COMEDK 2023 Evening Shift Question Paper with Solutions

To find the expression for the charge on the metal ball, we need to consider the forces acting on the ball when it is just suspended in the liquid due to the electric field. The forces involved are:

1. The weight of the ball (gravitational force), $F_{\text{gravity}} = mg$, where $m = \text{mass of the ball} = \rho V = \rho \left(\frac{4}{3}\pi r^3\right)$, and $g$ is the acceleration due to gravity.
2. The buoyant force due to the liquid, which can be given by Archimedes' principle, $F_{\text{buoyant}} = \text{Volume of displaced liquid} \times g \times \text{density of liquid} = V\sigma g = \left(\frac{4}{3}\pi r^3\right)\sigma g$.
3. The electric force in the upward direction which keeps the ball suspended, $F_{\text{electric}} = qE$, where $E$ is the electric field strength.

For the ball to be just suspended, the net force acting on it must be zero. This implies:

$F_{\text{electric}} = F_{\text{gravity}} - F_{\text{buoyant}}$

Substituting the expressions we have for each:

$qE = \rho \left(\frac{4}{3}\pi r^3\right) g - \sigma \left(\frac{4}{3}\pi r^3\right) g$

Isolating $q$, we find:

$q = \frac{\left(\rho \left(\frac{4}{3}\pi r^3\right) g - \sigma \left(\frac{4}{3}\pi r^3\right) g \right)}{E}$

Simplifying the expression:

$q = \frac{\left(\frac{4}{3}\pi r^3\right) (\rho - \sigma) g}{E}$

So, the correct expression for the charge on the metal ball, as per the given options, is option B:

$q=\frac{\left[4 \pi r^3(\rho-\sigma) g\right]}{3 E}$

To determine the energy of the radiation emitted when an electron in hydrogen transitions from the $3^{\text{rd}}$ excited state to the ground state, first, we need to understand what is meant by "ground state" and "excited states" in the context of a hydrogen atom.

The energy levels of an electron in a hydrogen atom are given by the formula:

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

where $E_n$ is the energy of the n-th energy level (with the ground state being $n = 1$) and 13.6 eV is the energy of the electron in the ground state. The ground state (n=1) has an energy of $-13.6$ eV. The $3^{\text{rd}}$ excited state would correspond to $n = 4$, since the excited states start counting after the ground state $n=1$ (first excited state is $n=2$, second is $n=3$, and so forth).

Therefore, the energy of the $3^{\text{rd}}$ excited state ($n=4$) is:

$E_4 = -\frac{13.6 \ eV}{4^2} = -\frac{13.6 \ eV}{16} = -0.85 \ eV$

To find the energy of the radiation emitted when the electron transitions from the $n = 4$ level to the ground state ($n = 1$), we calculate the difference in energy between these two levels:

$\Delta E = E_{final} - E_{initial} = E_1 - E_4 = (-13.6 \ eV) - (-0.85 \ eV) = -13.6 \ eV + 0.85 \ eV = -12.75 \ eV$

The negative sign indicates that energy is released, so the magnitude of the energy is 12.75 eV. Thus, the energy of the radiation emitted is 12.75 eV.

This corresponds to Option B.

To find the root mean square speed ($v_{rms}$) of the molecules at a different temperature and pressure, we can use the relations from the kinetic theory of gases. For a given mass of gas, the root mean square speed is directly proportional to the square root of the absolute temperature ($T$) and inversely proportional to the square root of the pressure ($P$), keeping in mind that the amount of gas does not change. Mathematically, this relationship can be expressed as:

$v_{rms} \propto \sqrt{\frac{T}{P}}$

Given that the initial $v_{rms}$ is $120 \, \text{m/s}$ at $88^\circ C$ (or $361 K$, since $T_{K} = T_{^\circ C} + 273$) and $1$ atmospheric pressure, and we want to find the $v_{rms}$ at $127^\circ C$ (or $400 K$) and $2$ atmospheric pressures, we can set up the following proportion :

$\frac{v_{rms,1}}{v_{rms,2}} = \sqrt{\frac{T_1/P_1}{T_2/P_2}}$

Using the given data:

$120 / v_{rms,2} = \sqrt{\frac{361 / 1}{400 / 2}}$

$v_{rms,2} = 120 \times \sqrt{\frac{400 / 2}{361 / 1}}$

$v_{rms,2} = 120 \times \sqrt{\frac{400}{2 \times 361}}$

$v_{rms,2} = 120 \times \sqrt{\frac{400}{722}}$

$v_{rms,2} = 120 \times \sqrt{\frac{2}{1.805}}$

$v_{rms,2} = 120 \times \sqrt{1.107694}$

$v_{rms,2} \approx 120 \times 1.052$

$v_{rms,2} \approx 126.24 \, \text{m/s}$

Hence, the correct answer is Option C: 126.3 m/s.

