JEE Main 2015 10th April Morning Shift Question Paper with Solutions

The capacitance $u$ is to be expressed in terms of the electronic charge $e$, Bohr radius $a_0$, Planck's constant $h$, and the speed of light $c$. We use dimensional analysis to find the correct expression.

First, recall the dimensions of each quantity in terms of mass (M), length (L), time (T), and current (I):

• Electronic charge $e$ has dimension $[I T]$.
• Bohr radius $a_0$ has dimension $[L]$.
• Planck's constant $h$ has dimension $[M L^2 T^{-1}]$.
• Speed of light $c$ has dimension $[L T^{-1}]$.

The capacitance $u$ has the dimension $[M^{-1} L^{-2} T^4 I^2]$. We assume $u$ is proportional to $e^a \times a_0^b \times h^c \times c^d$, so the dimensional equation is:

$[M^{-1} L^{-2} T^4 I^2] = [I T]^a \times [L]^b \times [M L^2 T^{-1}]^c \times [L T^{-1}]^d$

Expanding the right side:

$[I T]^a = I^a T^a$

$[L]^b = L^b$

$[M L^2 T^{-1}]^c = M^c L^{2c} T^{-c}$

$[L T^{-1}]^d = L^d T^{-d}$

Combining these, the right side becomes:

$M^c \times L^{b + 2c + d} \times T^{a - c - d} \times I^a$

Equating dimensions with the left side:

For mass (M): $c = -1$

For current (I): $a = 2$

For time (T): $a - c - d = 4$

Substituting $a = 2$ and $c = -1$:

$2 - (-1) - d = 4$

$2 + 1 - d = 4$

$3 - d = 4$

$-d = 1$

$d = -1$

For length (L): $b + 2c + d = -2$

Substituting $c = -1$ and $d = -1$:

$b + 2(-1) + (-1) = -2$

$b - 2 - 1 = -2$

$b - 3 = -2$

$b = 1$

Thus, the exponents are $a = 2$, $b = 1$, $c = -1$, $d = -1$. The expression for capacitance is:

$u \propto e^2 \times a_0^1 \times h^{-1} \times c^{-1} = \frac{e^2 a_0}{h c}$

Comparing with the options:

• Option A: $u = \frac{e^2 a_0}{h c}$ matches.
• Option B: $u = \frac{h c}{e^2 a_0}$ is the reciprocal.
• Option C: $u = \frac{e^2 c}{h a_0}$ does not match.
• Option D: $u = \frac{e^2 h}{c a_0}$ does not match.

Hence, the correct answer is Option A.

A block of mass $m = 10$ kg rests on a horizontal table with coefficient of friction $\mu = 0.05$. A bullet of mass $50$ g, which is $0.05$ kg, moving with speed $v$, embeds itself in the block. After the collision, the combined system moves and stops after traveling $2$ m. We need to find the height $H$ from which a freely falling object would acquire speed $\frac{v}{10}$, neglecting energy losses and using $g = 10$ m/s².

First, consider the inelastic collision between the bullet and the block. Conservation of momentum applies:

Initial momentum = Momentum of bullet + Momentum of block = $(0.05) \times v + (10) \times 0 = 0.05v$ kg·m/s.

After collision, the combined mass is $10 + 0.05 = 10.05$ kg. Let the common velocity be $V$.

Final momentum = $10.05 \times V$ kg·m/s.

Equating initial and final momentum:

$0.05v = 10.05V$

Solving for $V$:

$V = \frac{0.05v}{10.05} = \frac{5v}{1005} = \frac{v}{201}$

So, $V = \frac{v}{201}$ m/s.

After the collision, the block with embedded bullet slides on the table with initial velocity $V$ and stops after $s = 2$ m due to friction. The friction force causes deceleration.

The normal force $N$ equals the weight of the combined system:

$N = (10.05) \times 10 = 100.5 \text{ N}$

Friction force $f = \mu N = 0.05 \times 100.5 = 5.025$ N.

