JEE Main 2013 23rd April Morning Shift Question Paper with Solutions

The given formula for the time period of oscillation is $t = \sqrt{r^{2b} s^c d^{a/2}}$. We can simplify this expression by distributing the square root. Since $\sqrt{x^y} = x^{y/2}$, we rewrite the formula as:

$t = \left( r^{2b} s^c d^{a/2} \right)^{1/2} = r^{2b \cdot \frac{1}{2}} \cdot s^{c \cdot \frac{1}{2}} \cdot d^{\frac{a}{2} \cdot \frac{1}{2}} = r^b \cdot s^{c/2} \cdot d^{a/4}$

The problem states that the time period $t$ is directly proportional to $\sqrt{\frac{d}{s}}$. Expressing this proportionality mathematically:

$t \propto \sqrt{\frac{d}{s}} = \left( \frac{d}{s} \right)^{1/2} = d^{1/2} \cdot s^{-1/2}$

This proportionality implies that when the radius $r$ is held constant (as the observation does not involve varying $r$), the exponents of $d$ and $s$ in the expression for $t$ must match those in the proportionality expression. Therefore, comparing the exponents in $t = r^b \cdot s^{c/2} \cdot d^{a/4}$ and $d^{1/2} \cdot s^{-1/2}$, we set up the following equations:

For the exponent of $d$:

$\frac{a}{4} = \frac{1}{2}$

Solving for $a$:

$a = 4 \times \frac{1}{2} = 2$

For the exponent of $s$:

$\frac{c}{2} = -\frac{1}{2}$

Solving for $c$:

$c = 2 \times \left(-\frac{1}{2}\right) = -1$

Substituting $a = 2$ and $c = -1$ back into the expression for $t$:

$t = r^b \cdot s^{-1/2} \cdot d^{2/4} = r^b \cdot s^{-1/2} \cdot d^{1/2}$

Now, we must find $b$. Since $t$ has the dimension of time ($T$), we use dimensional analysis. The dimensions of the variables are:

• $[t] = T$
• $[r] = L$ (length)
• $[s] =$ surface tension = force per unit length. Force has dimensions $M L T^{-2}$, so $[s] = \frac{M L T^{-2}}{L} = M T^{-2}$
• $[d] =$ density = mass per unit volume = $M L^{-3}$

The dimensions of the right-hand side are:

$[r^b] = L^b$

$[s^{-1/2}] = \left( M T^{-2} \right)^{-1/2} = M^{-1/2} T^{1}$

$[d^{1/2}] = \left( M L^{-3} \right)^{1/2} = M^{1/2} L^{-3/2}$

Multiplying these together:

$[r^b] \cdot [s^{-1/2}] \cdot [d^{1/2}] = L^b \cdot M^{-1/2} T^{1} \cdot M^{1/2} L^{-3/2} = L^{b - \frac{3}{2}} M^{-1/2 + 1/2} T^{1} = L^{b - \frac{3}{2}} M^{0} T^{1}$

This must equal the dimension of $t$, which is $T$. Therefore, the exponent of $L$ must be zero (since there is no length dimension in time), and the exponent of $T$ is already 1, which matches. Setting the exponent of $L$ to zero:

$b - \frac{3}{2} = 0$

Solving for $b$:

$b = \frac{3}{2}$

Thus, the value of $b$ is $\frac{3}{2}$. Comparing with the options:

A. $\frac{3}{4}$

B. $\sqrt{3}$

C. $\frac{3}{2}$

D. $\frac{2}{3}$

Hence, the correct answer is Option C.

The man crouches and pushes the ground with a constant force $F$, raising his centre of gravity by $0.5 \text{ m}$ before his feet leave the ground. After take-off the centre of gravity rises an additional $1 \text{ m}$.

After take-off no external force acts other than gravity, so the kinetic energy at take-off equals the gain in gravitational potential energy during the aerial phase: $\frac{1}{2}mv^{2} = mg \times 1$. With $m = 70 \text{ kg}$ and $g = 10 \text{ m/s}^{2}$, this gives $\frac{1}{2}(70)v^{2} = 700$, hence $v^{2} = 20 \text{ m}^{2}\text{/s}^{2}$ and $v = 2\sqrt{5} \text{ m/s}$.

During the push phase the net upward force $F_{\text{net}} = F - mg$ does work over $0.5 \text{ m}$, producing the take-off kinetic energy: $F_{\text{net}} \times 0.5 = 700$, so $F_{\text{net}} = 1400 \text{ N}$. Because the man starts from rest and accelerates uniformly, the velocity increases from zero to $v = 2\sqrt{5} \text{ m/s}$ over this distance, meaning the maximum velocity occurs at the instant of take-off.

The maximum power delivered by the muscles equals the net force times the velocity at take-off: $P_{\max} = F_{\text{net}} \times v = 1400 \times 2\sqrt{5} = 2800\sqrt{5} \approx 6.26 \times 10^{3} \text{ Watts}$.

