WBJEE 2022 Paper 2 (Physics & Chemistry) Question Paper with Solutions

Intensity at any point on the screen is given by

$I = 4{I_0}{\cos ^2}{\phi \over 2}$

where, ${I_0}$ is the intensity of either wave and $\phi$ is the phase difference between two waves.

Phase difference $(\phi ) = {{2\pi } \over \lambda } \times$ path difference

$= {{2\pi } \over \lambda } \times \lambda = 2\pi$

$\therefore$ $I = 4{I_0}{\cos ^2}\left( {{{2\pi } \over 2}} \right)$

$= 4{I_0}{\cos ^2}(\pi ) = 4{I_0} = I$

$\Rightarrow {I_0} = {I \over 4}$ ...... (i)

When path difference is ${\lambda \over 4}$, then

$\phi = {{2\pi } \over \lambda } \times {\lambda \over 4} = {\pi \over 2}$

$\therefore$ $I = 4{I_0}{\cos ^2}\left( {{\pi \over 4}} \right) = 2{I_0} = 2 \times {I \over 4} = {I \over 2}$

Given, resistance per unit length of loop

$= \lambda \,\Omega /m$

(a) Current in the loop, $I = {E \over R}$

$I = {{Blv} \over R}$ ($\because$ emf $E = Blv$) ......... (i)

We know that,

${R \over {2(a + b)}} = \lambda \Rightarrow R = \lambda (2a + 2b)$ .......... (ii)

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

$I = {{Bbv} \over {\lambda (2a + 2b)}}$

So, option (a) is correct.

(b) Using right hand thumb rule, direction of current in the loop will be clockwise, looking from the top.

But as the flux is changing, so according to Lenz's law, direction of current will be anti-clockwise.

So, option (b) is incorrect.

(c) As the arm PS and QR has same potential, then ${V_P} - {V_S} = {V_Q} - {V_R}$ is also same.

So, option (c) is correct.

(d) As the arm SR is parallel to the velocity. So, there will not be any current or induction.

So, option (d) is correct.

Given, initial speed, $v = \sqrt 3 \,{v_e}$

We know that, ${v_e} = \sqrt {{{2GM} \over R}}$

At the surface of earth,

Total energy $= PE + KE$

$= - {{GMm} \over R} + {1 \over 2}m{v^2}$

$= - {{GMm} \over R} + {1 \over 2}m{\left( {\sqrt 3 \sqrt {{{2GM} \over R}} } \right)^2}$

$= - {{GMm} \over R} + {1 \over 2}3m \times \,{{2GM} \over R}$

$= -{{GMm} \over R} + {{3GMm} \over R} = {{2GMm} \over R}$

At infinity, total energy $= PE + KE$

$= 0 + {1 \over 2}\,m{v^2}$

From the principle of conservation of energy,

${{2GMm} \over R} = {1 \over 2}m{v^2}$

$\Rightarrow {{4GM} \over R} = {v^2}$

$\Rightarrow v = \sqrt {{{4GM} \over R}}$

$= \sqrt {2 \times {{2GM} \over R}} = \sqrt 2 \,{v_e}$

At any moment, the phase difference between current and voltage is given by

$\tan \phi = {{{X_L}} \over R}$

Since, voltage leads current by ${\pi \over 4}$, thus given circuit will be RL series circuit

$\therefore$ $\tan {\pi \over 4} = {{{X_L}} \over R} \Rightarrow R = {X_L} = \omega L$

From option (b), $R = 1\,k\Omega = 1000\,\Omega$

$\therefore$ ${X_L} = \omega L$

$= 100 \times 10$ [$\because$ $\omega=100$ rad/s]

$= 1000\,\Omega$

Thus, the condition is satisfied. Hence, option (b) is correct.

The given process BC is isothermal and AB is isochoric.

