COMEDK 2024 Evening Shift Question Paper with Solutions

To determine the nuclear radius of $^{125}\mathrm{Fe}$, we need to use the relationship between nuclear radius and mass number. The nuclear radius ($R$) of a nucleus is given by the empirical formula:

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

where:

$R_0$ is the proportionality constant (approximately 1.2-1.3 fermi) and $A$ is the mass number (number of nucleons).

Given the nuclear radius of $^{27}\mathrm{Al}$ (Aluminium-27) is 3.6 fermi, we can use this to find $R_0$:

$R_{^{27}\mathrm{Al}} = R_0 \cdot 27^{1/3}$

We know:

$R_{^{27}\mathrm{Al}} = 3.6 \text{ fermi}$

Therefore:

$3.6 = R_0 \cdot 27^{1/3}$

$R_0 = \frac{3.6}{27^{1/3}}$

We can solve for $R_0$:

$27^{1/3} = 3$

$R_0 = \frac{3.6}{3} = 1.2 \text{ fermi}$

Now, using this $R_0$, we find the radius of $^{125}\mathrm{Fe}$ (Iron-125):

$R_{^{125}\mathrm{Fe}} = R_0 \cdot 125^{1/3}$

Substitute $R_0$:

$R_{^{125}\mathrm{Fe}} = 1.2 \cdot 125^{1/3}$

We need to find $125^{1/3}$:

$125^{1/3} = 5$

Therefore:

$R_{^{125}\mathrm{Fe}} = 1.2 \cdot 5 = 6 \text{ fermi}$

Converting fermi to meters (since 1 fermi = $10^{-15}$ meters):

$6 \text{ fermi} = 6 \times 10^{-15} \text{ meters}$

The correct option is:

Option C

$6 \times 10^{-15} \mathrm{~m}$

To calculate the new resistance after the wire is stretched, we need to use the fact that resistance is proportional to the length and inversely proportional to the cross-sectional area of the wire. The formula for resistance is given by:

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

where $\rho$ is the resistivity of the material, $L$ is the length, and $A$ is the cross-sectional area.

Initially, the length of the wire is $10 \text{ m}$ and the resistance is $10 \Omega$. When the length is increased by $25 \%$, the new length becomes:

$L_{\text{new}} = 10 \text{ m} + (0.25 \times 10 \text{ m}) = 12.5 \text{ m}$

Let’s denote the initial resistance as $R_{\text{initial}} = 10 \Omega$ and the initial length as $L_{\text{initial}} = 10 \text{ m}$.

The volume of the wire remains constant during stretching, so:

$L_{\text{initial}} \times A_{\text{initial}} = L_{\text{new}} \times A_{\text{new}}$

This implies:

$10 \text{ m} \times A_{\text{initial}} = 12.5 \text{ m} \times A_{\text{new}}$

So,

$A_{\text{new}} = \frac{A_{\text{initial}} \times 10}{12.5} = 0.8 \times A_{\text{initial}}$

The new resistance is calculated as:

$R_{\text{new}} = \rho \frac{L_{\text{new}}}{A_{\text{new}}}$

Using the relationship between initial and final conditions, we get:

$R_{\text{new}} = \rho \frac{L_{\text{new}}}{A_{\text{new}}} = \rho \frac{L_{\text{new}}}{0.8 \times A_{\text{initial}}} = 1.25 \times \rho \frac{L_{\text{new}}}{A_{\text{initial}}}$

Since $R_{\text{initial}} = \rho \frac{L_{\text{initial}}}{A_{\text{initial}}}$, we can write:

$R_{\text{new}} = 1.25 \times R_{\text{initial}} \times \frac{L_{\text{new}}}{L_{\text{initial}}}$

Substitute the known values:

$R_{\text{new}} = 1.25 \times 10 \Omega \times \frac{12.5 \text{ m}}{10 \text{ m}}$

Simplify:

$R_{\text{new}} = 1.25 \times 10 \Omega \times 1.25$

$R_{\text{new}} = 1.5625 \times 10 \Omega$

$R_{\text{new}} = 15.625 \Omega$

Rounding it to one decimal place, $R_{\text{new}} \approx 15.6 \Omega$.

Therefore, the new resistance is Option B: $15.6 \Omega$.

Given, refractive index of the lens $𝜇=1.4$.

Both centres of curvature ${𝐶}_1$ and ${𝐶}_2$ lie to the right of the lens. According to the Cartesian sign convention, both radii of curvature are positive. Since the lens is thicker at the centre and tapers to zero at the edges, the first surface has a smaller radius of curvature than the second surface, giving ${𝑅}_1=+4\text{ cm}$ and ${𝑅}_2=+8\text{ cm}$.

