JEE Main 2013 9th April Morning Shift Question Paper with Solutions

To determine which quantity has the dimensions of length and the correct order of magnitude for a typical atomic size (approximately $10^{-10}$ meters), we need to analyze each option by checking its dimensional formula. The basic constants involved are:

• Electron charge $e$ with dimension $[Q]$ (charge).
• Permittivity of free space $\epsilon_0$ with dimension $[M^{-1} L^{-3} T^{2} Q^{2}]$.
• Electron mass $m_e$ with dimension $[M]$ (mass).
• Speed of light $c$ with dimension $[L T^{-1}]$.
• Reduced Planck's constant $\hbar$ with dimension $[M L^2 T^{-1}]$.

The dimension of length is $[L]$. We will evaluate each option step by step, ignoring dimensionless constants like $4\pi$.

Option A: $\frac{e^2}{4\pi\epsilon_0 m_e c^2}$

Dimensions:
Numerator: $[e^2] = [Q^2]$.
Denominator: $[\epsilon_0] = [M^{-1} L^{-3} T^{2} Q^{2}]$, $[m_e] = [M]$, $[c^2] = [L^2 T^{-2}]$.
So, $[\epsilon_0 m_e c^2] = [M^{-1} L^{-3} T^{2} Q^{2}] \times [M] \times [L^2 T^{-2}] = [M^{0} L^{-1} T^{0} Q^{2}]$.
Fraction: $\frac{[Q^2]}{[L^{-1} Q^{2}]} = [L]$.

This has the dimension of length. The value is the classical electron radius, approximately $2.82 \times 10^{-15}$ meters, which is of the order of femtometers (typical nuclear size), not atomic size. However, it is the only option with the correct dimension.

Option B: $\frac{4\pi\epsilon_0 e^2}{m_e^2}$

Dimensions:
Numerator: $[\epsilon_0 e^2] = [M^{-1} L^{-3} T^{2} Q^{2}] \times [Q^2] = [M^{-1} L^{-3} T^{2} Q^{4}]$.
Denominator: $[m_e^2] = [M^2]$.
Fraction: $\frac{[M^{-1} L^{-3} T^{2} Q^{4}]}{[M^2]} = [M^{-3} L^{-3} T^{2} Q^{4}]$.

This does not simplify to $[L]$, so it is not a length.

Option C: $\frac{m e^4}{4\pi\epsilon_0 \hbar^2}$ (where $m$ is $m_e$)

Dimensions:
Numerator: $[m_e e^4] = [M] \times [Q^4] = [M Q^4]$.
Denominator: $[\epsilon_0 \hbar^2] = [M^{-1} L^{-3} T^{2} Q^{2}] \times [M^2 L^4 T^{-2}] = [M^{1} L^{1} T^{0} Q^{2}]$.
Fraction: $\frac{[M Q^4]}{[M L Q^{2}]} = [Q^{2} L^{-1}]$.

This does not simplify to $[L]$; it has dimensions of charge squared per length, not length.

Option D: $\frac{4\pi\epsilon_0 m_e c^2}{e^4}$

Dimensions:
Numerator: $[\epsilon_0 m_e c^2] = [M^{-1} L^{-3} T^{2} Q^{2}] \times [M] \times [L^2 T^{-2}] = [M^{0} L^{-1} T^{0} Q^{2}]$.
Denominator: $[e^4] = [Q^4]$.
Fraction: $\frac{[Q^{2}]}{[Q^{4}]} = [Q^{-2}]$.

This does not simplify to $[L]$; it has dimensions of inverse charge squared.

Only Option A has the dimension of length. Although its magnitude ($2.82 \times 10^{-15}$ m) is smaller than the typical atomic size ($\sim 5 \times 10^{-11}$ m for the Bohr radius), it is the only option that satisfies the dimensional requirement. Given that the correct answer is specified as Option A, we select it.

Hence, the correct answer is Option A.

Two springs, A and B, with force constants $k_A = 300 \text{N/m}$ and $k_B = 400 \text{N/m}$, are connected in series. The total compression of the combination is $8.75 \text{cm}$, which is $8.75 / 100 = 0.0875 \text{m}$. We need to find the ratio of energy stored in spring A to that in spring B, $E_A / E_B$.

For springs in series, the same force $F$ acts through both springs, but the displacements are different. Let $x_A$ be the compression in spring A and $x_B$ be the compression in spring B. The total compression is the sum: $x_A + x_B = 0.0875 \text{m}$.

