fields, implying they move from a region of weaker magnetic field to a stronger magnetic field, not the other way around.

Taking these points into account, the most accurate answer would be:

Options B: A, C Only
A) They indeed align along the direction of an external magnetic field, and C) Their susceptibility is indeed a little more than zero, indicating a weak attraction to magnetic fields. Statement B can be considered incorrect not in its entirety that they are attracted, but in the use of the term "strongly," and D is certainly incorrect regarding their movement in magnetic fields.

To determine the percentage error in Young's modulus, we need to first understand the propagation of errors in the given formula.

Given the equation:

$\mathrm{Y}=49000 \frac{\mathrm{M}}{\mathrm{l}} \frac{\mathrm{dyne}}{\mathrm{cm}^2}$

where:

• $M$ is the mass (with its value given as $500 \mathrm{~g}$)
• $l$ is the extension (with its value given as $2 \mathrm{~cm}$)

The errors in the measurements are determined by the smallest scale divisions on the graph paper, which are:

• $5 \mathrm{~g}$ for the load axis
• $0.02 \mathrm{~cm}$ for the extension axis

To find the percentage error in Young's modulus ($Y$), we need to compute the relative errors in the measurements $M$ and $l$, and then propagate these errors through the given formula.

The relative error in $M$ is:

$\frac{\Delta M}{M} = \frac{5 \mathrm{~g}}{500 \mathrm{~g}} = 0.01$

The relative error in $l$ is:

$\frac{\Delta l}{l} = \frac{0.02 \mathrm{~cm}}{2 \mathrm{~cm}} = 0.01$

Since $Y$ is proportional to $M$ and inversely proportional to $l$, the overall percentage error in $Y$ is the sum of the percentage errors in $M$ and $l$:

$\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta l}{l} = 0.01 + 0.01 = 0.02$

To express this as a percentage, we multiply by 100:

$\text{Percentage error in } Y = 0.02 \times 100 = 2\%$

Thus, the percentage error in Young's modulus $Y$ is:

Option A: 2%

To solve for the area of the surface, we need to understand the relationship between the force exerted by the light, the light energy flux, and the area of the surface. The pressure exerted by the light on a non-reflecting surface is given by the formula:

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

where $P$ is the pressure, $F$ is the force, and $A$ is the area. The pressure due to the light can also be related to the energy flux $I$ by the relationship:

$P = \frac{I}{c}$

where $I$ is the light energy flux and $c$ is the speed of light in a vacuum ($3 \times 10^8 \mathrm{~m/s}$).

Given that the average force $F$ is $2.4 \times 10^{-4} \mathrm{~N}$ and the light energy flux $I$ is $360 \mathrm{~W/cm}^2$, we first convert the flux to $\mathrm{W/m}^2$:

$360 \mathrm{~W/cm}^2 = 360 \times 10^4 \mathrm{~W/m}^2$

Now we can use the relation between pressure and energy flux:

$P = \frac{I}{c} = \frac{360 \times 10^4}{3 \times 10^8} = 1.2 \mathrm{~N/m}^2$

Next, we use the pressure formula to find the area of the surface:

$P = \frac{F}{A} \implies A = \frac{F}{P}$

Substituting the values we have:

$A = \frac{2.4 \times 10^{-4} \mathrm{~N}}{1.2 \mathrm{~N/m}^2} = 2 \times 10^{-4} \mathrm{~m}^2$

The area of the surface is:

$A = 2 \times 10^{-4} \mathrm{~m}^2 = 0.02 \mathrm{~m}^2$

Hence, the correct option is:

Option D $0.02 \mathrm{~m}^2$

$$\begin{aligned} & \frac{v_1}{v_2}=\sqrt{\frac{r_2}{r_1}} \quad \text{.... (i)}\\ & m_1 v_1 r_1=h \\ & m_2 v_2 r_2=32 \\ & \Rightarrow \frac{v_1}{v_2}=\frac{1}{3} \frac{m_2}{m_1} \frac{r_2}{r_1} \quad \text{.... (ii)} \end{aligned}$$

