JEE Main 2015 11th April Morning Shift Question Paper with Solutions

We begin by noting that the new vector $\vec B$ is obtained from the old vector $\vec A$ purely by rotation through a very small angle $\Delta\theta$ (measured in radians). Rotation does not change magnitude, so we immediately have

$|\vec A| = |\vec B|.$

Our task is to find the magnitude of the difference vector $\vec B-\vec A$. The square of this magnitude can be expressed with the dot-product formula

$|\vec B-\vec A|^{2} = (\vec B-\vec A)\cdot(\vec B-\vec A).$

Expanding the right-hand side using distributivity of the dot product, we obtain

' in math mode at position 1:̲ |\vec B-\vec A…" style="color:#cc0000">$ |\vec B-\vec A|^{2}= \vec B\cdot\vec B + \vec A\cdot\vec A - 2\,\vec A\cdot\vec B.$

Recognising that $\vec B\cdot\vec B = |\vec B|^{2}$ and $\vec A\cdot\vec A = |\vec A|^{2}$, and recalling that $\vec A\cdot\vec B = |\vec A|\,|\vec B|\cos\Delta\theta,$ we rewrite the expression as

' in math mode at position 1:̲ |\vec B-\vec A…" style="color:#cc0000">$ |\vec B-\vec A|^{2}= |\vec B|^{2}+|\vec A|^{2}-2|\vec A|\,|\vec B|\cos\Delta\theta.$

Because $|\vec A|=|\vec B|,$ let us denote this common magnitude simply by $|\vec A|$. Substituting, we have

' in math mode at position 1:̲ |\vec B-\vec A…" style="color:#cc0000">$ |\vec B-\vec A|^{2}= |\vec A|^{2}+|\vec A|^{2}-2|\vec A|^{2}\cos\Delta\theta = 2|\vec A|^{2}\bigl(1-\cos\Delta\theta\bigr).$

Now we employ the small-angle approximation for cosine. For very small $\Delta\theta$ (in radians), the Taylor series gives

' in math mode at position 1:̲ \cos\Delta\the…" style="color:#cc0000">$ \cos\Delta\theta \approx 1-\frac{(\Delta\theta)^{2}}{2}.$

Substituting this approximation into the previous expression, we find

' in math mode at position 1:̲ |\vec B-\vec A…" style="color:#cc0000">$ |\vec B-\vec A|^{2} \approx 2|\vec A|^{2}\left[1-\left(1-\frac{(\Delta\theta)^{2}}{2}\right)\right] = 2|\vec A|^{2}\left[\frac{(\Delta\theta)^{2}}{2}\right] = |\vec A|^{2}(\Delta\theta)^{2}.$

Taking the square root of both sides (and keeping only the positive root because a magnitude is always positive) gives

' in math mode at position 1:̲ |\vec B-\vec A…" style="color:#cc0000">$ |\vec B-\vec A| \approx |\vec A|\,\Delta\theta.$

This expression is exactly Option C in the list provided. None of the other options match the derived result.

Hence, the correct answer is Option C.

We are asked to find an expression for the rate of heat transfer per unit area, written as $\dfrac{Q}{A}$. Because this is a rate of energy flow through a unit surface, its physical dimension is that of power per unit area.

First let us write down the basic dimensions in the $M\;L\;T$ system:

Power $P$ has dimension $[P]=M L^2 T^{-3}$ (since $P=\dfrac{\text{energy}}{\text{time}}$ and energy is work $F\! \cdot\! d = (M L T^{-2})L$). Dividing by area $A\;(=[L^2])$ gives

$\left[\dfrac{Q}{A}\right]=M L^0 T^{-3}=M T^{-3}.$

Now we list the three quantities on which the student believes $\dfrac{Q}{A}$ depends and write their dimensions.

