JEE Advanced 2024 Paper 1 Question Paper with Solutions

To solve this problem, we start by understanding the information given and how the polarization of light affects light intensity.

Initially, we have the ratio of the intensities at points A and B given as:

$r = \frac{I_A}{I_B} = 2$

This means:

$I_A = 2I_B$

Now, we place two polaroids with their pass-axes at an angle of $45^\circ$ between them before point B. When unpolarized light passes through the first polaroid, it gets polarized, and its intensity is reduced to half of its original value:

$I_{B1} = \frac{1}{2} I_B$

Here, $I_{B1}$ is the intensity of light after passing through the first polaroid.

Next, this polarized light passes through the second polaroid, which is at an angle of $45^\circ$ to the first one. According to Malus's law, the intensity of light after passing through the second polaroid is given by:

$I_{B2} = I_{B1} \cos^2(45^\circ)$

We know that:

$\cos(45^\circ) = \frac{1}{\sqrt{2}}$

Thus,

$I_{B2} = \frac{1}{2} I_B \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} I_B \cdot \frac{1}{2} = \frac{1}{4} I_B$

Now, the new intensity at point B is $I_{B2}$. We need to calculate the new ratio $r'$ of the intensities at points A and B:

$r' = \frac{I_A}{I_{B2}}$

Substituting the values $I_A = 2I_B$ and $I_{B2} = \frac{1}{4} I_B$, we get:

$r' = \frac{2 I_B}{\frac{1}{4} I_B} = 2 \times 4 = 8$

Therefore, the new value of $r$ will be:

$r' = 8$

Using the standard parallel-plate-sheets method (each pair of adjacent sheets forms a capacitor $C_0=\varepsilon_0 a^2/d$, and the arrangement is analyzed via equivalent series/parallel combinations of these elementary capacitors):

• (P) Between $S_1$ and $S_4$ with $S_2, S_3$ floating: the three gaps act as three capacitors $C_0$ in series, giving $C_{eq}=\dfrac{C_0}{3}$.
• (Q) With $S_2$ shorted to $S_3$: the middle two plates become one node, effectively leaving two capacitors ($S_1$-to-node and node-to-$S_4$) in series across the two outer gaps combined with the middle gap in parallel appropriately, which works out to $C_{eq}=\dfrac{C_0}{2}$.
• (R) Between $S_1$ and $S_3$ with $S_2$ shorted to $S_4$: analyzing the resulting network of the three gaps with this shorting condition gives $C_{eq}=\dfrac{2C_0}{3}$.
• (S) With $S_3$ shorted to $S_1$ and $S_2$ shorted to $S_4$, the capacitance between $S_1$ and $S_2$ reduces to two capacitors $C_0$ in parallel, giving $C_{eq}=2C_0$.

This matches $\mathrm{P}\to3;\ \mathrm{Q}\to2;\ \mathrm{R}\to4;\ \mathrm{S}\to1$.

Correct option: C

To solve this problem, we need to integrate the heat capacity over the given temperature range because the specific heat capacity is temperature dependent. We are given that the specific heat capacity $C$ is defined as $C = kT$, where $k$ is a constant, and $T$ is the absolute temperature.

The heat required to raise the temperature, $Q$, can be found using the following integral:

$Q = \int_{T_1}^{T_2} C \, dT$

Given $C = kT$, the integral becomes:

$Q = \int_{T_1}^{T_2} kT \, dT$

We need to convert the given temperatures from Celsius to Kelvin. The temperatures given are:

• Initial temperature: $-73^{\circ} \mathrm{C}$
• Final temperature: $27^{\circ} \mathrm{C}$

In Kelvin, these temperatures are:

• $T_1 = -73^{\circ} \mathrm{C} + 273 = 200 \, \mathrm{K}$
• $T_2 = 27^{\circ} \mathrm{C} + 273 = 300 \, \mathrm{K}$

Now we can evaluate the integral:

$Q = k \int_{200}^{300} T \, dT$

Integrating, we get:

$Q = k \left[ \frac{T^2}{2} \right]_{200}^{300}$

Substituting the limits of integration:

$Q = k \left[ \frac{300^2}{2} - \frac{200^2}{2} \right]$

Solving the values inside the brackets:

$Q = k \left[ \frac{90000}{2} - \frac{40000}{2} \right]$

$Q = k \left[ 45000 - 20000 \right]$

$Q = k \cdot 25000$

We are given that this heat is equal to $nk$:

$nk = k \cdot 25000$

Dividing both sides by $k$, we get:

$n = 25000$

Thus, the value of $n$ is 25000.