Using the lens maker's formula,

$\frac{1}{𝑓}=(𝜇-1)(\frac{1}{{𝑅}_1}-\frac{1}{{𝑅}_2})$

$\frac{1}{𝑓}=(1.4-1)(\frac{1}{4}-\frac{1}{8})=0.4\times \frac{1}{8}=\frac{1}{20}{\text{ cm}}^{-1}$

$\Rightarrow 𝑓=+20\text{ cm}$

Since the focal length is positive, the lens behaves as a converging lens of focal length $20\text{ cm}$.

Hence, the correct option is (B).

When the temperature of a wire increases, the Young's modulus of elasticity typically decreases. This is because, at higher temperatures, the atomic vibrations within the material increase, leading to a decrease in the stiffness and the ability of the material to sustain elastic deformations. Therefore, when the temperature is doubled, the Young's modulus of elasticity will decrease.

Hence, the correct answer is:

Option A: Will decrease

The electrical conductivity of a material is given by

$𝜎=\frac{𝑛{𝑒}^2𝜏}{𝑚}$

where $𝑛$ is the number density of charge carriers and $𝜏$ is the relaxation time.

As temperature increases, the amplitude of lattice vibrations increases, which causes more frequent collisions and therefore a decrease in the relaxation time $𝜏$.

Simultaneously, thermal energy breaks covalent bonds, causing the number density $𝑛$ of free charge carriers to increase exponentially. The effect of the substantial increase in $𝑛$ dominates over the decrease in $𝜏$, resulting in an overall increase in conductivity.

Hence, the correct option is (B).

To understand why the biconvex lens of glass with a refractive index of 1.5 behaves like a plane sheet of paper when dipped in a liquid, we must consider the behavior of light as it passes through different mediums and the concept of the refractive index.

The refractive index (or index of refraction) of a material determines how much the speed of light is reduced inside the material. When light moves from one medium to another, it bends at the interface between the two materials. This phenomenon is described by Snell's law:

$n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$

where:

$n_1$ (n_1) is the refractive index of the first medium,

$n_2$ (n_2) is the refractive index of the second medium,

$\theta_1$ (θ_1) is the angle of incidence, and

$\theta_2$ (θ_2) is the angle of refraction.

When the refractive index of the liquid is equal to that of the glass (1.5 in this case), the lens will not bend the light rays as they pass from the glass to the liquid. Therefore, the lens will no longer converge or diverge light; it will act as though it were a plane sheet of glass with no curvature.

Thus, for the biconvex lens to behave as a plane sheet of paper, the refractive index of the liquid must be equal to that of the glass. Hence, the correct answer is:

Option C: Equal to that of glass

The capillary rise $ℎ$ is given by

$ℎ=\frac{2𝑇\cos 𝜃}{𝑟𝜌𝑔}$

$⟹\cos 𝜃=\frac{ℎ𝑟𝑔}{2𝑇}𝜌$

Since $ℎ$, $𝑟$, and $𝑇$ are the same for all three liquids,

$\cos 𝜃\propto 𝜌$

Given the densities ${𝑃}_1<{𝑃}_2<{𝑃}_3$,

$\cos {𝜃}_1<\cos {𝜃}_2<\cos {𝜃}_3$

Since the liquids rise in the capillary ($ℎ>0$), the contact angle lies in the range $0\le 𝜃<\frac{𝜋}{2}$. As $\cos 𝜃$ is a strictly decreasing function in this interval,

$0\le {𝜃}_3<{𝜃}_2<{𝜃}_1<\frac{𝜋}{2}$

Hence, the correct option is (A).

The approximate frequency ranges for the given regions of the electromagnetic spectrum are:

Radiowaves have frequencies ranging from ${10}^4\text{ to }{10}^8\text{ Hz}$, matching (A) with (R).

Microwaves have frequencies ranging from ${10}^9\text{ to }{10}^{12}\text{ Hz}$, matching (B) with (S).

Infrared radiation has frequencies ranging from ${10}^{11}\text{ to }5\times {10}^{14}\text{ Hz}$, matching (C) with (Q).

X-rays have frequencies ranging from ${10}^{18}\text{ to }{10}^{20}\text{ Hz}$, matching (D) with (P).

Hence, the correct option is (B).

To solve this problem, we'll use the relationship between tangential acceleration, centripetal acceleration, and the net acceleration of an object moving in a circular path.