The feature of infrared waves that makes them useful for haze photography is Option B: Since it has large wavelength.

Infrared radiation has a longer wavelength than visible light. This characteristic allows it to penetrate atmospheric haze more effectively than the shorter wavelengths of visible light. Haze and fog are composed of small particles like water droplets which scatter visible light much more than infrared radiation. Since the longer wavelengths of infrared light are not scattered as much, they can pass through the haze and reach the camera. This ability to cut through atmospheric haze and fog enables photographers to capture clearer images with greater contrast and detail over long distances, which would not be possible with regular visible light photography.

The options A, C, and D do not accurately describe the feature that makes infrared wavelengths specifically useful for haze photography:

• Option A: Invisibility is not the key factor; the ability to penetrate atmospheric particulates like haze plays the critical role.
• Option C: The absorption by medium (like atmospheric gases) affects all electromagnetic waves to some extent, but it's not the reason why infrared is chosen for cutting through haze.
• Option D: Infrared waves actually have a lower frequency compared to visible light, not higher. Frequency inversely relates to the wavelength; hence, higher wavelength equates to a lower frequency in the electromagnetic spectrum.

To find the current drawn by the primary coil of the transformer, we can use the concept of power conservation in an ideal transformer. The power input to the primary coil is equal to the power output from the secondary coil. The formula for power is $P = IV$, where $I$ is current and $V$ is voltage.

Given that the transformer steps up the voltage from $22 \mathrm{~V}$ to $220 \mathrm{~V}$ and the load resistance on the secondary side is $110 \Omega$, we can first calculate the current in the secondary circuit using Ohm's law, $V = IR$, where $V$ is voltage, $I$ is current, and $R$ is resistance. Rearranging this, we get $I = \frac{V}{R}$.

The voltage across the secondary coil (and hence across the load resistance) is $220 \mathrm{~V}$, and the load resistance is $110 \Omega$. Thus, the current in the secondary circuit is:

$I_{\text{secondary}} = \frac{220 \mathrm{~V}}{110 \Omega} = 2 \mathrm{~A}$

Now, according to the principle of an ideal transformer (neglecting any losses), the power in the primary coil is equal to the power in the secondary coil. Thus,

$P_{\text{primary}} = P_{\text{secondary}}$

Applying the formula for power ($P = IV$) in both coils, we have:

$I_{\text{primary}} \times V_{\text{primary}} = I_{\text{secondary}} \times V_{\text{secondary}}$

We know $V_{\text{primary}} = 22 \mathrm{~V}$, $V_{\text{secondary}} = 220 \mathrm{~V}$, and we've calculated $I_{\text{secondary}} = 2 \mathrm{~A}$. Plugging these values into the equation, we get:

$I_{\text{primary}} \times 22 = 2 \times 220$

Solving for $I_{\text{primary}}$:

$I_{\text{primary}} = \frac{2 \times 220}{22} = \frac{440}{22} = 20 \mathrm{~A}$

Therefore, the current drawn by the primary coil of the transformer is 20 A.

The correct answer is Option A

20 A.

The magnetic permeability '$\mu$' of a material characterizes the ability of the material to allow the formation of magnetic fields within itself. In the context of different types of magnetic materials, the relative magnetic permeability ($\mu_r$, which is the ratio of the magnetic permeability of the material to the magnetic permeability of free space $\mu_0$), is a crucial parameter in distinguishing among diamagnetic, paramagnetic, and ferromagnetic substances.

For a paramagnetic substance, these materials are attracted by a magnet but do not retain magnetic properties when the external field is removed. Their magnetic dipoles tend to align with the magnetic field, thereby strengthening it. As a result, the relative magnetic permeability of a paramagnetic material is slightly greater than 1 ($\mu_r > 1$). However, this effect is typically quite weak.

Therefore, for the question about the magnetic permeability $\mu$ of a paramagnetic substance, the correct option is:

Option A: $\mu > 1$

This is because paramagnetic substances slightly enhance the magnetic field within themselves due to their magnetizable nature, hence making their magnetic permeability greater than 1, but not by a huge margin.

When two lenses are in contact, the power of the combination ($P_{\text{combined}}$) is given by the sum of the powers of the individual lenses ($P_1$ and $P_2$). The power of a lens is the reciprocal of its focal length in meters ($P = \frac{1}{f}$, where $P$ is in diopters and $f$ is in meters).

