COMEDK 2024 Morning Shift Question Paper with Solutions

Let ${𝑇}_1$ and ${𝑇}_2$ be the tensions in the strings making angles of ${60}^{\circ }$ and ${30}^{\circ }$ with the horizontal ceiling, respectively.

If $ℎ$ is the vertical distance of the stone below the ceiling, the lengths of the two strings are

${𝑙}_1=\frac{ℎ}{\sin {60}^{\circ }}=\frac{2ℎ}{\sqrt{\surd 3}}$

${𝑙}_2=\frac{ℎ}{\sin {30}^{\circ }}=2ℎ$

Since ${𝑙}_2>{𝑙}_1$, the string making an angle of ${30}^{\circ }$ with the ceiling is the longer string.

For the equilibrium of the stone, balancing forces in the horizontal direction gives

${𝑇}_1\cos {60}^{\circ }={𝑇}_2\cos {30}^{\circ }⟹{𝑇}_1\left(\frac{1}{2}\right)={𝑇}_2\left(\frac{\sqrt{\surd 3}}{2}\right)⟹{𝑇}_1=\sqrt{\surd 3}{𝑇}_2$

Balancing forces in the vertical direction gives

${𝑇}_1\sin {60}^{\circ }+{𝑇}_2\sin {30}^{\circ }=𝑚𝑔$

On substituting ${𝑇}_1=\sqrt{\surd 3}{𝑇}_2$, $𝑚=2\text{kg}$, and $𝑔=10{\text{m s}}^{-2}$,

$\left(\sqrt{\surd 3}{𝑇}_2\right)\left(\frac{\sqrt{\surd 3}}{2}\right)+{𝑇}_2\left(\frac{1}{2}\right)=2\times 10$

$\frac{3}{2}{𝑇}_2+\frac{1}{2}{𝑇}_2=20⟹2{𝑇}_2=20⟹{𝑇}_2=10\text{N}$

Hence, the correct option is (C).

To determine the electromotive force (emf) developed across the ends of the rotating rod, we can use the concept of motional emf generated in a rotating conductor in a magnetic field.

The formula for the motional emf in a rotating rod is given by:

$\varepsilon = \frac{1}{2} B \omega L^2$

where:

• $B$ is the magnetic field strength
• $\omega$ is the angular velocity
• $L$ is the length of the rotating rod

First, let's find the angular velocity $\omega$. The frequency $f$ is given as $100 \, \text{Hz}$. The relationship between angular velocity $\omega$ and frequency $f$ is:

$\omega = 2 \pi f$

Plugging in the values:

$\omega = 2 \pi \cdot 100 = 200 \pi \, \text{rad/s}$

The length of the rod $L$ is given as $2 \, \text{m}$ and the magnetic field strength $B$ is given as $2 \, \text{T}$.

Now, substituting these values into the motional emf formula:

$\varepsilon = \frac{1}{2} \cdot 2 \cdot 200 \pi \cdot (2)^2$

Simplifying this expression:

$\varepsilon = \frac{1}{2} \cdot 2 \cdot 200 \pi \cdot 4$

$\varepsilon = 800 \pi \, \text{V}$

Therefore, the emf developed across the ends of the rod is:

Option A: 800 $\pi$ volt

The focal length of a convex mirror is independent of the wavelength of the incident light. This is because the focal length of a mirror is determined by the curvature of the reflecting surface and not by the properties of the light used.

Therefore, even if the wavelength of the incident light is doubled, the focal length of the mirror will remain the same. As given, the original focal length of the convex mirror is $10 \mathrm{~cm}$. So, if the wavelength is doubled, the focal length will still be:

$10 \mathrm{~cm}$

Hence, the correct option is:

Option D: 10 cm

To determine the percentage drop in heat output, we first need to understand that the power output $P$ of an electrical device is proportional to the square of the voltage $V$. Mathematically, this relationship is represented as:

$P \propto V^2 \quad \text{or} \quad P = \frac{V^2}{R}$

where $R$ is the resistance of the heating unit. In this problem, we need to compare the power output at two different voltages, $400 \mathrm{~V}$ and $160 \mathrm{~V}$.

