JEE Main 2025 April 4 shift 1

The mean free time $\tau$ between two successive collisions is obtained from the relation $\tau = \frac{\lambda}{\bar{c}}$ where $\lambda$ is the mean free path and $\bar{c}$ is the average (root-mean-square) speed of the molecules. The collision frequency $f$, i.e. the number of collisions per second, is the reciprocal of the mean free time: $f = \frac{1}{\tau} = \frac{\bar{c}}{\lambda}$ Substituting the given values $\lambda = 3 \times 10^{-7}\,\text{m}$ and $\bar{c} = 600\,\text{m s}^{-1}$: $f = \frac{600}{3 \times 10^{-7}}\;\text{s}^{-1}$ $f = \frac{600}{3}\times 10^{7}\;\text{s}^{-1}$ $f = 200 \times 10^{7}\;\text{s}^{-1}$ $f = 2 \times 10^{9}\;\text{s}^{-1}$ Therefore, the collision frequency of oxygen molecules at 300 K and 1 atm is $2 \times 10^{9}\,\text{s}^{-1}$, which matches Option C.

The momentum change of the laser pulse upon reflection is $\Delta p = \frac{2E}{c}$.
By conservation of momentum, the mirror acquires an initial velocity $v = \frac{\Delta p}{m} = \frac{2E}{mc}$.
Applying energy conservation, the kinetic energy converts to potential energy at maximum deflection $\theta$: $\frac{1}{2}mv^2 = mgl(1-\cos\theta)$.
For small $\theta$, $1-\cos\theta \approx \frac{\theta^2}{2}$, so $\frac{1}{2}mv^2 = \frac{mgl\theta^2}{2}$, which simplifies to $v = \theta\sqrt{gl}$.
Equating the expressions for $v$, we get $\theta\sqrt{gl} = \frac{2E}{mc}$.
Solving for $\theta$ gives $\theta = \dfrac{2E}{mc\sqrt{gl}}$.

The shape of the liquid surface in a capillary tube depends upon the magnitude of the contact angle $\theta$ with the glass wall. \bullet If $\theta \lt 90^{\circ}$, the liquid wets glass. The meniscus is concave and $\cos\theta$ is positive. \bullet If $\theta = 90^{\circ}$, the liquid does not rise or fall. The meniscus is flat and $\cos\theta = 0$. \bullet If $\theta \gt 90^{\circ}$, the liquid does not wet glass. The meniscus is convex and $\cos\theta$ is negative. For the two liquids A and B we are given $K = \dfrac{\cos\theta_A}{\cos\theta_B}$. Case 1: $K \gt 0$ Both $\cos\theta_A$ and $\cos\theta_B$ have the same sign, so both menisci are of the same type (either both concave or both convex). Case 2: $K = 0$ This is possible only when $\cos\theta_A = 0$. Hence $\theta_A = 90^{\circ}$ (flat meniscus for liquid A) while liquid B can have any meniscus depending on $\theta_B$. Case 3: $K \lt 0$ The ratio is negative, so $\cos\theta_A$ and $\cos\theta_B$ have opposite signs. Exactly one liquid has $\theta \lt 90^{\circ}$ (concave) and the other has $\theta \gt 90^{\circ}$ (convex). Let $\cos\theta_A$ be positive and $\cos\theta_B$ be negative (the reverse choice still makes $K$ negative, but only one option \in the list matches a concave A and convex B). Hence \bullet liquid A: $\theta_A \lt 90^{\circ}$ \Rightarrow concave meniscus \bullet liquid B: $\theta_B \gt 90^{\circ}$ ⇒ convex meniscus This situation is described by Option C (statement 3): "K is negative, then liquid A has concave meniscus and liquid B has convex meniscus." Therefore, the correct statement is Option C (3).