The given powers of the lenses are:

• The first lens: $+1 \mathrm{D}$


• The second lens: $-2 \mathrm{D}$

We calculate the combined power as follows:

$P_{\text{combined}} = P_1 + P_2 = +1 \mathrm{D} + (-2 \mathrm{D}) = -1 \mathrm{D}$

The negative sign of the combined power indicates that the combination acts as a diverging lens. The focal length of the combined lens system can be found using the formula $f_{\text{combined}} = \frac{1}{P_{\text{combined}}}$:

$f_{\text{combined}} = \frac{1}{-1 \mathrm{D}} = -1 \mathrm{m} = -100 \mathrm{cm}$

The negative focal length confirms that the lens acts as a diverging lens. However, we usually refer to the magnitude of the focal length for lenses. Thus, the focal length of the combination is $100 \mathrm{cm}$, but since it's a diverging lens, the correct interpretation is:

Option A: diverging lens of focal length $100 \mathrm{cm}$

The de Broglie wavelength of a particle is given by the formula:

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

where $h$ is Planck's constant and $p$ is the momentum of the particle. The momentum $p$ can be expressed in terms of the mass $m$ and velocity $v$ of the particle as $p = m \cdot v$.

For the two particles in question, their de Broglie wavelengths are given by:

$\lambda_1 = \frac{h}{p_1} = \frac{h}{m_1 v_1},$

$\lambda_2 = \frac{h}{p_2} = \frac{h}{m_2 v_2}.$

To find the ratio $\frac{\lambda_1}{\lambda_2}$, we divide the equations:

$\frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{m_1 v_1}}{\frac{h}{m_2 v_2}} = \frac{m_2 v_2}{m_1 v_1}.$

Given that the original particle is at rest before decaying into two particles, by the conservation of momentum, the total momentum before decay must equal the total momentum after decay. Before the decay, the momentum is 0 (since the particle is at rest), and after the decay, it must still sum to 0. Thus, we have:

$m_1v_1 = m_2v_2.$

Substituting $m_1v_1 = m_2v_2$ into the equation for the ratio of the de Broglie wavelengths, we get:

$\frac{\lambda_1}{\lambda_2} = \frac{m_2 v_2}{m_1 v_1} = 1.$

Therefore, the ratio of their de Broglie wavelengths $\frac{\lambda_1}{\lambda_2}$ is 1 : 1. The correct option is:

Option B 1 : 1

Taking the potential of the common bottom wire as $0\text{ V}$, the potential at the left terminal of the $100\Omega$ resistor is $10\text{ V}$.

The node at the right terminal of the $100\Omega$ resistor is connected directly to the positive terminal of the $2\text{ V}$ battery, maintaining that node at $2\text{ V}$.

The potential difference across the $100\Omega$ resistor is

$𝑉=10\text{ V}-2\text{ V}=8\text{ V}$

Using Ohm's law, the current passing through the $100\Omega$ resistor is

$𝐼=\frac{𝑉}{𝑅}=\frac{8\text{ V}}{100\Omega }=0.08\text{ A}$

Hence, the correct option is (A).

To solve this problem, we can use the relation for an adiabatic process, which is given as:

$P V^\gamma = \text{constant}$

Where $P$ is the pressure, $V$ is the volume, and $\gamma$ (gamma) is the heat capacity ratio. In an adiabatic process, as the volume increases, the pressure decreases. We are given that the volume increases by $6.2\%$, so we can express the final volume $V_f$ in terms of the initial volume $V_i$ as:

$V_f = V_i + 0.062V_i = 1.062V_i$

Using the relation for an adiabatic process for both initial and final states, we have:

$P_i V_i^\gamma = P_f V_f^\gamma$

Substituting the expression for $V_f$, we get:

$P_i V_i^\gamma = P_f (1.062V_i)^\gamma$

$\frac{P_f}{P_i} = \left(\frac{V_i}{1.062V_i}\right)^\gamma$

$\frac{P_f}{P_i} = (1.062)^{-\gamma}$

Given that $\gamma = 1.4$, let's substitute that into the equation:

$\frac{P_f}{P_i} = (1.062)^{-1.4}$

We can now calculate the percentage change in pressure. However, remember that the percentage change is given by:

$\text{Percentage Change} = \left(\frac{P_f - P_i}{P_i}\right) \times 100\%$

But since we are finding $\frac{P_f}{P_i}$ directly, we need to adjust our formula to:

$\text{Percentage Change} = \left(\frac{P_f}{P_i} -1 \right) \times 100\%$

Substituting our value for $\frac{P_f}{P_i}$:

$\text{Percentage Change} = \left((1.062)^{-1.4} -1\right) \times 100\%$

Computing the value $(1.062)^{-1.4}$:

$= \left((1.062)^{-1.4} -1\right) \times 100\% \approx (-0.0868) \times 100\% = -8.68\%$

Since a negative sign indicates a decrease in pressure, the answer implies that the pressure decreases by approximately 8.68%. Thus, the correct answer is Option A: 8.68.

