JEE Main 2014 19th April Morning Shift Question Paper with Solutions

A bullet loses $\left(\frac{1}{n}\right)^{\text{th}}$ of its velocity when passing through one plank. This means if the initial velocity is $u$, after passing through one plank, the velocity becomes $u - \frac{u}{n} = u \left(1 - \frac{1}{n}\right) = u \left(\frac{n-1}{n}\right)$. The retardation is uniform, so we can use the equation of motion $v^2 = u^2 + 2as$, where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration (which is negative for retardation), and $s$ is the distance traveled (thickness of one plank).

For one plank, initial velocity $u$, final velocity $v = u \frac{n-1}{n}$, acceleration $a = -a$ (where $a > 0$ is the magnitude of retardation), and distance $s$. Substituting into the equation:

$\left(u \frac{n-1}{n}\right)^2 = u^2 + 2(-a)s$

$u^2 \frac{(n-1)^2}{n^2} = u^2 - 2as$

Bring all terms to one side:

$u^2 \frac{(n-1)^2}{n^2} - u^2 = -2as$

$u^2 \left( \frac{(n-1)^2}{n^2} - 1 \right) = -2as$

Simplify the expression inside the parentheses:

$\frac{(n-1)^2}{n^2} - 1 = \frac{(n-1)^2 - n^2}{n^2} = \frac{n^2 - 2n + 1 - n^2}{n^2} = \frac{-2n + 1}{n^2}$

So:

$u^2 \left( \frac{-2n + 1}{n^2} \right) = -2as$

Multiply both sides by $-1$:

$u^2 \left( \frac{2n - 1}{n^2} \right) = 2as$

Solve for $s$:

$s = \frac{u^2 (2n - 1)}{2a n^2}$

Now, to stop the bullet, the final velocity must be zero. Let $N$ be the number of planks required. The total distance covered is $N s$. Using the equation of motion again, with initial velocity $u$, final velocity $0$, acceleration $-a$, and distance $N s$:

$0^2 = u^2 + 2(-a)(N s)$

$0 = u^2 - 2a N s$

$2a N s = u^2$

$N = \frac{u^2}{2a s}$

Substitute the expression for $s$:

$N = \frac{u^2}{2a \left( \frac{u^2 (2n - 1)}{2a n^2} \right)} = \frac{u^2 \cdot 2a n^2}{2a u^2 (2n - 1)} = \frac{n^2}{2n - 1}$

Thus, the number of planks required to stop the bullet is $\frac{n^2}{2n - 1}$. Comparing with the options:

A. Infinite
B. $n$
C. $\frac{n^2}{2n-1}$
D. $\frac{2n^2}{n-1}$

The expression $\frac{n^2}{2n-1}$ matches option C. Therefore, the correct answer is option C.

Hence, the correct answer is Option C.

A heavy box is dragged along a rough horizontal floor. The coefficient of friction is given as $\mu = \frac{\sqrt{3}}{5}$. Person A pushes the box at an angle of 30° from the horizontal with a minimum force $F_A$, and person B pulls it at an angle of 60° from the horizontal with a minimum force $F_B$. We need to find the ratio $\frac{F_A}{F_B}$.

First, consider the forces acting on the box. The weight of the box is $mg$, where $m$ is the mass and $g$ is the acceleration due to gravity. The normal force $N$ is the force exerted by the floor perpendicular to the surface. The friction force opposes motion and is given by $f = \mu N$. The minimum force required to move the box occurs when the horizontal component of the applied force equals the maximum static friction force.