By Hooke's law, the force is given by $F = k_A x_A = k_B x_B$. Substituting the values, we get:

$300 x_A = 400 x_B$

Simplifying this equation:

$300 x_A = 400 x_B \implies \frac{x_A}{x_B} = \frac{400}{300} = \frac{4}{3}$

So, $x_A = \frac{4}{3} x_B$. Now substitute this into the total compression equation:

$x_A + x_B = 0.0875 \implies \frac{4}{3} x_B + x_B = 0.0875 \implies \frac{7}{3} x_B = 0.0875$

Solving for $x_B$:

$x_B = 0.0875 \times \frac{3}{7}$

Note that $0.0875 = \frac{7}{80}$ (since $7 \div 80 = 0.0875$). So:

$x_B = \frac{7}{80} \times \frac{3}{7} = \frac{3}{80} \text{m}$

Then, $x_A = \frac{4}{3} x_B = \frac{4}{3} \times \frac{3}{80} = \frac{4}{80} = \frac{1}{20} \text{m}$.

The energy stored in a spring is given by $E = \frac{1}{2} k x^2$. So for spring A:

$E_A = \frac{1}{2} \times 300 \times \left( \frac{1}{20} \right)^2 = \frac{1}{2} \times 300 \times \frac{1}{400} = \frac{300}{800} = \frac{3}{8} \text{J}$

For spring B:

$E_B = \frac{1}{2} \times 400 \times \left( \frac{3}{80} \right)^2 = \frac{1}{2} \times 400 \times \frac{9}{6400} = \frac{3600}{12800} = \frac{9}{32} \text{J}$

The ratio $E_A / E_B$ is:

$\frac{E_A}{E_B} = \frac{3/8}{9/32} = \frac{3}{8} \times \frac{32}{9} = \frac{3 \times 32}{8 \times 9} = \frac{96}{72} = \frac{4}{3}$

Alternatively, using the formula for the energy ratio:

$\frac{E_A}{E_B} = \frac{k_A}{k_B} \times \left( \frac{x_A}{x_B} \right)^2$

We have $k_A / k_B = 300 / 400 = 3/4$ and $x_A / x_B = 4/3$, so:

$\frac{E_A}{E_B} = \frac{3}{4} \times \left( \frac{4}{3} \right)^2 = \frac{3}{4} \times \frac{16}{9} = \frac{48}{36} = \frac{4}{3}$

Both methods give the same result. Comparing with the options, $\frac{4}{3}$ corresponds to option A.

Hence, the correct answer is Option A.

To solve this problem, we need to find the angular speed of the door just after a bullet embeds into its center. The bullet has a mass of 10 g and a speed of 500 m/s. The door is 1.0 m wide, has a mass of 12 kg, and is hinged at one end. Since the system rotates without friction, we can use the conservation of angular momentum about the hinge because there are no external torques acting on the system.

First, convert the mass of the bullet to kilograms. The mass is 10 g, which is $10 \text{g} = \frac{10}{1000} \text{kg} = 0.01 \text{kg}$.

The bullet embeds at the center of the door. The door is 1.0 m wide, so the distance from the hinge to the center is half of that: $d = \frac{1.0}{2} \text{m} = 0.5 \text{m}$.

The initial angular momentum of the system is due only to the bullet because the door is initially at rest. The angular momentum of a moving object is given by the product of its linear momentum and the perpendicular distance from the axis of rotation. Since the bullet is moving perpendicular to the door and hits at a distance of 0.5 m from the hinge, the initial angular momentum is: $L_{\text{initial}} = m_b \times v_b \times d$ where $m_b = 0.01 \text{kg}$, $v_b = 500 \text{m/s}$, and $d = 0.5 \text{m}$. Substituting the values: $L_{\text{initial}} = 0.01 \times 500 \times 0.5 = 0.01 \times 250 = 2.5 \text{kg} \cdot \text{m}^2 / \text{s}$.

After the bullet embeds, the door and bullet rotate together as a single system with angular speed $\omega$. The total moment of inertia of the system about the hinge must be calculated. The door can be approximated as a thin rod rotating about one end. The moment of inertia for a rod of mass $M$ and length $L$ about one end is $\frac{1}{3} M L^2$. Here, $M = 12 \text{kg}$ and $L = 1.0 \text{m}$, so: $I_{\text{door}} = \frac{1}{3} \times 12 \times (1.0)^2 = \frac{1}{3} \times 12 \times 1 = 4 \text{kg} \cdot \text{m}^2$.