Deceleration $a = \frac{f}{\text{total mass}} = \frac{5.025}{10.05} = 0.5$ m/s². Alternatively, $a = \mu g = 0.05 \times 10 = 0.5$ m/s².

Using the equation of motion: final velocity² = initial velocity² + 2 × acceleration × distance

$0^2 = V^2 + 2 \times (-0.5) \times 2$

$0 = V^2 - 2$

$V^2 = 2$

$V = \sqrt{2} \text{ m/s} \quad (\text{taking positive value})$

But $V = \frac{v}{201}$, so:

$\frac{v}{201} = \sqrt{2}$

$v = 201 \sqrt{2} \text{ m/s}$

Now, for the freely falling object: it starts from height $H$ with initial velocity 0 and acquires speed $\frac{v}{10}$ at the bottom. Using conservation of energy:

Potential energy loss = Kinetic energy gain

$m g H = \frac{1}{2} m \left( \frac{v}{10} \right)^2$

Cancel $m$ from both sides:

$g H = \frac{1}{2} \times \frac{v^2}{100}$

$10 H = \frac{v^2}{200}$

$H = \frac{v^2}{2000}$

Substitute $v = 201 \sqrt{2}$:

$H = \frac{(201 \sqrt{2})^2}{2000} = \frac{201^2 \times 2}{2000}$

Calculate $201^2$:

$201^2 = (200 + 1)^2 = 200^2 + 2 \times 200 \times 1 + 1^2 = 40000 + 400 + 1 = 40401$

So,

$H = \frac{40401 \times 2}{2000} = \frac{80802}{2000} = 40.401 \text{ m}$

Convert meters to kilometers (1 km = 1000 m):

$H = \frac{40.401}{1000} = 0.040401 \text{ km}$

The options are:

A. 0.2 km

B. 0.5 km

C. 0.4 km

D. None of these

Comparing $H \approx 0.0404$ km with the options:

0.0404 km is not close to 0.2 km, 0.5 km, or 0.4 km (which are 200 m, 500 m, and 400 m respectively, while 0.0404 km is 40.4 m).

Hence, the correct answer is Option D.

A block of mass $m = 0.1$ kg is attached to a spring with an unknown spring constant $k$. The block is compressed to a distance $x$ from its equilibrium position and released from rest. At this initial point, the block has zero kinetic energy because it is stationary, and the total energy is stored as spring potential energy, given by $E_i = \frac{1}{2} k x^2$. This energy is conserved until the collision occurs.

As the block moves towards the equilibrium position, it reaches a point where the displacement from equilibrium is $\frac{x}{2}$. At this position, the spring potential energy is $\frac{1}{2} k \left( \frac{x}{2} \right)^2 = \frac{1}{2} k \cdot \frac{x^2}{4} = \frac{1}{8} k x^2$. Let the velocity of the block at this point, just before collision, be $v$. The kinetic energy at this position is $\frac{1}{2} m v^2$.

By conservation of energy between the initial compressed position and the point at $\frac{x}{2}$, the initial total energy equals the sum of potential and kinetic energy at $\frac{x}{2}$:

$\frac{1}{2} k x^2 = \frac{1}{8} k x^2 + \frac{1}{2} m v^2$

Subtract $\frac{1}{8} k x^2$ from both sides:

$\frac{1}{2} k x^2 - \frac{1}{8} k x^2 = \frac{1}{2} m v^2$

$\frac{4}{8} k x^2 - \frac{1}{8} k x^2 = \frac{1}{2} m v^2$

$\frac{3}{8} k x^2 = \frac{1}{2} m v^2$

Multiply both sides by 8 to eliminate the denominator:

$3 k x^2 = 4 m v^2$

Rearrange to solve for $k x^2$:

$k x^2 = \frac{4 m v^2}{3} \quad \text{(Equation 1)}$

At the displacement $\frac{x}{2}$, the block collides with another block. After the collision, the first block comes to rest momentarily, and the second block moves with a velocity of 3 m/s. Let the mass of the first block be $m_1 = m = 0.1$ kg and the mass of the second block be $m_2$ (unknown). Before the collision, the second block is at rest, so its velocity is 0 m/s. After the collision, the first block has velocity 0 m/s, and the second block has velocity 3 m/s.