Start with person A, who pushes the box at 30° to the horizontal. When pushing, the vertical component of the force acts downward, increasing the normal force. Resolve $F_A$ into components:

• Horizontal component: $F_A \cos 30^\circ$
• Vertical component: $F_A \sin 30^\circ$ (downward)

The normal force $N_A$ is:

$N_A = mg + F_A \sin 30^\circ$

The friction force is:

$f_A = \mu N_A = \mu (mg + F_A \sin 30^\circ)$

At the point of motion, the horizontal component equals the friction force:

$F_A \cos 30^\circ = \mu (mg + F_A \sin 30^\circ)$

Substitute the trigonometric values: $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$:

$F_A \cdot \frac{\sqrt{3}}{2} = \mu \left( mg + F_A \cdot \frac{1}{2} \right)$

Given $\mu = \frac{\sqrt{3}}{5}$, substitute:

$F_A \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{5} \left( mg + \frac{F_A}{2} \right)$

Divide both sides by $\sqrt{3}$ (since $\sqrt{3} \neq 0$):

$\frac{F_A}{2} = \frac{1}{5} \left( mg + \frac{F_A}{2} \right)$

Multiply both sides by 5 to eliminate the denominator:

$5 \cdot \frac{F_A}{2} = mg + \frac{F_A}{2}$

$\frac{5F_A}{2} = mg + \frac{F_A}{2}$

Subtract $\frac{F_A}{2}$ from both sides:

$\frac{5F_A}{2} - \frac{F_A}{2} = mg$

$\frac{4F_A}{2} = mg$

$2F_A = mg$

$F_A = \frac{mg}{2}$

Now, consider person B, who pulls the box at 60° to the horizontal. When pulling, the vertical component acts upward, reducing the normal force. Resolve $F_B$ into components:

• Horizontal component: $F_B \cos 60^\circ$
• Vertical component: $F_B \sin 60^\circ$ (upward)

The normal force $N_B$ is:

$N_B = mg - F_B \sin 60^\circ$

The friction force is:

$f_B = \mu N_B = \mu (mg - F_B \sin 60^\circ)$

At the point of motion, the horizontal component equals the friction force:

$F_B \cos 60^\circ = \mu (mg - F_B \sin 60^\circ)$

Substitute the trigonometric values: $\cos 60^\circ = \frac{1}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$:

$F_B \cdot \frac{1}{2} = \mu \left( mg - F_B \cdot \frac{\sqrt{3}}{2} \right)$

Given $\mu = \frac{\sqrt{3}}{5}$, substitute:

$\frac{F_B}{2} = \frac{\sqrt{3}}{5} \left( mg - \frac{\sqrt{3} F_B}{2} \right)$

Multiply both sides by 10 to eliminate denominators:

$10 \cdot \frac{F_B}{2} = 10 \cdot \frac{\sqrt{3}}{5} \left( mg - \frac{\sqrt{3} F_B}{2} \right)$

$5F_B = 2\sqrt{3} \left( mg - \frac{\sqrt{3} F_B}{2} \right)$

Expand the right side:

$5F_B = 2\sqrt{3} \cdot mg - 2\sqrt{3} \cdot \frac{\sqrt{3} F_B}{2}$

Simplify the second term:

$2\sqrt{3} \cdot \frac{\sqrt{3}}{2} F_B = \sqrt{3} \cdot \sqrt{3} F_B = 3F_B$

So:

$5F_B = 2\sqrt{3} \, mg - 3F_B$

Add $3F_B$ to both sides:

$5F_B + 3F_B = 2\sqrt{3} \, mg$

$8F_B = 2\sqrt{3} \, mg$

Divide both sides by 2:

$4F_B = \sqrt{3} \, mg$

$F_B = \frac{\sqrt{3} \, mg}{4}$

Now, find the ratio $\frac{F_A}{F_B}$:

$\frac{F_A}{F_B} = \frac{\frac{mg}{2}}{\frac{\sqrt{3} \, mg}{4}} = \frac{mg}{2} \cdot \frac{4}{\sqrt{3} \, mg} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$

The ratio $\frac{F_A}{F_B} = \frac{2}{\sqrt{3}}$, which matches option B.

Hence, the correct answer is Option B.