The given details are:

Radius of the circular path, $r = 5 \, \text{m}$

Tangential acceleration, $a_t = 10 \, \text{m/s}^2$

Angle between net acceleration and centripetal acceleration, $\theta = 30^\circ$

First, let's denote the centripetal acceleration as $a_c$ and the net acceleration as $a_{net}$.

We know that the net acceleration is the vector sum of the centripetal acceleration and the tangential acceleration. Since the angle between the net acceleration and the centripetal acceleration is given as $30^\circ$, we can use trigonometric relationships to solve for the centripetal acceleration.

From the definition of net acceleration, we can establish:

$a_{\text{net}} = \sqrt{a_c^2 + a_t^2}$

Given that:

$\cos \theta = \frac{a_c}{a_{\text{net}}}$

Substituting the values we get:

$\cos 30^\circ = \frac{a_c}{\sqrt{a_c^2 + a_t^2}}$

Knowing $\cos 30^\circ = \frac{\sqrt{3}}{2}$, we substitute it into the equation:

$\frac{\sqrt{3}}{2} = \frac{a_c}{\sqrt{a_c^2 + (10 \, \text{m/s}^2)^2}}$

Rearranging and squaring both sides, we get:

$(\frac{\sqrt{3}}{2})^2 = \frac{a_c^2}{a_c^2 + 100}$

This simplifies to:

$\frac{3}{4} = \frac{a_c^2}{a_c^2 + 100}$

Cross-multiplying gives:

$3(a_c^2 + 100) = 4a_c^2$

$3a_c^2 + 300 = 4a_c^2$

$300 = a_c^2$

$a_c = \sqrt{300} = 10\sqrt{3} \, \text{m/s}^2$

Now, identifying that centripetal acceleration is given by:

$a_c = \frac{v^2}{r}$

Substituting values, we have:

$10\sqrt{3} = \frac{v^2}{5}$

Solving for $v$ gives:

$10\sqrt{3} \times 5 = v^2$

$v^2 = 50\sqrt{3}$

$v = \sqrt{50\sqrt{3}}$

Converting and simplifying:

$v = (\sqrt{50}) (\sqrt[4]{3})$

$v \approx 7.07 \cdot 1.32 = 9.3$

Thus, the instantaneous speed is $9.3 \, \text{m/s}$.

The correct answer is:

Option D: $9.3 \mathrm{~ms}^{-1}$

To design the parallel plate capacitor with the given specifications, we need to determine the minimum area of the plates. Let’s break down the given information and use it step by step.

The capacitor has a dielectric constant $\kappa = 5$ and a dielectric strength of $10^6 \mathrm{~V} \mathrm{~m}^{-1}$. The voltage rating is $2 \mathrm{~kV}$ and the maximum electric field allowed is $10 \%$ of its dielectric strength.

First, calculate the maximum electric field:

$E_{\text{max}} = 0.1 \times 10^6 \mathrm{~V} \mathrm{~m}^{-1} = 10^5 \mathrm{~V} \mathrm{~m}^{-1}$

Given the voltage rating, the separation between the plates (d) can be found using the relation:

$V = E_{\text{max}} \times d$

Rearranging for $d$:

$d = \frac{V}{E_{\text{max}}} = \frac{2000 \mathrm{~V}}{10^5 \mathrm{~V} \mathrm{~m}^{-1}} = 0.02 \mathrm{~m}$

Next, we use the formula for the capacitance of a parallel plate capacitor:

$C = \kappa \epsilon_0 \frac{A}{d}$

Where:

• $C = 60 \mathrm{~pF} = 60 \times 10^{-12} \mathrm{~F}$
• $\kappa = 5$
• $\epsilon_0 = 8.85 \times 10^{-12} \mathrm{~F m}^{-1}$
• $d = 0.02 \mathrm{~m}$
• $A = \text{Area of the plates}$

Rearranging for $A$:

$A = \frac{C \times d}{\kappa \times \epsilon_0}$

Substituting the values:

$A = \frac{60 \times 10^{-12} \mathrm{~F} \times 0.02 \mathrm{~m}}{5 \times 8.85 \times 10^{-12} \mathrm{~F m}^{-1}} = \frac{60 \times 0.02}{5 \times 8.85} \times 10^{-12 + 12} \mathrm{~m}^2 = \frac{1.2}{44.25} \mathrm{~m}^2$

Simplifying further, we get:

$A \approx 2.71 \times 10^{-2} \mathrm{~m}^2$

Therefore, the minimum area of the plates should be:

Option B: $2.7 \times 10^{-2} \mathrm{~m}^2$

For an unbiased p-n junction, the built-in barrier potential is ${𝑉}_0$.