Apply conservation of momentum. The momentum before the collision equals the momentum after the collision:

$m_1 v + m_2 \cdot 0 = m_1 \cdot 0 + m_2 \cdot 3$

$m_1 v = 3 m_2 \quad \text{(Equation 2)}$

For the collision to result in the first block stopping and the second block moving with 3 m/s, it must be an elastic collision where the masses are equal. This is a standard result for head-on elastic collisions between objects of equal mass. Thus, $m_1 = m_2$. Since $m_1 = 0.1$ kg, $m_2 = 0.1$ kg. Substitute into Equation 2:

$0.1 \cdot v = 3 \cdot 0.1$

$0.1 v = 0.3$

$v = \frac{0.3}{0.1} = 3 \text{ m/s}$

So, the velocity of the first block just before the collision is $v = 3$ m/s.

Now substitute $m = 0.1$ kg and $v = 3$ m/s into Equation 1:

$k x^2 = \frac{4 \times 0.1 \times (3)^2}{3}$

First, compute $(3)^2 = 9$:

$k x^2 = \frac{4 \times 0.1 \times 9}{3}$

$k x^2 = \frac{4 \times 0.9}{3}$

$k x^2 = \frac{3.6}{3}$

$k x^2 = 1.2$

The initial energy of the spring is $E_i = \frac{1}{2} k x^2$:

$E_i = \frac{1}{2} \times 1.2 = 0.6 \text{ J}$

Hence, the total initial energy of the spring is 0.6 J.

Therefore, the correct answer is Option A.

The formula for centripetal acceleration is $a = \frac{v^2}{r}$

Since $v^2$ is constant, the relationship becomes $a \propto \frac{1}{r}$

This represents a rectangular hyperbola.

Option (D) represents this correctly.

We are given a thin uniform square sheet with side length $a$, mass $m$, and we need to find the moment of inertia $I$ about one of its diagonals. The options are provided, and we will derive the moment of inertia step by step.

First, recall that for a thin square sheet, we can use the perpendicular axis theorem. This theorem states that for a planar body lying in the $xy$-plane, the moment of inertia about the $z$-axis (perpendicular to the plane) is the sum of the moments of inertia about the $x$-axis and $y$-axis (both in the plane and perpendicular to each other):

$I_z = I_x + I_y$

For a square sheet of side $a$ and mass $m$, the moment of inertia about an axis through the center and parallel to one side (say, the $x$-axis) is the same as that of a rectangular lamina. The formula for a rectangular lamina about an axis through its center and parallel to one side is $\frac{m}{12} \times (\text{length of the perpendicular side})^2$. Since the square has equal sides, both axes yield the same moment of inertia:

$I_x = \frac{m a^2}{12}$

$I_y = \frac{m a^2}{12}$

Here, $I_x$ is the moment of inertia about the $x$-axis (where the distance is measured along the $y$-direction), and $I_y$ is about the $y$-axis (distance along the $x$-direction). Adding these using the perpendicular axis theorem:

$I_z = I_x + I_y = \frac{m a^2}{12} + \frac{m a^2}{12} = \frac{2m a^2}{12} = \frac{m a^2}{6}$

So, $I_z = \frac{m a^2}{6}$.

Now, we need the moment of inertia about a diagonal. Consider the diagonal along the line $y = x$ in the coordinate system where the square is centered at the origin, with sides parallel to the axes. The vertices are at $\left(\frac{a}{2}, \frac{a}{2}\right)$, $\left(\frac{a}{2}, -\frac{a}{2}\right)$, $\left(-\frac{a}{2}, \frac{a}{2}\right)$, and $\left(-\frac{a}{2}, -\frac{a}{2}\right)$.