To calculate the torque ($\tau$) produced by a force on a wheel, the formula used is $\tau = rF \sin(\theta)$, where:

• $r$ is the distance from the axis to the point of application of the force, in meters,
• $F$ is the magnitude of the force, in Newtons,
• $\theta$ is the angle between the line of action of the force and the lever arm or radius.

In this case:

• The distance $r = 2$ cm $= 0.02$ meters (since $1$ cm $= 0.01$ m),
• The magnitude of the force $F = 200$ N,
• The angle $\theta$ is the difference between the angle formed by OP with the horizontal (x-axis) and the angle the force makes with the horizontal, which is $55^{\circ} - 25^{\circ} = 30^{\circ}$.

Thus, substituting these values into the torque formula:

$\tau = 0.02 \times 200 \times \sin(30^{\circ})$

Since $\sin(30^{\circ}) = 0.5$,

$\tau = 0.02 \times 200 \times 0.5 = 2 \text{ N m}$

Therefore, the torque produced by the force on the wheel is $2$ Newton-meters, which corresponds to Option C (2 N m).

The expression given is $\mathrm{P=P}_0 \exp \left(-\alpha t^2\right)$, where $\mathrm{P}$ is a physical quantity, $\alpha$ is a constant, and $t$ is time. Since the argument of the exponential function must be dimensionless (as the exponential function can only operate on dimensionless quantities), the term $-\alpha t^2$ must be dimensionless. This means that $\alpha$ must be chosen such that when it is multiplied by $t^2$ (where $t$ has dimensions of time, $[T]$), the result is a dimensionless quantity.

Let's consider the dimensions of each component:

• The dimensions of $\mathrm{P}$ and $\mathrm{P}_0$ are the same, as they are both representations of the physical quantity $\mathrm{P}$.
• The dimension of time, $t$, is $[T]$.
• The exponential function does not add or change dimensions; it requires its argument to be dimensionless.

To ensure the argument of the exponential $-\alpha t^2$ is dimensionless, $\alpha$ must have dimensions that cancel out the dimensions of $t^2$. Since the dimensions of $t^2$ are $[T]^2$, $\alpha$ must have dimensions that, when multiplied by $[T]^2$, result in a dimensionless quantity. Therefore, $\alpha$ must have the dimensions of $[T]^{-2}$.

So, the correct option is:

Option D: Have dimensions of $t^{-2}$.

To convert a galvanometer into a voltmeter, a high resistance needs to be added in series with the galvanometer. This is because a voltmeter must have a high resistance to ensure that it does not draw significant current from the circuit being measured, which can alter the voltage being measured. Therefore, out of the options given, we can immediately disregard the options mentioning a resistance in parallel with the galvanometer or resistances that are too low to make a sensible voltmeter.

The current sensitivity of the galvanometer is $10 \mu \mathrm{A} / \text{div}$, and it has 20 divisions. This means that the full-scale deflection current of the galvanometer is:

$20 \times 10 \mu \mathrm{A} = 200 \mu \mathrm{A} = 200 \times 10^{-6} \mathrm{A}$

To read up to $1 \mathrm{V}$, we require this full-scale current of $200 \mu \mathrm{A}$ to flow when a potential difference of $1 \mathrm{V}$ is applied across the modified galvanometer (now a voltmeter).

The internal resistance of the galvanometer, $R_g$, is $100 \Omega$. The voltage $V$ across the galvanometer and the series resistance $R_s$ together is given by Ohm's law, $V = I_t \times (R_g + R_s)$, where $I_t$ is the total current for full-scale deflection, which is 200 \mu \mathrm{A} = 200 \times 10^{-6} \mathrm{A}\). Set this equal to 1 V to find $R_s$.</p> <p>$1 = (200 \times 10^{-6}) \times (100 + R_s)$</p> <p>$1 = (200 \times 10^{-6} \times 100) + (200 \times 10^{-6} \times R_s)$</p> <p>$1 = 0.02 + (200 \times 10^{-6} \times R_s)$</p> <p>Subtracting 0.02 from both sides yields:</p> <p>$0.98 = 200 \times 10^{-6} \times R_s$</p> <p>Solving for $R_s$ gives:</p> <p>$R_s = \frac{0.98}{200 \times 10^{-6}} = \frac{0.98}{0.2 \times 10^{-3}} = 4900 \Omega$</p> <p>Hence, the correct answer is a4900 \Omega$resistor in series with the galvanometer to convert it into a voltmeter that can read up to$1 \mathrm{V}$$. This matches <strong>Option D:$4900 \Omega$ in series with the galvanometer.