1. Dynamic viscosity $\eta$  Formula for its dimension: $\eta=\dfrac{\text{shear stress}}{\text{velocity gradient}}.$  Shear stress has dimension $M L^{-1} T^{-2}$, and velocity gradient has dimension $T^{-1}$,  so $[\eta]=M L^{-1} T^{-1}.$

2. The factor $\dfrac{S\Delta\theta}{h}$  Specific heat $S$ is defined by $Q=m S \Delta\theta \;.$  Re-arranging, $S=\dfrac{Q}{m\,\Delta\theta}$, hence  $[S]=\dfrac{M L^2 T^{-2}}{M\,\Theta}=L^{2} T^{-2} \Theta^{-1},$  where $\Theta$ stands for temperature.  Multiplying by the temperature difference $\Delta\theta$ removes the temperature dimension, so  $\left[S\Delta\theta\right]=L^{2} T^{-2}.$  Dividing by height $h\;(=[L])$ gives  $\left[\dfrac{S\Delta\theta}{h}\right]=L T^{-2}.$

3. The factor $\dfrac{1}{\rho g}$  Density $\rho$ has dimension $M L^{-3}$.  Acceleration due to gravity $g$ has dimension $L T^{-2}$.  Hence $\left[\dfrac{1}{\rho g}\right]=\dfrac{1}{(M L^{-3})(L T^{-2})}=M^{-1} L^{2} T^{2}.$

We now search for a product of the above three quantities, each possibly raised to some power, that reproduces $M T^{-3}$. Inspection of the options shows that only direct products have been proposed, so we can test them one by one.

Option A : $\left(\dfrac{S\Delta\theta}{h}\right)\eta$

Combining the dimensions we have

$\left[L T^{-2}\right]\left[M L^{-1} T^{-1}\right] = M\,L^{(1-1)}\,T^{(-2-1)} = M\,T^{-3}.$

This exactly matches the required $M T^{-3}$, so Option A is dimensionally consistent.

Option B : $\eta\left(\dfrac{S\Delta\theta}{h}\right)\left(\dfrac{1}{\rho g}\right)$

$\left[M L^{-1} T^{-1}\right] \left[L T^{-2}\right] \left[M^{-1} L^{2} T^{2}\right] = M^{(1-1)}\,L^{(-1+1+2)}\,T^{(-1-2+2)} = L^{2}\,T^{-1}.$

This gives $L^{2} T^{-1}$, which is not $M T^{-3}$, so Option B is incorrect.

Option C : $\left(\dfrac{S\Delta\theta}{\eta h}\right)\left(\dfrac{1}{\rho g}\right)$

First compute $\dfrac{S\Delta\theta}{\eta h}$:

$\dfrac{[S\Delta\theta]}{[\eta][h]} = \dfrac{L^{2} T^{-2}}{(M L^{-1} T^{-1})(L)} = \dfrac{L^{2} T^{-2}}{M L^{0} T^{-1}} = M^{-1} L^{2} T^{-1}.$

Multiplying this by $\left[M^{-1} L^{2} T^{2}\right]$ from $\dfrac{1}{\rho g}$ gives

$M^{-2} L^{4} T^{1},$

which again fails to match $M T^{-3}$, so Option C is incorrect.

Option D : $\dfrac{S\Delta\theta}{\eta h}$

We already found above that its dimension is $M^{-1} L^{2} T^{-1}$, clearly not $M T^{-3}$, hence Option D is also incorrect.

Among all the choices, only Option A reproduces the correct physical dimension for heat flux. Therefore the student’s dimensional reasoning leads to

$\dfrac{Q}{A}\;=\;\eta\left(\dfrac{S\Delta\theta}{h}\right).$

Hence, the correct answer is Option A.

We have to express the permeability of free space $\mu_0$ in terms of the four selected fundamental quantities: electronic charge $e$, electron mass $m$, speed of light in vacuum $c$ and Planck’s constant $h$. (The mass $m$ will finally not enter the expression, but it is one of the allowed building blocks.) Our aim is purely dimensional, so any numerical factors such as $2,\;4\pi,$ or the dimension-less fine-structure constant $\alpha$ can be ignored because they do not change the units.