Torque (moment of the force) about the chosen origin is defined as $\vec{\tau} = \vec{r} \times \vec{F} \quad -(1)$ where $\vec{r}$ is the position vector of the point of application of the force $\vec{F}$. Therefore expression E ( $\vec{\tau} = \vec{r} \times \vec{F}$ ) is directly correct. Using Newton's second law $\vec{F} = \dfrac{d\vec{p}}{dt}$ and substituting \in $(1)$ gives $\vec{\tau} = \vec{r} \times \dfrac{d\vec{p}}{dt} \quad -(2)$ so expression C is also correct. The orbital angular momentum of the particle is defined as $\vec{L} = \vec{r} \times \vec{p} \quad -(3)$ Differentiating $(3)$ with respect to time (for a fixed origin) yields $\dfrac{d\vec{L}}{dt} = \dfrac{d}{dt}(\vec{r} \times \vec{p})$ By the rotational analogue of Newton's second law, $\dfrac{d\vec{L}}{dt} = \vec{\tau}$, hence $\vec{\tau} = \dfrac{d}{dt}(\vec{r} \times \vec{p}) \quad -(4)$ Thus expression B is correct. For a rigid body of constant moment of inertia $I$ about a fixed axis, the relation $\vec{\tau}_{\text{net}} = I \vec{\alpha} \quad -(5)$ holds, where $\vec{\alpha}$ is the angular acceleration. Hence expression D is correct under these usual JEE assumptions. Check expression A: $\vec{r} \times \vec{L}$ has dimensions $\bigl( \text{m}\bigr)\times \bigl(\text{kg m}^2\!/\text{s}\bigr)=\text{kg m}^3\!/\text{s}$, whereas torque has dimensions $\text{kg m}^2\!/\text{s}^2$. The two are not equal, and there is no general mechanical principle equating them. Hence expression A is incorrect. Therefore the correct expressions are B, C, D and E. Option C is the correct choice.

In Young's double-slit experiment, the fringe width $\beta$ is given by the standard formula $\beta = \frac{\lambda D}{d}$ where $\lambda$ = wavelength of light, $D$ = distance between the slits and the screen, $d$ = separation between the two slits. Case 1: Initial slit separation $d_1 = 0.2\ \text{mm}$. Fringe width $\beta_1 = \frac{\lambda D}{d_1}$ $-(1)$ Case 2: New slit separation $d_2 = 0.4\ \text{mm} = 2d_1$. Fringe width $\beta_2 = \frac{\lambda D}{d_2} = \frac{\lambda D}{2d_1} = \frac{1}{2}\left(\frac{\lambda D}{d_1}\right)$ Using $-(1)$, $\beta_2 = \frac{1}{2}\beta_1$ $-(2)$ The percentage change \in fringe width is $\frac{\beta_2 - \beta_1}{\beta_1}\times 100\% = \frac{\frac{1}{2}\beta_1 - \beta_1}{\beta_1}\times 100\%$ $= \left(-\frac{1}{2}\right)\times 100\% = -50\%$ The negative sign shows a decrease. Hence, the fringe width decreases by $50\%$. Therefore, the percentage change (magnitude) is $50\%$ ⇒ Option C.

The instantaneous current of a sinusoidal a.c. source is generally written as$i = I_0 \sin(\omega t)$where \bullet $I_0$ is the peak (maximum) current. \bullet $\omega$ is the angular frequency related to the ordinary frequency $f$ by$\omega = 2\pi f$. Comparing the given expression$i = 100\sqrt{2}\,\sin(100\pi t)$with$i = I_0 \sin(\omega t)$we identify $I_0 = 100\sqrt{2}\quad\text{and}\quad\omega = 100\pi\ \text{rad s}^{-1}$. Step 1: RMS (root-mean-square) value For a pure sine wave, the RMS current is related to the peak current by$I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$Applying this formula: $I_{\text{rms}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100\ \text{A}$. Step 2: Frequency Using the relation$f = \frac{\omega}{2\pi}$substitute $\omega = 100\pi$: $f = \frac{100\pi}{2\pi} = 50\ \text{Hz}$. Result RMS current $= 100\ \text{A}$ and frequency $= 50\ \text{Hz}$. These match Option C.