This maximum power is delivered at the moment of take-off, corresponding to option (B).

We have a boy of mass 20 kg standing on a cart of mass 80 kg. The cart is free to move and there is negligible friction between the cart and the ground. Initially, the boy is 25 m from a wall. The boy walks 10 m towards the wall on the cart. We need to find the final distance of the boy from the wall.

Since there is no external force acting on the system (friction is negligible), the center of mass of the system (boy + cart) remains stationary. This is a key principle we will use.

Set up a coordinate system with the wall at position $x = 0$. Initially, the boy is at $x_{\text{boy, initial}} = 25$ m. Let the initial position of the center of mass of the cart be $X$. The center of mass of the entire system ($x_{\text{cm}}$) is given by:

$x_{\text{cm}} = \frac{m_{\text{boy}} \cdot x_{\text{boy, initial}} + m_{\text{cart}} \cdot X}{m_{\text{boy}} + m_{\text{cart}}} = \frac{20 \cdot 25 + 80 \cdot X}{100} = \frac{500 + 80X}{100} = 5 + 0.8X$

Now, the boy walks 10 m towards the wall relative to the cart. Since the wall is to the left (negative x-direction), the boy moves $-10$ m relative to the cart. However, as the boy moves, the cart may also move. Let the displacement of the cart's center of mass be $D$ (positive if to the right). The displacement of the boy relative to the ground is then the displacement relative to the cart plus the displacement of the cart:

Displacement of boy relative to ground $= -10 + D$

Thus, the final position of the boy is:

$x_{\text{boy, final}} = x_{\text{boy, initial}} + (-10 + D) = 25 - 10 + D = 15 + D$

The final position of the cart's center of mass is:

$x_{\text{cart, final}} = X + D$

The center of mass remains unchanged, so:

$x_{\text{cm}} = \frac{m_{\text{boy}} \cdot x_{\text{boy, final}} + m_{\text{cart}} \cdot x_{\text{cart, final}}}{m_{\text{boy}} + m_{\text{cart}}} = \frac{20 \cdot (15 + D) + 80 \cdot (X + D)}{100}$

Set this equal to the initial center of mass:

$\frac{20(15 + D) + 80(X + D)}{100} = 5 + 0.8X$

Multiply both sides by 100 to clear the denominator:

$20(15 + D) + 80(X + D) = 100(5 + 0.8X)$

Expand both sides:

Left side: $20 \cdot 15 + 20 \cdot D + 80 \cdot X + 80 \cdot D = 300 + 20D + 80X + 80D = 300 + 100D + 80X$

Right side: $100 \cdot 5 + 100 \cdot 0.8X = 500 + 80X$

So:

$300 + 100D + 80X = 500 + 80X$

Subtract $80X$ from both sides:

$300 + 100D = 500$

Subtract 300 from both sides:

$100D = 200$

Divide by 100:

$D = 2$

So the cart moves 2 m to the right. The final position of the boy is:

$x_{\text{boy, final}} = 15 + D = 15 + 2 = 17 \text{ m}$

Therefore, the final distance of the boy from the wall is 17 m.

Hence, the correct answer is Option D.

A projectile of mass M has a horizontal range of 4 km. The range formula for a projectile is given by $R = \frac{u^2 \sin 2\theta}{g}$, where u is the initial velocity and θ is the angle of projection. So, we have:

$R = \frac{u^2 \sin 2\theta}{g} = 4 \text{km}$

At the highest point of its trajectory, the projectile explodes into two parts: one of mass $\frac{M}{4}$ and the other of mass $\frac{3M}{4}$. The heavier part (mass $\frac{3M}{4}$) falls vertically downward with zero initial speed, meaning its velocity immediately after the explosion is zero in both horizontal and vertical directions.

Since the explosion is an internal force, it does not affect the center of mass motion. Before the explosion, at the highest point, the entire projectile has only horizontal velocity $u_x = u \cos \theta$ and zero vertical velocity. Therefore, the center of mass continues with velocity $(u_x, 0)$ after the explosion.

Let $\vec{v_1} = (v_{1x}, v_{1y})$ be the velocity of the lighter part (mass $\frac{M}{4}$) and $\vec{v_2} = (0, 0)$ be the velocity of the heavier part (mass $\frac{3M}{4}$) after the explosion. The center of mass velocity is:

$\vec{v_{cm}} = \frac{ \left( \frac{M}{4} \right) \vec{v_1} + \left( \frac{3M}{4} \right) \vec{v_2} }{ M } = \frac{1}{4} \vec{v_1} + \frac{3}{4} \vec{v_2}$

Setting $\vec{v_{cm}} = (u_x, 0)$ and substituting $\vec{v_2} = (0, 0)$:

$(u_x, 0) = \frac{1}{4} (v_{1x}, v_{1y}) + \frac{3}{4} (0, 0)$

This simplifies to:

$(u_x, 0) = \left( \frac{v_{1x}}{4}, \frac{v_{1y}}{4} \right)$

Equating components:

$\frac{v_{1x}}{4} = u_x \quad \Rightarrow \quad v_{1x} = 4u_x$

$\frac{v_{1y}}{4} = 0 \quad \Rightarrow \quad v_{1y} = 0$

So, the lighter part has horizontal velocity $4u_x$ and zero vertical velocity immediately after the explosion.