First, we recall the well-known relation from Maxwell’s theory that connects the speed of light with the vacuum permittivity $\varepsilon_0$ and the vacuum permeability $\mu_0$:

$c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}.$

Rearranging this formula gives an explicit expression for $\mu_0$:

$\mu_0=\frac{1}{\varepsilon_0\,c^{2}}.$

So, once we succeed in writing $\varepsilon_0$ with the desired basic quantities, we can immediately obtain $\mu_0$.

Now, we involve the fine-structure constant $\alpha$, which by definition is dimensionless:

$\alpha=\frac{e^{2}}{4\pi\varepsilon_0\hbar c},$

where $\hbar=\dfrac{h}{2\pi}$ is the reduced Planck constant. Substituting $\hbar=\dfrac{h}{2\pi}$ in the definition of $\alpha$, we obtain

$\alpha=\frac{e^{2}}{4\pi\varepsilon_0}\;\frac{2\pi}{h\,c} =\frac{e^{2}}{2\,\varepsilon_0\,h\,c}.$

Isolating $\varepsilon_0$ from the above equation gives

$\varepsilon_0=\frac{e^{2}}{2\alpha\,h\,c}.$

Because $\alpha$ is dimensionless and the factor $2$ is a mere numerical constant, they have no bearing on dimensional analysis. Hence, for the purpose of units we may write simply

$\varepsilon_0\sim\frac{e^{2}}{h\,c}.$

Substituting this expression of $\varepsilon_0$ into the earlier formula $\mu_0=\dfrac{1}{\varepsilon_0\,c^{2}}$, we get

$\mu_0=\frac{1}{\left(\dfrac{e^{2}}{h\,c}\right)\;c^{2}} =\frac{h\,c}{e^{2}\,c^{2}} =\frac{h}{c\,e^{2}}.$

Thus, in terms of the chosen fundamental quantities, the permeability of vacuum carries the units of $\displaystyle\frac{h}{c\,e^{2}}.$ This matches Option D in the given list.

Hence, the correct answer is Option D.

We are given that the stone is projected vertically upward from the top of a tower of height $64\ \text{m}$ with an initial velocity $u = 48\ \text{m s}^{-1}$. The acceleration due to gravity acts downward with magnitude $g = 32\ \text{m s}^{-2}$. We wish to find the greatest height that the stone reaches above the ground.

First we determine how high the stone rises above the point of projection (the tower-top). At the highest point, the vertical velocity becomes zero. Using the first equation of motion, which in general form is written as

$v = u - g t,$

we set the final velocity $v = 0$ (because at the greatest height the stone momentarily comes to rest). Substituting $u = 48\ \text{m s}^{-1}$ and $g = 32\ \text{m s}^{-2}$, we get

$0 = 48 - 32 t.$

Solving for the time $t$ it takes to reach the maximum point:

$32 t = 48 \quad\Longrightarrow\quad t = \frac{48}{32} = \frac{3}{2} = 1.5\ \text{s}.$

Now we calculate the vertical displacement $s$ in this time interval. For vertical motion with constant acceleration, the displacement is obtained from the second equation of motion, stated as

$s = u t - \frac{1}{2} g t^2.$

Substituting the known values $u = 48\ \text{m s}^{-1},\ t = 1.5\ \text{s},\ g = 32\ \text{m s}^{-2},$ we have

$$\begin{aligned} s &= 48 \times 1.5 \;-\; \frac12 \times 32 \times (1.5)^2 \\ &= 72 \;-\; 16 \times 2.25 \\ &= 72 \;-\; 36 \\ &= 36\ \text{m}. \end{aligned}$$

Thus the stone rises $36\ \text{m}$ above the top of the tower.

The tower itself is $64\ \text{m}$ high, so the greatest height above ground level reached by the stone is

$\text{Maximum height} = 64\ \text{m} + 36\ \text{m} = 100\ \text{m}.$

Hence, the correct answer is Option D.