The moment of inertia about an axis is given by $I = \int r^2 dm$, where $r$ is the perpendicular distance from a point to the axis. For the axis $y - x = 0$, the distance $r$ from a point $(x, y)$ is:

$r = \frac{|x - y|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y|}{\sqrt{2}}$

Thus,

$I_d = \int r^2 dm = \int \left( \frac{|x - y|}{\sqrt{2}} \right)^2 dm = \int \frac{(x - y)^2}{2} dm$

Due to the symmetry of the square, $|x - y|^2 = (x - y)^2$, so we can write:

$I_d = \frac{1}{2} \int (x - y)^2 dm$

Expanding the square:

$(x - y)^2 = x^2 - 2xy + y^2$

So,

$I_d = \frac{1}{2} \int (x^2 - 2xy + y^2) dm = \frac{1}{2} \left[ \int x^2 dm - 2 \int xy dm + \int y^2 dm \right]$

Now, evaluate each integral:

• $\int x^2 dm$ is the moment of inertia about the $y$-axis, $I_y = \frac{m a^2}{12}$.
• $\int y^2 dm$ is the moment of inertia about the $x$-axis, $I_x = \frac{m a^2}{12}$.
• $\int xy dm$ is the product of inertia $I_{xy}$. Due to the symmetry of the square about both axes, for every point $(x, y)$, there are symmetric points $(x, -y)$, $(-x, y)$, and $(-x, -y)$, making the integral of $xy$ over the entire mass zero. Thus, $I_{xy} = 0$.

Substituting these values:

$I_d = \frac{1}{2} \left[ \frac{m a^2}{12} - 2 \cdot 0 + \frac{m a^2}{12} \right] = \frac{1}{2} \left[ \frac{m a^2}{12} + \frac{m a^2}{12} \right] = \frac{1}{2} \left[ \frac{2m a^2}{12} \right] = \frac{1}{2} \cdot \frac{m a^2}{6} = \frac{m a^2}{12}$

Therefore, the moment of inertia about the diagonal is $I = \frac{m a^2}{12}$.

Comparing with the options:

• A. $I = \frac{m a^2}{24}$
• B. $\frac{m a^2}{24} < I < \frac{m a^2}{12}$
• C. $I > \frac{m a^2}{12}$
• D. $I = \frac{m a^2}{12}$

Hence, the correct answer is Option D.

A uniform solid cylinder of mass $m$ is being pulled horizontally with a force $F$ applied at its center. The cylinder rolls without slipping and has a linear acceleration $a$. We need to find the value of $F$ in terms of $m$ and $a$.

Since the cylinder is rolling without slipping, both translational and rotational motions occur. We will apply Newton's second law for translation and the torque equation for rotation.

Consider the forces acting on the cylinder. The applied force $F$ acts horizontally at the center. Friction acts at the point of contact with the ground. Let $f$ be the friction force. Since the cylinder is accelerating to the right and rolling clockwise, friction opposes the applied force and acts to the left. This friction provides the torque for rotation.

Apply Newton's second law for translational motion in the horizontal direction. The net force equals mass times acceleration:

$F - f = m a \quad \text{(Equation 1)}$

Now, consider rotation about the center of the cylinder. The applied force $F$ acts at the center, so its torque is zero. The friction force $f$ acts at the rim, so its torque is $f \times R$, where $R$ is the radius. This torque causes angular acceleration.

The moment of inertia $I$ of a solid cylinder about its central axis is $I = \frac{1}{2} m R^2$. Let $\alpha$ be the angular acceleration. The torque equation is:

$\text{Torque} = I \alpha$

$f R = \left( \frac{1}{2} m R^2 \right) \alpha \quad \text{(Equation 2)}$

Simplify Equation 2 by dividing both sides by $R$:

$f = \frac{1}{2} m R \alpha \quad \text{(Equation 2a)}$

Since the cylinder rolls without slipping, the linear acceleration $a$ and angular acceleration $\alpha$ are related by:

$a = R \alpha \quad \text{(Equation 3)}$

Solve Equation 3 for $\alpha$:

$\alpha = \frac{a}{R}$

Substitute $\alpha = \frac{a}{R}$ into Equation 2a:

$f = \frac{1}{2} m R \times \frac{a}{R}$

Simplify by canceling $R$:

$f = \frac{1}{2} m a \quad \text{(Equation 4)}$

Now substitute Equation 4 into Equation 1:

$F - \frac{1}{2} m a = m a$

Solve for $F$:

$F = m a + \frac{1}{2} m a$

$F = \left(1 + \frac{1}{2}\right) m a$

$F = \frac{3}{2} m a$

Comparing with the options, $\frac{3}{2} m a$ corresponds to Option A.

Hence, the correct answer is Option A.

Consider a very long cylindrical galaxy of length $L$ and radius $R$ (with $R \ll L$), with mass uniformly distributed. A star orbits the galaxy in a plane perpendicular to the galaxy's axis and passing through its center, at a distance $r$ from the axis (where $r > R$ since the star is outside the galaxy). We need to find the relationship between the orbital time period $T$ and $r$.

Due to the cylindrical symmetry and the condition $R \ll L$, we can approximate the galaxy as an infinitely long cylinder for gravitational effects at distances $r$ (where $r \ll L$). The mass per unit length $\lambda$ is constant and given by the total mass divided by $L$. For an infinitely long cylinder, the gravitational field outside the cylinder depends only on $\lambda$.

Using Gauss's law for gravity, consider a cylindrical Gaussian surface of radius $r$ and length $l$, coaxial with the galaxy. The mass enclosed by this surface is $\lambda l$. The gravitational flux through the curved surface is $g \cdot 2\pi r l$ (since the field is radial and uniform), and through the ends is zero (as the field is parallel to the ends). Gauss's law states:

$\int \mathbf{g} \cdot d\mathbf{A} = -4\pi G M_{\text{enc}}$

Substituting the values:

$g \cdot 2\pi r l = -4\pi G (\lambda l)$

Solving for the magnitude of the gravitational field $g$ (ignoring the negative sign for direction):

$g = \frac{2G\lambda}{r}$

This gravitational force provides the centripetal force for the star's circular orbit. Let $m$ be the mass of the star and $v$ its orbital speed. The centripetal force is:

$F_c = \frac{m v^2}{r}$

The gravitational force is:

$F_g = m g = m \cdot \frac{2G\lambda}{r}$

Setting $F_g = F_c$:

$m \cdot \frac{2G\lambda}{r} = \frac{m v^2}{r}$

Canceling $m$ and $r$ (assuming $m \neq 0$ and $r \neq 0$):

$\frac{2G\lambda}{r} = \frac{v^2}{r}$

Multiplying both sides by $r$:

$2G\lambda = v^2$

Thus:

$v^2 = 2G\lambda$

Since $\lambda$ is constant for the galaxy, $v$ is constant and independent of $r$.

The orbital period $T$ is the time for one complete orbit, which is the circumference divided by the speed:

$T = \frac{2\pi r}{v}$

As $v$ is constant, $T$ is proportional to $r$:

$T \propto r$

Comparing with the options:

A. $T \propto \sqrt{r}$

B. $T \propto r$

C. $T \propto r^2$

D. $T^2 \propto r^3$

Option B matches the derived relationship.

Hence, the correct answer is Option B.

The pressure difference ($\Delta P$) across a curved liquid surface is determined by the surface tension ($T$) and the two principal radii of curvature ($R_1$ and $R_2$) is $\Delta P = T \left[ \frac{1}{R_1} + \frac{1}{R_2} \right]$

This excess pressure is the same as the excess pressure inside a drop or air bubble.

So,  $R_1 = R_2 = R$

$\Delta P = T \left[ \frac{1}{R} + \frac{1}{R} \right]$

$\Delta P = \frac{2T}{R}$

The Reynolds number is a dimensionless quantity used to predict flow patterns in fluid dynamics. It is given by the formula:

$\text{Re} = \frac{\rho v d}{\eta}$

where:

• $\rho$ is the density of the fluid,
• $v$ is the flow velocity,
• $d$ is the diameter of the pipe,
• $\eta$ is the dynamic viscosity.