From (i) & (ii)

$$\begin{aligned} & \sqrt{\frac{r_2}{r_1}}=\frac{1}{3} \frac{m_2}{m_1} \frac{r_2}{r_1} \\ & \frac{3 m_1}{m_2}=\sqrt{\frac{r_2}{r_1}} \\ & \frac{T_1}{T_2}=\left(\frac{r_1}{r_2}\right)^{3 / 2}=\left(\frac{m_2}{3 m_1}\right)=\frac{1}{27}\left(\frac{m_2}{m_1}\right)^3 \end{aligned}$$

When two conducting spheres of radii $a$ and $b$ are connected by a conducting wire, they come to the same potential because conductors in contact share charges until their potentials become equal. The potential of a charged sphere is given by $V = \frac{kQ}{R}$, where $V$ is the potential, $k$ is Coulomb's constant, $Q$ is the charge on the sphere, and $R$ is the radius of the sphere.

For the two spheres at the same potential, we have:

$\frac{kQ_a}{a} = \frac{kQ_b}{b}$

Where:

• $Q_a$ and $Q_b$ are the charges on the spheres with radii $a$ and $b$, respectively.
• $k$ cancels out from both sides as it is a constant.

From the above equation, to find the ratio of charges $\frac{Q_a}{Q_b}$, we rearrange it as follows:

$\frac{Q_a}{Q_b} = \frac{a}{b}$

Therefore, the ratio of the charges of the two spheres respectively is $\frac{a}{b}$.

So, the correct answer is Option C: $\frac{a}{b}$.

$$\begin{aligned} & f_o=\frac{v}{2 I} \quad \Rightarrow \quad f_{o_7}=8 \frac{v}{2 l} \\ & f_c=\frac{v}{4 I} \quad \Rightarrow \quad f_{c_7}=15 \frac{v}{4 I} \\ & \frac{f_{c_7}}{f_{o_7}}=15 \frac{v}{4 I} \frac{2 l}{8 v}=\frac{30}{32}=\frac{15}{16} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & (\vec{P}+\vec{Q})=\sqrt{2} A \\ & R=A \\ & \theta=90^{\circ} \end{aligned}\\ &\text { Resultant }=\sqrt{3} A \end{aligned}$$

To find the density of the sphere, we need to calculate its volume using the measured diameter. Then, using the mass of the sphere, we can calculate the density using the formula for density, which is $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$. Let's start by finding the accurate measurement of the diameter using the given vernier calipers readings.

The least count (LC) of the vernier calipers can be calculated using the formula:

$\text{LC} = \frac{\text{Value of one main scale division (MSD)}}{\text{Number of vernier scale divisions (VSD) that match with the main scale}}$

Given that 9 divisions of the main scale are equal to 10 divisions of the vernier scale and the shortest division on the main scale is equal to $1\, \mathrm{mm}$, we find:

$\text{LC} = \frac{1\, \mathrm{mm}}{10} = 0.1\, \mathrm{mm} = 0.01\, \mathrm{cm}$

For the main scale reading (MSR) of $2\, \mathrm{cm}$ and the second division of the vernier scale coinciding with a division on the main scale, the vernier scale reading (VSR) can be expressed as $\text{VSR} = 2 \times \text{LC}$.

So, $\text{VSR} = 2 \times 0.01\, \mathrm{cm} = 0.02\, \mathrm{cm}$.

The total measurement of the diameter (D) can be found by adding MSR and VSR:

$D = \text{MSR} + \text{VSR}$

$D = 2\, \mathrm{cm} + 0.02\, \mathrm{cm} = 2.02\, \mathrm{cm}$

Now, we can calculate the volume (V) of the sphere using its diameter with the formula $V = \frac{4}{3}\pi r^3$, where $r$ is the radius of the sphere. Remembering that the radius is half of the diameter, $r = \frac{D}{2} = \frac{2.02\, \mathrm{cm}}{2} = 1.01\, \mathrm{cm}$.