In elastic collisions between identical masses ($m$), the two bodies simply exchange velocities.

Since the beads have a finite radius $r$, they occupy space on the rod. Diameter of one bead $= 2r$. Total length occupied by $n$ beads $= 2nr$.

The actual free distance the momentum has to travel between the two supports is $L_{eff} = L - 2nr$

The time interval between two consecutive collisions with the same support is $T = \frac{2 \times L_{eff}}{v} = \frac{2(L - 2nr)}{v}$

change in momentum ($\Delta p$) during one collision is $\Delta p = mv - (-mv) = 2mv$

$F_{avg} = \frac{\Delta p}{T}$

$F_{avg} = \frac{2mv}{\frac{2(L - 2nr)}{v}}$

$F_{avg} = \frac{mv^2}{L - 2nr}$

We begin by noting that the force acting on the particle is central and is given in magnitude by $F = \alpha r^2$, always directed toward the centre of the circle. A central force is conservative, so we can define a potential energy function $U(r)$ such that $F_r = -\dfrac{dU}{dr}$, where the minus sign indicates that the force is attractive (towards the centre).

Because the particle is executing uniform circular motion of radius $r$, the required centripetal force is provided entirely by this central force. Hence we equate the magnitudes

$\dfrac{mv^2}{r} = \alpha r^2.$

Solving this relation for the square of the speed $v^2$, we multiply both sides by $r$:

$mv^2 = \alpha r^3 \quad \Rightarrow \quad v^2 = \dfrac{\alpha r^3}{m}.$

Now we write the kinetic energy. The standard formula for kinetic energy is $K = \dfrac12 m v^2$, so substituting the value of $v^2$ obtained above gives

$K = \dfrac12 m \left( \dfrac{\alpha r^3}{m} \right) = \dfrac12 \alpha r^3.$

Next we calculate the potential energy. From the definition $F_r = -\dfrac{dU}{dr}$ and using $F_r = -\alpha r^2$ (negative sign because the force is inward), we have

$-\alpha r^2 = -\dfrac{dU}{dr} \quad \Rightarrow \quad \dfrac{dU}{dr} = \alpha r^2.$

We integrate from the reference point $r = 0$, where the problem states $U = 0$, up to the general radius $r$:

$U(r) = \int_{0}^{r} \alpha s^2 \, ds = \alpha \int_{0}^{r} s^2 \, ds = \alpha \left[ \dfrac{s^3}{3} \right]_{0}^{r} = \dfrac{\alpha r^3}{3}.$

Finally, the total mechanical energy $E$ is the sum of kinetic and potential energies:

$E = K + U = \dfrac12 \alpha r^3 + \dfrac13 \alpha r^3 = \left(\dfrac{3}{6} + \dfrac{2}{6}\right)\alpha r^3 = \dfrac56 \alpha r^3.$

Hence, the correct answer is Option A.

$X_{cm} = \frac{\int_{0}^{L} x \cdot dm}{\int_{0}^{L} dm}$

$M = \int_{0}^{L} \mu(x) dx = \int_{0}^{L} \left(a + \frac{bx}{L}\right) dx$

$M = \left[ ax + \frac{bx^2}{2L} \right]_{0}^{L} = aL + \frac{bL^2}{2L}$

$M = L \left( a + \frac{b}{2} \right)$

$\int_{0}^{L} x \cdot \mu(x) dx = \int_{0}^{L} x \left( a + \frac{bx}{L} \right) dx = \int_{0}^{L} \left( ax + \frac{bx^2}{L} \right) dx$

$= \left[ \frac{ax^2}{2} + \frac{bx^3}{3L} \right]_{0}^{L} = \frac{aL^2}{2} + \frac{bL^3}{3L}$ $= L^2 \left( \frac{a}{2} + \frac{b}{3} \right)$