Let the potential energy be zero at the center $O$. The initial height of the particle at point $X$ relative to $O$ is $h_X = R \cos \theta$

At point $Y$, where $OY$ makes an angle $\phi$ with the horizontal, the height relative to $O$ is $h_Y = R \sin \phi$

By the Law of Conservation of Energy between points $X$ and $Y$:

$mgR \cos \theta = mgR \sin \phi + \frac{1}{2}mv^2$

$v^2 = 2gR(\cos \theta - \sin \phi) \quad \dots (1)$

At point $Y$, the forces acting on the particle along the radial direction are the normal reaction $N$ and the radial component of weight $mg \sin \phi$. The centripetal force equation is:  $mg \sin \phi - N = \frac{mv^2}{R}$

The normal reaction vanishes at point $Y$ ($N = 0$). Therefore, $mg \sin \phi = \frac{mv^2}{R}$

$v^2 = gR \sin \phi \quad \dots (2)$

Equating $(1)$ and $(2)$: $gR \sin \phi = 2gR(\cos \theta - \sin \phi)$

$\sin \phi = 2 \cos \theta - 2 \sin \phi$

$3 \sin \phi = 2 \cos \theta$

$\sin \phi = \frac{2}{3} \cos \theta$

For translational motion of the center of mass $O$: $f = M a_0 \quad \dots (1)$, where $a_0$ is the linear acceleration of the center of mass.

For rotational motion about the center of mass $O$: $\tau = I \alpha \implies f \cdot R = \left( \frac{1}{2} M R^2 \right) \alpha$

$f = \frac{1}{2} M R \alpha \quad \dots (2)$

The condition for no slipping at point $P$ is: $a = a_0 + R \alpha$

$a = \frac{f}{M} + \frac{2f}{M}$

$a = \frac{3f}{M}$

$f = \frac{Ma}{3}$

To solve this problem, we need to find the angular momentum of a ball at the highest point of its trajectory with respect to the point from which it was thrown. The ball has a mass of 160 g, is thrown at an angle of 60° to the horizontal with a speed of 10 m/s, and gravity is given as 10 m/s². Angular momentum is defined as the cross product of the position vector (from the reference point) and the linear momentum vector, so we'll calculate both at the highest point.

First, convert the mass to kilograms since SI units are required. The mass is 160 g, which is 160 divided by 1000, so mass $m = 0.160$ kg.

The initial velocity is 10 m/s at 60° to the horizontal. We resolve this into horizontal and vertical components:

Horizontal component: $u_x = u \cos \theta = 10 \cos 60^\circ$

Since $\cos 60^\circ = \frac{1}{2}$, $u_x = 10 \times \frac{1}{2} = 5$ m/s.

Vertical component: $u_y = u \sin \theta = 10 \sin 60^\circ$

Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $u_y = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ m/s.

At the highest point of the trajectory, the vertical component of velocity becomes zero ($v_y = 0$) because gravity acts downward, and the horizontal component remains unchanged since there is no horizontal acceleration. Therefore, the velocity at the highest point is purely horizontal: $v = v_x = u_x = 5$ m/s.

The linear momentum $\vec{p}$ at the highest point is mass times velocity:

$\vec{p} = m \vec{v} = 0.160 \times 5 = 0.8$ kg m/s in the horizontal direction.

So, $\vec{p} = (0.8, 0)$ kg m/s, where the first component is horizontal and the second is vertical.

Next, we find the position vector $\vec{r}$ of the ball at the highest point relative to the point of projection. This requires the horizontal distance traveled ($x$) and the maximum height ($H$).

The maximum height $H$ is given by the formula:

$H = \frac{u_y^2}{2g}$

Substituting the values: $H = \frac{(5\sqrt{3})^2}{2 \times 10} = \frac{25 \times 3}{20} = \frac{75}{20} = 3.75$ m.

The time taken to reach the highest point $t$ is the time when vertical velocity becomes zero:

$t = \frac{u_y}{g} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}$ seconds.

The horizontal distance $x$ covered in this time is:

$x = u_x \times t = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}$ m.

So, the position vector $\vec{r} = \left( \frac{5\sqrt{3}}{2}, 3.75 \right)$ m.

Angular momentum $\vec{L}$ with respect to the point of projection is:

$\vec{L} = \vec{r} \times \vec{p}$

In 2D, the magnitude of the cross product is $|\vec{L}| = |r_x p_y - r_y p_x|$, where $r_x$ and $r_y$ are the components of $\vec{r}$, and $p_x$ and $p_y$ are the components of $\vec{p}$.

Here, $r_x = \frac{5\sqrt{3}}{2}$, $r_y = 3.75$, $p_x = 0.8$, and $p_y = 0$.