Given:

• Time to fill the bucket, $t = 5$ minutes = $5 \times 60 = 300$ seconds,
• Volume of bucket, $V = 15$ litres = $15 \times 10^{-3}$ m$^3$ (since 1 litre = $10^{-3}$ m$^3$),
• Diameter of tap, $d = \frac{2}{\sqrt{\pi}}$ cm,
• Density of water, $\rho = 10^3$ kg/m$^3$,
• Viscosity of water, $\eta = 10^{-3}$ Pa.s.

First, convert the diameter to meters because SI units are required. Since 1 cm = 0.01 m,

$d = \frac{2}{\sqrt{\pi}} \times 10^{-2} \text{ m} = \frac{0.02}{\sqrt{\pi}} \text{ m}.$

Next, calculate the cross-sectional area $A$ of the tap. Since the tap is circular,

$A = \pi \left( \frac{d}{2} \right)^2.$

Substitute $d$:

$\frac{d}{2} = \frac{0.02}{2\sqrt{\pi}} = \frac{0.01}{\sqrt{\pi}} \text{ m},$

so,

$A = \pi \left( \frac{0.01}{\sqrt{\pi}} \right)^2 = \pi \times \frac{(0.01)^2}{\pi} = \pi \times \frac{0.0001}{\pi} = 0.0001 \text{ m}^2 = 10^{-4} \text{ m}^2.$

Now, find the volume flow rate $Q$, which is volume divided by time:

$Q = \frac{V}{t} = \frac{15 \times 10^{-3}}{300} = \frac{0.015}{300} = 0.00005 \text{ m}^3/\text{s} = 5 \times 10^{-5} \text{ m}^3/\text{s}.$

The flow velocity $v$ is given by $Q = A \times v$, so

$v = \frac{Q}{A} = \frac{5 \times 10^{-5}}{10^{-4}} = \frac{5 \times 10^{-5}}{1 \times 10^{-4}} = 5 \times 10^{-1} = 0.5 \text{ m/s}.$

Now, substitute all values into the Reynolds number formula:

$\text{Re} = \frac{\rho v d}{\eta} = \frac{(10^3) \times (0.5) \times \left( \frac{0.02}{\sqrt{\pi}} \right)}{10^{-3}}.$

First, compute the numerator:

$\rho v d = 1000 \times 0.5 \times \frac{0.02}{\sqrt{\pi}} = 500 \times \frac{0.02}{\sqrt{\pi}} = \frac{10}{\sqrt{\pi}}.$

Then, divide by $\eta$:

$\text{Re} = \frac{\frac{10}{\sqrt{\pi}}}{10^{-3}} = \frac{10}{\sqrt{\pi}} \times \frac{1}{10^{-3}} = \frac{10}{\sqrt{\pi}} \times 10^3 = \frac{10000}{\sqrt{\pi}}.$

Now, compute the numerical value. Using $\pi \approx 3.1416$, $\sqrt{\pi} \approx \sqrt{3.1416} \approx 1.77245$, so

$\text{Re} = \frac{10000}{1.77245} \approx 5641.89.$

Comparing with the options:

• A. 5500
• B. 550
• C. 1100
• D. 11000

The value 5641.89 is closest to 5500, with a difference of approximately 141.89, while the other options are farther away (550 is too small, 1100 is smaller, and 11000 is much larger). Therefore, the Reynolds number is close to 5500.

Hence, the correct answer is Option A.

(a → b): V↑,T↑ → Isobaric expansion → horizontal line rightward in P ⁣− ⁣V 

(b → c): V↑,T↓ → Adiabatic expansion → curve downwards (right)

(c → d): V↓,T↓ → Isobaric compression → horizontal line leftward

(d → a): V↓,T↑ → Adiabatic compression → curve upwards (left)

Option (A) correctly represents this behaviour.