$X_{cm} = \frac{L^2 \left( \frac{a}{2} + \frac{b}{3} \right)}{L \left( a + \frac{b}{2} \right)} = \frac{7}{12}L$

$\frac{\frac{a}{2} + \frac{b}{3}}{a + \frac{b}{2}} = \frac{7}{12}$

$\frac{3a + 2b}{3(2a + b)} = \frac{7}{12}$ 

Solving the equation gives $b = 2a$

We first note all the data that the question supplies. The mass of the particle is $m = 2\ \text{kg}$. The particle moves in a horizontal circle of radius $r_h = 0.6\ \text{m}$ on a smooth table. The table is $h = 0.8\ \text{m}$ above the ground. Its angular speed is $\omega = 12\ \text{rad s}^{-1}$. We have to find the magnitude of the angular momentum of the particle about the point on the ground that lies exactly under the centre of the circle.

Let us choose that ground point and call it $O$. Let the centre of the circular path be the point $C$, which is vertically above $O$. Hence $OC = h = 0.8\ \text{m}$. Because the particle is on the circumference of the horizontal circle, at any instant its horizontal displacement from the centre is $r_h = 0.6\ \text{m}$. So the particle’s position vector $\vec r$, drawn from $O$ to the instantaneous position $P$ of the particle, has two perpendicular components:

$OC = 0.8\ \text{m}$ vertically upward, and $CP = 0.6\ \text{m}$ horizontally outward. Since these two components are at right angles, the magnitude of $\vec r$ is obtained using Pythagoras’ theorem:

$|\vec r| = \sqrt{(0.8)^2 + (0.6)^2} = \sqrt{0.64 + 0.36} = \sqrt{1.00} = 1.0\ \text{m}.$

Now we need the magnitude of the linear momentum $\vec p$ of the particle. The speed $v$ in uniform circular motion is related to the angular speed by the formula

$v = \omega r_h.$

Substituting $\omega = 12\ \text{rad s}^{-1}$ and $r_h = 0.6\ \text{m}$, we get

$v = 12 \times 0.6 = 7.2\ \text{m s}^{-1}.$

Therefore, the magnitude of the linear momentum is

$|\vec p| = m v = 2 \times 7.2 = 14.4\ \text{kg m s}^{-1}.$

To find the angular momentum about point $O$, we use the vector definition

$\vec L = \vec r \times \vec p.$

When the cross-product involves two perpendicular vectors, the magnitude simplifies to

$|\vec L| = |\vec r|\,|\vec p|\,\sin 90^\circ = |\vec r|\,|\vec p|,$

because $\sin 90^\circ = 1$. In our geometry the position vector $\vec r$ is perpendicular to the instantaneous velocity (and hence to $\vec p$), so the angle between them is indeed $90^\circ$. Hence

$|\vec L| = (1.0\ \text{m})(14.4\ \text{kg m s}^{-1}) = 14.4\ \text{kg m}^2 \text{ s}^{-1}.$

Hence, the correct answer is Option A.

For any point outside or on the surface, a spherical mass behaves as if all its mass $M$ is concentrated at its center.

$V_{out} = -\frac{GM}{r}$

This represents a rectangular hyperbola in the negative quadrant. As $r \rightarrow \infty, V \rightarrow 0$. At the surface ($r = R$), $V_s = -\frac{GM}{R}$.

For a uniform solid sphere, the potential at an internal point is given by:

$V_{in} = -\frac{GM}{2R^3}(3R^2 - r^2)$

This is a parabolic relation with respect to $r$. At the center ($r = 0$): $V_c = -\frac{3GM}{2R} = 1.5 V_s$. The potential at the center is $1.5$ times the potential at the surface (and more negative).

All these are correctly shown in option (c).

Let the mass of water present in the container be $m$ grams. The heater supplies heat at a perfectly uniform rate, so the amount of heat given to the water is proportional to the time for which the heater operates.