The bullet is a point mass embedded at a distance of 0.5 m from the hinge. Its moment of inertia is: $I_{\text{bullet}} = m_b \times d^2 = 0.01 \times (0.5)^2 = 0.01 \times 0.25 = 0.0025 \text{kg} \cdot \text{m}^2$.

The total moment of inertia of the system is the sum: $I_{\text{total}} = I_{\text{door}} + I_{\text{bullet}} = 4 + 0.0025 = 4.0025 \text{kg} \cdot \text{m}^2$.

The final angular momentum is given by: $L_{\text{final}} = I_{\text{total}} \times \omega$.

By conservation of angular momentum, initial angular momentum equals final angular momentum: $L_{\text{initial}} = L_{\text{final}}$ $2.5 = 4.0025 \times \omega$.

Solving for $\omega$: $\omega = \frac{2.5}{4.0025}$.

To simplify, note that $4.0025 = \frac{1601}{400}$ because $4.0025 = 4 + \frac{25}{10000} = 4 + \frac{1}{400} = \frac{1600}{400} + \frac{1}{400} = \frac{1601}{400}$. Also, $2.5 = \frac{5}{2}$. So: $\omega = \frac{\frac{5}{2}}{\frac{1601}{400}} = \frac{5}{2} \times \frac{400}{1601} = \frac{5 \times 400}{2 \times 1601} = \frac{2000}{3202} = \frac{1000}{1601} \approx 0.6246 \text{rad/s}$.

This value is approximately 0.625 rad/s. Comparing with the options, 0.625 rad/s corresponds to option B. Additionally, the moment of inertia of the bullet is very small compared to that of the door (0.0025 kg·m² vs 4 kg·m²), so neglecting it gives $I_{\text{total}} \approx 4 \text{kg} \cdot \text{m}^2$ and $\omega = \frac{2.5}{4} = 0.625 \text{rad/s}$, which matches exactly.

Hence, the correct answer is Option B.

Hypotenuse ($H$): The total distance from the wall attachment point to the center of the sphere is the length of the string plus the radius. $H = 8\text{ cm} + 5\text{ cm} = 13\text{ cm}$
Base ($B$): The horizontal distance from the wall to the center of the sphere is simply the radius. $B = 5\text{ cm}$
Vertical Height ($V$): Using the Pythagorean theorem: $V = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}$

$\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{V}{H} = \frac{12}{13}$

For vertical equilibrium, the vertical component of the tension must balance the weight: $T \cos\theta = W$

$T \left( \frac{12}{13} \right) = W$

$T = \frac{13W}{12}$

Mass of removed part, $m = \frac{\text{Volume of removed part}}{\text{Volume of whole sphere}} \times M = \frac{\frac{4}{3}\pi (R/2)^3}{\frac{4}{3}\pi R^3} \times M = \frac{M}{8}$

The gravitational field $E$ at a distance $x$ from a point mass (or a uniform sphere at a far point) is $E = \frac{GM}{x^2}$.

Since $P$ is at a distance $x$ from the center, $E_{total} = \frac{GM}{x^2}$.

$E_{removed} \approx \frac{G(M/8)}{x^2}$

$E_{net} = E_{total} - E_{removed}$

$E_{net} \approx \frac{GM}{x^2} - \frac{GM}{8x^2}$

$E_{net} \approx \left(1 - \frac{1}{8}\right) \frac{GM}{x^2} = \mathbf{\frac{7}{8} \frac{GM}{x^2}}$

Tension in Brass wire ($F_b$): It supports only the bottom mass $2M$, hence, $F_b = 2Mg$
Tension in Steel wire ($F_s$): It supports both masses $M$ and $2M$, hence, $F_s = Mg + 2Mg = 3Mg$

$\frac{F_s}{F_b} = \frac{3Mg}{2Mg} = \frac{3}{2}$

$\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$

$\frac{\Delta L_s}{\Delta L_b} = \left( \frac{F_s}{F_b} \right) \times \left( \frac{L_s}{L_b} \right) \times \left( \frac{r_b}{r_s} \right)^2 \times \left( \frac{Y_b}{Y_s} \right)$

$\frac{\Delta L_s}{\Delta L_b} = \left( \frac{3}{2} \right) \times (a) \times \left( \frac{1}{b} \right)^2 \times \left( \frac{1}{c} \right)$

$\frac{\Delta L_s}{\Delta L_b} = \frac{3a}{2b^2c}$

Let us understand both statements step by step. We start by recalling the formula for the height $h$ to which a liquid rises in a capillary tube. The formula is:

$h = \frac{2S \cos \theta}{\rho g r}$

where $S$ is the surface tension of the liquid, $\theta$ is the angle of contact, $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $r$ is the radius of the capillary tube.