To solve this problem, we need to determine how the average force exerted by a single molecule on the walls of a closed container depends on the temperature $T$ for an ideal gas. Specifically, we are told that this force varies as $T^q$, and we must find the exponent $q$.

First, recall that the pressure $P$ exerted by an ideal gas is given by the kinetic theory equation:

$P = \frac{1}{3} \rho \overline{v^2}$

where $\rho$ is the density (mass per unit volume) and $\overline{v^2}$ is the mean square speed of the molecules. Since $\rho = \frac{N m}{V}$, where $N$ is the number of molecules, $m$ is the mass of one molecule, and $V$ is the volume, we can write:

$P = \frac{1}{3} \frac{N m}{V} \overline{v^2}$

From the kinetic theory of gases, the average kinetic energy per molecule is related to the temperature $T$ by:

$\frac{1}{2} m \overline{v^2} = \frac{3}{2} k T$

where $k$ is Boltzmann's constant. Solving for $\overline{v^2}$:

$\overline{v^2} = \frac{3 k T}{m}$

Substitute this into the pressure equation:

$P = \frac{1}{3} \frac{N m}{V} \cdot \frac{3 k T}{m}$

Simplify by canceling $m$ and the factor of 3:

$P = \frac{N k T}{V}$

This is the ideal gas law in terms of the number of molecules.

Now, pressure is force per unit area, so the total force on a wall of area $A$ is $F_{\text{total}} = P A$. Substituting the expression for $P$:

$F_{\text{total}} = \frac{N k T}{V} A$

This total force is the result of all molecules colliding with the wall. However, we need the average force exerted by a single molecule. Since the molecules are identical and the gas is isotropic, the total average force on the wall is the sum of the average forces from each molecule. Therefore:

$F_{\text{total}} = N \overline{F}$

where $\overline{F}$ is the average force per molecule. Equating the two expressions for $F_{\text{total}}$:

$N \overline{F} = \frac{N k T}{V} A$

Divide both sides by $N$:

$\overline{F} = \frac{k T A}{V}$

For a closed container, both the area $A$ of the wall and the volume $V$ are constant. Therefore, the average force per molecule $\overline{F}$ is proportional to $T$:

$\overline{F} \propto T$

This implies that $\overline{F}$ depends on $T$ as $T^q$ with $q = 1$.

Hence, the correct answer is Option C.

Using the trigonometric identity $\sin 2\theta = 2 \sin \theta \cos \theta$, we can rewrite $y(t)$:

$y = a(2 \sin \omega t \cos \omega t)$

$\sin \omega t = \frac{x}{a}$

Using $\cos \theta = \sqrt{1 - \sin^2 \theta}$: $\cos \omega t = \sqrt{1 - \left(\frac{x}{a}\right)^2}$

$y = 2a \left(\frac{x}{a}\right) \sqrt{1 - \frac{x^2}{a^2}}$

$y = 2x \sqrt{1 - \frac{x^2}{a^2}}$

$y^2 = 4x^2 \left(1 - \frac{x^2}{a^2}\right)$

$y^2 = 4x^2 - \frac{4x^4}{a^2}$

When $x = 0 \implies y = 0$. (Passes through the origin). When $x = \pm a \implies y = 0$

As $\omega t$ goes from $0$ to $\pi/2$, $x$ moves from $0$ to $a$.

In the same interval, $2\omega t$ goes from $0$ to $\pi$, meaning $y$ completes a half-sine wave ($0 \rightarrow a \rightarrow 0$). This forms a loop in the first and fourth quadrants. As $\omega t$ continues to $\pi$, $x$ returns to $0$ and $y$ goes from $0 \rightarrow -a \rightarrow 0$.

Shape: This specific frequency ratio of $1:2$ with zero phase difference results in a horizontal figure-eight shape.

The problem involves a simple harmonic oscillator with an angular frequency of 2 rad/s, subjected to an external force $F = \sin t$ N. The oscillator starts at rest at its equilibrium position when $t = 0$. We need to find its position at later times, which should be proportional to one of the given options.