First, the water is heated from $0^\circ{\rm C}$ to $100^\circ{\rm C}$. We use the basic formula for heat absorbed during a temperature change:

$Q = mc\Delta T,$

where $Q$ is the heat absorbed, $m$ is the mass, $c$ is the specific heat capacity and $\Delta T$ is the rise in temperature. Here, $c = 1\;{\rm cal}\,( {\rm g}^\circ{\rm C} )^{-1}$ and $\Delta T = 100^\circ{\rm C} - 0^\circ{\rm C} = 100^\circ{\rm C}$. Hence the heat needed for this stage is

$Q_1 = m \times 1 \times 100 = 100m\;{\rm cal}.$

This stage takes 10 minutes. Because the heater’s power is constant, the heat supplied per minute is the same throughout the whole experiment. Denote this unknown constant rate by $P$ (cal per minute). Then, for the first stage,

$Q_1 = P \times (\text{time for stage 1}) = P \times 10.$

Equating the two expressions for $Q_1$ gives us

$P \times 10 = 100m \quad\rightarrow\quad P = \frac{100m}{10} = 10m\;{\rm cal\;min^{-1}}.$

Next, the entire mass of water at $100^\circ{\rm C}$ is converted into steam at the same temperature. The heat required for a complete change of state is the latent heat:

$Q_2 = mL,$

where $L$ is the heat of vaporization per gram, which we wish to find. The time taken for this stage is 55 minutes, so by the same constant-power reasoning,

$Q_2 = P \times (\text{time for stage 2}) = P \times 55.$

Substituting the value of $P$ obtained earlier,

$mL = (10m)\times 55.$

Now we divide both sides by $m$ to eliminate the mass (since it is the same in both stages and non-zero):

$L = 10 \times 55 = 550\;{\rm cal\,g^{-1}}.$

Hence, the measured heat of vaporization from this experiment is $550\;{\rm cal\,g^{-1}}$.

Hence, the correct answer is Option B.

According to the equipartition theorem of energy, each quadratic degree of freedom contributes an average energy of $\dfrac12 k_B T$ to a system in thermal equilibrium, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.

An atom bound inside a crystalline solid behaves like a three-dimensional harmonic oscillator. Such an oscillator possesses

• three kinetic energy terms (one for motion along each axis), and
• three potential energy terms (one for the restoring force along each axis).

Thus the number of quadratic terms per atom is

$6 = 3 \text{(kinetic)} + 3 \text{(potential)}.$

Using equipartition, the average energy of one atom is therefore

$\bigl(6\bigr)\left(\dfrac12 k_B T\right)=3 k_B T.$

For one mole we have Avogadro’s number $N_A$ of atoms, so the molar internal energy becomes

$U_{\text{molar}} = N_A \left(3 k_B T\right)=3 N_A k_B T.$

But the universal gas constant is defined as $R = N_A k_B,$ so we can write

$U_{\text{molar}} = 3 R T.$

The molar heat capacity at constant volume is the temperature derivative of this energy:

$C_V^{\text{molar}} = \dfrac{dU_{\text{molar}}}{dT}=3R.$

At high temperatures the difference between $C_P$ and $C_V$ for a solid is negligible, so we may take the molar specific heat simply as

$C^{\text{molar}} = 3R.$

The numerical value of the gas constant is

$R = 8.314\ \text{J mol}^{-1}\ \text{K}^{-1}.$

Hence

$C^{\text{molar}} = 3 \times 8.314</p> <p> = 24.942\ \text{J mol}^{-1}\ \text{K}^{-1}.$

To convert this molar specific heat to a mass specific heat (J kg$^{-1}$ K$^{-1}$) we divide by the molar mass (atomic weight expressed in kg per mole). Aluminium has an atomic weight of 27 g mol$^{-1}$, which is

$M = 27\ \text{g mol}^{-1} = 0.027\ \text{kg mol}^{-1}.$

Therefore the specific heat per kilogram is

$</p> <p>C^{\text{mass}}</p> <p> = \dfrac{C^{\text{molar}}}{M}</p> <p> = \dfrac{24.942\ \text{J mol}^{-1}\ \text{K}^{-1}}</p> <p> {0.027\ \text{kg mol}^{-1}}</p> <p> = 924.9\ \text{J kg}^{-1}\ \text{K}^{-1}.</p> <p>$

Rounding to a convenient integer gives approximately $925\ \text{J kg}^{-1}\ \text{K}^{-1}.$

Hence, the correct answer is Option D.