Statement-1 claims that when the temperature of the liquid is raised, the height $h$ increases, assuming the density $\rho$ and the angle of contact $\theta$ remain constant.

Statement-2 states that the surface tension $S$ of a liquid decreases with a rise in temperature.

We know that Statement-2 is a well-established fact. The surface tension of liquids generally decreases as temperature increases because the increased thermal energy reduces the cohesive forces between molecules at the surface.

Now, let us analyze Statement-1 using the formula. Given that $\rho$, $\theta$, $g$, and $r$ are constant (as per the problem and since the capillary tube doesn't change), the height $h$ is directly proportional to the surface tension $S$. Therefore:

$h \propto S$

If the temperature rises, Statement-2 tells us that $S$ decreases. Since $h$ is proportional to $S$, a decrease in $S$ would lead to a decrease in $h$. However, Statement-1 claims that $h$ increases with temperature. This is a contradiction.

Therefore, Statement-1 is false because the height $h$ should decrease, not increase, when temperature rises (given that $\rho$ and $\theta$ are constant).

In summary:

• Statement-1 is false.
• Statement-2 is true.

Hence, the correct answer is Option B.

We are given two temperature scales: Kelvin (K) and a linear scale Y. The freezing and boiling points of water are known on both scales:

• On Kelvin scale: freezing point = 273 K, boiling point = 373 K
• On Y scale: freezing point = -160°Y, boiling point = -50°Y

We need to find the temperature on the Y scale that corresponds to 340 K. Since both scales are linear, the relationship between them can be expressed as:

$T_Y = m \cdot T_K + c$

where $T_Y$ is the temperature in °Y, $T_K$ is the temperature in K, $m$ is the slope, and $c$ is the intercept.

To find $m$ and $c$, we use the two known points:

• Point 1: (273 K, -160°Y)
• Point 2: (373 K, -50°Y)

The slope $m$ is calculated as the change in Y divided by the change in K:

$m = \frac{\Delta T_Y}{\Delta T_K} = \frac{(-50) - (-160)}{373 - 273} = \frac{-50 + 160}{100} = \frac{110}{100} = 1.1$

So, $m = 1.1$.

Now, substitute one point into the equation to find $c$. Using point 1 (273 K, -160°Y):

$-160 = 1.1 \cdot 273 + c$

First, compute $1.1 \cdot 273$:

$1.1 \cdot 273 = 1.1 \cdot (200 + 73) = 1.1 \cdot 200 + 1.1 \cdot 73 = 220 + 80.3 = 300.3$

So,

$-160 = 300.3 + c$

Solving for $c$:

$c = -160 - 300.3 = -460.3$

Thus, the equation is:

$T_Y = 1.1 \cdot T_K - 460.3$

Now, find $T_Y$ when $T_K = 340$ K:

$T_Y = 1.1 \cdot 340 - 460.3$

Compute $1.1 \cdot 340$:

$1.1 \cdot 340 = 1.1 \cdot (300 + 40) = 1.1 \cdot 300 + 1.1 \cdot 40 = 330 + 44 = 374$

Then,

$T_Y = 374 - 460.3 = -86.3$

Therefore, 340 K corresponds to -86.3°Y on the Y scale.

Comparing with the options:

• A: -73.7°Y
• B: -233.7°Y
• C: -86.3°Y
• D: -106.3°Y

Hence, the correct answer is Option C.

The problem involves identifying which lines a, b, and c correspond to monoatomic (M), diatomic (D), and polyatomic (P) gases on a graph of temperature change ($\Delta T$) versus heat supplied (Q) for an isobaric process. The initial state is the same for all gases, and vibrational degrees of freedom are ignored.

In an isobaric process, the heat supplied Q is related to the change in temperature $\Delta T$ by the equation:

$Q = n C_P \Delta T$

where n is the number of moles and $C_P$ is the molar specific heat at constant pressure. Rearranging for $\Delta T$:

$\Delta T = \frac{Q}{n C_P}$

Since n is the same for all gases (same initial state), $\Delta T$ is inversely proportional to $C_P$. Therefore, for a given Q, a higher $C_P$ results in a smaller $\Delta T$. On the graph with $\Delta T$ on the y-axis and Q on the x-axis, the slope of each line ($\Delta T$ per unit Q) is inversely proportional to $C_P$. Thus, a gas with a larger $C_P$ will have a flatter slope, and a gas with a smaller $C_P$ will have a steeper slope.