Let's denote the initial power output at $400 \mathrm{~V}$ as $P_1$ and the power output at $160 \mathrm{~V}$ as $P_2$. Given that the initial power output is $500 \mathrm{~W}$, we can write:

$P_1 = 500 \mathrm{~W}$

Since the power is proportional to the square of the voltage:

$P_1 = k \cdot V_1^2$

and

$P_2 = k \cdot V_2^2$

where $k$ is a proportionality constant. We can divide these equations to find the ratio of $P_2$ to $P_1$:

$\frac{P_2}{P_1} = \frac{V_2^2}{V_1^2}$

Substituting the given voltage values:

$\frac{P_2}{P_1} = \frac{(160)^2}{(400)^2} = \frac{25600}{160000} = \frac{16}{100} = 0.16$

This means that the power output at $160 \mathrm{~V}$ is $0.16$ times the power output at $400 \mathrm{~V}$. To find $P_2$, we multiply $P_1$ by $0.16$:

$P_2 = 0.16 \cdot 500 \mathrm{~W} = 80 \mathrm{~W}$

The percentage drop in heat output is then given by:

$\text{Percentage Drop} = \left( \frac{P_1 - P_2}{P_1} \right) \times 100$

Substituting the values:

$\text{Percentage Drop} = \left( \frac{500 \mathrm{~W} - 80 \mathrm{~W}}{500 \mathrm{~W}} \right) \times 100 = \left( \frac{420 \mathrm{~W}}{500 \mathrm{~W}} \right) \times 100 = 84\%$

Thus, the percentage drop in heat output is 84%. The correct answer is:

Option C - 84%.

Let ${𝐶}_0$ be the capacitance of the capacitor without dielectric.

The initial charge on the charged capacitor is

${𝑞}_0={𝐶}_0{𝑈}_0=78{𝐶}_0$

For the capacitor with the dielectric, the capacitance at potential $𝑈$ is $𝐶=𝜖{𝐶}_0=2𝑈{𝐶}_0$. The charge on this capacitor at voltage $𝑈$ is

${𝑞}_1=𝐶𝑈=2{𝐶}_0{𝑈}^2$

The charge remaining on the first capacitor at the common potential $𝑈$ is

${𝑞}_2={𝐶}_0𝑈$

By conservation of charge,

${𝑞}_1+{𝑞}_2={𝑞}_0⟹2{𝐶}_0{𝑈}^2+{𝐶}_0𝑈=78{𝐶}_0$

$\Rightarrow 2{𝑈}^2+𝑈-78=0$

Factoring the quadratic equation,

$(𝑈-6)(2𝑈+13)=0$

Since the potential $𝑈>0$,

$𝑈=6\text{V}$

Hence, the correct option is (A).

The distance of the displaced charge $𝑞$ from either of the fixed charges $-𝑞$ is $𝑟=\sqrt{\surd {𝑑}^2+{𝑥}^2}$.

The attractive force exerted by each charge $-𝑞$ on $𝑞$ has magnitude

${𝐹}_1=\frac{1}{4𝜋{𝜀}_0}\frac{{𝑞}^2}{{𝑟}^2}=\frac{{𝑞}^2}{4𝜋{𝜀}_0({𝑑}^2+{𝑥}^2)}$

The components of the two forces perpendicular to the $𝑥$-axis cancel by symmetry, while the components directed towards the origin add up to give the net restoring force:

$𝐹=-2{𝐹}_1\cos 𝜃=-2\left[\frac{{𝑞}^2}{4𝜋{𝜀}_0({𝑑}^2+{𝑥}^2)}\right]\frac{𝑥}{\sqrt{\surd {𝑑}^2+{𝑥}^2}}=-\frac{{𝑞}^2𝑥}{2𝜋{𝜀}_0({𝑑}^2+{𝑥}^2{)}^{3/2}}$

Since $𝑥\ll 𝑑$, this simplifies to

$𝐹\approx -\left(\frac{{𝑞}^2}{2𝜋{𝜀}_0{𝑑}^3}\right)𝑥$

The effective spring constant is $𝑘=\frac{{𝑞}^2}{2𝜋{𝜀}_0{𝑑}^3}$. The time period of oscillation is

$𝑇=2𝜋\sqrt{\surd \frac{𝑚}{𝑘}}=2𝜋\sqrt{\surd \frac{2𝜋{𝜀}_0𝑚{𝑑}^3}{{𝑞}^2}}=\sqrt{\surd \frac{8{𝜀}_0𝑚{𝜋}^3{𝑑}^3}{{𝑞}^2}}$

Hence, the correct option is (D).

