WBJEE 2023 Paper 2 (Physics & Chemistry) Question Paper with Solutions

"W" means the question is withdrawn. Hence, marks will be awarded to all appeared candidates.

In a SHM,

Acceleration, $f=-\omega^2 x$

Time period, $T=\frac{2 \pi}{\omega}$

$\therefore \quad|f T|=\left|-\omega^2 x \times \frac{2 \pi}{\omega}\right|=2 \pi \omega x$

$\therefore \quad|f T| \propto x$

Hence, $|f T|$ vs $x$ graph is a straight line.

Given, $\frac{d H}{d t}=$ constant $(K)$ ..... (i)

Where, $H=m s \theta$

$\therefore$ From Eq. (i),

$$\begin{aligned} & m s \frac{d \theta}{d t}=K \\ \therefore \quad & \frac{d \theta}{d t} \propto \frac{1}{s} \end{aligned}$$

i.e., slope of $\theta$ vs $t$ graph $\propto \frac{1}{s}$

As (Slope)$_B$ > (Slope)$_A$

$\therefore(\text { Specific heat })_A>(\text { Specific heat })_B$

The output, $Y=\overline{\overline{\bar{A} \cdot B}+\bar{C}}$

Let $D=\bar{A} \cdot B$

$$\begin{aligned} & \therefore \quad Y=\overline{\bar{D}+\bar{C}}=\overline{\overline{D \cdot C}}=D \cdot C=\bar{A} \cdot B \cdot C \\ & \therefore \quad Y=\bar{A} \cdot B \cdot C \end{aligned}$$

This is AND gate with input $\bar{A}, B$ and $C$

For $Y=1 ; \bar{A}=1, B=1, C=1 \Rightarrow$ one combination.

$\therefore$ The combination of $A, B$ and $C$ which gives $Y=0$ is

$2^n-1=2^3-1=7$

Let $u$, and $v$ are the speed at inlet and outlet of the pump.

$\therefore$ Applying Bernoulli's theorem at points 1 and 2.

$p_0+\frac{F}{A}+\frac{1}{2} \rho u^2=p_0+\frac{1}{2} \rho v^2$

$\therefore \quad \frac{F}{A}=\frac{1}{2} \rho\left(v^2-u^2\right)$ ..... (i)

By continuity equation, $A u=a v$

$\Rightarrow \quad u=\frac{a v}{A}$ ..... (ii)

From Eqs. (i) and (ii), we get

$$\begin{aligned} \frac{F}{A}=\frac{1}{2} \rho\left[v^2-\frac{a^2 v^2}{A^2}\right] & =\frac{1}{2} \rho v^2\left(\frac{A^2-a^2}{A^2}\right) \\ \text { As, } A \gg a \Rightarrow \frac{A^2-a^2}{A^2} & =1, \\ v & =\sqrt{2 F / \rho A} \end{aligned}$$

Let the collision of two balls

$\therefore$ By conservation of momentum,

$m v_0=m v_1+\frac{m}{2} v_2$

$\therefore \quad v_0=v_1+\frac{v_2}{2}$ ..... (i)

For elastic collision, $e=1=\frac{v_2-v_1}{v_0-0}$

$\therefore \quad v_0=v_2-v_1$ .... (ii)

From Eqs. (i) and (ii), we get

$$\begin{aligned} \frac{3 v_2}{2} & =2 v_0 \\ \Rightarrow \quad & v_2=\frac{4 v_0}{3} \end{aligned}$$

Similarly, the velocity of $3 \mathrm{rd}$ ball, $v_3=\left(\frac{4}{3}\right)^2 v_0$

$\therefore$ Velocity of $n$th ball, $v_n=\left(\frac{4}{3}\right)^{n-1} v_0$

Initially switch $K$ is closed,

Energy stored in capacitor $A, E_A=\frac{1}{2} C V^2$

Energy stored in capacitor $B, E_B=\frac{1}{2} C V^2$

$\therefore$ Total energy stored in two capacitors,

$E_i=E_A+E_B=\frac{1}{2} C V^2+\frac{1}{2} C V^2=C V^2$

When switch $K$ is open, and a dielectric $(K=3)$ is inserted in both capacitors, then