From A to B, we have

${p_0} \to 2{p_0}$

as $p \propto T$

Hence, ${T_B} = 2{T_A} = 2{T_0}$

${Q_{AB}} = \Delta {U_{AB}} = {W_{AB}} = {C_V}\Delta T + 0$

$\Rightarrow {Q_{AB}} = {5 \over 2}R(2{T_0} - {T_0}) = {5 \over 2}{p_0}{V_0}$

($\because$ ${C_V} = {5 \over 2}R$ for diatomic gas and $pV = RT$)

From B to C, we have,

${Q_{BC}} = \Delta {U_{BC}} + {W_{BC}} = 0 + R(2{T_0})\ln {{2{V_0}} \over {{V_0}}}$

$= 2R{T_0}\ln 2$

$= 2\ln 2{p_0}{V_0}$

Hence, ${Q_{total}} = {Q_{AB}} + {Q_{BC}}$

$= (2.5 + 2\ln 2){p_0}{V_0}$

$= 3.9\,{p_0}{V_0}$

According to the Hooke's law, for deformation of string of length $l$,

${{{F_1}} \over A} \propto {{\Delta {l_1}} \over l}$ ....... (i)

and in second case,

${{{F_2}} \over A} \propto {{\Delta {l_2}} \over l}$ ...... (ii)

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

${{{F_1}} \over {{F_2}}} = {{\Delta {l_1}} \over {\Delta {l_2}}} = {{{l_1} - l} \over {{l_2} - l}}$

$\Rightarrow ({F_2} - {F_1})l = {F_2}{l_1} - {F_1}{l_2}$

$\Rightarrow l = {{{F_2}{l_1} - {F_1}{l_2}} \over {{F_2} - {F_1}}}$

Given, energy of photons $= 10.2$ eV

Energy for nth state,

${E_n} = - 13.6\,{Z^2} \times \left( {{1 \over {n_i^2}} - {1 \over {n_f^2}}} \right)$

$= - 13.6 \times {(2)^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$ ($\because$ $Z = 2$)

$= - 13.6 \times 4 \times \left( {{{4 - 1} \over {16}}} \right)$

$= - 3.4 \times {3 \over 4}$

$= - 10.2$ eV

or, $E = 10.2$ eV

So, option (a) is correct.

(b) Number of lines in emission spectra

$= {{n(n - 1)} \over 2} = {{4 \times 3} \over 2} = 6$ [$\because$ $n = 4$]

So, option (b) is correct.

(c) For smallest wavelength of He⁺ spectrum the final state should be ${n_f} = \infty$

So, option (c) is incorrect.

(d) He⁺ electron cannot jump from $n = 2$ to $n = 3$ as the energy of photon is less than required.

So, option (d) is incorrect.

Considering the semi-circular wire of radius r. Force acting on it is shown below

Let T be the force exerted by each spring.

$\therefore$ $T + T = IBl$

$2T = IBl$

$2T = IB\,.\,\pi r$ ($\because$ $l = \pi r$)

$\therefore$ $T = {{IB\pi r} \over 2}$

Consider the figure given below.

Due to grounding, charge on outer sphere = 0

Now, potential due to inner sphere,

$V = {{kQ} \over r} = {{kQ} \over {{{3R} \over 2}}}$

$= {{2kQ} \over {3R}}$

or, ${1 \over {4\pi {\varepsilon _0}}}{{2Q} \over {3R}}$

Given, $S = \beta {k_B}A$

where, dimensional formula of $S = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]$

Dimensional formula of ${k_B} = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]$

Dimensional formula of $A = [{L^2}]$

Thus, $[{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}] = \beta [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][{L^2}]$

$\Rightarrow \beta = [{L^{ - 2}}]$

Consider the diagram given below.

When the charge q is displaced momentarily, it starts oscillating. The net force acting on q is given by ${F_{net}} = 2F\sin \theta$

$= - {{2\,kq \times q} \over {{{(\sqrt {{y^2} + {a^2}} )}^2}}}\,.\,{y \over {\sqrt {{y^2} + {a^2}} }} = - {{2k{q^2}y} \over {{{(\sqrt {{y^2} + {a^2}} )}^{{3 \over 2}}}}}$

For $a > > y$, we have

${F_{net}} = - {{2k{q^2}y} \over {{a^3}}}$

or, ${F_{net}} \propto - y$

de-Broglie wavelength of a charged particle having energy $\varepsilon$ is given by

$\lambda = {h \over {\sqrt {2m\varepsilon } }}$

Thus, ${\lambda _e} = {h \over {\sqrt {2{m_e}{\varepsilon _1}} }}$

${\lambda _\alpha } = {h \over {\sqrt {2{m_\alpha }{\varepsilon _2}} }}$

${\lambda _p} = {h \over {\sqrt {2{m_p}{\varepsilon _3}} }}$

Since, ${\lambda _e} = {\lambda _\alpha } = {\lambda _p}$ and ${m_\alpha } = {m_p} = {m_e}$