When a reverse bias $𝑉$ is applied, the external voltage aids the built-in electric field, increasing the effective barrier potential to ${𝑉}_0+𝑉$.

When a forward bias $𝑉$ is applied, the external voltage opposes the built-in electric field, reducing the effective barrier potential to ${𝑉}_0-𝑉$.

In the given graphs, graph B has the highest barrier potential, corresponding to the reverse-biased diode. Graph C has the lowest barrier potential, corresponding to the forward-biased diode. Graph A has an intermediate barrier potential, corresponding to the unbiased diode.

Therefore, A is an unbiased diode, B is a reverse-biased diode, and C is a forward-biased diode. This matching is given in column I.

Hence, the correct option is (D).

To determine the direction in which a negative charge particle will be deflected when moving in a magnetic field, we can use the right-hand rule for the Lorentz force. However, since the particle is negatively charged, we'll need to consider the direction of force opposite to what the right-hand rule would normally give for a positive charge.

Let's break this down step-by-step:

1. Direction of Velocity: The particle is moving upward.

2. Direction of Magnetic Field: The magnetic field is pointing towards the north.

According to the right-hand rule for a positive charge:

1. Point your thumb in the direction of the velocity (upward).

2. Point your fingers in the direction of the magnetic field (north).

3. Your palm will point in the direction of the force on a positive charge (west).

Since we are dealing with a negative charge, the force will be in the opposite direction of what the right-hand rule gives us. Therefore, the force will point to the east.

So, the negative charge particle will be deflected towards the east.

The correct answer is:

Option C: East

Let the bottom-left junction be $𝐴$, the bottom-right junction be $𝐵$, and the top junction be $𝐶$. The battery maintains a potential difference of ${𝑉}_{𝐴𝐵}=20\text{V}$ across terminals $𝐴$ and $𝐵$.

Between nodes $𝐴$ and $𝐶$, the $15𝜇\text{F}$ and $5𝜇\text{F}$ capacitors are in parallel:

${𝐶}_{𝐴𝐶}=15𝜇\text{F}+5𝜇\text{F}=20𝜇\text{F}$

Between nodes $𝐶$ and $𝐵$, the two $10𝜇\text{F}$ capacitors are in parallel:

${𝐶}_{𝐶𝐵}=10𝜇\text{F}+10𝜇\text{F}=20𝜇\text{F}$

Since ${𝐶}_{𝐴𝐶}$ and ${𝐶}_{𝐶𝐵}$ are connected in series across the $20\text{V}$ source, the potential difference across nodes $𝐶$ and $𝐵$ is:

${𝑉}_{𝐶𝐵}={𝑉}_{𝐴𝐵}\times \frac{{𝐶}_{𝐴𝐶}}{{𝐶}_{𝐴𝐶}+{𝐶}_{𝐶𝐵}}=20\times \frac{20}{20+20}=10\text{V}$

Each $10𝜇\text{F}$ capacitor is connected directly between nodes $𝐶$ and $𝐵$. Therefore, the charge acquired by each $10𝜇\text{F}$ capacitor is:

$𝑞=𝐶{𝑉}_{𝐶𝐵}=10𝜇\text{F}\times 10\text{V}=100𝜇\text{C}=1\times {10}^{-4}\text{C}$

Hence, the correct option is (D).

The correct answer is Option B: voltage is ahead of current in phase.

Here's why:

In an inductive circuit, the inductor opposes changes in current. When an AC voltage is applied, the inductor resists the flow of current initially. As the current starts to increase, the inductor builds up a magnetic field around itself, storing energy. This magnetic field opposes the change in current, causing the current to lag behind the voltage.

The relationship between voltage and current in an inductive circuit can be represented by a phase angle (φ), where:

φ = 90°

This means that the voltage leads the current by 90 degrees. This phase difference is a fundamental characteristic of inductive circuits.

Let's look at the other options:

Option A: voltage and current are in the same phase. This is incorrect because, as explained above, the voltage leads the current in an inductive circuit.

Option C: the phase between voltage and current depends upon the value of inductance. While the inductance value does influence the magnitude of the impedance and the amount of current, it doesn't change the fundamental phase relationship. The voltage will always lead the current by 90 degrees in an ideal inductor.

Option D: voltage lags behind current in phase. This is incorrect. The voltage leads the current in an inductive circuit.