Energy stored in capacitor $A$,

$E_A=\frac{1}{2}(3 C) V^2=\frac{3}{2} C V^2$

Energy stored in capacitor B,

$E_p=\frac{(C V)^2}{2 \times 3 C}=\frac{1}{6} C V^2$

$\therefore$ Total energy stored in two capacitors,

$E_f=E_A+E_B=\left(\frac{3}{2}+\frac{1}{6}\right) C V^2=\frac{5}{3} C V^2$

As the component of velocity is along $\vec{B}$ as well as perpendicular to $\vec{B}$. Hence, the trajectory of the particle is helical path.

Also,

$$\begin{aligned} & v_z=4 \mathrm{~m} / \mathrm{s}, z=2 \mathrm{~m} \\ &\therefore \quad z=v_z \cdot {t} \\ & \therefore \quad t=\frac{z}{v_z}=\frac{2}{4}=\frac{1}{2} \mathrm{~s} \\ \end{aligned}$$

The work required to move a charge $(q)$ from point $A$ to $B$ is

$$\begin{aligned} W_{\mathrm{ext}} & =q\left(V_B-V_A\right)=q\left(\frac{k p}{a^2}-0\right) \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{P q}{a^2} \quad\left(\because k=\frac{1}{4 \pi \varepsilon_0}\right) \end{aligned}$$

The acceleration due to gravity at height $H$ from the surface of planet is

$g_H=\frac{G M}{(R+H)^2}$ ..... (i)

The acceleration due to gravity at depth $H$ below the surface of planet is

$g_H=\frac{G M}{R^3}(R-H)$ .... (ii)

From Eq. (i) and (ii), we get

$$\begin{aligned} & \frac{G M}{(R+H)^2}=\frac{G M}{R^3}(R-H) \\ & \Rightarrow \quad H^2+H R-R^2=0 \\ & \Rightarrow \quad H=\frac{-R \pm \sqrt{R^2-4 \times 1 \times\left(-R^2\right)}}{2 \times 1} \\ & =\frac{-R \pm \sqrt{5} R}{2} \\ & \Rightarrow \quad H=\left(\frac{-1 \pm \sqrt{5}}{2}\right) R \\ & \Rightarrow \quad H \propto R \\ \end{aligned}$$

Hence, $H$ vs $R$ graph is straight line.

The top view of the cylinder and the dipole is

Hence, the dipole will experience a net force and a net torque in clockwise direction.

The given circuit is

Reading of voltmeter, $V=8 \mathrm{~V}$

$\therefore$ Potential difference across resistance $20 \Omega$

$V^{\prime}=10-8=2 \mathrm{~V}$

$\therefore$ Current through $20 \Omega$ resistor,

$I=\frac{V^{\prime}}{R_1^0}=\frac{2}{20}=\frac{1}{10} \mathrm{~A}$

Current through $160 \Omega$ resistor,

$I_2=\frac{V}{R_2}=\frac{8}{160}=\frac{1}{20} \mathrm{~A}$

$\therefore$ Current through voltmeter,

$$\begin{aligned} & I_1=\left(\frac{160}{160+R_V}\right) I \\ & \Rightarrow \quad I-I_2=\left(\frac{160}{160+R_V}\right) I \\ & \Rightarrow \quad \frac{1}{20}=\left(\frac{160}{160+R_V}\right) \times \frac{1}{10} \\ & \therefore \quad R_V=160 \Omega \\ \end{aligned}$$

The radius of the circular path of moving electron is

$r_n=m v_n / B q$ .... (i)

By Bohr's quantisation,

$m v_n r_n=\frac{n h}{2 \pi}=n\hbar$ .... (ii)

From Eqs (i) and (ii),

$B q r_n \cdot r_n=n \hbar \Rightarrow r_n^2=n \hbar / B q \Rightarrow r_n \propto \sqrt{n}$