Therefore, $V = \frac{4}{3}\pi (1.01\, \mathrm{cm})^3$,

Calculating the volume,

$V = \frac{4}{3}\pi (1.01)^3\, \mathrm{cm}^3 \approx \frac{4}{3} \times 3.1416 \times 1.0303\, \mathrm{cm}^3 = \frac{4}{3} \times 3.1416 \times 1.0303\, \mathrm{cm}^3 \approx 4.3434\, \mathrm{cm}^3$

Now to find the density ($\rho$) using the mass ($M = 8.635\, \mathrm{g}$) and volume ($V = 4.3434\, \mathrm{cm}^3$) calculated,

$\rho = \frac{M}{V} = \frac{8.635\, \mathrm{g}}{4.3434\, \mathrm{cm}^3} \approx 1.988\, \mathrm{g}/\mathrm{cm}^3$

Comparing this result to the given options, the closest value is:

Option B $2.0\, \mathrm{g}/\mathrm{cm}^3$

To determine the ratio of the kinetic energies of two particles after a stationary particle breaks into them, we can use the principle of conservation of momentum and the definition of kinetic energy.

Initially, since the particle is stationary, its total momentum is zero. Therefore, after the break, the total momentum of the two parts must also be zero. Mathematically, this can be written as:

$m_A v_A + m_B v_B = 0$

Solving for one of the velocities, we get:

$v_A = -\frac{m_B v_B}{m_A}$

The kinetic energies of the two parts are:

$K_A = \frac{1}{2} m_A v_A^2$

$K_B = \frac{1}{2} m_B v_B^2$

We need to find the ratio $\left(K_B: K_A\right)$. Substituting the expressions for $K_A$ and $K_B$, we get:

$\frac{K_B}{K_A} = \frac{\frac{1}{2} m_B v_B^2}{\frac{1}{2} m_A v_A^2} = \frac{m_B v_B^2}{m_A v_A^2}$

Using the expression we derived for $v_A$, we substitute it into the ratio:

$\frac{K_B}{K_A} = \frac{m_B v_B^2}{m_A \left( \frac{m_B v_B}{m_A} \right)^2} = \frac{m_B v_B^2}{m_A \frac{m_B^2 v_B^2}{m_A^2}} = \frac{m_B v_B^2 m_A^2}{m_A m_B^2 v_B^2} = \frac{m_A}{m_B}$

Thus, the ratio of their kinetic energies $\left(K_B: K_A\right)$ is:

$m_B: m_A$

Therefore, the correct answer is:

Option D: $m_B: m_A$

To solve for the ratio of the kinetic energies of a proton and an electron with the same de-Broglie wavelength, let us first recall the relationship between kinetic energy, momentum, and the de-Broglie wavelength.

The de-Broglie wavelength $\lambda$ is given by:

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

where $h$ is Planck’s constant and $p$ is the momentum.

Since the proton and the electron have the same de-Broglie wavelength, their momenta must be equal:

$\lambda_{\text{electron}} = \lambda_{\text{proton}} \implies \frac{h}{p_{\text{e}}} = \frac{h}{p_{\text{p}}} \implies p_{\text{e}} = p_{\text{p}}$

Next, the kinetic energy $K$ of a particle is related to its momentum $p$ and mass $m$ by the following formula:

$K = \frac{p^2}{2m}$

Given that the momentum $p$ is the same for both the proton and the electron, we can write the kinetic energies as:

$K_{\text{e}} = \frac{p^2}{2m_{\text{e}}}$

$K_{\text{p}} = \frac{p^2}{2m_{\text{p}}}$

The ratio of the kinetic energies is therefore:

$\frac{K_{\text{e}}}{K_{\text{p}}} = \frac{\frac{p^2}{2m_{\text{e}}}}{\frac{p^2}{2m_{\text{p}}}} = \frac{m_{\text{p}}}{m_{\text{e}}}$

Given that the mass of the proton $m_{\text{p}}$ is 1836 times the mass of the electron $m_{\text{e}}$, we have:

$\frac{m_{\text{p}}}{m_{\text{e}}} = 1836$

Thus, the ratio of the kinetic energies is:

$\frac{K_{\text{e}}}{K_{\text{p}}} = 1836$

Therefore, the correct answer is:

Option C

$1: 1836$

To determine the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus, we need to use the concept of mass-energy equivalence provided by Einstein's famous equation:

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

where:

• E is the binding energy.
• $\Delta m$ is the mass defect (difference in mass).
• $c$ is the speed of light in vacuum, which is approximately $3 \times 10^8 \, \text{m/s}$.

Given:

• Binding energy, $E = 18 \times 10^8 \, \text{J}$.

We need to find $\Delta m$, so rearranging the equation:

$\Delta m = \frac{E}{c^2}$

Substitute the given values:

$\Delta m = \frac{18 \times 10^8}{(3 \times 10^8)^2}$

Calculate the value:

$\Delta m = \frac{18 \times 10^8}{9 \times 10^{16}}$

$\Delta m = \frac{18}{9} \times 10^{-8}$

$\Delta m = 2 \times 10^{-8} \, \text{kg}$

To convert the mass from kilograms to micrograms ($\mu g$), we use the conversion factor $1 \, \text{kg} = 10^9 \, \mu g$:

$\Delta m = 2 \times 10^{-8} \, \text{kg} \times 10^9 \, \frac{\mu g}{\text{kg}}$

$\Delta m = 2 \times 10^{1} \, \mu g$

$\Delta m = 20 \, \mu g$

Therefore, the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus is 20 $\mu g$.

Option A (20 $\mu g$) is the correct answer.

$V_T=\left(\frac{3}{1+2}\right) \times 2=2 \mathrm{~V}$

To solve this problem, we need to understand the relationship between the current amplitude in a series LCR circuit at resonance and the resistance $R$. At resonance, the impedance $Z$ of the series LCR circuit is equal to the resistance $R$, and thus:

$Z = R$

The amplitude of the current $I_0$ at resonance is given by Ohm's law:

$I_0 = \frac{V_0}{R}$

where $V_0$ is the amplitude of the voltage supplied.

Now, if the resistance $R$ is halved while keeping the voltage amplitude $V_0$, capacitance $C$, and inductance $L$ the same, the new resistance becomes:

$R_{new} = \frac{R}{2}$

The new current amplitude $I_0'$ at resonance is given by the modified Ohm's law:

$I_0' = \frac{V_0}{R_{new}} = \frac{V_0}{\frac{R}{2}} = \frac{2V_0}{R} = 2 \times I_0$

Therefore, the current amplitude at resonance will be doubled. Hence, the correct answer is:

Option D: double

The electric flux through a surface is given by the formula:

$\Phi = \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} = |\overrightarrow{\mathrm{E}}||\overrightarrow{\mathrm{A}}|\cos\theta$

where $\overrightarrow{\mathrm{E}}$ is the electric field, $\overrightarrow{\mathrm{A}}$ is the area vector (with magnitude equal to the area of the surface and direction perpendicular to the surface, defined by the unit vector $\hat{n}$), and $\theta$ is the angle between $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{A}}$. However, when using unit vectors to describe the directions of $\overrightarrow{\mathrm{E}}$ and $\hat{n}$, the dot product can be used to simplify the calculation as follows:

$\Phi = \overrightarrow{\mathrm{E}} \cdot \left(\overrightarrow{\mathrm{A}}\right) = (\overrightarrow{\mathrm{E}} \cdot \hat{n})A$