Thus, we can say

${1 \over {\sqrt {{m_\alpha }{\varepsilon _2}} }} = {1 \over {\sqrt {{m_p}{\varepsilon _3}} }} = {1 \over {\sqrt {{m_e}{\varepsilon _1}} }}$

$\therefore$ ${\varepsilon _1} > {\varepsilon _3} > {\varepsilon _2}$

Here, all the three capacitors are connected in parallel. So, equivalent capacitance will be

${C_{eq}} = C + C + C = 3C$

Electric field due to line charge is given by

$E = {\lambda \over {2\pi {\varepsilon _0}d}}$

where, $\lambda$ = linear charge density.

$\lambda = {q \over l} \Rightarrow q = \lambda l$

Electric force due to one wire on another wire,

${F_E} = q\,.\,E$

$= \lambda l\,.\,{\lambda \over {2\pi {\varepsilon _0}d}}$

$\Rightarrow {F_E} = {{{\lambda ^2}l} \over {2\pi {\varepsilon _0}d}}$

Now, current due to moving charged wire is given as

$I = {{dq} \over {dt}} = {d \over {dt}}(\lambda dl)$ [$\because$ $v = {{dl} \over {dt}}$]

$= \lambda v$

Magnetic force on one wire due to another wire,

${F_B} = {{{\mu _0}{I_1}{I_2}l} \over {2\pi d}}$

Since, both forces balance each other, thus

${F_E} = {F_B}$

$\Rightarrow {{{\lambda ^2}l} \over {2\pi {\varepsilon _0}d}} = {{{\mu _0}(\lambda v)(\lambda v)l} \over {2\pi d}}$ [$\because$ ${I_1} = {I_2} = I$]

$\Rightarrow {1 \over {{\varepsilon _0}{\mu _0}}} = {v^2}$

or, $v = {1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }} = c$

Given, change in flux, $d\phi = 5$ Wb

Emf due to change in flux is given as

$E = {{d\phi } \over {dt}} \Rightarrow E\,.\,dt = 5$

Now, it is given that, $R = 100\,\Omega$

Thus, current, $I = {V \over R} = {E \over R}$ ..... (i)

Also, $I = {{dq} \over {dt}}$ ....... (ii)

where, $dq$ is the charge flowing per unit time.

Thus, from Eqs. (i) and (ii), we have

${E \over R} = {{dq} \over {dt}} \Rightarrow {{E\,.\,dt} \over R} = dq$

$\Rightarrow {5 \over {100}} = dq \Rightarrow dq = 0.05\,C$

Load F can be given as

$F = \left( {{{AY} \over L}} \right)\Delta x$

where, A = cross-sectional area

Y = Young's modulus

L = length

$\Delta$x = elongation in length.

i.e., F versus $\Delta$x graph is a straight line of slope ${{YA} \over L}$.

${(slope)_A} > {(slope)_B}$

${\left( {{{YA} \over L}} \right)_A} > {\left( {{{YA} \over L}} \right)_B}$

or, ${(A)_A} > {(A)_B}$

As we know that, they are of same material.

Hence, ${Y_B} = {Y_A}$

So, option (a) and (d) are correct.

For the given isotherm, the temperature of the curve EF is greater than the curve CD and AB.

As we know that, most probable speed of the molecule is given as

${v_p} = \sqrt {{{2RT} \over M}}$

Thus, the ${v_p}$ at 3 and 4 will be same.

Hence, ${v_p}$ at 3 = ${v_p}$ at 4 > ${v_p}$ at 2 > ${v_p}$ at 1.