So, $L = \left| \left( \frac{5\sqrt{3}}{2} \times 0 \right) - (3.75 \times 0.8) \right| = |0 - 3.75 \times 0.8| = | -3.0 | = 3.0$ kg m²/s.

Alternatively, using exact fractions:

$r_y = 3.75 = \frac{15}{4}$ m (since $3.75 = \frac{15}{4}$),

$p_x = 0.8 = \frac{4}{5}$ kg m/s,

So, $L = \left| -\left( \frac{15}{4} \times \frac{4}{5} \right) \right| = \left| -\frac{15 \times 4}{4 \times 5} \right| = \left| -\frac{15}{5} \right| = | -3 | = 3$ kg m²/s.

The magnitude is 3.0 kg m²/s, and the direction is perpendicular to the plane (negative sign indicates direction, but magnitude is positive).

Comparing with the options:

A. 1.73 kg m² s⁻¹

B. 3.46 kg m² s⁻¹

C. 3.0 kg m² s⁻¹

D. 6.0 kg m² s⁻¹

Hence, the correct answer is Option C.

We need to match each event from List-I with the appropriate time interval from List-II. Let's recall the approximate values for each event.

Starting with (a) the rotation period of Earth: This is the time for one complete rotation, which is one day. One day has 24 hours. Converting to seconds: 24 hours × 3600 seconds per hour = 86400 seconds. In scientific notation, this is 8.64 × 10$^4$ seconds. Since 8.64 is greater than 5, we round up to the next power of 10, giving an order of magnitude of 10$^5$ seconds. Therefore, (a) matches with (i).

Next, (b) the revolution period of Earth: This is the time for one orbit around the Sun, which is one year. One year has approximately 365 days. Converting to seconds: 365 days × 24 hours/day × 3600 seconds/hour = 365 × 86400 seconds. Calculating: 365 × 86400 = 31,536,000 seconds. In scientific notation, this is 3.1536 × 10$^7$ seconds. The order of magnitude is 10$^7$ seconds. Therefore, (b) matches with (ii).

Now, (c) the period of a light wave: Light waves are electromagnetic waves, and for visible light, the frequency is around 10$^{14}$ Hz to 10$^{15}$ Hz. Taking a typical frequency of 5 × 10$^{14}$ Hz, the period T is the reciprocal of frequency: T = 1 / f = 1 / (5 × 10$^{14}$) = 2 × 10$^{-15}$ seconds. The order of magnitude is 10$^{-15}$ seconds. Therefore, (c) matches with (iii).

Finally, (d) the period of a sound wave: Sound waves are audible in the frequency range of 20 Hz to 20,000 Hz. A typical frequency is 1000 Hz (1 kHz). The period T is the reciprocal: T = 1 / 1000 = 0.001 seconds = 10$^{-3}$ seconds. Even though periods can range from 5 × 10$^{-2}$ seconds (for 20 Hz) to 5 × 10$^{-5}$ seconds (for 20,000 Hz), the typical order for common sounds (like 1000 Hz) is 10$^{-3}$ seconds. Therefore, (d) matches with (iv).

Summarizing the matches:

• (a) matches with (i) 10$^5$ s
• (b) matches with (ii) 10$^7$ s
• (c) matches with (iii) 10$^{-15}$ s
• (d) matches with (iv) 10$^{-3}$ s

Comparing with the options:

• Option A: (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
• Option B: (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
• Option C: (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
• Option D: (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

Our matches correspond to Option A.

Hence, the correct answer is Option A.

The gravitational field is given by $\vec{g} = (5\hat{i} + 12\hat{j})$ N kg$^{-1}$. We need to find the change in gravitational potential energy ($\Delta U$) for a particle of mass $m = 2$ kg when it is moved from the origin $(0, 0)$ to the point $(7 \text{ m}, -3 \text{ m})$.

The change in gravitational potential energy is given by $\Delta U = -m \int_{\text{initial}}^{\text{final}} \vec{g} \cdot d\vec{r}$. Since $\vec{g}$ is constant, the integral simplifies to a path-independent calculation.