In summary, in an inductive circuit, the voltage leads the current by 90 degrees due to the inductor's property of opposing changes in current.

To determine the total resistance to the motion of the scooter, we can start by using the work-energy principle. According to this principle, the work done by the resistive forces to stop the scooter is equal to the initial kinetic energy of the scooter.

The initial kinetic energy (KE) of the scooter can be given by:

$KE = \frac{1}{2} m v^2$

where

• $m$ is the mass of the scooter
• $v$ is the initial velocity of the scooter ($7 \mathrm{~ms}^{-1}$).

We are given that the scooter comes to a stop after traveling $10 \mathrm{~m}$, and the work done by the resistive force $F$ over this distance can be represented as:

$Work = F \cdot d$

where

• $d$ is the distance covered by the scooter before coming to a stop ($10 \mathrm{~m}$).

Therefore, the work done by the resistive force is equal to the initial kinetic energy:

$F \cdot d = \frac{1}{2} m v^2$

From Newton's second law, we also know that the weight of the scooter $W$ is given by:

$W = m g$

thus, $m = \frac{W}{g}$

Substituting $m$ into the kinetic energy equation, we get:

$F \cdot 10 = \frac{1}{2} \left(\frac{W}{g}\right) \left(7\right)^2$

Simplifying this equation:

$F \cdot 10 = \frac{1}{2} \left(\frac{W}{g}\right) \cdot 49$

$F = \frac{1}{2} \left(\frac{49 W}{10 g}\right)$

$F = \frac{49 W}{20 g}$

Given that gravity $g = 9.8 \mathrm{~ms}^{-2}$, we can substitute this value into the equation:

$F = \frac{49 W}{20 \cdot 9.8}$

$F = \frac{49 W}{196}$

$F = \frac{W}{4}$

Therefore, the total resistance to the motion of the scooter is $\frac{1}{4} W$.

Thus, the correct answer is:

Option B: $\frac{1}{4} W$

The coefficient of volume expansion ($\beta$) is given as $49 \times 10^{-5} \mathrm{~K}^{-1}$. The formula to determine the change in volume due to thermal expansion is:

$\Delta V = V \beta \Delta T$

Here:

• $\Delta V$ is the change in volume.
• $V$ is the original volume.
• $\beta$ is the coefficient of volume expansion.
• $\Delta T$ is the change in temperature.

Given that the temperature change ($\Delta T$) is $50^{\circ} \mathrm{C}$, we have:

$\Delta V = V \cdot (49 \times 10^{-5} \mathrm{~K}^{-1}) \cdot 50$

$\Delta V = V \cdot (49 \times 10^{-5} \times 50)$

$\Delta V = V \cdot (2450 \times 10^{-5})$

$\Delta V = V \cdot 0.0245$

This indicates that the volume increases by 2.45%. Since density ($\rho$) is inversely proportional to volume ($V$), a 2.45% increase in volume will result in a 2.45% decrease in density.

Thus, the percentage change in density is 2.45%. Hence, the correct option is:

Option D: 2.45

Nodes 2 and 3 are connected by a wire of zero resistance, as are nodes 4 and 5.

With terminals 1 and 6 open, the two resistors of resistance $𝑅$ (between nodes 2 and 5, and between nodes 3 and 4) are connected in parallel between points 2 and 4.

Therefore, the equivalent resistance between points 2 and 4 is

${𝑅}_{24}=\frac{𝑅\times 𝑅}{𝑅+𝑅}=0.5𝑅$

Hence, the correct option is (D).

Given, mass of the cylinder $𝑀=2\text{ kg}$, radius $𝑅=0.2\text{ m}$, and initial angular velocity ${𝜔}_0=5{\text{ rad s}}^{-1}$.

The moment of inertia of the solid cylinder about its own axis is

${𝐼}_{𝑐}=\frac{1}{2}𝑀{𝑅}^2=\frac{1}{2}\times 2\times (0.2{)}^2=0.04{\text{ kg m}}^2$

The initial angular momentum of the cylinder is

${𝐿}_{𝑐}={𝐼}_{𝑐}{𝜔}_0=0.04\times 5=0.2{\text{ kg m}}^2{\text{ s}}^{-1}$

For the particle of mass $𝑚=1\text{ kg}$ moving with velocity $𝑣=5{\text{ m s}}^{-1}$, its angular momentum about the cylinder's axis just before striking is

${𝐿}_{𝑝}=𝑚𝑣𝑅=1\times 5\times 0.2=1.0{\text{ kg m}}^2{\text{ s}}^{-1}$

Since the particle's angular momentum aids the cylinder's rotation, the total initial angular momentum about the axis is