Given expression, $\overline A (A + B) + (B + AA)(A + \overline B )$

$= \overline A (A + B) + (B + A)(A + \overline B )$

$= A\overline A + \overline A B + AB + B\overline B + AA + A\overline B$

$= A + A\overline B + B(\overline A + \overline A )$ [$\because$ $A\overline A = B\overline B = 0$ and $A\,.\,A = A$]

$= A + A\overline B + B(\overline A + A)$

$= A(1 + \overline B ) + B(1)$ [$\because$ $= A + \overline A = 1$]

$= A + B$ [$\because$ $1 + \overline B = 1$]

The angular momentum is quantised to even multiple of ${h \over {2\pi }}$. Hence, the quantum numbers that are allowed are n = 2 and n = 4 ........ . So,

$E = 13.6\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

$= 13.6\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$

$= 13.6\left( {{1 \over 4} - {1 \over {16}}} \right)$

$= 2.55$ eV

Now, given that, $hc = 1242$ MeV-nm

Hence, $\lambda = {{hc} \over {2.55\,eV}} = {{1242\,eV\, - nm} \over {2.55\,eV}}$

= 487 nm

According to the question, given situation can be represented by the figure below.

From the Lorentz force, we have

$F = q(v \times B)$

Hence, force is acting directly towards the centre.

Thus, both angular momentum ($\mathbf{L}$) and torque ($\mathbf{T}$) will have the same direction. i.e. $\mathbf{T}$ || $\mathbf{L}$.

According to Gauss's theorem

Net flux through the surface $= {{{\mathrm{q_{enclosed}}}} \over {{\varepsilon _0}}}$

Here, ${\mathrm{q_{enclosed}}} = 0$

So, ${\phi _{\mathrm{curved\,surface}}} + {\phi _{\mathrm{circular\,surface}}} = 0$

${\phi _{\mathrm{curved\,surface}}} = - \,{\phi _{\mathrm{circular\,surface}}}$

$= - E\,.\,S = - ES\cos 180^\circ$

$= ES$ ($\because$ $\cos 180^\circ = - 1$)

$= E\,.\,\pi {R^2} = \pi {R^2}E$

As, electrostatic field is conservative in nature.

So, work done is path independent.

So, option (a) and (c) are correct.

At position A, direction of $\overrightarrow p$ is along + ve Y-axis.

At position A, direction of $\overrightarrow L$ is along + ve Z-axis because $\overrightarrow L$ = $\overrightarrow r$ $\times$ $\overrightarrow p$ and $\overrightarrow r$ is along + ve X-axsis.

Thus, $\overrightarrow \alpha$ = $\overrightarrow p$ $\times$ $\overrightarrow L$

$= \widehat i \times \widehat k$

$= - \widehat j$ or $-$ve Y-axis.

Given $p{V^3} = C,\,{C_V} = {{3R} \over 2},\,n = 1$ mol

For polytropic process,

$p{V^x} = C$

$\therefore$ $x = 3$

$C = {C_V} + {R \over {1 - x}} = {{3R} \over 2} + {R \over {1 - 3}} = R$

Heat supplied $(Q) = nC\Delta T$

$= 1 \times R \times ({T_2} - {T_1}) = R({T_2} - {T_1})$

Now, As $pV = nRT$ ...... (i)

$\therefore$ $pV = RT$ ($\because$ $n = 1$)

$\therefore$ $R = {{pV} \over T} = {{{p_1}{T_1}} \over {{T_1}}}$

Also, $p{V^3} = C$

$\Rightarrow \left( {{{RT} \over V}} \right){V^3} = C$ [from Eq. (i)]

$\Rightarrow T{V^2} = C \Rightarrow {T_1}V_1^2 = {T_2}V_2^2$

$\therefore$ ${T_2} = {{{T_1}V_1^2} \over {V_2^2}}$

Heat supplied $(Q) = {{{p_1}{V_1}} \over {{T_1}}}\left[ {{{{T_1}V_1^2} \over {V_2^2}} - {T_1}} \right] = {p_1}{V_1}\left[ {{{V_1^2} \over {V_2^2}} - 1} \right]$

According to the question,

$A = \sqrt {{A^2} + {A^2} + 2AA\cos \delta }$

$\Rightarrow \cos \delta = - {1 \over 2}$

or, $\delta = 120^\circ$

or, $= {{2\pi } \over 3}$

Given, trajectory of the particle,

$\overrightarrow r = b\cos \omega t\widehat i + b\sin \omega t\widehat j$

It's component can be written as

$x = b\cos \omega t$ ..... (i)

$y = b\sin \omega t$ ...... (ii)

Both Eqs. (i) and (ii) together represent the circle in x-y plane. Thus, the trajectory of the particle in x-y plane is circle.