To find the time the helium nucleus is in the tube, we can use the kinematic equations that describe motion under uniform acceleration. The relevant equation in this setting, given the initial velocity ($v_{i}$), final velocity ($v_{f}$), displacement ($s$), and acceleration ($a$), is:

$v_{f} = v_{i} + at$

However, we don't directly know the acceleration or the time, so we'll use another equation that relates these variables without the need for acceleration. This equation is:

$s = v_{i}t + \frac{1}{2}at^2$

Given that we know the initial and final velocities and the displacement, but still lack the acceleration, we can use a different form that makes use of both initial and final velocities:

$s = \frac{v_{i} + v_{f}}{2} \cdot t$

This equation comes from averaging the initial and final velocities in a uniformly accelerated motion to find the average velocity, then multiplying by the time to find the displacement.

We know from the problem that:

• The length of the tube, which is the displacement $s = 4$ m,


• The initial velocity $v_{i} = 2000$ m/s,


• And the final velocity $v_{f} = 8000$ m/s.

Plugging these values into our formula, we have:

$4 = \frac{2000 + 8000}{2} \cdot t$

This simplifies to:

$4 = \frac{10000}{2} \cdot t$

Which further simplifies to:

$4 = 5000t$

Solving for $t$ gives:

$t = \frac{4}{5000} = 0.0008$ s

Hence, the time the particle is in the tube is $0.0008$ seconds or $8 \times 10^{-4}$ s, which corresponds to Option C.

The force constant of a uniform spring is inversely proportional to its length, so $𝑘\propto \frac{1}{𝑙}$.

Let the original spring have length $𝐿$ and spring constant $𝑘$. When cut into lengths in the ratio $1:3:4$, the lengths of the three parts are

${𝑙}_1=\frac{1}{8}𝐿,{𝑙}_2=\frac{3}{8}𝐿,{𝑙}_3=\frac{4}{8}𝐿=\frac{1}{2}𝐿$

The corresponding spring constants are

${𝑘}_1=8𝑘,{𝑘}_2=\frac{8}{3}𝑘,{𝑘}_3=2𝑘$

When connected in series, the equivalent spring constant ${𝑘}^{\prime }$ satisfies

$\frac{1}{{𝑘}^{\prime }}=\frac{1}{{𝑘}_1}+\frac{1}{{𝑘}_2}+\frac{1}{{𝑘}_3}=\frac{1}{8𝑘}+\frac{3}{8𝑘}+\frac{4}{8𝑘}=\frac{8}{8𝑘}=\frac{1}{𝑘}⟹{𝑘}^{\prime }=𝑘$

When connected in parallel, the equivalent spring constant ${𝑘}^{\prime \prime }$ is

${𝑘}^{\prime \prime }={𝑘}_1+{𝑘}_2+{𝑘}_3=8𝑘+\frac{8}{3}𝑘+2𝑘=\frac{38}{3}𝑘$

Therefore, the ratio of the spring constants is

$\frac{{𝑘}^{\prime }}{{𝑘}^{\prime \prime }}=\frac{𝑘}{\frac{38}{3}𝑘}=\frac{3}{38}$

This gives ${𝑘}^{\prime }:{𝑘}^{\prime \prime }=3:38$.

Hence, the correct option is (B).

When a neutron undergoes a head-on elastic collision with a lead nucleus, we can analyze the process using the principles of conservation of momentum and kinetic energy, which apply since it's an elastic collision. The question involves calculating the fractional change in the kinetic energy of the neutron after the collision.

Let's denote:

• $m_n$ as the mass of the neutron,
• $m_p$ as the mass of the lead nucleus, where given it is 206 times the mass of the neutron, so $m_p = 206m_n$,
• $v_i$ as the initial velocity of the neutron before the collision,
• $v_f$ as the final velocity of the neutron after the collision,
• $u_i = 0$ as the initial velocity of the lead nucleus (since it's considered stationary before the collision),
• $u_f$ as the final velocity of the lead nucleus after the collision.

Since momentum is conserved in an elastic collision, the total initial momentum equals the total final momentum, which can be expressed as:

$m_n v_i + m_p u_i = m_n v_f + m_p u_f$

Given $u_i = 0$, we can simplify this to:

$m_n v_i = m_n v_f + m_p u_f$

Because the collision is elastic, kinetic energy is also conserved. The change in kinetic energy we are interested in involves calculating the kinetic energy before and after the collision, specifically for the neutron, and then finding the fractional change.