The explosion occurs at the highest point, which is at a horizontal distance of half the range from the firing point. Since the range is 4 km, this distance is:

$\frac{R}{2} = \frac{4}{2} = 2 \text{km}$

The lighter part is now projected horizontally from this point with initial velocity $4u_x$ and from a height H, where H is the maximum height of the original projectile. The maximum height H is given by:

$H = \frac{u^2 \sin^2 \theta}{2g}$

The time taken for the lighter part to fall to the ground from height H with zero initial vertical velocity is found using the equation of motion:

$H = \frac{1}{2} g t^2 \quad \Rightarrow \quad t = \sqrt{\frac{2H}{g}}$

In this time, the horizontal distance covered by the lighter part from the explosion point is:

$d = v_{1x} \cdot t = 4u_x \cdot \sqrt{\frac{2H}{g}}$

Substituting $H = \frac{u^2 \sin^2 \theta}{2g}$:

$d = 4u_x \cdot \sqrt{ \frac{2}{g} \cdot \frac{u^2 \sin^2 \theta}{2g} } = 4u_x \cdot \sqrt{ \frac{u^2 \sin^2 \theta}{g^2} } = 4u_x \cdot \frac{u \sin \theta}{g}$

Let $u_y = u \sin \theta$ (the initial vertical component of velocity for the original projectile). Then:

$d = 4u_x \cdot \frac{u_y}{g}$

From the original range equation:

$R = \frac{2 u_x u_y}{g} = 4 \text{km}$

So:

$\frac{u_x u_y}{g} = \frac{R}{2} = \frac{4}{2} = 2 \text{km}$

Substituting this into the expression for d:

$d = 4 \cdot 2 = 8 \text{km}$

This is the horizontal distance traveled by the lighter part from the explosion point. The total horizontal range from the firing point is the distance to the explosion point plus d:

$\text{Total range} = 2 \text{km} + 8 \text{km} = 10 \text{km}$

Hence, the horizontal range of the lighter part is 10 km.

So, the answer is Option C.

The angular momentum of a particle about a point is given by the cross product of its position vector and its linear momentum, that is, $\vec{L} = \vec{r} \times \vec{p}$. Here, $\vec{p} = m \vec{v}$, where $m$ is the mass and $\vec{v}$ is the velocity vector.

First, we are given the position vector as a function of time: $\vec{r}(t) = 5\hat{i} - 2t^2\hat{j}$. The mass $m = 2$ kg.

To find the velocity $\vec{v}(t)$, we differentiate the position vector with respect to time:

$\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(5)\hat{i} + \frac{d}{dt}(-2t^2)\hat{j}$

The derivative of the constant 5 is 0, and the derivative of $-2t^2$ is $-4t$. So,

$\vec{v}(t) = 0 \cdot \hat{i} - 4t \hat{j} = -4t \hat{j} \text{ m/s}$

Now, the linear momentum $\vec{p} = m \vec{v} = 2 \times (-4t \hat{j}) = -8t \hat{j} \text{ kg m/s}$.

Next, we compute the angular momentum $\vec{L} = \vec{r} \times \vec{p}$:

$\vec{L} = (5\hat{i} - 2t^2\hat{j}) \times (-8t \hat{j})$

Expanding the cross product:

$\vec{L} = 5\hat{i} \times (-8t \hat{j}) + (-2t^2\hat{j}) \times (-8t \hat{j})$

Let's evaluate each term separately. For the first term:

$5\hat{i} \times (-8t \hat{j}) = 5 \times (-8t) \times (\hat{i} \times \hat{j}) = -40t \times \hat{k} = -40t \hat{k}$

because $\hat{i} \times \hat{j} = \hat{k}$. For the second term:

$(-2t^2\hat{j}) \times (-8t \hat{j}) = (-2t^2) \times (-8t) \times (\hat{j} \times \hat{j}) = 16t^3 \times \vec{0} = 0$

since the cross product of any vector with itself is zero. Therefore,

$\vec{L} = -40t \hat{k} + 0 = -40t \hat{k} \text{ kg m}^2\text{s}^{-1}$

Now, we need the angular momentum at $t = 2$ s. Substituting $t = 2$:

$\vec{L} = -40 \times 2 \hat{k} = -80 \hat{k} \text{ kg m}^2\text{s}^{-1}$

Comparing with the given options, $-80\hat{k}$ corresponds to option A.

Hence, the correct answer is Option A.

A body of mass $m$ is tied to a spring and whirled in a horizontal plane with constant angular velocity. The spring elongates due to the centrifugal force acting on the body. Let the original length of the spring be $L$ cm. In the first scenario, with angular velocity $\omega$, the elongation is 1 cm. Therefore, the total length of the spring becomes $L + 1$ cm, which is the radius of the circular path.

The centrifugal force is given by $m \omega^2 r$, where $r$ is the radius. This force is balanced by the spring force, which is $k \times \text{elongation}$, where $k$ is the spring constant. For the first case:

$k \times 1 = m \omega^2 (L + 1)$

Simplifying, we get:

$k = m \omega^2 (L + 1) \quad \text{(Equation 1)}$

When the angular velocity is doubled to $2\omega$, the elongation becomes 5 cm. The total length is now $L + 5$ cm. The force balance equation becomes:

$k \times 5 = m (2\omega)^2 (L + 5)$

Simplifying the right side:

$5k = m \cdot 4\omega^2 (L + 5)$

$5k = 4 m \omega^2 (L + 5) \quad \text{(Equation 2)}$

Substitute the expression for $k$ from Equation 1 into Equation 2:

$5 \left[ m \omega^2 (L + 1) \right] = 4 m \omega^2 (L + 5)$

Divide both sides by $m \omega^2$ (assuming $m \neq 0$ and $\omega \neq 0$):

$5 (L + 1) = 4 (L + 5)$

Expand both sides:

$5L + 5 = 4L + 20$

Subtract $4L$ from both sides:

$5L - 4L + 5 = 20$

$L + 5 = 20$

Subtract 5 from both sides:

$L = 15$

Thus, the original length of the spring is 15 cm.

Hence, the correct answer is Option A.

We have a composite wire made of copper and steel joined end to end. The copper wire is 1.0 m long and the steel wire is 0.5 m long, both with the same cross-sectional area. The composite wire is stretched by a load that causes the copper wire to extend by 1 mm. We need to find the total extension of the composite wire.

Given:

• Length of copper wire, $L_c = 1.0 \text{m}$
• Length of steel wire, $L_s = 0.5 \text{m}$
• Extension of copper wire, $\Delta L_c = 1 \text{mm} = 0.001 \text{m}$
• Young's modulus of copper, $Y_c = 1.0 \times 10^{11} \text{Nm}^{-2}$
• Young's modulus of steel, $Y_s = 2.0 \times 10^{11} \text{Nm}^{-2}$

Since the wires are joined end to end, the same force $F$ acts on both wires. The cross-sectional area $A$ is the same for both.

Using Hooke's law for the copper wire:

$Y_c = \frac{F \cdot L_c}{A \cdot \Delta L_c}$

Solving for the force $F$:

$F = \frac{Y_c \cdot A \cdot \Delta L_c}{L_c}$

Substituting the values:

$F = \frac{(1.0 \times 10^{11}) \cdot A \cdot (0.001)}{1.0}$

$F = (1.0 \times 10^{11}) \cdot A \cdot 0.001$

$F = 1.0 \times 10^{11} \times 10^{-3} \cdot A$

$F = 1.0 \times 10^{8} A \text{ Newtons}$

Now, for the steel wire, the same force $F$ is applied. Using Hooke's law:

$Y_s = \frac{F \cdot L_s}{A \cdot \Delta L_s}$

Solving for the extension of steel $\Delta L_s$:

$\Delta L_s = \frac{F \cdot L_s}{A \cdot Y_s}$

Substituting the values:

$\Delta L_s = \frac{(1.0 \times 10^{8} A) \cdot (0.5)}{A \cdot (2.0 \times 10^{11})}$

The $A$ cancels out:

$\Delta L_s = \frac{1.0 \times 10^{8} \cdot 0.5}{2.0 \times 10^{11}}$

$\Delta L_s = \frac{5.0 \times 10^{7}}{2.0 \times 10^{11}}$

$\Delta L_s = \frac{5.0}{2.0} \times 10^{7-11}$

$\Delta L_s = 2.5 \times 10^{-4} \text{m}$

Converting to millimeters (since 1 m = 1000 mm):

$\Delta L_s = 2.5 \times 10^{-4} \times 1000 = 0.25 \text{mm}$

The total extension of the composite wire is the sum of the extensions of the copper and steel parts:

$\Delta L_{\text{total}} = \Delta L_c + \Delta L_s = 1 \text{mm} + 0.25 \text{mm} = 1.25 \text{mm}$

Hence, the correct answer is Option D.

First, recall the formula for the height $h$ to which a liquid rises in a capillary tube:

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

Here, $S$ is the surface tension of the liquid (water), $\theta$ is the angle of contact, $\rho$ is the density of water, $g$ is the acceleration due to gravity, and $r$ is the radius of the capillary tube. The radius $r$, density $\rho$, gravity $g$, and surface tension $S$ remain constant. Only $\theta$ changes when the tube is waxed.