The equation of motion for a forced harmonic oscillator is given by Newton's second law. The general form is $m \frac{d^2x}{dt^2} + kx = F_{\text{ext}}$, where $m$ is the mass and $k$ is the spring constant. The natural angular frequency $\omega$ is related to $k$ and $m$ by $\omega^2 = \frac{k}{m}$. Given $\omega = 2$ rad/s, we have $\omega^2 = 4$, so $\frac{k}{m} = 4$. Substituting, the equation becomes:

$\frac{d^2x}{dt^2} + 4x = \frac{F_{\text{ext}}}{m}$

The external force is $F_{\text{ext}} = \sin t$ N, so:

$\frac{d^2x}{dt^2} + 4x = \frac{\sin t}{m}$

Since $m$ is a constant, we can write the right-hand side as $\frac{1}{m} \sin t$. The solution to this differential equation will give the position $x(t)$.

The general solution is the sum of the homogeneous solution (free oscillation) and a particular solution (due to the external force). First, solve the homogeneous equation:

$\frac{d^2x_h}{dt^2} + 4x_h = 0$

The characteristic equation is $r^2 + 4 = 0$, which has roots $r = \pm 2i$. Thus, the homogeneous solution is:

$x_h = A \cos 2t + B \sin 2t$

where $A$ and $B$ are constants to be determined from initial conditions.

Next, find a particular solution $x_p$ for the non-homogeneous equation. The forcing term is $\frac{1}{m} \sin t$. Since the driving frequency (1 rad/s) is different from the natural frequency (2 rad/s), there is no resonance. Assume a particular solution of the form:

$x_p = C \cos t + D \sin t$

where $C$ and $D$ are constants to be determined. Compute the second derivative:

$\frac{d^2x_p}{dt^2} = -C \cos t - D \sin t$

Substitute $x_p$ and its second derivative into the differential equation:

$(-C \cos t - D \sin t) + 4(C \cos t + D \sin t) = \frac{1}{m} \sin t$

Simplify the left-hand side:

$(-C + 4C) \cos t + (-D + 4D) \sin t = 3C \cos t + 3D \sin t$

Set this equal to the right-hand side:

$3C \cos t + 3D \sin t = 0 \cdot \cos t + \frac{1}{m} \sin t$

Equate coefficients of like trigonometric functions. For $\cos t$:

$3C = 0 \implies C = 0$

For $\sin t$:

$3D = \frac{1}{m} \implies D = \frac{1}{3m}$

Thus, the particular solution is:

$x_p = \frac{1}{3m} \sin t$

The general solution is the sum of $x_h$ and $x_p$:

$x(t) = A \cos 2t + B \sin 2t + \frac{1}{3m} \sin t$

Apply the initial conditions. At $t = 0$, the oscillator is at its equilibrium position, so $x(0) = 0$:

$x(0) = A \cos 0 + B \sin 0 + \frac{1}{3m} \sin 0 = A \cdot 1 + B \cdot 0 + 0 = A$

Set equal to zero:

$A = 0$

So the solution simplifies to:

$x(t) = B \sin 2t + \frac{1}{3m} \sin t$

The oscillator is at rest at $t = 0$, so the initial velocity is zero: $\frac{dx}{dt}(0) = 0$. Compute the derivative:

$\frac{dx}{dt} = 2B \cos 2t + \frac{1}{3m} \cos t$

Evaluate at $t = 0$:

$\frac{dx}{dt}(0) = 2B \cos 0 + \frac{1}{3m} \cos 0 = 2B \cdot 1 + \frac{1}{3m} \cdot 1 = 2B + \frac{1}{3m}$

Set equal to zero:

$2B + \frac{1}{3m} = 0 \implies 2B = -\frac{1}{3m} \implies B = -\frac{1}{6m}$

Substitute $B$ back into the solution:

$x(t) = -\frac{1}{6m} \sin 2t + \frac{1}{3m} \sin t$

Factor out $\frac{1}{m}$: