JEE Main 2021 24th February Morning Shift Question Paper with Solutions

A $\to$ B = isothermal process

B $\to$ C = isobaric process

C $\to$ A = isochoric process

also, V₂ = 2V₁

work done by gas in the complete cycle ABCA is -

$\Rightarrow$ w = w_AB + w_BC + w_CA .....(1)

$\Rightarrow$ w_CA = 0, as isochoric process

$\Rightarrow$ w_AB = 2P₁V₁ ln$\left( {{{{v_2}} \over {{v_1}}}} \right)$ = 2 nRT ln(2)

$\Rightarrow$ w_BC = P₂(V₁ $-$ V₂) = P₂ (V₁ $-$ 2V₁) = $-$P₂V₁ = $-$nRT

$\Rightarrow$ Now put the value of w_AB, w_BC and w_CA in equation, we get

$\Rightarrow$ w = 2nRT ln(2) $-$ nRT + 0

$\Rightarrow$ w = nRT [2ln (2) $-$ 1]

$\Rightarrow$ w = nRT [ln (2) $-$ ${{1 \over 2}}$]

Given, initial amplitude = A

Velocity at mean position, v = A$\omega$

Applying conservation of momentum at mean position, we get

M₁v₁ = M₂v₂

MA$\omega$ = (M + m)v'

$\Rightarrow v' = {{MA\omega } \over {M + m}} = {{MA\sqrt {{k \over M}} } \over {M + m}}$

$\therefore$ $v' = A'\omega ' = A'\sqrt {{k \over {M + m}}}$

$\Rightarrow A' = {{MA\sqrt {{k \over M}} } \over {M + m}} \times \sqrt {{{M + m} \over k}}$

$A' = \sqrt {{M \over {M + m}}} A$

Given, $\mu$_r = $\varepsilon$_r = 2

where, $\mu$_r is relative permeability, $\varepsilon$_r is relative permittivity.

Speed of electromagnetic wave v is given by

$v = {c \over n}$

where, n = refractive index = $\sqrt {{\mu _r}{\varepsilon _r}} = \sqrt 4 = 2$

$\Rightarrow v = {{3 \times {{10}^8}} \over 2}$ = 15 $\times$ 10⁷ m/s

$\because$ x $\times$ 10⁷ = 15 $\times$ 10⁷

$\Rightarrow$ x = 15

where, k is Boltzmann constant,

T is temperature and x is displacement.

We know that, ${{{x^2}} \over {\alpha kT}}$ is a dimensionless quantity.

$\therefore$ $\left[ {{{{x^2}} \over {\alpha kT}}} \right] = [{M^0}{L^0}{T^0}] \Rightarrow [\alpha ] = {{[{x^2}]} \over {[k][T]}}$

$\Rightarrow [\alpha ] = {{[{L^2}]} \over {[k][T]}}$ ..... (i)

Since, dimensions of k are

$[k] = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]$ ...... (ii)

Dimensions of temperature are

$[T] = [K]$ ..... (iii)

Substituting Eqs. (ii) and (iii) in Eq. (i), we get

$[\alpha ] = {{[{L^2}]} \over {[{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}$

$[\alpha ] = [{M^{ - 1}}{T^2}]$

According to dimensional analysis,

$[W] = [\alpha {\beta ^2}]$

$\Rightarrow [{\beta ^2}] = {{[W]} \over {[\alpha ]}}$

$\Rightarrow [{\beta ^2}] = {{{M^1}{L^2}{T^{ - 2}}]} \over {[{M^{ - 1}}{T^2}]}} = [{M^2}{L^2}{T^{ - 4}}]$

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

From the graph for first line, the slope is negative and intercept is positive.

So, equation of line is

$v = - mt + c$

$\Rightarrow {a_1} = {{dv} \over {dt}} = - m$

Similarly, for second line, the slope is positive and intercept is negative, so equation of line is

$v = mt - c$

$\Rightarrow {a_2} = {{dv} \over {dt}} = m$

$\therefore$ The corresponding acceleration-time graph as shown below


Hence, option (a) is correct.

The circuit can be shown as below


The current through the circuit,

$I = {{{\varepsilon _1} - {\varepsilon _2}} \over {{r_1} + {r_2}}} = {{6 - 4} \over {10}} = {1 \over 5}A$

$\therefore$ Potential difference across points X and Y is

${V_{XY}} = {E_2} + I{r_2} = 4 + {1 \over 5} \times 8 = 5.6V$

Given, L = 2 $\times$ 10^$-$4 H, R = 6.28 $\Omega$, f₀ = 10 MHz = 10 $\times$ 10⁶ Hz

$\therefore$ Quality factor $= {\omega _0}{L \over R} = 2\pi {f_0}{L \over R}$

$= 2\pi \times 10 \times {10^6} \times {{2 \times {{10}^{ - 4}}} \over {6.28}}$

$= 2 \times {10^3} = 2000$

We know that,

$Y = 3K(1 - 2\sigma )$

$\Rightarrow \sigma = {1 \over 2}\left( {1 - {Y \over {3K}}} \right)$ ..... (i)

Also, $Y = 2\eta (1 + \sigma )$

$\Rightarrow \sigma = {Y \over {2\eta }} - 1$ .... (ii)

On comparing Eqs. (i) and (ii), we get

$\left( {1 - {Y \over {3K}}} \right){1 \over 2} = {Y \over {2\eta }} - 1$

On solving, we get

$K = {{\eta Y} \over {9\eta - 3Y}}$ N/m²

The graph for given equation is shown below


At maximum height, the slope of tangent drawn,

$\tan \theta = {{dy} \over {dx}} = {{2x} \over 4} = {x \over 2}$ [$\because$ $y = {{{x^2}} \over 4}$]

$\Rightarrow 0.5 = {x \over 2}$ ($\because$ $\mu$ = tan$\theta$)

$\Rightarrow$ x = 1 m

$\therefore$ $y = {{{x^2}} \over 4} = {1 \over 4}$ = 0.25 m = 25 cm

For convex mirror, the focal length (f) and radius of curvature (r) are related as $f = + {r \over 2}$.

Given, $i = {\alpha _0}t + \beta {t^2}$

where, $\alpha$₀ = 20 A/s, $\beta$ = 8 A/s²

We know that, $i = {{dq} \over {dt}}$

$\Rightarrow {{dq} \over {dt}} = i = {\alpha _0}t + \beta {t^2} = 20t + 8{t^2}$

$\Rightarrow dq = (20t + 8{t^2})dt$

On integrating both sides, we get

$\int\limits_0^q {dq = \int\limits_0^{15} {(2t + 8{t^2})dt} }$

$q = \left[ {{{20{t^2}} \over 2} + {{8{t^3}} \over 3}} \right]_0^{15} = 10 \times {(15)^2} + {8 \over 3} \times {(15)^3}$

$\therefore$ q = 11250 C

Given, resistance, R = 2 k$\Omega$ = 2 $\times$ 10³ $\Omega$

In Zener breakdown,

$i = {V \over R} = {5 \over {2 \times {{10}^3}}} = 2.5 \times {10^{ - 3}}$

$\because$ $x \times {10^{ - 4}} = 25 \times {10^{ - 4}}$

$\therefore$ x = 25

The given situation is shown below


The gravitational force between these two stars provide the required centripetal force for rotation in a circle about their common centre.

Assuming 2 m at origin, the centre of mass of the system lies at

$x = {{2m \times 0 + m \times d} \over {2m + m}} = {d \over 3}$

Hence, ${F_G} = {F_C}$

where, F_G is gravitational force between them and F_C is centripetal force.

$\Rightarrow {{G{m_1}{m_2}} \over {{r^2}}} = 2m{\omega ^2}x$

$\Rightarrow {{G(2m)(m)} \over {{d^2}}} = 2m{\omega ^2} \times {d \over 3} \Rightarrow {\omega ^2} = {{3Gm} \over {{d^3}}}$

$\Rightarrow \omega = \sqrt {{{3Gm} \over {{d^3}}}}$

We know that,

$\omega = {{2\pi } \over T}$

$\therefore$ $T = {{2\pi } \over \omega }$

$\sqrt {{{3Gm} \over {{d^3}}}}$

$= 2\pi \sqrt {{{{d^3}} \over {3Gm}}}$ [using Eq. (i)]

Given, I₀ = 100 lumens

When analyser is rotated through an angle $\theta$, the intensity of light will becomes

I = I₀ cos²$\theta$ = 100 $\times$ cos²30$^\circ$

= 100 $\times$ ${\left( {{{\sqrt 3 } \over 2}} \right)^2}$ = 75 lumens

We know that, $\gamma = 3\alpha$ .... (i)

where, $\alpha$ is the coefficient of linear expansion and $\gamma$ is the coefficient of volume expansion.

We know that,

${{\Delta V} \over V} = \gamma \Delta T$

$\Rightarrow {{\Delta V} \over V} = 3\alpha \Delta T$ [from Eq. (i)]

$\Delta V = 3{a^3}\alpha \Delta T$ [ $\because$ volume of cube = a³ ]

Given, coefficient of static friction, $\mu$_s = 0.2


Various forces acting on block are shown below


Frictional force $\le$ mg

$\Rightarrow$ N $\times$ 0.2 $\le$ 5

$\Rightarrow$ N $\le$ 25

$\therefore$ Magnitude of horizontal force, F = N = 25 N

In transition A, hydrogen atom comes from higher energy state n = $\infty$ to lower energy state n = 1. Hence, transition A represents series limit of Lyman series. In transition B, hydrogen atom comes from higher energy state n = 5 to lower energy state n = 2. Hence, transition B represents 3rd line of Balmer series. In transition C, hydrogen atom comes from higher energy state n = 5 to lower energy state n = 3. Hence, transition C represents 2nd line of Paschen series.

Hence, option (c) is correct.

Given, amplitude $\propto$ width of slit

$\Rightarrow$ A₂ = 3A₁

We know that,

${{{I_{\max }}} \over {{I_{\min }}}} = {{{{(\sqrt {{I_1}} + \sqrt {{I_2}} )}^2}} \over {{{(\sqrt {{I_1}} - \sqrt {{I_2}} )}^2}}}$

$\because$ Intensity, I $\propto$ A²

$\therefore$ ${{{I_{\max }}} \over {{I_{\min }}}} = {{{{({A_1} + {A_2})}^2}} \over {{{(|{A_1} - {A_2}|)}^2}}}$

$= {\left( {{{{A_1} + 3{A_1}} \over {|{A_1} - 3{A_1}|)}}} \right)^2} = {\left( {{{4{A_1}} \over {2{A_1}}}} \right)^2} = {4 \over 1}$

$\therefore$ ${I_{\max }}:{I_{\min }} = 4:1$

Given, period of revolution of first satellite,

T₁ = 1h

Period of revolution of second satellite,

T₂ = 8h

$\therefore$ ${{{T_1}} \over {{T_2}}} = {1 \over 8}$

We know that, $\omega = {{2\pi } \over T}$

$\Rightarrow \omega \propto {1 \over T}$

$\because$ ${{{\omega _1}} \over {{\omega _2}}} = {{{T_2}} \over {{T_1}}} \Rightarrow {{{\omega _1}} \over {{\omega _2}}} = {8 \over 1}$

or ${\omega _1}:{\omega _2} = 8:1$

Given, m = 1 kg, R = 1 m

We know that,

$F = {{G{m_1}{m_2}} \over {{r^2}}}$

$\because$ ${F_1} = {{Gmm} \over {{{(2R)}^2}}} = {{G{m^2}} \over {4{R^2}}}$

and ${F_2} = {{Gmm} \over {{{(\sqrt 2 R)}^2}}} = {{G{m^2}} \over {2{R^2}}}$

Net force on one particle,

${F_{net}} = {F_1} + {F_2}\cos 45^\circ + {F_2}\cos 45^\circ$

$= {F_1} + 2{F_2}\cos 45^\circ$

$= {{G{m^2}} \over {4{R^2}}} + 2\left( {{{G{m^2}} \over {2{R^2}}}} \right).{1 \over {\sqrt 2 }}$

$= {{G{m^2}} \over {4{R^2}}} + {{G{m^2}} \over {\sqrt 2 {R^2}}}$

$= {{G{m^2}} \over {{R^2}}}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right]$

As the gravitational force provides the necessary centripetal force, so

${F_{net}} = {F_C} = {{m{v^2}} \over R}$

Here, F_C = centripetal force.

$\Rightarrow {{G{m^2}} \over {{R^2}}}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right] = {{m{v^2}} \over R}$

$\Rightarrow v = {1 \over 2}\sqrt {{{Gm} \over R}(1 + 2\sqrt 2 )}$

$\Rightarrow v = {1 \over 2}\sqrt {G(1 + 2\sqrt 2 )}$

As we know, $\lambda = {h \over p} = {h \over {\sqrt {2mK} }}$

If linear momenta of two photons are equal, then their wavelengths is also equal.

Also, if the wavelength is decreased, then the momentum and energy of photon will increase.

Hence, option (d) is correct.

According to Pascal's law,

${{{F_1}} \over {{A_1}}} = {{{F_2}} \over {{A_2}}}$

Initially, ${{100g} \over {{A_1}}} = {{mg} \over {{A_2}}}$ .... (i)

Finally, ${{Mg} \over {16{A_1}}} = {{mg} \over {\left( {{{{A_2}} \over {16}}} \right)}}$ ..... (ii)

On dividing Eqs. (i) by (ii), we get

${{100 \times 16} \over M} = {1 \over {16}}$

$\therefore$ M = 25600 kg

Let M and R be the mass and radius of four bodies. Then, as per question, their moment of inertia are

I₁ = ${{M{R^2}} \over 2}$,

I₂ = ${{M{R^2}} \over 2}$,

I₃ = ${{M{R^2}} \over 2}$,

I₄ = ${2 \over 5}M{R^2}$

$\therefore$ I₁ = I₂ = I₃ > I₄

Given, emitter current, I_E = 4 mA

Collector current, I_C = 3.5 mA

Current gain in common base amplifier,

$\alpha = {{{I_C}} \over {{I_E}}}$

$\Rightarrow \alpha = {{3.5} \over 4} = {7 \over 8}$

Also, current gain in common emitter amplifier,

$\beta = {\alpha \over {1 - \alpha }}$

$\Rightarrow \beta = {{7/8} \over {1 - 7/8}}$

$\beta = 7$

Given, C₁ = C₂ = C

When both capacitors are connected in series, their equivalent capacitance will be

${1 \over {{C_s}}} = {1 \over C} + {1 \over C} = {2 \over C}$

$\Rightarrow {C_s} = {C \over 2}$

When both capacitors are connected in parallel, their equivalent capacitance will be

C_p = C + C = 2C

$\therefore$ The ratio of equivalent capacitance in series and parallel combination is

${{{C_s}} \over {{C_p}}} = {{C/2} \over {2C}} = {1 \over 4}$

$\therefore$ C_s : C_p = 1 : 4

We know that, in isothermal process, $\Delta$T = 0

In isochoric process, $\Delta$V = 0

In adiabatic process, $\Delta$Q = 0

In isobaric process, $\Delta$p = 0

So, the correct match is,

A $\to$ 2, B $\to$ 3, C $\to$ 4, D $\to$ 1

Given, audio signal,

V_m = 20 sin 2$\pi$(1500t) .... (i)

Carrier signal, V_c = 80 sin 2$\pi$(100000t) .... (ii)

We know that, modulation index,

${m_f} = {{{A_m}} \over {{A_c}}}$

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

A_m = 20, A_c = 80

Percentage of modulation index,

${m_f} = {{{A_m}} \over {{A_c}}} \times 100 = {{20} \over {80}} \times 100 = 25\%$

The situation is shown below


Using conservation of linear momentum in y-direction,

p_i = p_f

As, p_i = 0

and p_f = mv₁ sin30$^\circ$ $-$ mv₂ sin30$^\circ$

$\Rightarrow$ 0 = m $\times$ ${1 \over 2}$v₁ $-$ m $\times$ ${1 \over 2}$v₂

$\Rightarrow$ v₁ = v₂ or v₁ : v₂ = 1 : 1

Since, v₁ : v₂ = x : y (given)

$\therefore$ x = 1

In a transformer,

${{{N_p}} \over {{N_s}}} = {{{V_p}} \over {{V_s}}}$

where, N_p = number of turns in primary circuit, N_s = number of turns in secondary circuit = 24, V_p = potential of primary circuit = 220 V and V_s = potential of secondary circuit = 12 V

$\Rightarrow {{{N_p}} \over {24}} = {{220} \over {12}}$

$\Rightarrow {N_p} = 440$

We can replace $-$ Q charge at origin by + Q and $-$ 2Q. Now, due to + Q charge at every corner of cube, electric field at centre of cube is zero. So, net electric field at centre is only due to $-$ 2Q charge at origin. Vector form of electric field strength,

$E = {{Kqr} \over {{r^3}}}$

Here, position vector, $r = {a \over 2}(\widehat x + \widehat y + \widehat z)$

$\Rightarrow \left| r \right| = \sqrt {{{\left( {{a \over 2}} \right)}^2} + {{\left( {{a \over 2}} \right)}^2} + {{\left( {{a \over 2}} \right)}^2}} = {{\sqrt 3 a} \over 2}$

$\Rightarrow E = {1 \over {4\pi {\varepsilon _0}}} \times {{( - 2Q)a/2} \over {{{\left( {{a \over 2}\sqrt 3 } \right)}^3}}}\,.\,(\widehat x + \widehat y + \widehat z)$

$E = {{-2Q(\widehat x + \widehat y + \widehat z)} \over {3\sqrt 3 \pi {a^2}{\varepsilon _0}}}$

For a reaction, A(g) $\to$ B(g)

Given, K_p (equilibrium constant) = 100

Temperature = 300 K

Pressure = 1 atm

Formula used, $\Delta$G$^\circ$ = $-$ RT ln K_p .... (i)

Here, $\Delta$G$^\circ$ = standard Gibb's free energy

R = gas constant = 8.31 J mol^$-$1 K^$-$1

Put value in Eq. (i), we get

$\Delta$G$^\circ$ = $-$ R (300) ln 100

$\Delta$G$^\circ$ = $-$ R (300) (2) ln (10)

$\because$ ln (10) = 2.3

$\Delta$G$^\circ$ = $-$ R(300) (2) (2.3)

$\Delta$G$^\circ$ = $-$ 1380 R

Hence, $\Delta$G$^\circ$ = $-$ xR

$\therefore$ x = 1380

Composition of Gun metal $\to$ 90% Cu + 8% Sn + 2% Zn

Note :

Alloys	Composition
Brass	60% Cu + 40% Zn
Bronze	75-90% Cu + 10-25% Sn
Gun-metal	90% Cu + 8% Sn + 2% Zn
Aluminium bronze	90% Cu + 10% Al
Monel metal	30% Cu + 67% Ni + 3% (Fe + Mn)
German Silver	56% Cu + 24% Zn + 20% Ni
Manganin	78-86% Cu + 13-18% Mn + 1-4% Ni
Bell metal	80% Cu + 20% Sn
Constantan	60% Cu + 40% Ni

Methane (CH₄) gas is evolved due to anaerobic degradation of vegetation which causes global warming and cancer. It is a heat trapping gas that forces the planet to warm drastically and quickly.

On moving left to right in a period of the periodic table, Ionisation energy (I.E.) increases due to increase in effective nuclear charge i.e. Z_eff.

But due to extra stability of fully filled and half-filled electronic configuration of Mg and P required ionisation enthalpy is more from neighbouring elements.

i.e. First ionisation enthalpy order is Al < Mg < Si < S < P.

An amphoteric compound is a molecule or ion that can react with both as an acid or as a base.

BeO = Amphoteric

BaO = Basic

Be(OH)₂ = Amphoteric

Sr(OH)₂ = Basic

Both beryllium compound BeO and Be(OH)₂ are amphoteric in nature while compound BaO and Sr(OH)₂ are basic in nature, they form alkaline solution in H₂O.

Coordination number is the number of nearest neighbours of a central atom in the structure.

bcc has a coordination number of 8 and contains 2 atoms per unit cell.

This is because each atom touches four atoms in the layer above it, four in the layer below it and none in its own layer.

Given, weight of compound A = 4.5 g

Molecular weight of compound A = 90 g/mol

Volume of solution (in mL) = 250 mL

Now, molarity is defined as number of moles of solute or compound A divided by volume of solution (in L).

$M = {{Number\,of\,moles\,of\,solute\,(n)} \over {Volume\,of\,solution}}$

$= {{{{4.5} \over {90}}} \over {{{250} \over {1000}}}} = 0.2$ = 2 $\times$ 10^$-$1 M

$\therefore$ $n = {{Weight\,of\,solute\,(compound\,A)} \over {Molecular\,weight\,of\,solute\,(compound\,A)}}$

Hence, x $\times$ 10^$-$1 $\mu$

x = 2

The two half reaction, one separately are as follows

${S_8} + 16{e^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }}$ (Reduction)

${S_8} + 12{H_2}O\buildrel {} \over \longrightarrow 4{S_2}O_3^{2 - } + 24{H^ + } + 16{e^ - }$
_____________________________________

$2{S_8} + 12{H_2}O\buildrel {} \over \longrightarrow \mathop {8{S^{2 - }} + 4{S_2}O_3^{2 - }}\limits_{} + 24{H^ + }$

For balancing in basic medium, Add an equal number of OH^$-$ that of H⁺, we get

$2{S_8} + 12{H_2}O + 24O{H^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }} + 4{S_2}O_3^{2 - } + 24{H_2}O$

$2{S_8} + 24O{H^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }} + 4{S_2}O_3^{2 - } + 12{H_2}O$

or ${S_8} + 12O{H^ - }\buildrel {} \over \longrightarrow 4{S^2} + 2{S_2}O_3^{2 - } + 6{H_2}O$ .... (i)

On comparing (i) with

${S_8} + aO{H^ - }(aq)\buildrel {} \over \longrightarrow b{S^2}(aq) + c{S_2}O_3^{2 - } + d{H_2}O$

We get, a = 12; b = 4; c = 2; d = 6

Step 1 : In step 1, NaNO₂ + HCl , 0-5°C used for diazotisation reaction. It will form diazonium salt. Further, it will react with KCN to form cyanobenzene.


Step 2 : In step 2, SnCl₂ and HCl is a Stephen’s reduction reagent. Cyanobenzene reduced to benzaldehyde by SnCl₂/HCl.

Al₂O₃ (aluminium oxide) was leached with alkali to get Na[Al(OH)₄] (X) (sodium aluminate). The solution of sodium aluminate when passed through gas Y, i.e. carbon dioxide (CO₂) forms Al₂O₃.xH₂O (Z).

Complete reactions are as follows

Given, mass of Li³⁺ = 8.3 times of mass of proton formula,

De-Broglie wavelength, $\lambda = {h \over {\sqrt {2mqV} }}$

Here, h = Planck's constant = 6.624 $\times$ 10^$-$34 J-s

m = Mass of atom

q = Charge (or number of electrons)

${\lambda _{Li}} = {h \over {\sqrt {2{m_{Li}}(3e)v} }}$ .... (i)

${\lambda _P} = {h \over {\sqrt {2{m_P}(e)v} }}$ .... (ii)

Now, Eq. (i) divided by Eq. (ii), we get,

${{{\lambda _{Li}}} \over {{\lambda _P}}} = \sqrt {{{{m_P}(e)v} \over {{m_{Li}}(3e)v}}}$

We know that m_Li = 8.3 m_p

${{{\lambda _{Li}}} \over {{\lambda _P}}} = \sqrt {{{{m_P} \times e \times V} \over {8.3\,{m_P} \times 3e \times V}}} = \sqrt {{1 \over {8.3 \times 3}}} = {1 \over 5} = 0.2$ = 2 $\times$ 10^$-$1

Hence, x $\times$ 10^$-$1, x = 2

Isostructural compounds are those compounds which have same structure as well as same hybridisation.

Formula for find hybridisation : H = (Lone pair + Sigma bond + coordinate bond )

If number of sigma bond ($\sigma$), co-ordinate bond and lone pair are same for given pairs, they are isostructural.