To find the velocities after the collision, we use the formula derived for velocities in a one-dimensional elastic collision involving two masses:

The velocity of the neutron ($v_f$) after the collision is given by:

$v_f = \frac{(m_n - m_p) v_i}{m_n + m_p}$

Substituting $m_p = 206m_n$, we get:

$v_f = \frac{(m_n - 206m_n) v_i}{m_n + 206m_n}$

$v_f = \frac{-205m_n}{207m_n} v_i$

$v_f = -\frac{205}{207} v_i$

The fractional change in kinetic energy of the neutron is given by the change in kinetic energy relative to the initial kinetic energy. The kinetic energy ($KE$) formula is $KE = \frac{1}{2}mv^2$, so the fractional change in kinetic energy for the neutron is:

$\Delta KE = \frac{\frac{1}{2}m_n v_f^2 - \frac{1}{2}m_n v_i^2}{\frac{1}{2}m_n v_i^2}$

Substituting $v_f = -\frac{205}{207} v_i$, we have:

$\Delta KE = \frac{\frac{1}{2}m_n (-\frac{205}{207} v_i)^2 - \frac{1}{2}m_n v_i^2}{\frac{1}{2}m_n v_i^2}$

=$\frac{\frac{1}{2}m_n (\frac{205}{207})^2 v_i^2 - \frac{1}{2}m_n v_i^2}{\frac{1}{2}m_n v_i^2}$

=$\frac{(\frac{205}{207})^2 - 1}{1}$

Calculating this gives us a fractional change in the kinetic energy:

=$\left(\frac{205^2 - 207^2}{207^2}\right)$

=$\left(\frac{42025 - 42849}{42849}\right)$

=$\frac{-824}{42849}$

This fraction simplifies approximately to a fractional decrease, and when calculated, it yields a value around -0.01923, which is about a 2% decrease in kinetic energy. Hence, the correct answer is:

Option B: 2% decrease.

The stated value of an alternating voltage, ${𝑉}_{\text{rms}}=220\text{V}$, represents its root-mean-square value.

The peak value of this ac voltage is:

${𝑉}_0=\sqrt{\surd 2}{𝑉}_{\text{rms}}=\sqrt{\surd 2}\times 220\text{V}\approx 311\text{V}$

In contrast, $220\text{V}$ dc maintains a constant value of $220\text{V}$. Because the peak value of ac exceeds the corresponding value of dc, $220\text{V}$ ac is more dangerous.

Hence, the correct option is (A).

The magnetic field at the center of a current-carrying circular coil is given by the formula:

$B = \frac{\mu_0 I}{2R}$

where,

• $B$ is the magnetic field at the center of the coil,
• $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$),
• $I$ is the current passing through the coil, and
• $R$ is the radius of the coil.

Let's determine the magnetic field before and after the radius of the coil is doubled.

Initially, let the magnetic field be $B_1$ with coil radius $R$. According to the formula:

$B_1 = \frac{\mu_0 I}{2R}$

After the radius of the coil is doubled, the new radius is $2R$. Let the new magnetic field be $B_2$. Thus,

$B_2 = \frac{\mu_0 I}{2(2R)} = \frac{\mu_0 I}{4R}$

The ratio of the initial magnetic field to the final magnetic field would be:

$\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I}{4R}}$

By simplifying the equation, we cancel out $\mu_0$, $I$, and $R$ as they are common in both the numerator and the denominator, leaving:

$\frac{B_1}{B_2} = \frac{2R}{R} \times \frac{4}{2} = 2$

This means the ratio of the magnetic field before the doubling of the radius to after is 2:1. Therefore, the correct answer is:

Option B: 2 : 1

To find the ratio of the radii $R_P : R_Q$ of two nuclei with mass numbers $A_P = 27$ and $A_Q = 125$ respectively, we use the empirical formula for the radius of a nucleus, which is given by:

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

where $R$ is the radius, $A$ is the mass number of the nucleus, and $R_0$ is a constant. Using this formula, we can express the radii of nuclei P and Q as:

$R_P = R_0 A_P^{1/3}$

$R_Q = R_0 A_Q^{1/3}$

To find the ratio $R_P : R_Q$, we divide the expression for $R_P$ by the expression for $R_Q$:

$\frac{R_P}{R_Q} = \frac{R_0 A_P^{1/3}}{R_0 A_Q^{1/3}} = \frac{A_P^{1/3}}{A_Q^{1/3}}$

Substituting the given values of $A_P = 27$ and $A_Q = 125$, we get:

$\frac{R_P}{R_Q} = \frac{27^{1/3}}{125^{1/3}}$

Since $27 = 3^3$ and $125 = 5^3$, the expression simplifies to:

$\frac{R_P}{R_Q} = \frac{3}{5}$

Therefore, the ratio of their radii $R_P : R_Q$ is $3 : 5$, which corresponds to Option B.