Let $d\vec{r} = dx\,\hat{i} + dy\,\hat{j}$. Then, $\vec{g} \cdot d\vec{r} = (5\hat{i} + 12\hat{j}) \cdot (dx\,\hat{i} + dy\,\hat{j}) = 5\,dx + 12\,dy$.

So, the integral becomes:

' in math mode at position 1:̲ \int_{(0,0)}^{…" style="color:#cc0000">$ \int_{(0,0)}^{(7,-3)} \vec{g} \cdot d\vec{r} = \int_{(0,0)}^{(7,-3)} (5\,dx + 12\,dy)$

We can compute this by integrating along the x-axis first and then the y-axis. From $(0,0)$ to $(7,0)$, $y = 0$ (so $dy = 0$), and then from $(7,0)$ to $(7,-3)$, $x = 7$ (so $dx = 0$).

First segment: from $(0,0)$ to $(7,0)$:

' in math mode at position 1:̲ \int_{x=0}^{7}…" style="color:#cc0000">$ \int_{x=0}^{7} 5\,dx + \int_{y=0}^{0} 12\,dy = \int_{0}^{7} 5\,dx + 0 = 5 \left[ x \right]_{0}^{7} = 5 \times (7 - 0) = 35$

Second segment: from $(7,0)$ to $(7,-3)$:

' in math mode at position 1:̲ \int_{x=7}^{7}…" style="color:#cc0000">$ \int_{x=7}^{7} 5\,dx + \int_{y=0}^{-3} 12\,dy = 0 + \int_{0}^{-3} 12\,dy = 12 \left[ y \right]_{0}^{-3} = 12 \times (-3 - 0) = 12 \times (-3) = -36$

Adding both segments:

' in math mode at position 1:̲ \int_{(0,0)}^{…" style="color:#cc0000">$ \int_{(0,0)}^{(7,-3)} \vec{g} \cdot d\vec{r} = 35 + (-36) = -1 \text{ J kg}^{-1}$

Alternatively, using a straight-line path parameterization: let $x = 7t$, $y = -3t$, where $t$ goes from 0 to 1. Then $dx = 7\,dt$, $dy = -3\,dt$.

' in math mode at position 1:̲ \vec{g} \cdot …" style="color:#cc0000">$ \vec{g} \cdot d\vec{r} = 5(7\,dt) + 12(-3\,dt) = 35\,dt - 36\,dt = -dt$

' in math mode at position 1:̲ \int_{0}^{1} (…" style="color:#cc0000">$ \int_{0}^{1} (-1)\,dt = -1 \left[ t \right]_{0}^{1} = -1 \times (1 - 0) = -1 \text{ J kg}^{-1}$

Both methods give the same result: $-1$ J kg$^{-1}$.

Now, the change in potential energy is:

' in math mode at position 1:̲ \Delta U = -m …" style="color:#cc0000">$ \Delta U = -m \times \left( \int \vec{g} \cdot d\vec{r} \right) = - (2 \text{ kg}) \times (-1 \text{ J kg}^{-1}) = -2 \times (-1) = 2 \text{ J}$

Hence, the change in gravitational potential energy is 2 J.

Comparing with the options:

A. 71 J

B. $13\sqrt{58}$ J

C. 2 J

D. 1 J

So, the correct answer is Option C.

The velocity of water near the surface is given as 18 km h$^{-1}$, and the river depth is 5 m. To find the shearing stress between horizontal layers of water, we use Newton's law of viscosity, which states that the shearing stress (τ) is equal to the coefficient of viscosity (η) multiplied by the velocity gradient (dv/dy). The formula is τ = η × (dv/dy).

First, we need to convert the velocity from km h$^{-1}$ to m s$^{-1}$ because SI units are required. We know that 1 km = 1000 m and 1 hour = 3600 seconds. So, 18 km h$^{-1}$ = 18 × (1000 m) / (3600 s) = 18 × (1000 ÷ 3600) m s$^{-1}$. Simplifying 1000 ÷ 3600 gives 5/18, so 18 × (5/18) = 5 m s$^{-1}$. Therefore, the velocity at the surface is 5 m s$^{-1}$.

At the bottom of the river (depth = 5 m), the velocity of water is zero due to the no-slip condition. The velocity gradient dv/dy is the change in velocity per unit depth. Assuming a linear velocity profile from the bottom to the surface, dv/dy = (velocity at surface - velocity at bottom) ÷ depth = (5 m s$^{-1}$ - 0 m s$^{-1}$) ÷ 5 m = 5 ÷ 5 = 1 s$^{-1}$.

The coefficient of viscosity η is given as 10$^{-2}$ poise. Poise is a CGS unit, so we convert it to SI units (Pa·s or N s m$^{-2}$). We know that 1 poise = 0.1 Pa·s. Therefore, η = 10$^{-2}$ poise = 10$^{-2}$ × 0.1 Pa·s = 10$^{-3}$ Pa·s = 10$^{-3}$ N s m$^{-2}$.

Now, substituting into the formula for shearing stress: τ = η × (dv/dy) = (10$^{-3}$ N s m$^{-2}$) × (1 s$^{-1}$) = 10$^{-3}$ N m$^{-2}$.

Comparing with the options, 10$^{-3}$ N m$^{-2}$ corresponds to option B. Hence, the correct answer is Option B.

We are given that a large number of liquid drops, each of radius $r$, coalesce to form a single big drop of radius $R$. The energy released during coalescence is converted into the kinetic energy of the big drop. We need to find the speed of the big drop, given the surface tension $T$ and density $\rho$.

First, recall that the surface energy of a liquid drop is given by the surface tension multiplied by the surface area. For a single small drop of radius $r$, the surface area is $4\pi r^2$, so its surface energy is $T \times 4\pi r^2$.

Let the number of small drops be $n$. The volume of one small drop is $\frac{4}{3}\pi r^3$, and the volume of the big drop is $\frac{4}{3}\pi R^3$. Since volume is conserved during coalescence:

$n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$

Canceling $\frac{4}{3}\pi$ from both sides:

$n r^3 = R^3$

Solving for $n$:

$n = \frac{R^3}{r^3}$

The total surface energy before coalescence is the sum of the surface energies of all small drops:

$\text{Initial surface energy} = n \times T \times 4\pi r^2 = \frac{R^3}{r^3} \times T \times 4\pi r^2 = T \times 4\pi \frac{R^3}{r}$

The surface energy after coalescence is that of the big drop:

$\text{Final surface energy} = T \times 4\pi R^2$

The energy released is the difference between the initial and final surface energies:

$E_{\text{released}} = \text{Initial surface energy} - \text{Final surface energy} = T \times 4\pi \frac{R^3}{r} - T \times 4\pi R^2$

Factor out $T \times 4\pi$:

$E_{\text{released}} = T \times 4\pi \left( \frac{R^3}{r} - R^2 \right)$

Factor $R^2$ from the expression inside the parentheses:

$E_{\text{released}} = T \times 4\pi R^2 \left( \frac{R}{r} - 1 \right)$

Note that $\frac{R}{r} - 1 = \frac{R - r}{r}$. Also, observe that $\frac{1}{r} - \frac{1}{R} = \frac{R - r}{R r}$. Therefore:

$R - r = R r \left( \frac{1}{r} - \frac{1}{R} \right)$

Substitute this into the expression:

$E_{\text{released}} = T \times 4\pi R^2 \times \frac{R - r}{r} = T \times 4\pi R^2 \times \frac{R r \left( \frac{1}{r} - \frac{1}{R} \right)}{r}$

Simplify by canceling $r$ in the numerator and denominator:

$E_{\text{released}} = T \times 4\pi R^2 \times R \left( \frac{1}{r} - \frac{1}{R} \right) = T \times 4\pi R^3 \left( \frac{1}{r} - \frac{1}{R} \right)$

This energy is converted into the kinetic energy of the big drop. The kinetic energy is given by $\frac{1}{2} M v^2$, where $M$ is the mass of the big drop and $v$ is its speed.

The mass $M$ is the density times volume:

$M = \rho \times \frac{4}{3}\pi R^3$

Set the energy released equal to the kinetic energy:

$T \times 4\pi R^3 \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{1}{2} M v^2$

Substitute $M$:

$T \times 4\pi R^3 \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{1}{2} \times \rho \times \frac{4}{3}\pi R^3 \times v^2$

Divide both sides by $\pi R^3$ (assuming $R \neq 0$):

$T \times 4 \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{1}{2} \times \rho \times \frac{4}{3} \times v^2$

Simplify the right side:

$\frac{1}{2} \times \frac{4}{3} = \frac{4}{6} = \frac{2}{3}$

So:

$4T \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{2}{3} \rho v^2$

Solve for $v^2$. Multiply both sides by $\frac{3}{2}$:

$v^2 = 4T \left( \frac{1}{r} - \frac{1}{R} \right) \times \frac{3}{2} \times \frac{1}{\rho}$

Simplify:

$v^2 = \frac{4 \times 3}{2} \times \frac{T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{12}{2} \times \frac{T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right) = 6 \times \frac{T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right)$

Therefore:

$v^2 = \frac{6T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right)$

Take the square root of both sides:

$v = \sqrt{ \frac{6T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right) }$

Comparing with the options, this matches option B.

Hence, the correct answer is Option B.

The pressure difference between the two points is determined by the height difference $\Delta h = 5\text{ cm}$ in the manometer:

$\Delta P = \rho g \Delta h = 1\text{ g/cc} \times 10^3\text{ cm/s}^2 \times 5\text{ cm} = 5 \times 10^3\text{ dyne/cm}^2$

Using the equation of continuity $A_1v_1 = A_2v_2$: $6v_A = 10v_B \implies v_A = \frac{5}{3}v_B$

Substituting this into Bernoulli's equation, $\frac{1}{2}\rho(v_A^2 - v_B^2) = P_B - P_A$:

$\frac{1}{2} \times 1 \left[ \left(\frac{5}{3}v_B\right)^2 - v_B^2 \right] = 5 \times 10^3$

$\frac{1}{2} \left( \frac{16}{9}v_B^2 \right) = 5 \times 10^3 \implies v_B = \frac{3}{4} \times 10^2\text{ cm/s}$

$\text{Volume flow rate} = v_B A_B = \left(\frac{3}{4} \times 10^2\right) \times (10 \times 10^{-2}) = 7.5\text{ cc/s}$

The sphere cools by radiation since it is in a vacuum. The net power radiated by a black body is given by the Stefan-Boltzmann law. The net power radiated per unit area is $\sigma (T^4 - T_0^4)$, where $T$ is the temperature of the sphere and $T_0$ is the surrounding temperature. The surface area of the sphere is $4\pi R^2$, so the total net power radiated is:

$P_{\text{net}} = 4\pi R^2 \sigma (T^4 - T_0^4)$

This power loss equals the rate of decrease of the internal energy of the sphere. Therefore:

$-\frac{dU}{dt} = 4\pi R^2 \sigma (T^4 - T_0^4)$

The internal energy $U$ depends on the temperature. Given the specific heat per unit mass is $\alpha T^3$, the heat capacity per unit mass is $c = \alpha T^3$. For a mass $M$, the change in internal energy for a temperature change $dT$ is:

$dU = M c dT = M \alpha T^3 dT$

Thus, the rate of change of internal energy with time is:

$\frac{dU}{dt} = M \alpha T^3 \frac{dT}{dt}$

Substituting into the power equation:

$- M \alpha T^3 \frac{dT}{dt} = 4\pi R^2 \sigma (T^4 - T_0^4)$

Rearranging for $dt$:

$dt = -\frac{M \alpha T^3}{4\pi R^2 \sigma (T^4 - T_0^4)} dT$

To find the time taken for the sphere to cool from $3T_0$ to $2T_0$, integrate from $T = 3T_0$ to $T = 2T_0$. Since the temperature decreases, the negative sign is handled by reversing the limits:

$t = \int_{2T_0}^{3T_0} \frac{M \alpha T^3}{4\pi R^2 \sigma (T^4 - T_0^4)} dT$

Factor the denominator using $T^4 - T_0^4 = (T^2)^2 - (T_0^2)^2 = (T^2 - T_0^2)(T^2 + T_0^2)$. However, a substitution simplifies the integral. Let $u = T^4 - T_0^4$, then $du = 4T^3 dT$, so $T^3 dT = \frac{du}{4}$.

Change the limits: when $T = 2T_0$, $u = (2T_0)^4 - T_0^4 = 16T_0^4 - T_0^4 = 15T_0^4$. When $T = 3T_0$, $u = (3T_0)^4 - T_0^4 = 81T_0^4 - T_0^4 = 80T_0^4$.

Substitute:

$t = \int_{15T_0^4}^{80T_0^4} \frac{M \alpha}{4\pi R^2 \sigma} \cdot \frac{1}{u} \cdot \frac{du}{4} = \frac{M \alpha}{4\pi R^2 \sigma} \cdot \frac{1}{4} \int_{15T_0^4}^{80T_0^4} \frac{du}{u}$

The integral of $\frac{du}{u}$ is $\ln|u|$, and since $u > 0$, it is $\ln u$:

$t = \frac{M \alpha}{16\pi R^2 \sigma} \left[ \ln u \right]_{15T_0^4}^{80T_0^4} = \frac{M \alpha}{16\pi R^2 \sigma} \left( \ln(80T_0^4) - \ln(15T_0^4) \right)$

Simplify using logarithm properties:

$t = \frac{M \alpha}{16\pi R^2 \sigma} \ln \left( \frac{80T_0^4}{15T_0^4} \right) = \frac{M \alpha}{16\pi R^2 \sigma} \ln \left( \frac{80}{15} \right)$

Reduce $\frac{80}{15}$ by dividing numerator and denominator by 5:

$\frac{80}{15} = \frac{16}{3}$

Thus:

$t = \frac{M \alpha}{16\pi R^2 \sigma} \ln \left( \frac{16}{3} \right)$

Comparing with the options, this matches Option B.

Hence, the correct answer is Option B.

The work done on the gas is $W = -P \cdot \Delta V$

$W = -100 \cdot (1 - 2) = 100 \text{ J}$

This 100 J represents energy transferred into the gas through mechanical work.

In the second step, the gas is heated at a constant volume. 

Energy supplied ($Q$) = 150 J. Work done ($W$) = 0 J (since volume does not change)

$\Delta U_2 = 150 \text{ J}$

$\Delta U_{total} = \text{Work done on gas} + \text{Heat supplied}$

$\Delta U_{total} = 100 \text{ J} + 150 \text{ J} = 250 \text{ J}$

The gas molecule has a kinetic energy equivalent to 0 °C. Since temperature in kinetic theory is given in Kelvin, convert 0 °C to Kelvin: 0 °C = 273 K. The average translational kinetic energy of a gas molecule is given by $\frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature. Therefore, the kinetic energy $K$ at the surface is:

$K = \frac{3}{2} k_B \times 273$

As the molecule rises straight up without colliding with other molecules, it will lose kinetic energy and gain potential energy. At the maximum height, its kinetic energy becomes zero, and all the initial kinetic energy is converted to potential energy. Assuming the potential energy at the Earth's surface is zero, by conservation of energy:

Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy

$K + 0 = 0 + U_{\text{max}}$

So,

$U_{\text{max}} = K = \frac{3}{2} k_B \times 273$

The potential energy at height $h$ above the Earth's surface is $M g h$, where $M$ is the mass of the molecule, $g$ is the acceleration due to gravity, and $h$ is much less than the Earth's radius (as given). Therefore:

$U_{\text{max}} = M g h$

Equating the expressions for potential energy:

$M g h = \frac{3}{2} k_B \times 273$

Solving for $h$:

$h = \frac{\frac{3}{2} k_B \times 273}{M g}$

Simplify the fraction:

$h = \frac{3}{2} \times \frac{273 k_B}{M g}$

Multiply the constants:

$h = \frac{3 \times 273 k_B}{2 M g}$

Calculate $3 \times 273$:

$3 \times 273 = 819$

So,

$h = \frac{819 k_B}{2 M g}$

Comparing with the given options, this matches option B. Hence, the correct answer is Option B.