The speed of sound in an ideal gas is defined by the formula $v = \sqrt{\gamma \frac{p}{\rho}}$, where $\gamma$ is the adiabatic index and $\frac{p}{\rho}$ is constant.
Given $\frac{p}{\rho}$ is the same for all gases, the ratio of speeds is $v_{He} : v_{CH_4} : v_{CO_2} = \sqrt{\gamma_{He}} : \sqrt{\gamma_{CH_4}} : \sqrt{\gamma_{CO_2}}$.
For Helium (monatomic), the degrees of freedom $f=3$, so $\gamma_{He} = 1 + \frac{2}{f} = \frac{5}{3}$.
For polyatomic molecules like $CH_4$ and $CO_2$, the degrees of freedom are $f=6$, yielding $\gamma = 1 + \frac{2}{6} = \frac{4}{3}$.
Substituting these values, we get $v_{He} : v_{CH_4} : v_{CO_2} = \sqrt{\frac{5}{3}} : \sqrt{\frac{4}{3}} : \sqrt{\frac{4}{3}}$.

The required physical quantity is the ratio of electric flux $\Phi_E$ to magnetic flux $\Phi_B$. We must find the dimensions of each flux, then divide them. Electric flux Electric field $E$ is force per unit charge. Force has dimensions $MLT^{-2}$, and charge has dimensions $AT$. Therefore, $E : \; \frac{MLT^{-2}}{AT}=MLT^{-3}A^{-1}$ Electric flux is $\Phi_E = E \times \text{area}$, and area has dimension $L^{2}$. Thus $\Phi_E : \; (MLT^{-3}A^{-1})(L^{2}) = ML^{3}T^{-3}A^{-1}$ Magnetic flux Magnetic induction $B$ is obtained from $F = qvB$, giving $B = F/(qv)$. $B : \; \frac{MLT^{-2}}{(AT)(LT^{-1})}=MT^{-2}A^{-1}$ Magnetic flux is $\Phi_B = B \times \text{area}$, so $\Phi_B : \; (MT^{-2}A^{-1})(L^{2}) = ML^{2}T^{-2}A^{-1}$ Ratio of the two fluxes $\frac{\Phi_E}{\Phi_B} : \; \frac{ML^{3}T^{-3}A^{-1}}{ML^{2}T^{-2}A^{-1}} = L^{3-2}\,T^{-3-(-2)}\,A^{-1-(-1)} = L^{1}T^{-1}$ Hence the dimensional formula is $M^{0}L^{1}T^{-1}A^{0}$, giving $P = 0,\; Q = 1,\; R = -1,\; S = 0$ Therefore, $Q = 1$ and $R = -1$, which corresponds to Option D.

The mirror formula and the lateral magnification formula for a spherical mirror are $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\qquad -(1)$ $m=-\frac{v}{u}\qquad\qquad\!\!\!\! -(2)$ The object is kept \in front of the mirror, so by the Cartesian sign convention \bullet object distance $u$ is always negative, \bullet for a convex mirror the focal length $f$ is positive and the image distance $v$ is positive (virtual, behind the mirror). Since the given magnification $\dfrac{1}{2}$ is positive (erect image), the mirror must be convex. Given data for the first position: $u_1=-40\text{ cm},\qquad m_1=\frac{1}{2}$ From $-(2)$ : $-\frac{v_1}{u_1}=m_1\Longrightarrow v_1=-m_1u_1=-\frac{1}{2}\times (-40)=+20\text{ cm}$ Substituting $u_1$ and $v_1$ \in $-(1)$ to find the focal length: $\frac{1}{f}=\frac{1}{20}+\frac{1}{-40}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}$ $\Rightarrow\; f=+40\text{ cm}$ For the new position the required magnification is $m_2=\frac{1}{3}\quad(\text{still positive, erect image})$ Using $-(2)$ again: $-\frac{v_2}{u_2}=m_2\Longrightarrow v_2=-m_2u_2=-\frac{1}{3}u_2$ $-(3)$ Insert $v_2$ from $-(3)$ into the mirror formula $-(1)$: $\frac{1}{f}=\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{-\dfrac{u_2}{3}}+\frac{1}{u_2}=-\frac{3}{u_2}+\frac{1}{u_2}=-\frac{2}{u_2}$ With $f=+40\text{ cm}$, we get $\frac{1}{40}=-\frac{2}{u_2}\;\;\Longrightarrow\;\;u_2=-80\text{ cm}$ Comparison of object positions: Initial distance $|u_1|=40\text{ cm}$ Final distance $|u_2|=80\text{ cm}$ The object must therefore be shifted farther from the mirror by $|u_2|-|u_1|=80-40=40\text{ cm}$ Case 1: Moving the object 40 cm away from the mirror produces the required magnification $\dfrac{1}{3}$. Hence, the correct option is Option A (40 cm away from the mirror).