Now, we determine $C_P$ for each gas type, ignoring vibrational degrees of freedom:

• For a monoatomic gas (M), the degrees of freedom f = 3 (translational only). The molar specific heat at constant volume is $C_V = \frac{f}{2} R = \frac{3}{2} R$. Then, $C_P = C_V + R = \frac{3}{2} R + R = \frac{5}{2} R$.
• For a diatomic gas (D), f = 5 (3 translational + 2 rotational). So, $C_V = \frac{5}{2} R$, and $C_P = \frac{5}{2} R + R = \frac{7}{2} R$.
• For a polyatomic gas (P), we consider f = 6 (3 translational + 3 rotational, assuming non-linear structure, which is standard when vibrational modes are ignored). Thus, $C_V = \frac{6}{2} R = 3R$, and $C_P = 3R + R = 4R$.

Comparing the $C_P$ values:

$C_P(\text{M}) = \frac{5}{2}R = 2.5R$

$C_P(\text{D}) = \frac{7}{2}R = 3.5R$

$C_P(\text{P}) = 4R = 4.0R$

So, $C_P(\text{M}) < C_P(\text{D}) < C_P(\text{P})$.

Since $\Delta T$ is inversely proportional to $C_P$, the slope of the line ($\Delta T$ vs Q) is largest for the gas with the smallest $C_P$ and smallest for the gas with the largest $C_P$:

• Monoatomic gas (M) has the smallest $C_P$ (2.5R), so it has the steepest slope.
• Diatomic gas (D) has a medium $C_P$ (3.5R), so it has a medium slope.
• Polyatomic gas (P) has the largest $C_P$ (4.0R), so it has the flattest slope.

Therefore, the lines on the graph, ordered from steepest to flattest slope, correspond to M, D, P. The problem asks for lines a, b, and c respectively. Given the correct answer is Option B, which is M, D, P, this means:

• Line a corresponds to monoatomic gas (M).
• Line b corresponds to diatomic gas (D).
• Line c corresponds to polyatomic gas (P).

Hence, the correct answer is Option B.

The temperature of a real gas does not always decrease during free expansion. Depending on the gas and conditions, it may increase, decrease, or remain unchanged.

In an ideal gas, intermolecular forces are neglected, so the internal energy does not contain intermolecular potential energy. Hence, the internal energy of an ideal gas depends only on molecular kinetic energy (translational, rotational, and vibrational depending on the molecule and temperature). For a real gas, intermolecular forces exist, so its internal energy contains both kinetic and potential energy components.

The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$, where $L$ is the length and $g$ is the acceleration due to gravity. For the shorter pendulum of length 1 m, the time period is $T_1 = 2\pi \sqrt{\frac{1}{g}}$. For the longer pendulum of length 4 m, the time period is $T_2 = 2\pi \sqrt{\frac{4}{g}} = 2\pi \cdot \sqrt{4} \cdot \sqrt{\frac{1}{g}} = 2\pi \cdot 2 \cdot \sqrt{\frac{1}{g}} = 4\pi \sqrt{\frac{1}{g}}$. Comparing $T_1$ and $T_2$, we see $T_2 = 2 \cdot (2\pi \sqrt{\frac{1}{g}}) = 2 T_1$.

The angular frequency $\omega$ is related to the time period by $\omega = \frac{2\pi}{T}$. So for the shorter pendulum, $\omega_1 = \frac{2\pi}{T_1}$. For the longer pendulum, $\omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{2 T_1} = \frac{\pi}{T_1}$. The difference in angular frequencies is $\omega_1 - \omega_2 = \frac{2\pi}{T_1} - \frac{\pi}{T_1} = \frac{\pi}{T_1}$.

The phase difference between the two pendulums at time $t$ is $\Delta \theta = (\omega_1 - \omega_2) t = \left( \frac{\pi}{T_1} \right) t$. They are in phase when this phase difference is an integer multiple of $2\pi$, so $\Delta \theta = 2\pi n$ for some integer $n$. Substituting, we get $\frac{\pi}{T_1} t = 2\pi n$. Solving for $t$, divide both sides by $\pi$: $\frac{t}{T_1} = 2n$, so $t = 2n T_1$.

The time $t$ is also the time taken by the shorter pendulum to complete $N$ oscillations, which is $t = N T_1$. Setting this equal to the expression above, $N T_1 = 2n T_1$. Dividing both sides by $T_1$ (assuming $T_1 \neq 0$), we get $N = 2n$. The smallest positive integer $N$ occurs when $n = 1$, giving $N = 2$. This means after 2 oscillations of the shorter pendulum, they are again in phase for the first time after the initial displacement.

Verifying with the time: at $t = 2 T_1$, the shorter pendulum completes exactly 2 oscillations. The longer pendulum has a time period of $2 T_1$, so in time $2 T_1$, it completes exactly one oscillation ($\frac{t}{T_2} = \frac{2 T_1}{2 T_1} = 1$). Both pendulums return to their starting positions with the same velocity direction, confirming they are in phase.

Hence, the correct answer is Option A.

The displacements of the particle due to the two sound waves are given by:

$y_1 = 0.05\cos(0.50\pi x - 100\pi t)$

$y_2 = 0.05\cos(0.46\pi x - 92\pi t)$

where $y_1$, $y_2$, and $x$ are in meters, and $t$ is in seconds. We need to find the speed of sound in the medium.

A general wave equation traveling in the positive x-direction is written as $y = A \cos(kx - \omega t)$, where $k$ is the wave number and $\omega$ is the angular frequency. The wave number $k$ is related to the wavelength $\lambda$ by $k = \frac{2\pi}{\lambda}$, and the angular frequency $\omega$ is related to the frequency $f$ by $\omega = 2\pi f$. The speed of sound $v$ is given by the formula $v = f \lambda$. Substituting the relations, we get:

$v = f \lambda = \left( \frac{\omega}{2\pi} \right) \times \left( \frac{2\pi}{k} \right) = \frac{\omega}{k}$

Therefore, the speed can be directly calculated as $v = \frac{\omega}{k}$ for each wave. Since both waves are traveling in the same medium, the speed should be identical for both.

For the first wave $y_1$, the argument is $0.50\pi x - 100\pi t$. Comparing with the general form, we have:

Wave number $k_1 = 0.50\pi$ rad/m,

Angular frequency $\omega_1 = 100\pi$ rad/s.

Thus, the speed $v_1 = \frac{\omega_1}{k_1} = \frac{100\pi}{0.50\pi}$.

Simplifying, the $\pi$ cancels out:

$v_1 = \frac{100}{0.50} = \frac{100}{\frac{1}{2}} = 100 \times 2 = 200 \text{ m/s}$

For the second wave $y_2$, the argument is $0.46\pi x - 92\pi t$. Comparing with the general form, we have:

Wave number $k_2 = 0.46\pi$ rad/m,

Angular frequency $\omega_2 = 92\pi$ rad/s.

Thus, the speed $v_2 = \frac{\omega_2}{k_2} = \frac{92\pi}{0.46\pi}$.

Simplifying, the $\pi$ cancels out:

$v_2 = \frac{92}{0.46} = \frac{92}{\frac{46}{100}} = 92 \times \frac{100}{46} = \frac{92 \times 100}{46}$

Dividing 92 by 46:

$\frac{92}{46} = 2, \quad \text{so} \quad 2 \times 100 = 200 \text{ m/s}$

Alternatively, $92 \div 46 = 2$, so $92 = 46 \times 2$, and thus:

$\frac{92 \times 100}{46} = \frac{46 \times 2 \times 100}{46} = 2 \times 100 = 200 \text{ m/s}$

Both waves give the same speed of 200 m/s, which is consistent since they are in the same medium. Therefore, the speed of sound in the medium is 200 m/s.

Now, comparing with the options:

A. 92 m/s

B. 200 m/s

C. 100 m/s

D. 332 m/s

Hence, the correct answer is Option B.

An engine approaches a hill with a constant speed. The whistle is blown when the engine is 0.9 km away from the hill. Convert this distance to meters: 0.9 km = 900 meters. The echo is heard by the driver after 5 seconds, and the speed of sound is 330 m/s. We need to find the speed of the engine, denoted as $v$ m/s.

When the whistle is blown, the sound travels towards the hill at 330 m/s. The hill is stationary. The echo is the sound reflecting off the hill and returning to the engine. During the 5 seconds, the engine is moving towards the hill, so the distance covered by the sound includes the initial distance to the hill and the reduced distance on the return due to the engine's movement.

Consider the total distance traveled by the sound in 5 seconds. Since the speed of sound is 330 m/s, the total distance covered by sound is:

$\text{Distance} = \text{Speed} \times \text{Time} = 330 \times 5 = 1650 \text{ meters}.$

This distance is the sum of the distance from the engine's initial position to the hill and the distance from the hill back to the engine's position when the echo is heard.

Let the initial position of the engine be point A, and the hill be point B. So, AB = 900 meters. In 5 seconds, the engine moves towards the hill with speed $v$ m/s, so the distance covered by the engine is $v \times 5$ meters. Let the position of the engine when the echo is heard be point D. Therefore, AD = $5v$ meters.

The distance from D to the hill (point B) is DB. Since D is between A and B, DB = AB - AD = 900 - 5v meters.

The sound travels from A to B (900 meters) and then from B to D (DB = 900 - 5v meters). So, the total distance traveled by sound is:

$\text{Total distance} = \text{AB} + \text{DB} = 900 + (900 - 5v) = 1800 - 5v \text{ meters}.$

But we know the total distance is 1650 meters. Set them equal:

$1800 - 5v = 1650$

Solve for $v$:

Subtract 1800 from both sides:

$-5v = 1650 - 1800$

$-5v = -150$

Divide both sides by -5:

$v = \frac{-150}{-5} = 30 \text{ m/s}.$

Hence, the speed of the engine is 30 m/s.

So, the answer is Option D.

The interaction potential energy ($U$) between two point dipoles $\vec{p}_1$ and $\vec{p}_2$ separated by a distance $r$ is given by:

$U = \frac{1}{4\pi\epsilon_0 r^3} \left[ \vec{p}_1 \cdot \vec{p}_2 - 3(\vec{p}_1 \cdot \hat{r})(\vec{p}_2 \cdot \hat{r}) \right]$

In the axial configuration where both dipoles are collinear and point along the line joining them (i.e., $\vec{p}_1 \parallel \vec{p}_2 \parallel \vec{r}$):

$\vec{p}_1 \cdot \vec{p}_2 = p_1 p_2$, $\vec{p}_1 \cdot \hat{r} = p_1$, $\vec{p}_2 \cdot \hat{r} = p_2$

$U = \frac{1}{4\pi\epsilon_0 x^3} \left[ p_1 p_2 - 3(p_1)(p_2) \right]$

$U = -\frac{2 p_1 p_2}{4\pi\epsilon_0 x^3}$

$F = -\frac{dU}{dx}$

$F = -\frac{6 p_1 p_2}{4\pi\epsilon_0 x^4}$ (the $-ve$ sign indicates attraction)

Consider two identical balls, each with mass $m$ and charge $q$, suspended from a fixed support by strings of length $l$. At electrostatic equilibrium, the balls repel each other, causing the strings to make small angles with the vertical. We need to find how the separation $x$ between the balls depends on $l$.

For each ball, three forces act: the gravitational force $mg$ downward, the tension $T$ along the string, and the electrostatic repulsive force $F_e$ horizontally away from the other ball. The electrostatic force is given by Coulomb's law:

$F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2}$

where $x$ is the separation between the balls and $\epsilon_0$ is the permittivity of free space.

Since the setup is symmetric and the angles are small, let $\theta$ be the small angle each string makes with the vertical. The separation $x$ can be expressed in terms of $\theta$. Each ball is displaced horizontally by $l \sin\theta$ from the vertical, so the total separation is:

$x = 2 l \sin\theta$

Using the small-angle approximation, $\sin\theta \approx \theta$ (where $\theta$ is in radians), we get:

$x \approx 2 l \theta$

Resolving the forces for one ball at equilibrium:

• Vertically: The vertical component of tension balances the weight.

$T \cos\theta = mg$

Since $\theta$ is small, $\cos\theta \approx 1$, so:

$T \cdot 1 = mg \quad \Rightarrow \quad T = mg$

• Horizontally: The horizontal component of tension balances the electrostatic force.

$T \sin\theta = F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2}$

Using $\sin\theta \approx \theta$ and substituting $T = mg$:

$mg \theta = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2}$

From the separation equation $x \approx 2 l \theta$, solve for $\theta$:

$\theta \approx \frac{x}{2l}$

Substitute this expression for $\theta$ into the horizontal force equation:

$mg \left( \frac{x}{2l} \right) = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2}$

Multiply both sides by $2l$ to simplify:

$mg x = 2l \cdot \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2}$

Simplify the right side:

$mg x = \frac{2l}{4\pi\epsilon_0} \frac{q^2}{x^2} = \frac{l}{2\pi\epsilon_0} \frac{q^2}{x^2}$

Multiply both sides by $x^2$:

$mg x \cdot x^2 = \frac{l}{2\pi\epsilon_0} q^2$

$mg x^3 = \frac{l}{2\pi\epsilon_0} q^2$

Solve for $x^3$:

$x^3 = \frac{l}{2\pi\epsilon_0} \frac{q^2}{mg}$

The terms $\frac{q^2}{2\pi\epsilon_0 mg}$ are constant since $q$, $m$, $g$, and $\epsilon_0$ are constants. Let $k = \frac{q^2}{2\pi\epsilon_0 mg}$, so:

$x^3 = k l$

Taking the cube root of both sides:

$x = (k l)^{1/3} = k^{1/3} l^{1/3}$

Thus, $x$ is proportional to $l^{1/3}$.

Comparing with the options:

A. $l$

B. $l^2$

C. $l^{2/3}$

D. $l^{1/3}$

Hence, the correct answer is Option D.

In a meter bridge experiment, we have a wire of length 100 cm. The null point occurs when the ratio of the resistances equals the ratio of the lengths of the wire segments. Initially, resistance X is balanced against resistance Y, and the null point is at 40 cm from the end where X is connected. Since X is less than Y, we use the formula:

$\frac{X}{Y} = \frac{l_1}{100 - l_1}$

Here, $l_1 = 40$ cm. Substituting the values:

$\frac{X}{Y} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3}$

So, we have $\frac{X}{Y} = \frac{2}{3}$, which means $X = \frac{2}{3} Y$.

Now, we need to balance a resistance of 3X against Y. Let the new null point be at $l$ cm from the same end where X was connected. The balance condition becomes:

$\frac{3X}{Y} = \frac{l}{100 - l}$

Substitute $\frac{X}{Y} = \frac{2}{3}$ into the equation:

$\frac{3X}{Y} = 3 \times \frac{X}{Y} = 3 \times \frac{2}{3} = 2$

So, the equation simplifies to:

$2 = \frac{l}{100 - l}$

To solve for $l$, multiply both sides by $100 - l$:

$2 \times (100 - l) = l$

Which gives:

$200 - 2l = l$

Now, bring all terms involving $l$ to one side:

$200 = l + 2l$

$200 = 3l$

Dividing both sides by 3:

$l = \frac{200}{3} \approx 66.67 \text{ cm}$

This value is approximately 67 cm. Comparing with the options, 67 cm corresponds to option C.

Hence, the correct answer is Option C.

The letter 'A' is made of uniform wire with a resistance of 1.0 Ω per cm. The sides (legs) are each 20 cm long, and the cross piece is 10 cm long. The apex angle is 60°. We need to find the resistance between the ends of the legs, labeled as points A and B.

First, visualize the structure. The apex angle at the top is 60°, and the legs CA and CB are both 20 cm. The cross piece is horizontal and connects the midpoints of the legs. Let D be the midpoint of CA and E be the midpoint of CB. Since the legs are 20 cm, CD = DA = CE = EB = 10 cm. The cross piece DE is 10 cm long. Given the apex angle of 60° and CD = CE = 10 cm, triangle CDE is equilateral, so DE = 10 cm, which matches.

The circuit has the following segments with resistances (since resistance per cm is 1 Ω):

• AD (from A to D): 10 cm → 10 Ω
• DC (from D to C): 10 cm → 10 Ω
• CE (from C to E): 10 cm → 10 Ω
• EB (from E to B): 10 cm → 10 Ω
• DE (from D to E): 10 cm → 10 Ω

The circuit between A and B can be represented as:

• A is connected to D by 10 Ω.
• D is connected to E by 10 Ω (DE) and to C by 10 Ω (DC).
• E is connected to B by 10 Ω and to C by 10 Ω (CE).

To find the equivalent resistance between A and B, note that between D and E, there are two parallel paths:

1. Direct path DE: 10 Ω
2. Path via C: DC + CE = 10 Ω + 10 Ω = 20 Ω

The equivalent resistance between D and E is the parallel combination:

$\frac{1}{R_{DE}} = \frac{1}{10} + \frac{1}{20} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20}$

$R_{DE} = \frac{20}{3} \Omega \approx 6.6667 \Omega$

Now, the total path from A to B is: A to D (10 Ω), then D to E (20/3 Ω), then E to B (10 Ω). These are in series, so:

$R_{AB} = R_{AD} + R_{DE} + R_{EB} = 10 + \frac{20}{3} + 10 = 20 + \frac{20}{3} = \frac{60}{3} + \frac{20}{3} = \frac{80}{3} \Omega \approx 26.6667 \Omega$

Comparing with the options:

• A. 50.0 Ω
• B. 10 Ω
• C. 36.7 Ω
• D. 26.7 Ω

The value 26.6667 Ω is closest to 26.7 Ω.

Hence, the correct answer is Option D.