For an unwaxed glass capillary tube, water wets the surface, and the angle of contact $\theta$ is acute (less than 90°). Typically, $\theta$ is close to 0° for pure water and clean glass. Thus, $\cos \theta$ is positive, and $h$ is positive, meaning water rises in the tube.

When the inner wall is coated with wax, the surface becomes hydrophobic (water-repelling). Wax increases the angle of contact $\theta$ because water does not wet the surface as well. For wax, $\theta$ becomes obtuse (greater than 90°), often around 106° for paraffin wax. Therefore, $\theta$ increases compared to the unwaxed tube.

Now, examine the effect on $h$. Since $\theta$ increases from less than 90° to more than 90°, $\cos \theta$ changes:

• When $\theta < 90^\circ$, $\cos \theta > 0$ (positive).
• When $\theta > 90^\circ$, $\cos \theta < 0$ (negative).

Substituting into the formula:

For unwaxed tube: $h_{\text{unwaxed}} = \frac{2S \cos \theta_{\text{unwaxed}}}{\rho g r} > 0$ (since $\cos \theta_{\text{unwaxed}} > 0$).

For waxed tube: $h_{\text{waxed}} = \frac{2S \cos \theta_{\text{waxed}}}{\rho g r} < 0$ (since $\cos \theta_{\text{waxed}} < 0$).

A negative $h$ means capillary depression occurs—water falls below the surrounding level instead of rising. The question asks for the height "up to which water rises." In the waxed tube, water does not rise; it depresses. Therefore, compared to the positive rise in the unwaxed tube, the height of rise $h$ decreases (it becomes negative, so it is less than the positive value).

Thus:

• The angle of contact $\theta$ increases (from acute to obtuse).
• The height $h$ to which water rises decreases (from positive to negative).

Comparing with the options:

A. $\theta$ increases and $h$ increases → Incorrect, because $h$ decreases.

B. $\theta$ decreases and $h$ decreases → Incorrect, because $\theta$ increases.

C. $\theta$ increases and $h$ decreases → Correct.

D. $\theta$ decreases and $h$ increases → Incorrect, because $\theta$ increases and $h$ decreases.

Hence, the correct answer is Option C.

When the tube is turned vertical, let the mercury be displaced by a distance $x$. The new lengths of the air columns will be $(40 - x)$ for the lower section and $(40 + x)$ for the upper section.

Applying Boyle's Law ($P_1V_1 = P_2V_2$) for both air columns:

Upper air column: $P_1(40 + x) = P_0(40) \implies P_1 = \frac{76 \times 40}{40 + x}$

Lower air column: $P_2(40 - x) = P_0(40) \implies P_2 = \frac{76 \times 40}{40 - x}$

$P_2 = P_1 + 20$

$\frac{76 \times 40}{40 - x} = \frac{76 \times 40}{40 + x} + 20$

$\frac{152}{40 - x} = \frac{152}{40 + x} + 1$

$152 \left[ \frac{1}{40 - x} - \frac{1}{40 + x} \right] = 1$

$152 \left[ \frac{(40 + x) - (40 - x)}{1600 - x^2} \right] = 1$

$152 \left( \frac{2x}{1600 - x^2} \right) = 1 \implies 304x = 1600 - x^2$

$x^2 + 304x - 1600 = 0$

$x \approx \frac{-304 + 314.35}{2} = \frac{10.35}{2} \approx 5.18\text{ cm}$

The ratio of the coefficient of volume expansion of the glass container to that of the viscous liquid is given as 1 : 4. Let the coefficient of volume expansion of the glass container be $\gamma_g$ and that of the liquid be $\gamma_l$. Therefore, $\gamma_g / \gamma_l = 1/4$, which implies $\gamma_l = 4\gamma_g$.

Let the initial inner volume of the container at temperature $T_0$ be $V_0$. We need to find the fraction of $V_0$ that the liquid should occupy at $T_0$ so that the volume of the vacant space remains constant at all temperatures. Let this fraction be $k$, so the initial volume of the liquid is $k V_0$ and the initial vacant space is $(1 - k) V_0$.

When the temperature changes by $\Delta T$, the new volume of the container is $V_0 (1 + \gamma_g \Delta T)$. The new volume of the liquid is $k V_0 (1 + \gamma_l \Delta T) = k V_0 (1 + 4\gamma_g \Delta T)$, since $\gamma_l = 4\gamma_g$. The vacant space at the new temperature is the difference between the container volume and the liquid volume:

$\text{Vacant space} = V_0 (1 + \gamma_g \Delta T) - k V_0 (1 + 4\gamma_g \Delta T)$

This vacant space must remain constant and equal to the initial vacant space $(1 - k) V_0$ at all temperatures. Therefore, we set up the equation:

$V_0 (1 + \gamma_g \Delta T) - k V_0 (1 + 4\gamma_g \Delta T) = (1 - k) V_0$

Dividing both sides by $V_0$ (assuming $V_0 \neq 0$):

$1 + \gamma_g \Delta T - k (1 + 4\gamma_g \Delta T) = 1 - k$

Expanding the left side:

$1 + \gamma_g \Delta T - k - 4k\gamma_g \Delta T = 1 - k$

Subtracting 1 from both sides:

$\gamma_g \Delta T - k - 4k\gamma_g \Delta T = -k$

Adding $k$ to both sides:

$\gamma_g \Delta T - 4k\gamma_g \Delta T = 0$

Factoring out $\gamma_g \Delta T$:

$\gamma_g \Delta T (1 - 4k) = 0$

Since $\gamma_g \neq 0$ and $\Delta T$ can be arbitrary (not always zero), the term $(1 - 4k)$ must be zero:

$1 - 4k = 0$

Solving for $k$:

$4k = 1$

$k = \frac{1}{4}$

Thus, the liquid should occupy $\frac{1}{4}$ of the inner volume of the container at the initial temperature. This fraction corresponds to the ratio 1 : 4.

Verifying the solution: with $k = \frac{1}{4}$, the initial liquid volume is $\frac{V_0}{4}$ and the initial vacant space is $\frac{3V_0}{4}$. At a new temperature, the container volume is $V_0 (1 + \gamma_g \Delta T)$, and the liquid volume is $\frac{V_0}{4} (1 + 4\gamma_g \Delta T)$. The vacant space is:

$V_0 (1 + \gamma_g \Delta T) - \frac{V_0}{4} (1 + 4\gamma_g \Delta T) = V_0 + V_0\gamma_g \Delta T - \frac{V_0}{4} - V_0\gamma_g \Delta T = V_0 - \frac{V_0}{4} = \frac{3V_0}{4}$

This is constant and equal to the initial vacant space, confirming the solution.

Hence, the correct answer is Option B.

We are given:

• Mass of water, $m_w = 500$ g at 0°C
• Mass of ice, $m_i = 100$ g at 0°C
• Water equivalent of calorimeter, $W = 40$ g (meaning the calorimeter behaves like 40 g of water for heat absorption)
• Mass of steam, $m_s = 10$ g at 100°C
• Latent heat of ice, $L_i = 80$ cal/g
• Latent heat of steam, $L_s = 540$ cal/g

We need to find the final mass of water in the calorimeter after adding steam. The specific heat capacity of water is $c = 1$ cal/g°C.

First, calculate the total heat released by the steam when it condenses and cools to 0°C. The steam undergoes two processes:

1. Condensation: Steam at 100°C condenses to water at 100°C, releasing latent heat. $Q_{\text{condense}} = m_s \times L_s = 10 \times 540 = 5400 \text{ cal}$
2. Cooling: The condensed water cools from 100°C to 0°C, releasing sensible heat. $Q_{\text{cool}} = m_s \times c \times \Delta T = 10 \times 1 \times (100 - 0) = 1000 \text{ cal}$

Total heat available from steam: $Q_{\text{available}} = Q_{\text{condense}} + Q_{\text{cool}} = 5400 + 1000 = 6400 \text{ cal}$

Next, calculate the heat required to melt all the ice. The ice is at 0°C, and melting it to water at 0°C requires latent heat: $Q_{\text{melt all ice}} = m_i \times L_i = 100 \times 80 = 8000 \text{ cal}$

Since $Q_{\text{available}} = 6400$ cal is less than $Q_{\text{melt all ice}} = 8000$ cal, the available heat is insufficient to melt all the ice. Therefore, only part of the ice melts, and the final temperature remains at 0°C.

The heat available (6400 cal) is used solely to melt ice because the calorimeter and the existing water are at 0°C, and the temperature does not change (so they absorb no heat). The mass of ice melted is: $\text{Mass of ice melted} = \frac{Q_{\text{available}}}{L_i} = \frac{6400}{80} = 80 \text{ g}$

Thus, 80 g of ice melts into water, and the remaining ice is: $\text{Remaining ice} = 100 - 80 = 20 \text{ g}$

Now, calculate the total water in the calorimeter:

• Original water: 500 g
• Water from melted ice: 80 g
• Water from condensed steam: 10 g (since the steam condenses and cools to 0°C)

Total water mass: $\text{Total water} = 500 + 80 + 10 = 590 \text{ g}$

The calorimeter's water equivalent (40 g) does not add to the water mass; it only accounts for heat capacity. The remaining 20 g of ice is not water, so it is not included in the liquid water mass.

Hence, the water in the calorimeter is 590 g.

The options are: A. 580 g B. 590 g C. 600 g D. 610 g

So, the correct answer is Option B.

Let us analyze Statement-1 and Statement-2 step by step.

First, consider Statement-1: "The internal energy of a perfect gas is entirely kinetic and depends only on absolute temperature of the gas and not on its pressure or volume." A perfect gas, also known as an ideal gas, follows the ideal gas law and has no intermolecular forces. Therefore, the internal energy of an ideal gas is solely due to the kinetic energy of its molecules and does not include any potential energy. According to the kinetic theory of gases, this internal energy is a function of temperature alone and is independent of pressure or volume. Hence, Statement-1 is true.