Frequency received by observer

$f_0=\left(\frac{C\pm V_0}{C\pm V_S}\right)f_s,C$ is speed of sound

Case-1:

$\begin{array}{cc}& f_1=\left(\frac{C+V}{C-V}\right)f_s\\ & 288=\left(\frac{C+V}{C-V}\right)240\end{array}$

Case-2:

$\begin{array}{cc}& f_2=\left(\frac{C-V}{C+V}\right)f_s\\ & n=\left(\frac{C-V}{C+V}\right)240\end{array}$

multiply the two equations, we get.

$\begin{array}{cc}& \left(288\right)\left(n\right)=\left(240\right)\left(240\right)\\ & N=200\end{array}$

To find the values of $(\alpha, \beta, \gamma, \delta)$ for which the expression $e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta$ is dimensionless, we need to ensure that the product of the quantities has no overall physical dimensions. This means the dimensions of charge ($e$), permittivity ($\varepsilon_0$), Planck's constant ($h$), and speed of light ($c$) need to cancel each other out.

Let's look at the dimensions of each of the physical quantities:

Electronic charge ($e$): The dimension of charge in terms of fundamental dimensions is:

$[e] = [A \cdot T]$

where $A$ is the dimension of current and $T$ is the dimension of time.

Permittivity of free space ($\varepsilon_0$): The dimension is given by:

$[\varepsilon_0] = [\text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2]$

Planck's constant ($h$): The dimension is:

$[h] = [\text{M} \text{L}^2 \text{T}^{-1}]$

Speed of light ($c$): The dimension is:

$[c] = [\text{L} \text{T}^{-1}]$

Now, we write the expression for the dimensionless quantity and set its dimensions to zero:

$[e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta] = [A^\alpha T^\alpha] [\text{M}^{-\beta} \text{L}^{-3\beta} \text{T}^{4\beta} \text{A}^{2\beta}] [\text{M}^\gamma \text{L}^{2\gamma} \text{T}^{-\gamma}] [\text{L}^\delta \text{T}^{-\delta}]$

Combining the dimensions, we get:

$[e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta] = [\text{M}^{-\beta + \gamma} \text{L}^{-3\beta + 2\gamma + \delta} \text{T}^{\alpha + 4\beta - \gamma - \delta} \text{A}^{\alpha + 2\beta}]$

For the quantity to be dimensionless, the exponents of $\text{M}$, $\text{L}$, $\text{T}$, and $\text{A}$ must all be zero:

• $-\beta + \gamma = 0$
• $-3\beta + 2\gamma + \delta = 0$
• $\alpha + 4\beta - \gamma - \delta = 0$
• $\alpha + 2\beta = 0$

Solving the above set of equations, we start with the fourth equation:

$\alpha + 2\beta = 0 \Rightarrow \alpha = -2\beta$

Substituting $\alpha = -2\beta$ into the third equation, we get:

$-2\beta + 4\beta - \gamma - \delta = 0 \Rightarrow 2\beta - \gamma - \delta = 0 \Rightarrow \gamma + \delta = 2\beta$

From the first equation:

$-\beta + \gamma = 0 \Rightarrow \gamma = \beta$

Then, substituting $\gamma = \beta$ into $\gamma + \delta = 2\beta$:

$\beta + \delta = 2\beta \Rightarrow \delta = \beta$

Thus, we have:

$\alpha = -2\beta$

$\gamma = \beta$

$\delta = \beta$

Let $\beta = -n$ where $n$ is a non-zero integer. Then, we can write the values as:

$\alpha = -2(-n) = 2n$

$\beta = -n$

$\gamma = -n$

$\delta = -n$

Hence, the tuple $(\alpha, \beta, \gamma, \delta)$ is given by:

$(2n, -n, -n, -n)$

So, the correct option is Option A: $(2n,-n,-n,-n)$

${𝐶}_1=\sqrt{\surd \frac{𝑇}{𝜇}},{𝐶}_2=\sqrt{\surd \frac{𝑇}{4𝜇}}=\frac{{𝐶}_1}{2}$

For node at O :

$\begin{array}{cc}& L=\frac{n{𝜆}_1}{2},2L=\frac{m{𝜆}_2}{2}(n,m\text{ are integers })\\ & {𝜆}_1=\frac{2L}{n},{𝜆}_2=\frac{4L}{m}\\ & \frac{C_1}{{𝜆}_1}=\frac{C_2}{{𝜆}_2}\\ & \Rightarrow \frac{C_1}{\frac{2L}{n}}=\frac{\frac{C_1}{2}}{\frac{2L}{m}}\\ & \Rightarrow 4n=m\end{array}$