Given that $\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}$ and $\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$, and the area $A = 4 \mathrm{m}^2$, we can substitute them into our formula. Note that since $\overrightarrow{\mathrm{A}} = A\hat{n}$, the magnitude of the area vector is the area of the surface itself. First, let's find $\overrightarrow{\mathrm{E}} \cdot \hat{n}$:

$\overrightarrow{\mathrm{E}} \cdot \hat{n} = \left(\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}\right) \cdot \left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$

To compute the dot product, we multiply corresponding components and then add them up:

$\left(\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}\right) \cdot \left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right) = \frac{1}{6}(2\cdot2 + 6\cdot1 + 8\cdot1)$

$= \frac{1}{6}(4 + 6 + 8) = \frac{18}{6} = 3$

Then, the electric flux through the surface is:

$\Phi = (\overrightarrow{\mathrm{E}} \cdot \hat{n})A = 3 \times 4 \mathrm{m}^2$

$\Phi = 12 \mathrm{Vm}$

So, the electric flux for that surface is $12 \mathrm{Vm}$.

$$\begin{aligned} & A = 36 \times 10^{-4} \mathrm{~m}^2 \\ & I= 6 \times 10^{-2} \mathrm{~m} \\ & =6 \mathrm{~cm} \\ V & =\frac{6 \mathrm{~cm}}{1 \mathrm{sec}}=6 \mathrm{~cm} / \mathrm{s} \\ \varepsilon & =B / \mathrm{vn}^2=0.5 \times \frac{6}{100} \times \frac{6}{100} \\ & =18 \times 10^{-4} \mathrm{~V} \\ E & =\frac{n^2 \varepsilon^2}{R} t=100 \times \frac{18 \times 18 \times 10^{-4} \times 10^{-4}}{10^2} \times 1 \\ & =324 \times 10^{-10} \times 10^2 \\ & =3.24 \times 10^{-6} \end{aligned}$$

To find the ratio of specific heats at constant volume ($C_V$) of the gases, we need to understand the degrees of freedom each type of gas molecule has, as this determines their specific heat capacity at constant volume. Degrees of freedom refer to the number of independent ways in which a molecule can store energy.

A monoatomic gas molecule has 3 translational degrees of freedom. This is because it can move in three dimensions: x, y, and z. A diatomic gas, if we assume it to be rigid for this context, has 5 degrees of freedom: 3 translational like the monoatomic gas and 2 rotational since it can rotate around two axes perpendicular to the bond axis connecting the two atoms. Vibrational modes are not considered at room temperature for a rigid diatomic gas, as these require higher energy to become accessible.

The specific heat capacity at constant volume for a monoatomic gas is given by:

$C_{V, mono} = \frac{3}{2} R$

And for a diatomic gas, it's:

$C_{V, diatomic} = \frac{5}{2} R$

Where $R$ is the ideal gas constant.

To find the ratio of their specific heats at constant volume, we divide the specific heat of the monoatomic gas by that of the diatomic gas:

$\frac{C_{V, mono}}{C_{V, diatomic}} = \frac{\frac{3}{2} R}{\frac{5}{2} R} = \frac{3}{2} \times \frac{2}{5} = \frac{3}{5}$

Thus, the correct ratio of the specific heats at constant volume of the monoatomic to diatomic (rigid) gases is given by Option B: $\frac{3}{5}$.

$$\begin{aligned} \text { (1) } P_a V_a & =P_b V_b \quad \text{.... (i)}\\ P_c V_c & =P_d V_d \quad \text{.... (ii)} \end{aligned}$$

$$\begin{aligned} \text { (2) } P_a V_a^{\gamma-1} & =P_d V_d^{\gamma-1} \quad \text{.... (iii)}\\ P_b V_b^{\gamma-1} & =P_c V_c^{\gamma-1} \quad \text{.... (iv)} \end{aligned}$$