The correct option is C, $\mathrm{S} \mathrm{~m}^{-1}$.

Electrical conductivity is a measure of the ability of a material to conduct electric current. The SI unit of electrical conductivity is the siemens per meter ($\mathrm{S} \mathrm{~m}^{-1}$). It is the inverse of electrical resistivity (with resistivity measured in ohm-meters, $\Omega \mathrm{~m}$), and thus conductivity is expressed as the reciprocal of resistivity. Therefore, the formula to calculate conductivity ($\sigma$) from resistivity ($\rho$) is:

$\sigma = \frac{1}{\rho}$

So, when resistivity is in $\Omega \mathrm{~m}$, conductivity is given in $\mathrm{S} \mathrm{~m}^{-1}$, making option C the correct answer. The other units mentioned are related to electrical terms but do not correctly represent the unit of electrical conductivity:

• Option A: "S m" is not properly formatted; it lacks the exponent to denote per meter.
• Option B: $\Omega \mathrm{~m}$ is the unit of electrical resistivity, not conductivity.
• Option D: $\Omega \mathrm{~m}^{-1}$ is not a standard unit for any electrical property.

The wavelength of light changes when it moves from one medium to another due to a change in its speed. The speed of light in any medium is given by $v = \frac{c}{n}$, where $v$ is the speed of light in the medium, $c$ is the speed of light in vacuum, and $n$ is the refractive index of the medium.

The wavelength of light in any medium can be related to its wavelength in vacuum by the equation:

$\lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{n} = \frac{\lambda_{\text{air}}}{n}$

Given that the refractive index of air is approximately 1 (i.e., $n_{\text{air}} \approx 1$), we can use the given refractive index of the new medium ($n_{\text{medium}} = 1.5$) to find the new wavelength in that medium.

Let the wavelength of the light in air be $\lambda_{\text{air}} = 600 \, \mathrm{nm}$ and in the new medium let it be $\lambda_{\text{medium}}$. Substituting the given values into the equation, we get:

$\lambda_{\text{medium}} = \frac{600 \, \mathrm{nm}}{1.5} = 400 \, \mathrm{nm}$

Therefore, the wavelength of the light in the new medium with a refractive index of 1.5 will be $400 \, \mathrm{nm}$. Hence, the correct answer is Option D, 400 nm.

Gamma decay refers to the emission of gamma rays from a nucleus. Gamma rays are high-energy photons and do not carry any charge. They are typically emitted from the nucleus during radioactive decay when the nucleus transitions from a higher energy state to a lower energy state. Unlike alpha and beta decay, gamma decay does not involve the emission of particles such as protons or neutrons from the nucleus. Therefore, the emission of gamma rays does not change the composition of the nucleus in terms of its protons and neutrons.

Given this understanding:

• Option A - Incorrect. This describes alpha decay, where the mass number is reduced by 4 and the atomic number is reduced by 2, not gamma decay.
• Option B - Incorrect. This is more characteristic of beta decay, where a neutron is converted into a proton, increasing the atomic number by 1, but the mass number remains unchanged. Again, not gamma decay.
• Option C - Correct. During gamma decay, neither the mass number nor the atomic number of the atom is changed. Gamma decay only involves a change in the energy state of the nucleus, with no alteration in the number of protons or neutrons.
• Option D - Incorrect. This also describes alpha decay, not gamma decay.

Therefore, the correct answer is Option C: Mass number and atomic number are not changed during gamma decay.

The fringe width $𝛽$ in Young's double-slit experiment is given by

$𝛽=\frac{𝜆𝐷}{𝑑}$

For a constant slit separation $𝑑$ and screen distance $𝐷$,

$𝛽\propto 𝜆$

Given, ${𝛽}_1=3.2\times {10}^{-4}\text{ m}$, ${𝜆}_1=6400\text{ Å}$, and ${𝜆}_2=4800\text{ Å}$.

The new fringe width ${𝛽}_2$ is

${𝛽}_2={𝛽}_1\left(\frac{{𝜆}_2}{{𝜆}_1}\right)=3.2\times {10}^{-4}\times \frac{4800}{6400}=2.4\times {10}^{-4}\text{ m}$

Therefore, the change in fringe width is

$\Delta 𝛽={𝛽}_1-{𝛽}_2=(3.2-2.4)\times {10}^{-4}\text{ m}=0.8\times {10}^{-4}\text{ m}$

Hence, the correct option is (C).