The photoelectric equation is $eV_0 = h\nu - \phi$, where $V_0$ is the stopping potential, $\nu$ is the frequency of incident light and $\phi$ is the work function of the metal. From the above relation we see that $V_0$ depends only on the frequency $\nu$ (and the work function $\phi$), not on the intensity of light. Therefore, even if the intensity is increased, the stopping potential remains unchanged. Hence Assertion A is false. Intensity of light represents the number of photons incident per unit time. When the frequency $\nu$ is kept greater than the threshold frequency $\nu_0$ so that each photon has energy $h\nu \gt \phi$, an increase in the number of photons results in a proportional increase in the number of emitted photoelectrons. Thus the emission rate (photoelectric current for a given potential) increases. Reason R is therefore true. We conclude: Assertion A is false but Reason R is true. This corresponds to Option B.

The velocity of the particle is $\vec{v} = \frac{d\vec{r}}{dt} = a\omega(-\sin\omega t \hat{i} + \cos\omega t \hat{j})$.
The acceleration is $\vec{a} = \frac{d\vec{v}}{dt} = -a\omega^2(\cos\omega t \hat{i} + \sin\omega t \hat{j})$.
Substituting the given expression for $\vec{r}$, we obtain $\vec{a} = -\omega^2\vec{r}$.
By Newton's second law, the force is $\vec{F} = m\vec{a} = -m\omega^2\vec{r}$.
The negative sign indicates that the force $\vec{F}$ is directed opposite to the position vector $\vec{r}$.

Given: $T_1=\sqrt{\ 3}T_2$ Horizontal equilibrium $T_1\cos\theta_1=T_2\cos\theta_2$ Substitute: $\sqrt{3}T_2\cos\theta_1=T_2\cos\theta_2$​ Cancel $T_2$​: $\sqrt{3}\cos\theta_1=\cos\theta_2$ From the options We know: $\cos60^{\circ}=\frac{1}{2}$ $\cos30^{\circ}=\frac{\sqrt{3}}{2}$ Try: $\theta_1=60^{\circ\ }$ Then: $\sqrt{3}\cos60^{\circ}=\sqrt{3}\cdot\frac{1}{2}=\frac{\sqrt{3}}{2}$​​ So: $\cos\ \theta\ _2=\frac{\sqrt{\ 3}}{2}$ $\theta\ _2=30^{\circ\ }$ $T_1\sin\theta_1+T_2\sin\theta_2=mg$ Substitute: $\sqrt{3}T_2\cdot\frac{\sqrt{3}}{2}+T_2\cdot\frac{1}{2}=mg$ $\frac{3}{2}T_2+\frac{1}{2}T_2=mg\Rightarrow2T_2=mg$ $T_2 = \dfrac{mg}{2}$

The instantaneous current is given as $I(t) = 0.02\,t + 0.01$ ampere. Charge $Q$ transported \in a time interval is obtained from the definition $Q = \int_{t_1}^{t_2} I(t)\,dt$. Here $t_1 = 1\ \text{s}$ and $t_2 = 2\ \text{s}$, so we need $Q = \int_{1}^{2} \left(0.02\,t + 0.01\right) dt$. Integrate term by term: $\int 0.02\,t\,dt = 0.02 \times \frac{t^2}{2} = 0.01\,t^2$ $\int 0.01\,dt = 0.01\,t$. Thus the antiderivative is $0.01\,t^2 + 0.01\,t$. Evaluate between the limits: At $t = 2\ \text{s}$: $0.01 \times (2)^2 + 0.01 \times 2 = 0.04 + 0.02 = 0.06\ \text{C}$. At $t = 1\ \text{s}$: $0.01 \times (1)^2 + 0.01 \times 1 = 0.01 + 0.01 = 0.02\ \text{C}$. Charge that flows \in the interval is the difference: $Q = 0.06 - 0.02 = 0.04\ \text{C}$. The charge transported from $t = 1\ \text{s}$ to $t = 2\ \text{s}$ is $0.04\ \text{C}$. Hence, the correct option is Option D .