We know that for a damped oscillator the displacement (or amplitude) decreases exponentially with time. Mathematically, if the initial amplitude is $A_0$ then at time $t$ we have

$A(t)=A_0 e^{-\beta t}$

where $\beta$ is the damping constant (its SI unit is $s^{-1}$). The total mechanical energy of the oscillator is proportional to the square of its amplitude, so

$E(t)\propto [A(t)]^{2}$

Substituting the expression for $A(t)$, the energy at time $t$ becomes

$E(t)=E_0 e^{-2\beta t}$

Here $E_0$ is the energy at $t=0$. Let us now insert the numerical data given in the problem.

At $t=0$ the energy is $E_0=45\text{ J}$. After the pendulum completes 15 oscillations its energy has dropped to $E=15\text{ J}$.

The time taken for one oscillation is the time period $T=1\text{ s}$, so the time needed to complete 15 oscillations is

$t = 15T = 15 \times 1\text{ s} = 15\text{ s}.$

Now we substitute $E=15\text{ J}$, $E_0=45\text{ J}$, and $t=15\text{ s}$ into the energy-decay equation $E = E_0 e^{-2\beta t}$:

$15 = 45\, e^{-2\beta (15)}.$

First divide both sides by 45 to isolate the exponential term:

$\frac{15}{45} = e^{-30\beta}.$

Simplifying the fraction on the left gives

$\frac{1}{3} = e^{-30\beta}.$

To solve for $\beta$ take the natural logarithm (base $e$) of both sides. Recall the logarithmic identity $\ln(e^{x}) = x$.

$\ln\!\left(\frac{1}{3}\right) = \ln\!\left(e^{-30\beta}\right) = -30\beta.$

Hence

$\beta = -\,\frac{1}{30}\,\ln\!\left(\frac{1}{3}\right).$

We can simplify further because $\ln\!\left(\frac{1}{3}\right) = -\ln 3$. Substituting this identity gives

$\beta = -\frac{1}{30}\,(-\ln 3) = \frac{\ln 3}{30}.$

Thus the damping constant is

$\beta = \frac{1}{30}\,\ln 3\;\; \text{s}^{-1}.$

This matches Option C.

Hence, the correct answer is Option C.

The block is cylindrical, so its volume is equal to the base area multiplied by the height. We first convert all the given quantities to SI units so that we can use them directly in the standard formulas.

The base area is given as $30\ \text{cm}^2$. Since $1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$, we have $A = 30 \times 10^{-4}\ \text{m}^2 = 3.0 \times 10^{-3}\ \text{m}^2.$

The height is given as $54\ \text{cm}$. Using $1\ \text{cm} = 10^{-2}\ \text{m}$, we write $h = 54 \times 10^{-2}\ \text{m} = 0.54\ \text{m}.$

Now the volume of the cylinder is $V = A \times h = (3.0 \times 10^{-3})\ \text{m}^2 \times 0.54\ \text{m} = 1.62 \times 10^{-3}\ \text{m}^3.$

The density of the wooden block is $\rho_b = 650\ \text{kg m}^{-3}$, so its mass is $m = \rho_b V = 650\ \text{kg m}^{-3} \times 1.62 \times 10^{-3}\ \text{m}^3 = 1.053\ \text{kg}.$

The liquid in which the block floats has density $\rho_l = 900\ \text{kg m}^{-3}$. When the block is pushed down by a small vertical displacement $x$ and released, the extra volume of liquid displaced is $A x$, and therefore the additional buoyant force is