Now, consider Statement-2: "A perfect gas is heated keeping pressure constant and later at constant volume. For the same amount of heat the temperature of the gas at constant pressure is lower than that at constant volume." We need to compare the temperature change when the same amount of heat is supplied to the gas under constant pressure and constant volume conditions.

Let the heat supplied be $Q$. For a constant volume process, the heat supplied is given by $Q = n C_v \Delta T_v$, where $n$ is the number of moles, $C_v$ is the molar specific heat at constant volume, and $\Delta T_v$ is the temperature change at constant volume.

For a constant pressure process, the heat supplied is $Q = n C_p \Delta T_p$, where $C_p$ is the molar specific heat at constant pressure, and $\Delta T_p$ is the temperature change at constant pressure.

Since the same heat $Q$ is supplied in both cases:

$n C_v \Delta T_v = n C_p \Delta T_p$

Canceling $n$ (assuming $n \neq 0$):

$C_v \Delta T_v = C_p \Delta T_p$

For an ideal gas, $C_p$ and $C_v$ are related by $C_p = C_v + R$, where $R$ is the gas constant, and $C_p > C_v$ because at constant pressure, some heat is used to do work during expansion, whereas at constant volume, all heat goes into increasing internal energy.

Substituting $C_p = C_v + R$:

$C_v \Delta T_v = (C_v + R) \Delta T_p$

Rearranging for $\Delta T_v$:

$\Delta T_v = \frac{C_v + R}{C_v} \Delta T_p$

$\Delta T_v = \left(1 + \frac{R}{C_v}\right) \Delta T_p$

Since $\frac{R}{C_v} > 0$, the term $1 + \frac{R}{C_v} > 1$, so:

$\Delta T_v > \Delta T_p$

This means the temperature change at constant volume ($\Delta T_v$) is greater than the temperature change at constant pressure ($\Delta T_p$) for the same heat input. Therefore, for the same amount of heat, the temperature of the gas at constant pressure is lower than at constant volume (assuming the same initial temperature). Hence, Statement-2 is true.

Now, we must determine if Statement-2 is the correct explanation for Statement-1. Statement-1 states that internal energy depends only on temperature, which is a fundamental property of ideal gases derived from kinetic theory. Statement-2 describes the difference in temperature changes under constant pressure and constant volume heating, which arises because at constant pressure, part of the heat is used for work done in expansion, leaving less heat to increase temperature and internal energy. While Statement-2 relies on the fact that internal energy is a function of temperature only (so that at constant volume, all heat increases internal energy), it does not explain why internal energy is entirely kinetic and temperature-dependent in the first place. Thus, Statement-2 is true and consistent with Statement-1 but does not provide the explanation for it.

Therefore, both statements are true, but Statement-2 is not the correct explanation of Statement-1.

Hence, the correct answer is Option C.

The bob of the pendulum is made of iron, and it is oscillating over a horizontal coil carrying direct current. The direct current in the coil produces a constant magnetic field. Since the coil is horizontal, the magnetic field along its axis is vertical. Iron is a ferromagnetic material, so it is attracted to the magnetic field. The coil is below the pendulum, so the magnetic force on the iron bob acts downward, toward the coil.

The gravitational force on the bob is $mg$ downward. The additional magnetic force $F_m$ also acts downward. Therefore, the total downward force when the bob is at rest is $mg + F_m$. The effective acceleration due to gravity, $g_{\text{eff}}$, is given by:

$g_{\text{eff}} = g + \frac{F_m}{m}$

where $m$ is the mass of the bob. Since $F_m$ is positive and downward, $g_{\text{eff}} > g$.

The time period $T$ of a simple pendulum depends on the effective acceleration due to gravity:

$T = 2\pi\sqrt{\frac{l}{g_{\text{eff}}}}$

Because $g_{\text{eff}} > g$, we have:

$\frac{l}{g_{\text{eff}}} < \frac{l}{g}$

Taking square roots (all terms are positive):

$\sqrt{\frac{l}{g_{\text{eff}}}} < \sqrt{\frac{l}{g}}$

Multiplying both sides by $2\pi$:

$2\pi\sqrt{\frac{l}{g_{\text{eff}}}} < 2\pi\sqrt{\frac{l}{g}}$

So:

$T < 2\pi\sqrt{\frac{l}{g}}$

Now, consider damping. In air alone, damping occurs due to air resistance. With the coil present, the constant magnetic field and the motion of the iron bob induce eddy currents in the bob. By Lenz's law, these eddy currents oppose the motion of the bob, causing additional electromagnetic damping. Therefore, the total damping (air resistance plus electromagnetic damping) is larger than in air alone.

Hence, the time period is less than $2\pi\sqrt{\frac{l}{g}}$ and damping is larger than in air alone.