The fundamental frequency of an open organ pipe of length $𝐿$ is given by

$𝑓=\frac{𝑣}{2𝐿}$

where $𝑣$ is the speed of sound.

Let ${𝐿}_1=22\text{ cm}$ and ${𝐿}_2=22.5\text{ cm}$ be the lengths of pipes A and B respectively. The corresponding fundamental frequencies are ${𝑓}_1$ and ${𝑓}_2$, where ${𝑓}_1>{𝑓}_2$ since ${𝐿}_1<{𝐿}_2$.

The relation between the two frequencies is

${𝑓}_2={𝑓}_1\left(\frac{{𝐿}_1}{{𝐿}_2}\right)$

Given that the beat frequency is $2{\text{ s}}^{-1}$,

${𝑓}_1-{𝑓}_2=2$

${𝑓}_1(1-\frac{{𝐿}_1}{{𝐿}_2})=2$

${𝑓}_1\left(\frac{{𝐿}_2-{𝐿}_1}{{𝐿}_2}\right)=2$

On substituting ${𝐿}_1=22\text{ cm}$ and ${𝐿}_2=22.5\text{ cm}$,

${𝑓}_1\left(\frac{22.5-22}{22.5}\right)=2$

${𝑓}_1\left(\frac{0.5}{22.5}\right)=2⟹{𝑓}_1=2\times 45=90\text{ Hz}$

Hence, the correct option is (B).

To find the cross-sectional area of the stream $20 \mathrm{~cm}$ below the tap, we can use the equation of continuity which states that for an incompressible and steady flow of fluid, the product of cross-sectional area (A) and the velocity of flow (V) at any point is constant throughout the flow. We can pair this with the equation that determines the velocity of a freely falling object to address the change in velocity as the water falls.

First, let's denote:

• Initial cross-sectional area as $A_1 = 1 \mathrm{~cm}^2$
• Initial velocity as $V_1 = 2 \mathrm{~m/s}$
• Final cross-sectional area as $A_2$, which we need to find.
• Final velocity as $V_2$, which we can find using the formula for the velocity of a freely falling object.

The velocity of a freely falling object can be determined by the formula:

$V_2 = \sqrt{V_1^2 + 2gh}$

where

• $g = 10 \mathrm{~m/s^2}$ is the acceleration due to gravity,
• $h = 20 \mathrm{~cm} = 0.2 \mathrm{~m}$ is the height the water has fallen,

Let's calculate $V_2$:

$V_2 = \sqrt{(2)^2 + 2 \cdot 10 \cdot 0.2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \mathrm{~m/s}$

Now, using the equation of continuity $A_1V_1 = A_2V_2$, we can find $A_2$:

$1 \cdot 2 = A_2 \cdot 2\sqrt{2}$

$A_2 = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \mathrm{~cm^2}$

Since $\sqrt{2} \approx 1.414$, substituting this value gives us:

$A_2 \approx \frac{1.414}{2} \mathrm{~cm^2} \approx 0.707 \mathrm{~cm^2}$

Therefore, the cross-sectional area of the stream $20 \mathrm{~cm}$ below the tap is approximately $0.707 \mathrm{~cm^2}$, making the correct option C) 0.707 cm$^2$.

To understand how the stress in the legs of a man who grows into a giant changes, we first need to consider how his weight and the cross-sectional area of his legs will change if his height increases by a factor of 8, while his density remains constant.

Let's denote the original height of the man by $H$, and his new height by $H'$. Given that $H' = 8H$, we can say that the man's linear dimensions all increase by a factor of 8.

Assuming the man's shape and density remain constant, his volume will scale with the cube of the change in his linear dimensions. Since his volume is directly proportional to his weight (assuming density is constant), his weight $W'$ will increase by the cube of the linear scale factor:

$W' = 8^3W$

where $W$ is his original weight. This is because volume (and thus weight, given constant density) scales as the cube of height, and if height increases by a factor of 8, volume and weight increase by $8^3$.

Now, let's consider the stress in the legs, knowing that stress is defined as force (weight, in this context) per unit area. The critical area here is the cross-section of the man's legs.

The cross-sectional area $A$ of his legs will increase as the square of the linear dimension, since area scales with the square of the length. So, if his height increases by a factor of 8, the cross-sectional area of his legs will increase by $8^2$, i.e.,

$A' = 8^2A$

Stress is defined as the force divided by the area over which the force is applied ($\sigma = \frac{F}{A}$). The stress in his legs $σ'$ after the growth could thus be given by:

$\sigma' = \frac{W'}{A'}$

Substituting the expressions for $W'$ and $A'$ gives:

$\sigma' = \frac{8^3W}{8^2A} = 8\frac{W}{A}$