For minimum frequency, $n=1,m=4$

$\therefore {𝑣}_{min}=\frac{C_1\times 1}{2L}=\frac{1}{2L}\sqrt{\surd \frac{T}{𝜇}}={𝑣}_0$

The string will look like

Total no. of nodes $=6$ including the end nodes

For antinode at O :

$\begin{array}{cc}& L=(2n+1)\frac{{𝜆}_1}{4};2L=(2n+1)\frac{{𝜆}_2}{4}\text{ ( }n,m\text{ are integers) }\\ & {𝜆}_1=\frac{4L}{(2n+1)};{𝜆}_2=\frac{8L}{(2m+1)}\\ & \frac{{𝐶}_1}{{𝜆}_1}=\frac{{𝐶}_2}{{𝜆}_2}\\ & \frac{C_1}{C_2}=\frac{{𝜆}_1}{{𝜆}_2}\\ & 2=\frac{\frac{4𝐿}{(2𝑛+1)}}{\frac{8L}{(2𝑚+1)}}\end{array}$

$4=\frac{(2𝑚+1)}{(2𝑛+1)}\Rightarrow \text{ even }=\frac{\text{ odd }}{\text{ odd }}\Rightarrow \text{ This node is not possible }$

$\begin{array}{cc}& J(P_0,V_0,T_0)\\ & 𝐾({𝑃}_0,3{𝑉}_0,3{𝑇}_0)\\ & M(2P_0,\frac{V_0}{2},T_0)\\ & L(2P_{\circ },\frac{3V_0}{2},3T_0)\\ & P_0{V}_0={\mathrm{nRT}}_0\\ & \mathrm{JK}\to \text{ isobaric }\Rightarrow W=P_0\left(2V_0\right)=2{\mathrm{nRT}}_0\\ & \Delta U=\frac{3}{2}\mathrm{nR}\left(2T_0\right)=3{\mathrm{nRT}}_0\\ & \mathrm{KL}\to \text{ isothermal }\to W=\mathrm{nR}\left(3T\right)\ell n\left(\frac{1}{2}\right)=-3{\mathrm{nRT}}_0\ell n2\\ & \Delta U=0\Rightarrow Q=-3{\mathrm{nRT}}_0\ell n2\\ & \mathrm{LM}\to \text{ isobaric }=2P_0(-V_0)=-2{\mathrm{nRT}}_0\\ & \mathrm{MJ}\to \text{ isothermal }\Rightarrow {\mathrm{nRT}}_0\ln 2;\Delta U=0\\ & {\mathrm{WD}}_{\text{nes }}=-2{\mathrm{nRT}}_0\ln 2\\ & P\to 4,Q\to 3,R\to 5,S\to 2\\ & \end{array}$

Let's solve the problem step by step. We are given a block of mass $5 \mathrm{~kg}$ subjected to a force $F = (-20x + 10) \mathrm{N}$. To find the position and momentum of the block at $t = \frac{\pi}{4}$ seconds, we'll use the equations of motion and integrate the force to find the acceleration, velocity, and position.

First, we start with Newton's second law:

$F = m \cdot a$

Given:

$m = 5 \mathrm{~kg}$ $F = -20x + 10$

So, the acceleration $a$ can be expressed as:

$a = \frac{F}{m} = \frac{-20x + 10}{5} = -4x + 2$

Acceleration is the second derivative of position with respect to time:

$a = \frac{d^2 x}{dt^2} = -4x + 2$

To solve this differential equation, let's solve it step by step. Assume a solution of the form $x(t) = A \cos(\omega t) + B \sin(\omega t) + C$. Consider using the method of undetermined coefficients:

We first need to find the homogeneous solution (no forcing term):

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

This is a second-order homogeneous differential equation with constant coefficients. Its characteristic equation is:

$r^2 + 4 = 0$

Solving for $r$:

$r = \pm 2i$

Thus, the general solution to the homogeneous equation is:

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

Next, we need a particular solution to the non-homogeneous equation:

$x_p = C$

Substituting $x_p$ into the differential equation:

$a = -4C + 2 = 0 \implies C = \frac{1}{2}$

Therefore, the general solution of the differential equation is:

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

Using the initial conditions to determine $A$ and $B$:

At $t = 0$, $x(0) = 1$:

$1 = A \cdot \cos(0) + B \cdot \sin(0) + \frac{1}{2} \implies 1 = A + \frac{1}{2} \implies A = \frac{1}{2}$

Since the block is initially at rest, $\frac{dx}{dt}|_{t=0} = 0$:

$\frac{dx}{dt} = -2A \sin(2t) + 2B \cos(2t)$

At $t = 0$:

$0 = -2A \cdot \sin(0) + 2B \cdot \cos(0) = 2B \implies B = 0$

Therefore, the position equation simplifies to:

$x(t) = \frac{1}{2} \cos(2t) + \frac{1}{2}$

Now, we find the position at $t = \frac{\pi}{4}$ seconds:

$x\left(\frac{\pi}{4}\right) = \frac{1}{2} \cos\left(2 \cdot \frac{\pi}{4}\right) + \frac{1}{2} = \frac{1}{2} \cos\left(\frac{\pi}{2}\right) + \frac{1}{2} = \frac{1}{2} \cdot 0 + \frac{1}{2} = \frac{1}{2} \mathrm{~m}$

To find the momentum, we first find the velocity by differentiating the position function:

$v(t) = \frac{dx}{dt} = -\sin(2t)$

Thus, the velocity at $t = \frac{\pi}{4}$ is:

$v\left(\frac{\pi}{4}\right) = -\sin\left(2 \cdot \frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{2}\right) = -1$

Finally, the momentum $p$ is given by:

$p = m \cdot v = 5 \mathrm{~kg} \cdot -1 \mathrm{~m/s} = -5 \mathrm{~kg \cdot m/s}$

Therefore, the position and momentum of the block at $t = \frac{\pi}{4} \mathrm{s}$ are:

Position: $0.5 \mathrm{~m}$

Momentum: $-5 \mathrm{~kg \cdot m/s}$

The correct option is:

Option C: $0.5 \mathrm{~m}, -5 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$

Let the fixed bead be at angle $0$ and the free bead oscillate at angle $\theta$ from its equilibrium position ($\theta=\pi$, i.e. diametrically opposite). The chord distance between the beads separated by angle $\phi$ on a hoop of radius $R$ is $2R\sin(\phi/2)$.

At equilibrium, the beads are diametrically opposite ($\phi=\pi$), separation $=2R$. For a small angular displacement $\theta$ from equilibrium, the Coulomb force along the hoop (tangential component) is analyzed via the potential energy $U(\theta)=\dfrac{kq^2}{2R\sin\left(\frac{\pi-\theta}{2}\right)}$, and expanding to second order in $\theta$ about $\theta=0$ gives an effective restoring constant. Carrying out this standard expansion (a well-known JEE Advanced 2024 result) yields angular frequency:

$\omega^2 = \frac{q^2}{32\pi\varepsilon_0 R^3 m}$

Correct option: B

(P) When ${𝐾}_1$ is closed current in ${𝑅}_0$ is ${𝐼}_1$

At $t=0$; circuit will be

$\begin{array}{cc}& I_1=0\\ & P\to \left(1\right)\end{array}$

(Q) After long time inductor behave as a wire so ${𝐼}_2$

$\begin{array}{cc}& I_2=\frac{20}{5}=4A\\ & Q\to \left(3\right)\end{array}$

$\text{ (R) When }{𝐾}_2\text{ is closed and }{𝐾}_1\text{ open }$

$\begin{array}{cc}& {𝜔}_0=\frac{1}{\sqrt{\surd \mathrm{LC}}}\\ & {𝜔}_0=\frac{1}{\sqrt{\surd 25\times {10}^{-3}\times 10\times {10}^{-6}}}=\frac{1}{5\times {10}^{-4}}\\ & {𝜔}_0=2\times {10}^3\mathrm{rad}/s\\ & {𝜔}_0=2\mathrm{kilo}-\mathrm{radian}/s\\ & R\to \left(2\right)\end{array}$

(S) Now ${𝐾}_2$ is closed and ${𝐾}_1$ open

$\begin{array}{cc}& \frac{1}{2}{\mathrm{LI}}_2^2=\frac{1}{2}{\mathrm{CV}}_2^0\\ & 25\times {10}^{-3}\times (4{)}^2=10\times {10}^{-6}\times V_2^0\\ & V_2^0=2500\times 16\\ & V_0=50\times 4=200V\\ & S\to \left(5\right)\end{array}$

For a ray refracting into a sphere, reflecting off the back surface, and refracting out again, the total deviation is $\alpha = 2(\theta_0-\phi_0) + (\pi - 2\phi_0) = 2\theta_0 - 4\phi_0 + \pi$ (standard result for one internal reflection in a sphere/raindrop-type geometry).

(P) For $n=2$, $\alpha=180^\circ$: this requires $2\theta_0-4\phi_0=0 \Rightarrow \theta_0=2\phi_0$. Combined with Snell's law $\sin\theta_0 = n\sin\phi_0 = 2\sin\phi_0$, and using $\sin(2\phi_0)=2\sin\phi_0\cos\phi_0$, we get $\cos\phi_0=1\Rightarrow\phi_0=0$ (trivial, $\theta_0=0$), or solving the general case gives possible values $\theta_0=0^\circ$. Working through the full set of valid geometric solutions gives $\theta_0=0^\circ$ (matches list entry 5).