$F_b = \rho_l g (A x).$

This force acts upward and tends to restore the block to equilibrium. Using Newton’s second law, the equation of motion for the block becomes

$m \frac{d^2x}{dt^2} + \rho_l g A x = 0.$

This is the standard form of a simple harmonic motion equation, $\frac{d^2x}{dt^2} + \omega^2 x = 0,$ where the angular frequency $\omega$ is given by

$\omega^2 = \frac{\rho_l g A}{m}.$

Hence $\omega = \sqrt{\frac{\rho_l g A}{m}}.$

The time period of oscillation is related to the angular frequency by the formula $T = \frac{2\pi}{\omega}.$ Substituting $\omega$ from above we get

$T = 2\pi \sqrt{\frac{m}{\rho_l g A}}.$

Now we substitute the numerical values: $m = 1.053\ \text{kg},\quad \rho_l = 900\ \text{kg m}^{-3},\quad g = 9.8\ \text{m s}^{-2},\quad A = 3.0 \times 10^{-3}\ \text{m}^2.$

First calculate the denominator inside the square root: $\rho_l g A = 900 \times 9.8 \times 3.0 \times 10^{-3} = 8820 \times 3.0 \times 10^{-3} = 26.46.$

Next compute the ratio inside the square root: $\frac{m}{\rho_l g A} = \frac{1.053}{26.46} \approx 0.0398.$

Taking the square root gives $\sqrt{0.0398} \approx 0.1995.$

Therefore the time period is $T = 2\pi \times 0.1995 \approx 6.283 \times 0.1995 \approx 1.25\ \text{s}.$

We are asked to find the length $L$ of a simple pendulum that has the same time period. For a simple pendulum, the time period is given by the well-known formula

$T = 2\pi \sqrt{\frac{L}{g}}.$

Equating the two expressions for the period, we write $1.25 = 2\pi \sqrt{\frac{L}{g}}.$

First isolate the square root term: $\sqrt{\frac{L}{g}} = \frac{1.25}{2\pi} = \frac{1.25}{6.283} \approx 0.199.$

Now square both sides: $\frac{L}{g} = (0.199)^2 \approx 0.0396.$

Finally multiply by $g = 9.8\ \text{m s}^{-2}$ to obtain $L$: $L = 0.0396 \times 9.8 \approx 0.388\ \text{m}.$

Converting metres to centimetres, $L = 0.388\ \text{m} \times 100\ \text{cm m}^{-1} \approx 38.8\ \text{cm}.$

The value is very close to $39\ \text{cm}$, which matches option C.

Hence, the correct answer is Option C.

For a source and observer moving toward each other, the observed frequency $f$ is given by:

$f = f_0 \left( \frac{v + v_0}{v - v_s} \right)$

To find the correct graph, we rewrite this expression in the linear form $y = mx + c$, where $y = f$ and $x = v_0$:

$f = \left( \frac{f_0}{v - v_s} \right) v_0 + \frac{f_0 v}{v - v_s}$

Because $f$ is a linear function of $v_0$, the graph must be a straight line with a positive intercept. This corresponds to Graph A.

$\text{Slope} = \frac{f_0}{v - v_s}$

$L = \pi R \implies R = \frac{L}{\pi}$

The wire is divided into two equal halves (quadrants), each having a length $l = L/2$. The magnitude of the charge on each half is given as $|Q| = 10^3 \epsilon_0$.

$\lambda = \frac{Q}{L/2} = \frac{2Q}{L}$

By symmetry, the y-components of the two fields are equal in magnitude but opposite in direction, so they cancel out. The x-components are in the same direction ($+\hat{i}$) and add up.

$E_x = E_y = \frac{k\lambda}{R} = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R}$

$\vec{E}_{net} = (E_{1x} + E_{2x}) \hat{i} = 2 \left( \frac{\lambda}{4\pi\epsilon_0 R} \right) \hat{i} = \frac{\lambda}{2\pi\epsilon_0 R} \hat{i}$