For a body of mass $m$ projected from the earth's surface (radius $R$) let the minimum speed required to reach infinity with zero speed be $v_e$ (escape velocity). Gravitational potential energy at the earth's surface is $U_s = -\dfrac{GMm}{R}$, while at infinity it is $U_\infty = 0$ (this is the highest possible value of gravitational potential energy). Using conservation of mechanical energy, the total energy at the surface must equal the total energy at infinity: $\dfrac{1}{2} m v_e^2 + \left(-\dfrac{GMm}{R}\right) = 0 \quad -(1)$ Solving $(1)$ gives the escape velocity: $v_e = \sqrt{\dfrac{2GM}{R}}$ The kinetic energy that must be supplied is therefore $K = \dfrac{1}{2} m v_e^2 = \dfrac{1}{2} m \left(\dfrac{2GM}{R}\right) = \dfrac{GMm}{R}$ Using $g = \dfrac{GM}{R^2}$, we rewrite this as $K = m g R$ Assertion A claims the required kinetic energy is $\dfrac{1}{2} m g R$, which is not equal to the correct value $m g R$. Hence Assertion A is false. Reason R states that "the maximum potential energy of a body is zero when it is projected to infinity from the earth's surface." Because gravitational potential is defined to be zero at infinity and is negative everywhere else, this statement is true. Therefore, A is false but R is true. The correct option is Option A.

Simplify the expression: $Y = \overline{A}BC + \overline{A}C = \overline{A}C(B+1) = \overline{A}C$.
To realize $Y = \overline{A}BC + \overline{A}C$ directly, one needs to perform:

1. Inversion of $A$ (1 NOT gate).
2. ANDing for terms $\overline{A}BC$ (3-input AND) and $\overline{A}C$ (2-input AND).
3. ORing the results (2-input OR).
   Configuration A provides these specific gates.
   Configuration B uses a NOR gate, which is functionally complete and can emulate the necessary AND/OR logic via De Morgan's theorem, making it a valid configuration.
   Configuration C is invalid as it introduces a 3-input OR gate not required by the product-of-sums structure of the original expression.

Let the original length of the air column in the closed organ pipe be $L$. The pipe has uniform cross-section, so filling it with water to $\frac{1}{5}$ of its volume means water occupies $\frac{1}{5}$ of its length as well. Therefore, the new length of the air column is $L' = L - \frac{L}{5} = \frac{4L}{5}$ For a closed organ pipe, the fundamental (first harmonic) frequency is given by the relation $f = \frac{v}{4L}$ where $v$ is the speed of sound \in air. After adding water, the new fundamental frequency becomes $f' = \frac{v}{4L'} = \frac{v}{4\left(\frac{4L}{5}\right)} = \frac{5v}{16L}$ To compare the two frequencies, form the ratio $\frac{f'}{f} = \frac{\frac{5v}{16L}}{\frac{v}{4L}} = \frac{5v}{16L} \times \frac{4L}{v} = \frac{5}{4} = 1.25$ This shows the new frequency is $1.25$ \times the original. The percentage change is $\left(1.25 - 1\right) \times 100\% = 0.25 \times 100\% = 25\%$ Hence, the fundamental frequency increases by $25\%$. Correct option: Option A