Again, $\frac{m^2 v_n^2}{B q}=n \hbar \Rightarrow v_n^2=\frac{B q n \hbar}{m^2}$

$\therefore \quad v_n=\frac{\sqrt{B q \hbar \cdot n}}{m}$

$\therefore$ For minimum velocity of electron, $n=1$

$$\begin{aligned} \therefore \quad v_{\min } & =\sqrt{q B \hbar / m} \\ \text { Now, } E_n & =\mathrm{KE}+\mathrm{PE}=\mathrm{KE}-2(\mathrm{KE}) \\ & =-\mathrm{KE}=-\frac{1}{2} m v_n^2=-\frac{1}{2} m \cdot \frac{B q \hbar n}{m^2} \\ \therefore \quad & E_n \times n \end{aligned}$$

Also, $h v \propto n \Rightarrow v \propto n$

The time period of the satellite near surface of planet is

$$\begin{aligned} T^2 & =\frac{4 \pi^2}{G M} R^3=\frac{4 \pi^2}{G \cdot \frac{4}{3} \pi R^3 \cdot \rho} R^3 \\ \therefore \quad T^2 & =\frac{3 \pi}{\rho G} \end{aligned}$$

$\therefore$ For the earth, $T_e^2=\frac{3 \pi}{\rho_e G}$ .... (i)

For the moon, $T_m^2=\frac{3 \pi}{\rho_m G}$ ..... (ii)

As $T_e=T_m=90 \mathrm{~rev} / \mathrm{min}$

$\therefore$ From Eqs. (i) and (ii), we get

$$\begin{aligned} \quad \frac{3 \pi}{\rho_e G} & =\frac{3 \pi}{\rho_m G} \\ \therefore \quad \rho_e & =\rho_m \end{aligned}$$

By conservation of angular momentum,

$$\begin{aligned} & I \omega=\left(I+M R^2\right) \cdot \omega^{\prime} \\ \therefore & \quad\frac{\omega^{\prime}}{\omega}=\frac{I}{I+m R^2} \end{aligned}$$

$\therefore$ The fractional loss of angular velocity is

$$\begin{aligned} \frac{\omega-\omega^{\prime}}{\omega} & =1-\frac{\omega^{\prime}}{\omega}=1-\frac{I}{I+m R^2} \\ & =\frac{m R^2}{I+m R^2} \end{aligned}$$

The magnetic force on a current carrying wire is

$$\begin{aligned} & \mathbf{F} ={I l} \times \mathbf{B}={I} \lambda \times \mathbf{B} \\ \therefore \quad & F =I \lambda B \sin 90^{\circ}=I \lambda B \end{aligned}$$

Initially only gravity force acts on the magnet.

$\therefore$ The speed of magnet increases from zero to certain value. When the magnet reaches near the ring, then opposing magnetic force $F_m$ acts on the magnet. So, its speed decreases.

Similarly when magnet crosses the ring and goes away, then only gravity force acts and the speed of magnet increases.

Measurement of the object $=6.50 \mathrm{~cm}$

$\therefore$ Number of significant figures $=3$

Now, when the measured value is converted from one unit to another unit, then the number of significant figures remains same in both cases.

$\therefore$ The number of significant figures in $0.0650 \mathrm{~m}$ is 3.

$$\begin{aligned} & \tan 30^{\circ}=\frac{\sin r}{\sin i} \\ & \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{\sin r}{\sin i} \quad \text{.... (i)}\\ \end{aligned}$$

By Snell's law,

$\mu_x \sin i=\mu_y \sin r \Rightarrow \frac{c}{v_x} \sin i=\frac{c}{v_y} \sin r$

$\Rightarrow \quad \frac{v_y}{v_x}=\frac{\sin r}{\sin i}$ .... (ii)

From Eqs. (i) and (ii), we get

$\frac{v_y}{v_x}=\frac{1}{\sqrt{3}} \Rightarrow v_x=\sqrt{3} v_y$

The FBD of the liquid above the level is

$\therefore \quad \tan \theta=\frac{f}{g}=\frac{h}{a} \therefore h=\frac{f}{g} a$

$\therefore \quad i=-\frac{i_0}{T} t+i_0$ .... (i)

The charge flowing through the coil is

$$\begin{aligned} Q & =\text { Area of } i-t \text { graph } \\ Q & =\frac{1}{2} \times i_0 \times T \\ \therefore \quad i_0 & =2 Q / T \end{aligned}$$

$\therefore$ From Eq. (i),

$i=-\frac{1}{T} \cdot \frac{2 Q}{T} \cdot t+\frac{2 Q}{T}=\frac{2 Q}{T}\left(1-\frac{t}{T}\right)$

$\therefore$ Amount of heat generated in the coil is,

$$\begin{aligned} H & =\int_0^T i^2 R d t=\int_0^T\left(\frac{2 Q}{T}\right)^2\left(1-\frac{t}{T}\right)^2 R \cdot d t \\ & =\frac{4 Q^2 R}{T^2} \cdot \int_0^T\left(1-\frac{2}{T} t+\frac{t^2}{T^2}\right) d t \\ & =\frac{4 Q^2 R}{T^2}\left[t-\frac{2}{T} \cdot \frac{t^2}{2}+\frac{1}{T^2} \cdot \frac{t^3}{3}\right]_0^T=\frac{4 Q^2 R}{3 T} \end{aligned}$$

M = 0.4 kg, L = 4 m

$$\begin{aligned} W_{\mathrm{ext}} & =U_f-U_i=0-(-m g h) \\ & =m g h=\left(\frac{M}{L} \times 0.6\right) \times 10 \times 0.3 \end{aligned}$$

$$\begin{aligned} & =\frac{0.4}{4} \times 0.6 \times 3 \\ & =0.18 \mathrm{~J} \end{aligned}$$

$\vec{B}=\left(4 t^2-2 t+6\right) \hat{\mathbf{k}} \Rightarrow \frac{d B}{d t}=(8 t-2)$

By Maxwell's equation,

For $r < R$,

$$\begin{aligned} & & \oint E_{\text {ind }} \cdot d l & =\frac{d \phi_B}{d t}=-\left(\frac{d B}{d t}\right) A \\ \Rightarrow & & E_{\text {ind }}(2 \pi r) & =-(8 t-2) \cdot \pi R^2 \\ \therefore & & E_{\text {ind }} & =-(4 t-1) \cdot \frac{R^2}{r} \end{aligned}$$

$\Rightarrow E_{\text {ind }} \propto \frac{1}{r} \Rightarrow E_{\text {ind }}$ vs $r$ does not vary linearly.

As $E_{\text {ind }}$ is a non-conservative field. So, it will form a closed loop.

Now, $\left(\frac{d B}{d t}\right)_{t=0}=8 \times 0-2=-2$

$\therefore B$ will decrease. So, by Lenz's law the charge particle will move in anti-clockwise direction to oppose the decreasing magnetic field.

$\text { Now, }\left|E_{\text {ind }}\right|_{t=2 \mathrm{~s}}=\left.(4 t-1) \cdot \frac{R^2}{2 R}\right|_{t=2}=\frac{7 R}{2}$

$\therefore$ Acceleration of charge particle is

$a=\frac{q E_{\text {ind }}}{m}=\frac{q}{m} \cdot \frac{7 R}{2}=\frac{7 q R}{2 m}$

The direction of the magnetic field in an electromagnetic wave can be determined using the cross product of the wave vector and the electric field vector. In this case, the electric field is given by $\vec{E} = E_0(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \sin(kz - \omega t)$, and the wave is propagating in the $+\hat{\mathbf{k}}$ direction (along the z-axis).