$$\begin{aligned} & \text { (i) } \div \text { (iii) } \Rightarrow \frac{V_a}{V_a^{\gamma-1}}=\frac{P_b V_b}{P_d V_d{ }^{\gamma-1}} \text { (v) } \\ & \text { (ii) } \div \text { (iv) } \Rightarrow \frac{V_c}{V_c^{\gamma-1}}=\frac{P_d V_d}{P_b V_b{ }^{\gamma-1}} \text { (vi) } \\ & \text { (v) } \times \text { (vi) } \Rightarrow \frac{V_a}{V_a^{\gamma-1}} \frac{V_c}{V_c{ }^{\gamma-1}}=\frac{P_b V_b}{P_d V_d^{\gamma-1}} \times \frac{P_d V_d}{P_b V_b^{\gamma-1}} \\ & \Rightarrow V_a^{\gamma-2} V_c{ }^{\gamma-2}=V_d^{\gamma-1} V_b{ }^{\gamma-2} \\ & \Rightarrow \frac{V_a}{V_d}=\frac{V_b}{V_c} \end{aligned}$$

To determine the distance traveled by the tips of the second hand and the minute hand, we need to calculate the circumference of the circles they make. Let's start by calculating the distances traveled by both hands over a period of 30 minutes.

First, the circumference formula is given by:

$C = 2\pi r$

For the second hand:

The length of the second hand is $75 \space \text{cm}$. The second hand completes one full revolution every 60 seconds, so in 1 minute, the second hand travels:

$\text{Distance per minute} = 2\pi \times 75 \text{ cm}$

In 30 minutes, the second hand will travel:

$\text{Total distance traveled by second hand} = 30 \times 2\pi \times 75 \text{ cm}$

Now, let's calculate it with $\pi = 3.14$:

$\text{Total distance traveled by second hand} = 30 \times 2 \times 3.14 \times 75 \text{ cm}$

$\text{Total distance traveled by second hand} = 30 \times 471 \text{ cm} = 14130 \text{ cm}$

For the minute hand:

The length of the minute hand is $60 \text{ cm}$. The minute hand completes one full revolution every 60 minutes, so in 30 minutes, the minute hand travels:

$\text{Total distance traveled by minute hand} = 0.5 \times 2\pi \times 60 \text{ cm}$

Again, let's calculate it with $\pi = 3.14$:

$\text{Total distance traveled by minute hand} = 0.5 \times 2 \times 3.14 \times 60 \text{ cm}$

$\text{Total distance traveled by minute hand} = 0.5 \times 376.8 \text{ cm} = 188.4 \text{ cm}$

The difference in distance $x$ is:

$x = 14130 \text{ cm} - 1884 \text{ cm}$

$x = 13941.6 \text{ cm}$

Convert this distance into meters:

$x = 13941.6 \text{ cm} \times \frac{1 \text{ meter}}{100 \text{ cm}}$

$x \approx 139.4 \text{ meters}$

Thus, the value of $x$ in meters is nearly 139.4 meters, making the correct option:

Option C

The force exerted by the ball on the hand can be calculated using the formula derived from Newton's second law of motion, which is $F = \frac{\Delta p}{\Delta t}$, where $F$ is the force, $\Delta p$ represents the change in momentum, and $\Delta t$ is the time over which this change occurs.

The change in momentum, $\Delta p$, can be calculated as the difference between the final momentum, $p_f$, and the initial momentum, $p_i$. In this scenario, because the ball comes to a stop in the player's hand, its final velocity (and hence, its final momentum) is 0. Therefore, the change in momentum is equal to the initial momentum of the ball (since final momentum is zero).