Now,

${{d\overrightarrow r } \over {dt}} = \omega b( - \sin \omega t\widehat i + \cos \omega t\widehat j)$

${{{d^2}\overrightarrow r } \over {d{t^2}}} = - b{\omega ^2}(\cos \omega t\widehat i + \sin \omega t\widehat j)$

$= - {\omega ^2}(\overrightarrow r )$

Therefore, $\overrightarrow a = - {\omega ^2}\overrightarrow r$

Hence, option (c) and (d) are incorrect.

Energy of the particle in circular motion is given as ${1 \over 2}m{\omega ^2}{A^2}$ (where, A is the maximum displacement from the centre), which is constant.

Hence, ${E \over \omega }$ is also constant.

So, option (a) and (b) are correct.

Given, ${V_s} = 10\,V$

${V_z} = 6\,V$

Maximum current $({I_{z\,\max }}) = 40\,mA = 40 \times {10^{ - 3}}\,A$

Minimum current $({I_{z\,\min }}) = 10\,mA = 10 \times {10^{ - 3}}\,A$

Maximum value of series resistance R_S in the voltage regulator circuit is given as

${R_{\max }} = {{{V_s} - {V_z}} \over {{I_{z\,\min }}}}$

$= {{10 - 6} \over {10 \times {{10}^{ - 3}}}}$

$= 400\,\Omega$

Maximum height of projectile,

$h = {{{u^2}{{\sin }^2}\theta } \over {2g}}$ ...... (i)

At maximum height velocity of projectile,

$v = u\cos \theta$

$\therefore$ Momentum at highest point $= mu\cos \theta$

$\therefore$ Angular momentum about origin,

$p = mu\cos \theta \times h$

$= mu\cos \theta \times {{{u^2}{{\sin }^2}\theta } \over {2g}}$

$= {{m{u^3}\cos \theta {{\sin }^2}\theta } \over {2g}}$

$\Rightarrow p \propto {u^3}$

According to work-energy theorem,

change in kinetic energy = work done

= force $\times$ distance

Thus, $\mathrm{force = {{change\,in\,energy} \over {change\,in\,distance}}}$

At x = 10 m, we have

$\mathrm{F = {{20 - 40} \over {12 - 8}}\widehat i = - 5\widehat i\,N}$

Given, initial velocity, $u = 0$ m/s

Final velocity, $v = 100$ m/s

Mass of ball, $m = 50 \times {10^{ - 3}}$ kg

Time of contact $= 0.02\,s$

Impulse, $F = {{m(v - u)} \over t}$

$= {{50 \times {{10}^{ - 3}}(100 - 0)} \over {0.02}}$

$= {{5 \times 100} \over 2}$

$= 250\,N$

We know that, $E = - {{dV} \over {dr}}$

Thus, electric field from ($-$4 to $-$2),

$E = - \left( {{{10 - 0} \over {4 - ( - 2)}}} \right) = {{10} \over 2} = 5$ N/C

Electric field from ($-$2 to 2),

$E = - \left( {{{10 - 10} \over {2 - ( - 2)}}} \right) = 0$

Electric field from (2 to 7),

$E = - \left( {{{10 - 0} \over {2 - 7}}} \right) = - 2$ N/C

Moment of inertia about line $AX = m{a^2} + m{d^2}$

We know that,

$\sin \theta = {d \over a}$

$\Rightarrow {1 \over 2} = {d \over a}$ ($\because$ $\theta = 30^\circ$)

$\Rightarrow d = {a \over 2}$

$\therefore$ ${I_{Ax}} = m{a^2} + m{\left( {{a \over 2}} \right)^2}$

$= m{a^2} + {{m{a^2}} \over 4} = {5 \over 4}m{a^2}$

$U = A{p^2}V$, $A =$ constant

For adiabatic process, $dQ = 0$

$\therefore$ From first law of thermodynamics,

$dQ = dU + dW$

$\therefore$ $dU + dW = 0$ ($\because$ $dQ = 0$)

$\Rightarrow d(A{p^2}V) + pdV = 0$

$\Rightarrow A(2p\,.\,Vdp + {p^2}dV) + pdV = 0$

$\Rightarrow (A{p^2} + p)dV + 2ApVdp = 0$

$\Rightarrow (A{p^2} + p)dV = - 2ApVdp$

$\Rightarrow {{dV} \over { - 2AV}} = {{pdp} \over {p(Ap + 1)}}$