Comparing with the options:

Option A states $T < 2\pi\sqrt{\frac{l}{g}}$ but damping is smaller, which is incorrect.

Option B states $T = 2\pi\sqrt{\frac{l}{g}}$ and damping is larger, which is incorrect.

Option C states $T > 2\pi\sqrt{\frac{l}{g}}$ and damping is smaller, which is incorrect.

Option D states $T < 2\pi\sqrt{\frac{l}{g}}$ and damping is larger, which matches our conclusion.

Hence, the correct answer is Option D.

A sonometer wire of length 114 cm is fixed at both ends, and we need to place two bridges to divide it into three segments with fundamental frequencies in the ratio 1 : 3 : 4. For a string fixed at both ends, the fundamental frequency $f$ is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length of the segment, $T$ is the tension (constant throughout the wire), and $\mu$ is the linear mass density (also constant). Since $T$ and $\mu$ are constant, the frequency $f$ is inversely proportional to the length $L$, so $f \propto \frac{1}{L}$.

Given the frequency ratio $f_1 : f_2 : f_3 = 1 : 3 : 4$, the lengths will be inversely proportional to these frequencies. Therefore, the ratio of the lengths is $L_1 : L_2 : L_3 = \frac{1}{1} : \frac{1}{3} : \frac{1}{4}$. To simplify, multiply each term by the least common multiple (LCM) of the denominators 1, 3, and 4, which is 12:

$L_1 : L_2 : L_3 = \frac{1}{1} \times 12 : \frac{1}{3} \times 12 : \frac{1}{4} \times 12 = 12 : 4 : 3$

So, $L_1 : L_2 : L_3 = 12k : 4k : 3k$ for some constant $k$. The total length is the sum of the segments:

$12k + 4k + 3k = 19k = 114 \text{ cm}$

Solving for $k$:

$19k = 114 \implies k = \frac{114}{19} = 6$

Now, find the lengths:

$L_1 = 12k = 12 \times 6 = 72 \text{ cm}$

$L_2 = 4k = 4 \times 6 = 24 \text{ cm}$

$L_3 = 3k = 3 \times 6 = 18 \text{ cm}$

The segments must be adjacent and cover the entire wire. The bridges are placed such that the segments are formed consecutively. Starting from one end (say end A), the first bridge is at a distance $L_1 = 72$ cm from A. The second bridge is at $L_1 + L_2 = 72 + 24 = 96$ cm from A. The third segment from 96 cm to 114 cm is $L_3 = 18$ cm.

To verify, the fundamental frequencies are proportional to the reciprocals of the lengths:

$f_1 \propto \frac{1}{72}, \quad f_2 \propto \frac{1}{24}, \quad f_3 \propto \frac{1}{18}$

The ratio is:

$f_1 : f_2 : f_3 = \frac{1}{72} : \frac{1}{24} : \frac{1}{18}$

Simplify by multiplying each term by the LCM of 72, 24, and 18, which is 72:

$\frac{1}{72} \times 72 = 1, \quad \frac{1}{24} \times 72 = 3, \quad \frac{1}{18} \times 72 = 4$

So, $f_1 : f_2 : f_3 = 1 : 3 : 4$, which matches the given ratio.

Now, check the options:

• Option A: Bridges at 36 cm and 84 cm. Segments: 36 cm, 48 cm (84 - 36), 30 cm (114 - 84). Frequencies proportional to $\frac{1}{36}$, $\frac{1}{48}$, $\frac{1}{30}$. Ratios: $\frac{1}{36} : \frac{1}{48} : \frac{1}{30} = 20 : 15 : 24$ (after multiplying by 720, the LCM), which is not 1:3:4.
• Option B: Bridges at 24 cm and 72 cm. Segments: 24 cm, 48 cm (72 - 24), 42 cm (114 - 72). Frequencies proportional to $\frac{1}{24}$, $\frac{1}{48}$, $\frac{1}{42}$. Ratios: $\frac{1}{24} : \frac{1}{48} : \frac{1}{42} = 14 : 7 : 8$ (after multiplying by 336, the LCM), which is not 1:3:4.
• Option C: Bridges at 48 cm and 96 cm. Segments: 48 cm, 48 cm (96 - 48), 18 cm (114 - 96). Frequencies proportional to $\frac{1}{48}$, $\frac{1}{48}$, $\frac{1}{18}$. Ratios: $\frac{1}{48} : \frac{1}{48} : \frac{1}{18} = 3 : 3 : 8$ (after multiplying by 144, the LCM), which is not 1:3:4.
• Option D: Bridges at 72 cm and 96 cm. Segments: 72 cm, 24 cm (96 - 72), 18 cm (114 - 96). Frequencies proportional to $\frac{1}{72}$, $\frac{1}{24}$, $\frac{1}{18}$. Ratios: $\frac{1}{72} : \frac{1}{24} : \frac{1}{18} = 1 : 3 : 4$ (as shown above).

Hence, the correct answer is Option D.