The direction of the magnetic field $\vec{B}$ can be found using the right-hand rule for the cross product $\hat{\mathbf{v}} \times \vec{E}$, where $\hat{\mathbf{v}}$ is the direction of wave propagation. In this case, $\hat{\mathbf{v}} = \hat{\mathbf{k}}$. So, we calculate :

$ \hat{\mathbf{v}} \times \vec{E} = \hat{\mathbf{k}} \times (\hat{\mathbf{i}} + \hat{\mathbf{j}}) $

This cross product gives us the direction of the magnetic field :

• The cross product $\hat{\mathbf{k}} \times \hat{\mathbf{i}}$ yields $+\hat{\mathbf{j}}$.
• The cross product $\hat{\mathbf{k}} \times \hat{\mathbf{j}}$ yields $-\hat{\mathbf{i}}$.

So, combining these results, we get :

$ \hat{\mathbf{k}} \times (\hat{\mathbf{i}} + \hat{\mathbf{j}}) = -\hat{\mathbf{j}} + \hat{\mathbf{i}} = -\hat{\mathbf{i}} + \hat{\mathbf{j}} $

Therefore, the direction of the magnetic field $\vec{B}$ associated with the electromagnetic wave is $-\hat{\mathbf{i}} + \hat{\mathbf{j}}$, which corresponds to Option B.

The current through the circuit is

$$\begin{aligned} & I=\frac{9}{200+100}=\frac{3}{100} \mathrm{~A} \\ \therefore \quad & V_L=I \cdot R_L=\frac{3}{100} \times 100=3 \mathrm{~V} \end{aligned}$$

As $V_L< V_Z$. So, the Zener diode will not activate and $V_L=3 \mathrm{~V}$.

Given, gravitational potential,

$V=-\frac{G M}{r}+\frac{A}{r^2}$

By principle of homogeneity,

$$\begin{aligned} {\left[\frac{G M}{r}\right] } & =\left[\frac{A}{r^2}\right] \\ \therefore \quad[A] & =[G M r] \end{aligned}$$

$$\begin{aligned} & \text { Now, }[G]=\left[\frac{F r^2}{m_1 m_2}\right]=\frac{\left[M L T^{-2}\right]\left[\mathrm{L}^2\right]}{\left[\mathrm{M}^2\right]}=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \cdot \mathrm{T}^{-2}\right] \\ & \therefore \quad[A]=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2]}[\mathrm{M}][\mathrm{L}]=\left[\mathrm{L}^4 \mathrm{~T}^{-2}\right]\right. \end{aligned}$$

According to the question,

$\begin{gathered} {[A]=[G]^x[M]^y[c]^z} \\ \Rightarrow\left[\mathrm{L}^4 \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^x[M]^y\left[\mathrm{LT}^{-1}\right]^z \end{gathered}$

$$\begin{aligned} \therefore-x+y & =0 \Rightarrow x=y \quad \text{... (i)}\\ 3 x+z & =4 \quad \text{... (ii)}\\ -2 x-z & =-2 \quad \text{... (iii)} \end{aligned}$$

From Eqs. (i), (ii) and (iii), we get

$$\begin{aligned} & x=y=2, z=-2 \\ \therefore \quad & A=G^2 M^2 c^{-2}=\frac{G^2 M^2}{c^2} \end{aligned}$$

The velocities of the molecules of ideal gas are $1,3,5,5,6$ and $5 \mathrm{~m} / \mathrm{s}$.

$\therefore$ Average speed,

$\bar{v}=\frac{\sum v}{n}=\frac{1+3+5+5+6+5}{6} \Rightarrow \frac{25}{6}=4.1 \mathrm{~m} / \mathrm{s}$

Root mean square speed, $v_{\mathrm{rms}}=\sqrt{\frac{\Sigma v^2}{n}}$

$$\begin{aligned} \therefore v_{\mathrm{rms}} & =\sqrt{\frac{1^2+3^2+5^2+5^2+6^2+5^2}{6}} \\ & =\sqrt{\frac{121}{6}}=\frac{11 \sqrt{6}}{6}=4.5 \mathrm{~m} / \mathrm{s} \\ \therefore v_{\mathrm{rms}} & >\bar{v} \end{aligned}$$

$r_{\perp}=H_{\max }=\frac{u^2 \sin ^2 \theta}{2 g}, v=u \cos \theta$

Angular momentum at the maximum height about point $O$ is