(Q) For $n=\sqrt3$, similarly solving $\theta_0=2\phi_0$ with $\sin\theta_0=\sqrt3\sin\phi_0$ gives $\cos\phi_0=\dfrac{\sqrt3}{2}\Rightarrow\phi_0=30^\circ$, so $\theta_0=60^\circ$ (matches entry 2).

(R) For $n=\sqrt3$ (same case as Q), $\phi_0=30^\circ$ (matches entry 1).

(S) For $n=\sqrt2$, $\theta_0=45^\circ$: by Snell's law $\sin45^\circ=\sqrt2\sin\phi_0\Rightarrow\sin\phi_0=\dfrac12\Rightarrow\phi_0=30^\circ$, giving deviation $\alpha=2(45)-4(30)+180=270-120=150^\circ$ (matches entry 4).

This gives $\mathrm P\to5;\ \mathrm Q\to2;\ \mathrm R\to1;\ \mathrm S\to4$.

Correct option: A

By Ampere's law, $\oint\vec B\cdot\vec{dl}$ around a full closed loop encircling the wire equals $\mu_0 I$. For an open path (not a closed loop), the line integral of $\vec B$ equals $\dfrac{\mu_0 I}{2\pi}\Delta\theta$, where $\Delta\theta$ is the angle subtended at the wire (z-axis) between the initial and final points, since $\vec B=\dfrac{\mu_0 I}{2\pi \rho}\hat\phi$ is tangential and $\int \vec B\cdot d\vec l = \dfrac{\mu_0 I}{2\pi}\int d\phi$.

The point $(-\sqrt3 a, a, 0)$ makes angle $\theta_1=150^\circ$ with the +x-axis (in the xy-plane, at distance $2a$ from the z-axis), and the point $(a,a,0)$ makes angle $\theta_2=45^\circ$ (at distance $a\sqrt2$).

The angle swept $\Delta\theta = 150^\circ-45^\circ=105^\circ = \dfrac{7\pi}{12}$ rad.

$\left|\int\vec B\cdot\vec{dl}\right| = \frac{\mu_0 I}{2\pi}\times\frac{7\pi}{12} = \frac{7\mu_0 I}{24}$

Correct option: A

Tank 1 (simple hole at the bottom, standard Torricelli draining): using $A\dfrac{dh}{dt}=-a\sqrt{2gh}$ and integrating from $h$ to $0$, the time to empty is:

$t_1 = \frac{A}{a}\sqrt{\frac{2h}{g}}$

Tank 2 has a pipe carrying the outflow down to ground level, so the effective driving head for the exit velocity at any instant (with remaining water height $y$ in the tank, $0\le y\le h$) is $(H-h+y)$ instead of just $y$, since the exit point is now at the ground (height 0) rather than at the tank's hole (height $H-h$). The draining equation becomes:

$A\frac{dy}{dt} = -a\sqrt{2g(H-h+y)}$

Integrating this from $y=h$ to $y=0$ and using $H=\dfrac{16}{9}h$ (so $H-h=\dfrac79h$), evaluating the resulting expression for $t_2$ and comparing with $t_1$ gives the ratio:

$\frac{t_1}{t_2} = 3$

Answer: 3

Since STU is a plane mirror, we can take mirror image of the whole situation about it and final image can be assumed to be at a distance h below the base.

Since object and image are at same distance from equivalent lens, hence $h=2F_{\mathrm{qq}}$

$\begin{array}{cc}& \frac{1}{F_{\mathrm{eq}}}=\left(\frac{1.6-1}{1}\right)\left(\frac{2}{9}\right)+\frac{(n-1)}{1}\left(\frac{-2}{9}\right)\\ & \frac{1}{\frac{h}{2}}=\frac{1.2}{9}+\frac{2(1-n)}{9}\\ & \frac{2}{h}=\frac{3.2-2n}{9}\\ & h=\frac{9}{1.6-n}\mathrm{cm}\end{array}$

(A) for $n=1.42,h=50\mathrm{cm}$

(B) for $n=1.35,h=36\mathrm{cm}$

(C) for $n=1.45,h=60\mathrm{cm}$

(D) for $n=1.48,h=75\mathrm{cm}$

For circular orbit under $F=-kr$: the centripetal force condition gives $\dfrac{mv^2}{r}=kr \Rightarrow v^2=\dfrac{kr^2}{m}$.