$𝐿={𝐿}_{𝑐}+{𝐿}_{𝑝}=0.2+1.0=1.2{\text{ kg m}}^2{\text{ s}}^{-1}$

After the particle sticks to the surface of the cylinder, the total moment of inertia of the system becomes

$𝐼={𝐼}_{𝑐}+𝑚{𝑅}^2=0.04+1\times (0.2{)}^2=0.08{\text{ kg m}}^2$

By conservation of angular momentum, the final angular velocity $𝜔$ is

$𝜔=\frac{𝐿}{𝐼}=\frac{1.2}{0.08}=15.0{\text{ rad s}}^{-1}$

Hence, the correct option is (A).

To determine the additional kinetic energy required for a satellite to just escape into outer space, we need to understand the relationship between kinetic energy and gravitational potential energy in a circular orbit.

The total energy (E) of a satellite in a stable circular orbit around the Earth is given by:

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

Where:

• $m$ is the mass of the satellite
• $v$ is the orbital velocity of the satellite
• $G$ is the gravitational constant
• $M$ is the mass of the Earth
• $r$ is the radius of the orbit of the satellite

The kinetic energy (KE) of the satellite in orbit is:

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

For a satellite in a stable circular orbit, the kinetic energy is half of the magnitude of the gravitational potential energy, i.e.

$KE = \frac{GMm}{2r}$

We know the satellite's kinetic energy is $1.69 \times 10^{10} \mathrm{~J}$.

The total energy (E) will thus be:

$E = - KE$

$E = -1.69 \times 10^{10} \mathrm{~J}$

To just escape from Earth's gravity, the total energy should be zero because the satellite would have kinetic energy equal to the gravitational potential energy (but positive because it escapes the gravitational pull). Therefore, the additional kinetic energy required, $\Delta KE$, should be equal to the magnitude of the total energy:

$\Delta KE = 1.69 \times 10^{10} \mathrm{~J}$

Thus, the required additional kinetic energy for the satellite to escape into outer space is:

Option B: $1.69 \times 10^{10} \mathrm{~J}$

To find the approximate separation of the centers of charges in the water molecule, we need to consider the work done in rotating the dipole in an electric field. The work done $W$ to rotate a dipole of moment $\vec{p}$ by an angle $\theta$ in an electric field $\vec{E}$ is given by:

$W = pE \cos(\theta) - pE \cos(0)$

For a rotation by $180^{\circ}$, $\theta = 180^{\circ}$, and $\cos(180^{\circ}) = -1$:

$W = pE \cos(180^{\circ}) - pE \cos(0)$

Since $\cos(0) = 1$, the above equation simplifies to:

$W = pE(-1) - pE(1)$

$W = -pE - pE$

$W = -2pE$

Given that the work done is:

$5 \times 10^{-25} \text{ J}$

And the electric field strength is:

$E = 2.5 \times 10^4 \text{ N/C}$

We can find the dipole moment, $p$:

$5 \times 10^{-25} = -2p(2.5 \times 10^4)$

Solving for $p$:

$5 \times 10^{-25} = -5 \times 10^4 p$

$p = \frac{5 \times 10^{-25}}{-5 \times 10^4}$

$p = -1 \times 10^{-29} \text{ C m}$

Since the dipole moment $p$ is given by the product of the charge $q$ on the water molecule and the separation $d$ of the charges, we have:

$p = qd$

For a water molecule, the charge of each center (hydrogen or oxygen) can be approximated to the elementary charge $e$ (since a water molecule is neutral overall, but the centers have polarity):

$q = 1.6 \times 10^{-19} \text{ C}$

Therefore, we can solve for the separation $d$:

$d = \frac{p}{q}$

$d = \frac{1 \times 10^{-29}}{1.6 \times 10^{-19}}$

$d = \frac{1}{1.6} \times 10^{-10}$

$d \approx 0.625 \times 10^{-10} \text{ m}$

So, the approximate separation of the centers of charges is:

$0.625 \times 10^{-10} \text{ m}$

Hence, the correct option is:

Option B: $0.625 \times 10^{-10} \mathrm{~m}$

The latent heat of vaporisation, also known as the enthalpy of vaporisation, reflects the energy required to transform a substance from a liquid phase to a gas phase without changing its temperature. According to the first law of thermodynamics, the change in internal energy (ΔU) during the process of vaporisation can be calculated using the formula:

$ΔU = Q - W$

Where:

• $Q$ is the heat added (latent heat of vaporisation)
• $W$ is the work done in the process

In this scenario:

• The latent heat of vaporisation ($Q$) is $2240 \mathrm{~J}$.
• The work done ($W$) is $168 \mathrm{~J}$.