The initial momentum, $p_i$, of the ball can be calculated using the formula $p = mv$, where $m$ is the mass of the ball and $v$ is its velocity. Given that the mass of the ball is $150 \, \mathrm{g} = 0.15 \, \mathrm{kg}$ (converting grams to kilograms) and its velocity is $20 \, \mathrm{m/s}$, we have:

$p_i = (0.15 \, \mathrm{kg}) \times (20 \, \mathrm{m/s}) = 3 \, \mathrm{kg \cdot m/s}$

Since the change in momentum, $\Delta p$, equals the initial momentum ($p_i$) because the final momentum is 0, the force exerted can be found by substituting $\Delta p$ and $\Delta t$ into the first formula:

$F = \frac{3 \, \mathrm{kg \cdot m/s}}{0.1 \, \mathrm{s}} = 30 \, \mathrm{N}$

Therefore, the magnitude of force exerted by the ball on the hand of the player is 30 N, which corresponds to Option C.

Given that the bodies A, B, and C have equal kinetic energies, we can use the relationship between kinetic energy ($K.E.$) and linear momentum ($p$) to find the ratio of their momenta. Recall the formula for kinetic energy is $K.E. = \frac{1}{2}mv^2$ and the formula for momentum is $p = mv$, where $m$ is the mass and $v$ is the velocity of the object.

First, from the kinetic energy formula, we can express the velocity in terms of kinetic energy and mass:

$v = \sqrt{\frac{2 \cdot K.E.}{m}}.$

The momentum can then be rewritten using the velocity expression obtained from the kinetic energy equation:

$p = m\sqrt{\frac{2 \cdot K.E.}{m}} = \sqrt{2m \cdot K.E.}.$

Given that the kinetic energies are the same for all three bodies, we can ignore the kinetic energy term when comparing the ratios, simplifying our comparison to the square root of their masses:

$p \propto \sqrt{m}.$

Now, we calculate the ratio of their linear momenta using their masses. Note that the masses should be in consistent units for a valid comparison, so we'll use kilograms for all:

• Mass of A = $400 \mathrm{~g} = 0.4 \mathrm{~kg}$


• Mass of B = $1.2 \mathrm{~kg}$


• Mass of C = $1.6 \mathrm{~kg}$

Thus, the ratio of their momenta will be proportional to the square root of their masses:

$\text{Ratio of momenta} = \sqrt{0.4} : \sqrt{1.2} : \sqrt{1.6} = \sqrt{\frac{4}{10}} : \sqrt{\frac{12}{10}} : \sqrt{\frac{16}{10}} = \sqrt{\frac{2}{5}} : \sqrt{\frac{6}{5}} : \sqrt{\frac{8}{5}}.$

Simplifying these we get:

$\text{Ratio of momenta} = \frac{\sqrt{2}}{\sqrt{5}} : \frac{\sqrt{6}}{\sqrt{5}} : \frac{\sqrt{8}}{\sqrt{5}} = \sqrt{2} : \sqrt{6} : \sqrt{8}.$

Recognizing that $\sqrt{6}$ is equivalent to $\sqrt{2} \cdot \sqrt{3}$ and that $\sqrt{8}$ is equivalent to $\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{2} = 2\sqrt{2}$, we see this can also be expressed as:

$\sqrt{2} : \sqrt{2} \cdot \sqrt{3} : 2\sqrt{2}.$

Dividing through by $\sqrt{2}$ to simplify the ratio, the final ratio of their linear momenta is:

$1 : \sqrt{3} : 2,$

which matches Option A $1: \sqrt{3}: 2.$

$$\begin{aligned} & \frac{1}{2} m v_0^2=\frac{1}{4 \pi \epsilon_0} \frac{z(e)}{r} \\ & \frac{1}{2} \times 6.72 \times 10^{-27} v_0^2=9 \times 10^9 \times \frac{80 \times 2 \times 1.6 \times 1.6 \times 10^{-19} \times 10^{-19}}{4.5 \times 10^{-14}} \\ & v_0^2=\frac{2 \times 9 \times 80 \times 1.6 \times 1.6 \times 2}{6.72 \times 4.5} \times 10^{-38+14+27+9} \\ & v_0^2=243.8 \times 10^{12} \\ & v_0=15.6 \times 10^6 \\ & v_0=156 \times 10^5 \end{aligned}$$