First, let's analyze the given information. The radius of the circular path, $r$, is given as $\frac{5}{\pi}$ meters. The frequency of revolution, $f$, is numerically equal to the radius, so $f = r = \frac{5}{\pi}$ Hz.

The acceleration we need to find is the centripetal acceleration, which is given by the formula:

$a = \omega^2 r$

Here, $\omega$ (omega) is the angular velocity, which can be calculated from the frequency $f$ using the relationship:

$\omega = 2 \pi f$

Substitute $f = \frac{5}{\pi}$ into the equation for $\omega$:

$\omega = 2 \pi \left(\frac{5}{\pi}\right) = 10 \, \text{rad/s}$

Now, substitute $\omega = 10 \, \text{rad/s}$ and $r = \frac{5}{\pi}$ meters into the centripetal acceleration formula:

$a = \omega^2 r = (10)^2 \left(\frac{5}{\pi}\right)$

Thus, the acceleration $a$ is:

$a = 100 \left(\frac{5}{\pi}\right) = \frac{500}{\pi} \, \text{m/s}^2$

Therefore, the correct option is:

Option D: $\left(\frac{500}{\pi}\right) \mathrm{ms}^{-2}$

To find the current through the conductor at $220^{\circ} \mathrm{C}$, we need to understand the relationship between temperature and resistance, assuming that the resistance of the conductor increases linearly with temperature. This relationship is given by:

$R_T = R_0 (1 + \alpha T)$

Where:

• $R_T$ is the resistance at temperature $T$.
• $R_0$ is the resistance at $0^{\circ} \mathrm{C}$.
• $\alpha$ is the temperature coefficient of resistance.
• $T$ is the temperature in degrees Celsius.

Since the current $I$ is inversely proportional to the resistance $R$ (Ohm's Law: $I = \frac{V}{R}$), we'll first express the given currents $a$ and $b$ at the corresponding temperatures in terms of resistance:

At $0^{\circ} \mathrm{C}$:

$I_0 = a = \frac{V}{R_0}$

At $100^{\circ} \mathrm{C}$:

$I_{100} = b = \frac{V}{R_{100}} = \frac{V}{R_0 (1 + 100\alpha)}$

Dividing the equations for $b$ by the equation for $a$, we get:

$\frac{b}{a} = \frac{V / R_{100}}{V / R_0} = \frac{R_0}{R_0 (1 + 100\alpha)} = \frac{1}{1 + 100\alpha}$

Solving for $\alpha$, we get:

$1 + 100\alpha = \frac{a}{b}$

$100\alpha = \frac{a}{b} - 1$

$\alpha = \frac{1}{100} \left(\frac{a}{b} - 1\right)$

Now let's determine the current at $220^{\circ} \mathrm{C}$:

At $220^{\circ} \mathrm{C}$:

$I_{220} = \frac{V}{R_{220}} = \frac{V}{R_0 (1 + 220\alpha)}$

Substituting the value of $\alpha$ from above:

$I_{220} = \frac{V}{R_0 \left(1 + 220 \cdot \frac{1}{100} \left(\frac{a}{b} - 1\right)\right)}$

$I_{220} = \frac{V}{R_0 \left(1 + 2.2 \left(\frac{a}{b} - 1\right)\right)}$

$I_{220} = \frac{V}{R_0 \left(1 + 2.2 \frac{a}{b} - 2.2\right)}$

$I_{220} = \frac{V}{R_0 \left(\frac{b + 2.2 a - 2.2 b}{b}\right)}$

$I_{220} = \frac{V b}{R_0 (b + 2.2 a - 2.2 b)}$

$I_{220} = \frac{V b}{R_0 (2.2 a - 1.2 b)}$

Using $I_0 = \frac{V}{R_0}$ from above:

$I_{220} = \frac{b I_0}{2.2 a - 1.2 b}$

$I_{220} = \frac{b a}{2.2 a - 1.2 b}$

Multiplying numerator and denominator by 5, we get:

$I_{220} = \frac{5 a b}{11 a - 6 b}$

Therefore, the current through the conductor at $220^{\circ} \mathrm{C}$ is given by Option C:

$\frac{5 a b}{11 a - 6 b}$

The areal velocity of a planet is the rate at which its radius vector sweeps out area. It is given by:

$\mathrm{A} = \frac{1}{2} r^2 \omega$

where $r$ is the distance from the planet to the sun and $\omega$ is its angular velocity. The angular momentum of the planet is given by:

$L = I \omega = Mr^2 \omega$

where $I$ is the moment of inertia of the planet. Substituting the expression for $\omega$ from the areal velocity equation, we get:

$L = Mr^2 \left( \frac{2A}{r^2} \right) = 2MA$

Therefore, the angular momentum of the planet is 2MA. So the answer is Option C.

For aluminium, atomic mass $𝑀=27{\text{g mol}}^{-1}$ and atomic number $𝑍=13$.

The number of atoms in $𝑚=0.75\text{g}$ of aluminium is

$𝑁=\frac{𝑚}{𝑀}{𝑁}_{\text{A}}=\frac{0.75}{27}\times 6.023\times {10}^{23}\approx 1.673\times {10}^{22}$

The total magnitude of positive (or negative) charge is

$𝑄=𝑁𝑍𝑒=1.673\times {10}^{22}\times 13\times 1.602\times {10}^{-19}\text{C}\approx 3.48\times {10}^4\text{C}$

Given the diagonal of the square coin is $𝑑=14\text{mm}$, the side length $𝑠$ is

$𝑠=\frac{𝑑}{\sqrt{\surd 2}}=\frac{14}{\sqrt{\surd 2}}\text{mm}\approx 10\text{mm}={10}^{-2}\text{m}$

The dipole moment of the system is

$𝑝=𝑄\times 𝑠=(3.48\times {10}^4\text{C})\times \left({10}^{-2}\text{m}\right)=348\text{C m}$

Hence, the correct option is (D).

To analyze the potential difference across the capacitor, let's first understand the behavior of the circuit elements involved. An ideal diode allows current to flow only in one direction, blocking current in the reverse direction. A capacitor, on the other hand, stores energy in the form of an electric field and can charge and discharge based on the applied voltage.

The ac source has a voltage of $220 \mathrm{~V}$. The given voltage is typically an RMS (Root Mean Square) value, which is a form of averaging used for ac voltages. The peak voltage ($V_{\text{peak}}$) of the ac source can be calculated using the relationship:

$V_{\text{peak}} = V_{\text{RMS}} \times \sqrt{2}$

Substituting the given RMS value, we have:

$V_{\text{peak}} = 220 \mathrm{~V} \times \sqrt{2}$

$V_{\text{peak}} \approx 220 \mathrm{~V} \times 1.414$

$V_{\text{peak}} \approx 311 \mathrm{~V}$

Since the diode is ideal, during the positive half cycle of the ac voltage, the diode will conduct and allow current to flow, charging the capacitor to the peak voltage of the ac source, which is 311 V.

During the negative half cycle, the diode will be in reverse bias and will not conduct, thus isolating the capacitor and maintaining its charge. Therefore, the potential difference across the capacitor will be equal to the peak voltage of the ac source.

Hence, the potential difference across the capacitor is:

$\boxed{311 \mathrm{~V}}$

So, the correct answer is:

Option B

$311 \mathrm{~V}$

To determine the total mass of the gas in the box, we first need to understand the given problem and apply the principles of kinetic theory of gases. Let's summarize the given data:

• Side length of the cubical box, $L = 1 \ \text{m}$
• Pressure of the gas, $P = 100 \ \text{Nm}^{-2}$
• Observation time, $t = 1 \ \text{s}$
• Number of hits on the wall, $n = 500$

We are given that the atom travels with the root mean square (rms) speed (denoted by $v_{rms}$) parallel to one edge of the cube, and makes 500 hits on the wall in 1 second. From this, we can find the $v_{rms}$ as follows:

Each hit corresponds to the atom travelling across the side of the cube and bouncing back. The distance travelled between two consecutive hits is $2L$, as the atom goes to the wall and returns. Therefore, the total distance travelled in 1 second is:

$500 \times 2L = 500 \times 2 \times 1 = 1000 \ \text{m}$

So, the rms speed $v_{rms}$ is:

$v_{rms} = \frac{1000 \ \text{m}}{1 \ \text{s}} = 1000 \ \text{m/s}$

Now, using the formula from the kinetic theory of gases for pressure:

$P = \frac{1}{3} \rho v_{rms}^2$

where $\rho$ is the density of the gas. We can express density $\rho$ as:

$\rho = \frac{m}{V}$

Here, $m$ is the mass of the gas and $V$ is the volume of the box. For a cubical box, $V = L^3$, so:

$V = 1^3 \ \text{m}^3 = 1 \ \text{m}^3$

Substituting the values into the pressure equation:

$100 \ \text{Nm}^{-2} = \frac{1}{3} \left( \frac{m}{1} \right) (1000)^2$

Simplifying, we get:

$100 = \frac{1}{3} m \times 1000000$

$100 = \frac{1000000}{3} m$

$m = \frac{100 \times 3}{1000000}$

$m = 0.0003 \ \text{kg}$

Converting this mass to grams:

$m = 0.0003 \ \text{kg} \times 1000 \ \text{g/kg} = 0.3 \ \text{g}$

Therefore, the total mass of the gas in the box is:

Option B: 0.3 grams

The coefficient of mutual induction, commonly denoted by $M$, is defined by the relationship between the induced electromotive force (EMF) in one coil and the rate of change of current in another coil. According to Faraday's Law of Induction, the induced EMF ( $\mathcal{E}$ ) in a coil is given by:

$\mathcal{E} = -M \frac{dI}{dt}$

Where:

• $\mathcal{E}$ is the induced EMF in volts (V).
• $M$ is the mutual inductance.
• $\frac{dI}{dt}$ is the rate of change of current in amperes per second (A/s).

Rearranging this equation to solve for $M$, we get:

$M = - \frac{\mathcal{E}}{\frac{dI}{dt}}$

From this relationship, the units of $M$ can be derived. The unit of EMF ( $\mathcal{E}$ ) is volts (V), and the unit of $\frac{dI}{dt}$ is amperes per second (A/s). Thus, the unit of the coefficient of mutual induction is:

$M = \frac{\text{V}}{\text{A/s}} = \text{volt} \cdot \text{second} / \text{ampere}$

Therefore, the correct unit for the coefficient of mutual induction is:

Option A: volt. second / ampere

First, let's identify the voltage transformation ratio for the transformer. The primary voltage $V_p$ is $330 \mathrm{~V}$ and the secondary voltage $V_s$ is $33 \mathrm{~V}$. The voltage transformation ratio $k$ is given by:

$k = \frac{V_p}{V_s} = \frac{330}{33} = 10$

This means the primary voltage is 10 times the secondary voltage.

Next, we need to determine the current in the secondary coil, which can be found using Ohm's Law. The impedance of the device connected to the secondary coil is given as $110 \Omega$, and the voltage across it is $33 \mathrm{~V}$:

$I_s = \frac{V_s}{Z} = \frac{33}{110} = 0.3 \mathrm{~A}$

Now, using the current transformation relation in a transformer, which is inversely proportional to the voltage transformation ratio:

$I_p = \frac{I_s}{k} = \frac{0.3}{10} = 0.03 \mathrm{~A}$

The current drawn by the primary coil of the transformer is therefore $0.03 \mathrm{~A}$.

So, the correct answer is:

Option B:
0.03 A

In a single slit diffraction experiment, the width of the central maxima is given by the diffraction pattern formula derived from the Huygens-Fresnel principle. The position of the first minima in the diffraction pattern is given by:

$a \sin(\theta) = n \lambda$

where $a$ is the slit width, $\theta$ is the angle of the diffraction minimum, $n$ is the order of the minimum (for the first minimum, $n=1$), and $\lambda$ is the wavelength of the light used.

The angular width of the central maxima is approximately twice the angle to the first minimum:

$\Delta \theta = \frac{2 \lambda}{a}$

When projecting this onto a screen at a distance $D$, the width of the central maximum can be approximated by:

$\beta = 2D \tan(\theta) \approx 2D \left(\frac{\lambda}{a}\right)$

If the slit width $a$ is doubled, say to $2a$, then the new width of the central maximum $\beta'$ becomes:

$\beta' = 2D \left(\frac{\lambda}{2a}\right) = \frac{\beta}{2}$

Thus, if the slit width is doubled, the corresponding width of the central maxima will be:

Option C: $\frac{\beta}{2}$.