Substituting these values into the equation, we get:

$ΔU = 2240 \mathrm{~J} - 168 \mathrm{~J}$

Therefore,

$ΔU = 2072 \mathrm{~J}$

Hence, the increase in internal energy is:

Option B: 2072 J

To determine the power generated by the turbine, we need to consider the input energy, losses due to friction, and the rate at which water falls.

The potential energy of the falling water can be calculated using the formula:

$P.E = mgh$

where:

• $m$ is the mass of the water per second ($20 \, \mathrm{kg} \, \mathrm{s}^{-1}$)
• $g$ is the acceleration due to gravity ($9.8 \, \mathrm{m} \, \mathrm{s}^{-2}$)
• $h$ is the height from which the water falls ($80 \, \mathrm{m}$)

Substituting the values, the potential energy per second is:

$P.E = 20 \, \mathrm{kg} \, \mathrm{s}^{-1} \times 9.8 \, \mathrm{m} \, \mathrm{s}^{-2} \times 80 \, \mathrm{m}$

$P.E = 15680 \, \mathrm{J/s}$

This is the input energy per second, which is also the power input. However, because 20% of the input energy is lost due to frictional forces, only 80% of the input energy is effectively used by the turbine. Therefore, the power generated by the turbine $P$ is:

$P = 15680 \times 0.80$

$P = 12544 \, \mathrm{W}$

$P = 12.544 \, \mathrm{kW}$

Thus, the power generated by the turbine is approximately:

Option A: 12.8 kW.

Initially, the electrostatic force between the two charges separated by a distance $𝑟$ is given by Coulomb's law:

$𝐹=\frac{1}{4𝜋{𝜀}_0}\frac{{𝑞}^2}{{𝑟}^2}$

When $30\%$ of the charge of $N$ is transferred to $M$, the transferred charge is:

$\Delta 𝑞=30\%\times (-𝑞)=-0.3𝑞$

The new charges on $M$ and $N$ are:

${𝑞}_M=+𝑞+(-0.3𝑞)=+0.7𝑞$

${𝑞}_N=-𝑞-(-0.3𝑞)=-0.7𝑞$

The magnitude of the new force between them is:

${𝐹}^{\prime }=\frac{1}{4𝜋{𝜀}_0}\frac{\left|{𝑞}_M{𝑞}_N\right|}{{𝑟}^2}=\frac{1}{4𝜋{𝜀}_0}\frac{\left(0.7𝑞\right)\left(0.7𝑞\right)}{{𝑟}^2}=0.49\left(\frac{1}{4𝜋{𝜀}_0}\frac{{𝑞}^2}{{𝑟}^2}\right)=\frac{49}{100}𝐹$

Hence, the correct option is (C).

Given, ${𝑓}_{𝑜}=60\text{ cm}$, ${𝑓}_{𝑒}=5\text{ cm}$, ${𝑢}_{𝑜}=-300\text{ cm}$, and the least distance of distinct vision $𝐷=25\text{ cm}$.

For the objective lens, using the lens formula,

$\frac{1}{{𝑣}_{𝑜}}-\frac{1}{{𝑢}_{𝑜}}=\frac{1}{{𝑓}_{𝑜}}$

$\frac{1}{{𝑣}_{𝑜}}-\frac{1}{-300}=\frac{1}{60}⟹\frac{1}{{𝑣}_{𝑜}}=\frac{1}{60}-\frac{1}{300}=\frac{4}{300}=\frac{1}{75}$

$\therefore {𝑣}_{𝑜}=75\text{ cm}$

The magnification produced by the objective is

${𝑚}_{𝑜}=\frac{{𝑣}_{𝑜}}{{𝑢}_{𝑜}}=\frac{75}{-300}=-0.25$

For the eyepiece forming the final image at the least distance of distinct vision (${𝑣}_{𝑒}=-𝐷$),

${𝑚}_{𝑒}=1+\frac{𝐷}{{𝑓}_{𝑒}}=1+\frac{25}{5}=6$

The total magnification produced by the telescope is

$𝑚={𝑚}_{𝑜}\times {𝑚}_{𝑒}=(-0.25)\times 6=-1.5$

Hence, the correct option is (C).