For a simple pendulum executing small-angle oscillations, the restoring torque gives the angular equation of motion $\alpha = -\frac{g}{l}\,\theta$ where $\alpha$ is the angular acceleration, $g$ is the acceleration due to gravity and $l$ is the length of the pendulum. Case 1: First pendulum Length $= l_1$, angular displacement $= \theta_1$ Therefore $\alpha_1 = -\frac{g}{l_1}\,\theta_1$ $-(1)$ Case 2: Second pendulum Length $= l_2$, angular displacement $= \theta_2$ Therefore $\alpha_2 = -\frac{g}{l_2}\,\theta_2$ $-(2)$ The question states that both pendulums have the same angular acceleration, so $\alpha_1 = \alpha_2$ $-(3)$ Substituting $(1)$ and $(2)$ into $(3)$: $-\frac{g}{l_1}\,\theta_1 = -\frac{g}{l_2}\,\theta_2$ The factor $-g$ is common on both sides and cancels out, giving $\frac{\theta_1}{l_1} = \frac{\theta_2}{l_2}$ Cross-multiplying, we obtain the required relation $\theta_1\,l_2 = \theta_2\,l_1$ This matches Option D. Answer : Option D

Field due to one infinite positively charged sheet is $E=\frac{\sigma}{2\varepsilon_0}$ directed away from the sheet. There are two identical positive sheets. At points A and B, which lie between the sheets: Left sheet produces field to the right. Right sheet produces field to the left. Magnitudes are equal, so they cancel. So between the sheets only the charged sphere contributes. Outside a uniformly charged sphere, $E\propto\frac{1}{r^2}$ Point A is closer to the sphere than BBB, hence $E_A>E_B$ Now point C: It lies outside both sheets. Both sheet fields point left and add: $E_{\text{sheets}}=\frac{\sigma}{\varepsilon_0}$ Sphere field at C also points left, so it adds. Thus $E_C=\frac{\sigma}{\varepsilon_0}+E_{\text{sphere at C}}$ At point D: Again sheet contribution is $\frac{\sigma}{\varepsilon_0}$ Sphere field also adds. But D is closer to the sphere than C, so $E_{\text{sphere at D}}>E_{\text{sphere at C}}$ Therefore $E_D>E_C$

Electric field due to infinite non-conducting sheet is $E=\frac{\sigma}{2\varepsilon_0}$ Given $\sigma=100\mu C/m^2=10^{-4}C/m^2$ So $E=\frac{10^{-4}}{2(8.85\times10^{-12})}$ $=5.65\times10^6\text{ N/C}$ Force on each charge: F=qEF with $q=2\times10^{-6}C$ So $F=(2\times10^{-6})(5.65\times10^6)$ =11.3N The two equal and opposite charges form an electric dipole. Dipole moment p=qd where d=20cm=0.2m So $p=(2\times10^{-6})(0.2)=4\times10^{-7}$ Torque on dipole \in electric field: $\tau=pE\sin\theta$ with $\theta=30^{\circ}$ Thus $\tau=(4\times10^{-7})(5.65\times10^6)\cdot\frac{1}{2}​$ $=1.13Nm$

For a hydrogen-like ion, Bohr's radius of the electron in the $n^{\text{th}}$ orbit is given by $r_n = \frac{n^{2}a_0}{Z}$ where $a_0$ is the Bohr radius of the first orbit of the hydrogen atom and $Z$ is the atomic number of the nucleus. Both $Li^{2+}$ and $He^{+}$ have a single electron, so the same formula applies. Case 1: $Li^{2+}$ For lithium, $Z_{Li}=3$ and for the $5^{\text{th}}$ orbit, $n=5$. $r_{Li} = \frac{(5)^2\,a_0}{3}= \frac{25a_0}{3}$ Case 2: $He^{+}$ For helium, $Z_{He}=2$ and again $n=5$. $r_{He} = \frac{(5)^2\,a_0}{2}= \frac{25a_0}{2}$ Now compute the required ratio: $\frac{r_{Li}}{r_{He}} = \frac{\dfrac{25a_0}{3}}{\dfrac{25a_0}{2}} = \frac{2}{3}$ Hence, the ratio of the radii is $\dfrac{2}{3}$. Option D is correct.