The physical position of the centre of mass of a body is determined entirely by its geometry (shape and size) and the distribution of mass within it. It is a fixed, unique point relative to the body and does not depend on the choice of the coordinate system chosen to describe it.

The distribution of mass directly influences the location of the centre of mass, ruling out (A). The coordinates of the centre of mass change with the choice of origin, but the point itself is independent of the coordinate system, ruling out (B). Furthermore, for bodies such as a hollow sphere or a circular ring, the centre of mass lies in empty space outside the material of the body, ruling out (D).

Hence, the correct option is (C).

The correct option is:

Option B: The energy gap of $\mathrm{Sn}$ is zero volt while that of $\mathrm{Si}$ is $0.07 \mathrm{~V}$

Explanation:

Both $\mathrm{Sn}$ (tin) and $\mathrm{Si}$ (silicon) belong to the 14th group of the periodic table, and they share some common properties. However, the key difference explaining why $\mathrm{Sn}$ behaves as a metal while $\mathrm{Si}$ behaves as a semiconductor lies in their energy band structure.

In metals, the valence band and conduction band overlap, which means there is no energy gap ($E_g$) between them. This allows electrons to move freely within the material, contributing to high electrical conductivity.

In semiconductors, there is a small energy gap between the valence band and the conduction band. For silicon, this energy gap is approximately $0.07 \mathrm{~eV}$. This small but significant gap means that at room temperature, some electrons can gain enough thermal energy to jump from the valence band to the conduction band, allowing moderate electrical conductivity that can be manipulated through doping and other methods.

For tin, the energy gap is essentially zero because the valence and conduction bands overlap, which leads to metallic behavior with high electrical conductivity.

Thus, the primary reason why $\mathrm{Sn}$ is a metal and $\mathrm{Si}$ is a semiconductor is due to the difference in their energy gaps. The energy gap of $\mathrm{Sn}$ is effectively zero, while $\mathrm{Si}$ has a small but finite energy gap of $0.07 \mathrm{~eV}$.

To find the ratio of the maximum speed required by a rocket to escape the gravitational force of planet $\mathrm{X}$ to that of Earth, we must consider the escape velocity formula:

$v_e = \sqrt{\frac{2GM}{R}}$

where:

• $v_e$ is the escape velocity
• $G$ is the gravitational constant
• $M$ is the mass of the planet
• $R$ is the radius of the planet

Given:

• The mass of Earth ($M_{\text{Earth}}$) is nine times the mass of planet X ($M_X$),
• The radius of Earth ($R_{\text{Earth}}$) is four times the radius of planet X ($R_X$).

So, we have:

$M_{\text{Earth}} = 9M_X$

$R_{\text{Earth}} = 4R_X$

The escape velocity for planet X is:

$v_{e,X} = \sqrt{\frac{2GM_X}{R_X}}$

The escape velocity for Earth is:

$v_{e, \text{Earth}} = \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}}$

Substitute $M_{\text{Earth}} = 9M_X$ and $R_{\text{Earth}} = 4R_X$:

$v_{e, \text{Earth}} = \sqrt{\frac{2G(9M_X)}{4R_X}} = \sqrt{\frac{18GM_X}{4R_X}} = \sqrt{\frac{9GM_X}{2R_X}} = 3\sqrt{\frac{GM_X}{2R_X}}$

However, note that:

$v_{e,X} = \sqrt{\frac{2GM_X}{R_X}}$

So, the ratio of the escape velocity of planet X to Earth is:

$\frac{v_{e,X}}{v_{e, \text{Earth}}} = \frac{\sqrt{\frac{2GM_X}{R_X}}}{3\sqrt{\frac{GM_X}{2R_X}}}$

$= \frac{\sqrt{\frac{2GM_X}{R_X}}}{3 \cdot \sqrt{\frac{GM_X}{2R_X}}}$

$= \frac{1}{3} \cdot \sqrt{\frac{2GM_X}{R_X} \cdot \frac{2R_X}{GM_X}}$

$= \frac{1}{3} \sqrt{4}$

$= \frac{1}{3} \cdot 2$

$= \frac{2}{3}$

Therefore, the ratio of the maximum speed required by a rocket to escape the gravitational force of planet $\mathrm{X}$ to that of Earth is:

$\boxed{\frac{2}{3}}$

The binding energy of a nucleus is the energy required to disassemble the nucleus into its constituent protons and neutrons. The formula to calculate the binding energy is given by:

$\text{Binding Energy} = \left(\text{Total mass of individual nucleons} - \text{Mass of the nucleus}\right) \times 931.5 \, \text{MeV/u}$

For tritium ($^3_1\text{H}$), it consists of 1 proton and 2 neutrons. We are given the following masses:

Mass of tritium nucleus, $\text{m}_{\text{T}} = 3.01605 \, \text{u}$

Mass of a proton, $\text{m}_{\text{p}} = 1.00782 \, \text{u}$

Mass of a neutron, $\text{m}_{\text{n}} = 1.00866 \, \text{u}$

First, we calculate the total mass of individual nucleons:

$\text{Total mass of nucleons} = 1 \times 1.00782 \, \text{u} + 2 \times 1.00866 \, \text{u}$

$= 1.00782 \, \text{u} + 2.01732 \, \text{u}$

$= 3.02514 \, \text{u}$

Next, we calculate the mass defect (difference between the total mass of individual nucleons and the mass of the nucleus):

$\text{Mass defect} = \text{Total mass of nucleons} - \text{Mass of the nucleus}$

$= 3.02514 \, \text{u} - 3.01605 \, \text{u}$

$= 0.00909 \, \text{u}$

Now, we convert the mass defect into energy using the conversion factor 931.5 MeV/u:

$\text{Binding Energy} = 0.00909 \, \text{u} \times 931.5 \, \text{MeV/u}$

$\approx 8.5 \, \text{MeV}$

Therefore, the binding energy of the tritium nucleus is approximately 8.5 MeV. Hence, the correct option is:

Option A: 8.5 MeV

A coil typically offers a certain resistance to a direct current (DC), which is also known as its ohmic resistance. In the case of an alternating current (AC), the coil presents additional opposition to the flow of current, which is known as impedance. This impedance is due to both the ohmic resistance and the inductive reactance of the coil.

The inductive reactance $X_L$ of a coil in an AC circuit is given by the formula:

$X_L = 2 \pi f L$

where:

• $f$ is the frequency of the alternating current
• $L$ is the inductance of the coil

The total impedance $Z$ of the coil in an AC circuit, considering both the resistance $R$ and the inductive reactance $X_L$, is given by the formula:

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

Since impedance $Z$ takes into account both ohmic resistance and inductive reactance, the resistance offered by the coil to an alternating current will always be greater than its resistance to a direct current, assuming the inductance $L$ is not zero.

Therefore, the resistance offered by the coil to the alternating current will be:

Option B: Greater than $20 \Omega$

Let the speeds of the two objects be $v_1$ and $v_2$.

When the two objects are moving in the same direction, the relative speed is given by:

$|v_1 - v_2|$

According to the problem, this relative speed results in an increase in the distance between them by $6 \mathrm{~m}$ in one second. Therefore, we have:

$|v_1 - v_2| = 6$

When the two objects are moving towards each other, the relative speed is given by:

$v_1 + v_2$

In this case, the distance decreases by $8 \mathrm{~m}$ in one second, so we have:

$v_1 + v_2 = 8$

We now have two equations:

1. $|v_1 - v_2| = 6$

1. $v_1 + v_2 = 8$

We can solve these equations by considering the absolute value condition in two cases:

Case 1: $v_1 \ge v_2$

v_1 - v_2 = 6 \quad \text{(from |v_1 - v_2| = 6 )}

$v_1 + v_2 = 8$

Adding these two equations:

$(v_1 - v_2) + (v_1 + v_2) = 6 + 8$

$2v_1 = 14$

$v_1 = 7$

Substituting $v_1 = 7$ into $v_1 + v_2 = 8$:

$7 + v_2 = 8$

$v_2 = 1$

Case 2: $v_1 < v_2$

v_2 - v_1 = 6 \quad \text{(from |v_1 - v_2| = 6 )}

$v_1 + v_2 = 8$

Adding these two equations:

$(v_2 - v_1) + (v_1 + v_2) = 6 + 8$

$2v_2 = 14$

$v_2 = 7$

Substituting $v_2 = 7$ into $v_1 + v_2 = 8$:

$v_1 + 7 = 8$

$v_1 = 1$

From both cases, we find that the speeds of the objects are $7 \mathrm{~ms}^{-1}$ and $1 \mathrm{~ms}^{-1}$.

Therefore, the correct answer is:

Option B: $7 \mathrm{~ms}^{-1}$ and $1 \mathrm{~ms}^{-1}$

Fringe width in air is given by

$𝛽=\frac{𝜆𝐷}{𝑑}$

When the entire apparatus is immersed in a liquid of refractive index $𝜇$, the wavelength becomes ${𝜆}^{\prime }=\frac{𝜆}{𝜇}$, while $𝐷$ and $𝑑$ remain unchanged. The new fringe width ${𝛽}^{\prime }$ is

${𝛽}^{\prime }=\frac{{𝜆}^{\prime }𝐷}{𝑑}=\frac{𝛽}{𝜇}$

Given $𝛽=2.6\text{mm}$ and $𝜇=1.3$,

${𝛽}^{\prime }=\frac{2.6}{1.3}=2\text{mm}$

Hence, the correct option is (C).

The correct answer is Option D: zero. Here's why:

Phase change upon reflection occurs when light goes from a medium with a lower refractive index to a medium with a higher refractive index. This is because the reflected wave undergoes a 180° phase shift. However, in this case, the light is moving from a medium with a lower refractive index ($\mu$) to a medium with a higher refractive index ($2\mu$). Since there's no reflection involved in this scenario, there is no phase change.

Here's a breakdown of why phase change occurs during reflection:

When light waves travel from one medium to another, they undergo a change in speed due to the different refractive indices. At the boundary between the two media, a portion of the light is reflected back into the original medium. When the light is reflected from a denser medium (higher refractive index), the reflected wave undergoes a phase shift of 180°. This is because the electric field vector of the reflected wave is inverted.

In our case, the light is transmitted from the first medium to the second, not reflected back. Therefore, there is no phase change.

To solve for the minimum value of $\Delta \mathrm{t}$ that allows for constructive interference between successive pulses, we must determine the time it takes for a wave pulse to travel from one end of the string to the other and back. This time is essentially the period for which constructive interference will occur.

First, we find the wave speed, $v$, on the string. The wave speed is given by the formula:

$v = \sqrt{\frac{T}{\mu}}$

where:

1. $T$ is the tension in the string, which is given as $2.5 \mathrm{~N}$.
2. $\mu$ is the linear mass density of the string.

The linear mass density $\mu$ can be calculated using the formula:

$\mu = \frac{m}{L}$

Where:

1. $m$ is the mass of the string, given as $10^{-3} \mathrm{~kg}$.
2. $L$ is the length of the string, given as $25 \mathrm{~cm}$ (which we convert to meters as $0.25 \mathrm{~m}$).

Thus,

$\mu = \frac{10^{-3} \mathrm{~kg}}{0.25 \mathrm{~m}} = 4 \times 10^{-3} \mathrm{~kg/m}$

Now, substituting $\mu$ and $T$ back into the wave speed formula:

$v = \sqrt{\frac{2.5 \mathrm{~N}}{4 \times 10^{-3} \mathrm{~kg/m}}} = \sqrt{625 \mathrm{~m^2/s^2}} = 25 \mathrm{~m/s}$

Now that we have the wave speed, we can calculate the time it takes for the wave pulse to travel the length of the string and back (i.e., a round trip). The total distance for a round trip is $2L$, so the time interval $\Delta t$ is given by:

$\Delta t = \frac{2L}{v}$

Substituting the given values:

$\Delta t = \frac{2 \times 0.25 \mathrm{~m}}{25 \mathrm{~m/s}} = \frac{0.5 \mathrm{~m}}{25 \mathrm{~m/s}} = 0.02 \mathrm{~s} = 20 \mathrm{~ms}$

Thus, the minimum value of $\Delta \mathrm{t}$ for constructive interference between successive pulses is:

Option D: 20 ms