To solve this problem, we need to determine the amount of heat required to raise the temperature of the hydrogen gas in the cylinder by $20^{\circ} \text{C}$. We’ll start by using the given data and applying the necessary equations from thermodynamics. Let's break down the steps:

1. Calculate the number of moles of hydrogen gas (H₂) at STP:

At STP (Standard Temperature and Pressure), the volume of one mole of an ideal gas is 22.4 liters. Given the cylinder has a fixed capacity of 44.81 liters, we can find the number of moles, $n$:

$n = \frac{\text{Volume of gas}}{\text{Volume of one mole at STP}} = \frac{44.81 \, \text{L}}{22.4 \, \text{L/mol}} = 2 \, \text{mol}$

2. Use the specific heat capacity at constant volume (Cₜ):

For hydrogen gas (H₂), which is diatomic, the molar heat capacity at constant volume $C_V$ is:

$C_V = \frac{5}{2} R$

where $R = 8.31 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1}$. So:

$C_V = \frac{5}{2} \times 8.31 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1} = 20.775 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1}$

3. Calculate the heat required:

The heat required $Q$ to raise the temperature of the gas by $\Delta T = 20^{\circ} \text{C}$ (which is the same in Kelvin) is given by the formula:

$Q = n \cdot C_V \cdot \Delta T$

Plugging in the values:

$Q = 2 \, \text{mol} \times 20.775 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1} \times 20 \, \text{K} = 831 \, \text{J}$

Thus, the amount of heat needed to raise the temperature of the hydrogen gas in the cylinder by $20^{\circ} \text{C}$ is 831 J.

Therefore, the correct option is:

Option C: 831 J

The wavelength of the emitted photon in a hydrogen atom transition is given by the Rydberg formula:

$\frac{1}{𝜆}=𝑅(\frac{1}{{𝑛}_2^1}-\frac{1}{{𝑛}_2^2})$

For the transition from $𝑛=3$ to $𝑛=2$:

$\frac{1}{𝜆}=𝑅(\frac{1}{2^2}-\frac{1}{3^2})=𝑅(\frac{1}{4}-\frac{1}{9})=\frac{5𝑅}{36}$

For the transition from $𝑛=2$ to $𝑛=1$:

$\frac{1}{{𝜆}^{\prime }}=𝑅(\frac{1}{1^2}-\frac{1}{2^2})=𝑅(1-\frac{1}{4})=\frac{3𝑅}{4}$

Taking the ratio of the two expressions:

$\frac{{𝜆}^{\prime }}{𝜆}=\frac{5𝑅/36}{3𝑅/4}=\frac{5}{36}\times \frac{4}{3}=\frac{5}{27}$

$\therefore {𝜆}^{\prime }=\frac{5𝜆}{27}$

Hence, the correct option is (A).

To determine the potential of the bigger drop formed by the combination of 216 smaller drops, we need to consider the following principles of electrostatics and geometry:

1. Volume Conservation: The total volume of the bigger drop will be equal to the sum of the volumes of the 216 smaller drops. If each smaller drop has a radius $r$, then the volume of one small drop is:

$V_{\text{small}} = \dfrac{4}{3} \pi r^3$

Since there are 216 such drops, the total volume becomes:

$V_{\text{total}} = 216 \times \dfrac{4}{3} \pi r^3$

Let $R$ be the radius of the bigger drop. The volume of the bigger drop is:

$V_{\text{big}} = \dfrac{4}{3} \pi R^3$

By volume conservation, we have:

$\dfrac{4}{3} \pi R^3 = 216 \times \dfrac{4}{3} \pi r^3$

or,

$R^3 = 216 r^3$

Therefore,

$R = 6r$

2. Charge Conservation: The total charge on the bigger drop will be the sum of the charges on all the smaller drops. If each smaller drop has a potential of $200 \mathrm{~V}$ and radius $r$, then its charge $q$ can be given by:

$V_{\text{small}} = \dfrac{k q}{r}$

Given $V_{\text{small}} = 200 \mathrm{~V}$, we get:

$q = \dfrac{200 r}{k}$

For 216 such drops, the total charge $Q_{\text{total}}$ is:

$Q_{\text{total}} = 216 \times \dfrac{200 r}{k} = \dfrac{43200 r}{k}$

3. Potential of the Bigger Drop: The potential of the bigger drop can be calculated using its charge and radius. The formula for the potential of a spherical drop is:

$V_{\text{big}} = \dfrac{k Q_{\text{total}}}{R}$

Substitute the values:

$V_{\text{big}} = \dfrac{k \dfrac{43200 r}{k}}{6r} = \dfrac{43200 r}{6r} = 7200 \mathrm{~V}$