For any rigid body rolling without slipping down an inclined plane, conservation of mechanical energy gives $mgh \;=\;\frac12\,m v^{2}\;+\;\frac12\,I\omega^{2}$ Since the body rolls without slipping, $\omega=\dfrac{v}{R}$. Thus $mgh \;=\;\frac12\,m v^{2}\;+\;\frac12\,I\dfrac{v^{2}}{R^{2}} =\frac12\,m v^{2}\Bigl(1+\frac{I}{mR^{2}}\Bigr)$ Therefore the speed attained at the bottom is $v=\sqrt{\frac{2gh}{\,1+\dfrac{I}{mR^{2}}\,}}$ Case 1 : Circular ring For a thin ring about its symmetry axis, $I_{ring}=mR^{2}$, so $\dfrac{I}{mR^{2}}=1$. $v_{ring}=\sqrt{\frac{2gh}{1+1}} =\sqrt{\frac{2gh}{2}} =\sqrt{gh}$ Case 2 : Solid sphere For a solid sphere about a diameter, $I_{sphere}=\tfrac{3}{5}mR^{2}$, hence $\dfrac{I}{mR^{2}}=\tfrac{3}{5}$. $v_{sphere}=\sqrt{\frac{2gh}{1+\tfrac{3}{5}}} =\sqrt{\frac{2gh}{\tfrac{8}{5}}} =\sqrt{\frac{5gh}{4}}$ Required ratio: $\frac{v_{ring}}{v_{sphere}} =\sqrt{\frac{gh}{\tfrac{5gh}{4}}} =\sqrt{\frac{4}{5}} =\sqrt{\frac{x}{5}}$ Hence $x=4$. Thus, the ratio of their velocities at the bottom of the plane is $\sqrt{\dfrac{4}{5}}$, giving $x = 4$.

For a slab under a tangential (shearing) force $F$, \bullet the shearing (tangential) stress is $\tau = \dfrac{F}{A}$, where $A$ is the area of the face on which the force acts. \bullet the shearing strain is $\gamma = \tan\theta \approx \theta$ for small angles. \bullet shear modulus (rigidity modulus) is defined by $G = \dfrac{\text{shear stress}}{\text{shear strain}} = \dfrac{\tau}{\gamma}.$ Both slabs have square cross-section of side $l$ and thickness $d$. The force $F$ is applied on a narrow face of each slab; that face has dimensions $l \times d$, so its area is $A = l\,d$. Case 1: Material 1 Thickness $d_1$, narrow-face area $A_1 = l\,d_1$. Shear stress $\tau_1 = \dfrac{F}{A_1} = \dfrac{F}{l\,d_1}$. Let the small angle of deformation be $\theta_1$, so shear strain $\gamma_1 = \theta_1$. Hence $G_1 = \dfrac{\tau_1}{\gamma_1} = \dfrac{F}{l\,d_1\,\theta_1} \quad -(1)$. Case 2: Material 2 Thickness $d_2 = 2d_1$, narrow-face area $A_2 = l\,d_2 = l\,(2d_1) = 2l\,d_1$. Shear stress $\tau_2 = \dfrac{F}{A_2} = \dfrac{F}{2l\,d_1} = \dfrac{\tau_1}{2}$. Given angle $\theta_2 = 2\theta_1$, so shear strain $\gamma_2 = \theta_2 = 2\theta_1$. Therefore $G_2 = \dfrac{\tau_2}{\gamma_2} = \dfrac{\tau_1/2}{2\theta_1} = \dfrac{\tau_1}{4\theta_1} \quad -(2)$. Divide $(2)$ by $(1)$: $\dfrac{G_2}{G_1} = \dfrac{\tau_1/(4\theta_1)}{F/(l\,d_1\,\theta_1)} = \dfrac{F}{4l\,d_1\,\theta_1}\times \dfrac{l\,d_1\,\theta_1}{F} = \dfrac{1}{4}.$ Hence $G_2 = \dfrac{G_1}{4}$. Given $G_1 = 4 \times 10^9 \text{ N m}^{-2}$, $G_2 = \dfrac{4 \times 10^9}{4} = 1 \times 10^9 \text{ N m}^{-2}.$ The required value is $x = 1$.