JEE Main 2021 18th March Evening Shift Question Paper with Solutions

Radius, $r = {{mv} \over {qB}} = {{\sqrt {2mK} } \over {qB}}$

$\Rightarrow K = {{{r^2}{q^2}{B^2}} \over {2m}}$

$\therefore$ ${{{K_p}} \over {{K_\alpha }}} = {\left( {{{{r_p}} \over {{r_\alpha }}}} \right)^2} \times {\left( {{{{q_p}} \over {{q_\alpha }}}} \right)^2} \times {{{m_\alpha }} \over {{m_p}}}$

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

$= 4$

From Doppler effect, we know

$\lambda ' = \lambda \left[ {1 + {v \over c}} \right]$

here, $v = 286 \times {10^3}$ m/s

$c = 3 \times {10^8}$ m/s

$\therefore$ $\lambda ' - \lambda = {{\lambda v} \over c}$

$= {{630 \times {{10}^{ - 9}} \times 286 \times {{10}^3}} \over {3 \times {{10}^8}}}$

= 6 $\times$ 10^$-$10 m

General equation of SHM

x = A sin($\omega$'t $\pm$ $\phi$)

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

Given, $T = {\pi \over \omega }$

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

$\therefore$ Equation becomes,

x = a sin(2$\omega$t $\pm$ $\phi$)

Here coefficient of t is 2$\omega$.

you can see only option (D) has coefficient 2$\omega$.

At point P,

$\mu \sin 45^\circ = 1\sin r$

$\Rightarrow \sin r = {\mu \over {\sqrt 2 }}$

For light to come out of prism

sin r should be less than 1.

$\therefore$ $\sin r < 1$

$\Rightarrow {\mu \over {\sqrt 2 }} < 1$

$\Rightarrow \mu < \sqrt 2$

$\Rightarrow \mu < 1.414$

$\therefore$ The material which have refractive index less than 1.414, that material can emerge light out of the prism.

Here only red material have refractive index less than 1.414.

$\therefore$ Red light will come out of the prism.

$\alpha = {{{I_C}} \over {{I_E}}}$ & $\beta = {{{I_C}} \over {{I_B}}}$ & ${I_E} = {I_B} + {I_C}$

$\therefore$ ${{{I_E}} \over {{I_C}}} = {{{I_B}} \over {{I_C}}} + {{{I_C}} \over {{I_C}}}$

$\Rightarrow {1 \over \alpha } = {1 \over \beta } + 1 = {{1 + \beta } \over \beta }$

$\Rightarrow \alpha = {\beta \over {1 + \beta }}$

Magnetic energy, U = ${1 \over 2}$LI$_0^2$

Given : U = 25% of U₀.

$\Rightarrow$ ${1 \over 2}L{I^2} = {1 \over 4} \times {1 \over 2}LI_0^2$

$\Rightarrow {I^2} = {{I_0^2} \over 4} \Rightarrow I = {{{I_0}} \over 2}$

$\therefore$ $I = {I_0}(1 - {e^{ - t/\tau }})$

$\Rightarrow {{{I_0}} \over 2} = {I_0}(1 - {e^{ - t/\tau }})$

$\Rightarrow {1 \over 2} = {e^{ - t/\tau }}$

$\Rightarrow {e^{t/\tau }} = 2$

$\Rightarrow t = \tau \ln 2$

$\Rightarrow t = {L \over R}\ln 2$

$\therefore$ From figure,

L = $\pi$R

$\Rightarrow$ R = ${L \over \pi }$

Moment of inertia about center O,

I = MR² = M${\left( {{L \over \pi }} \right)^2}$ = ${{{M{L^2}} \over {{\pi ^2}}}}$

For calculation of Range from tower.

$d = \sqrt {2R{h_T}} + \sqrt {2R{h_R}}$

$\therefore$ h_T = height of tower

& h_R = height of receiver

For 1^st case : h_T = 20m, h_R = 0

$\because$ ${d_1} = \sqrt {2 \times 6400 \times {{10}^3} \times 20}$ = 16 km

For 2^nd case : h_T = 20m, h_R = 5m

$\because$ ${d_2} = \sqrt {2 \times 6400 \times {{10}^3} \times 20} + \sqrt {2 \times 6400 \times {{10}^3} \times 5}$

= 16 + 8 = 24 km

$\therefore$ % change in range

$= {{{d_2} - {d_1}} \over {{d_1}}} \times 100$

$= {{24 - 16} \over {16}} \times 100$

$= {8 \over {16}} \times 100$ = 50%

For objects to float

mg = 2$\omega$²R

$\omega$ = angular velocity of earth.

R = Radius of earth

$\omega = \sqrt {{g \over R}}$ ..... (1)

Duration of day = T

$T = {{2\pi } \over \omega }$ ..... (2)

$\Rightarrow T = 2\pi \sqrt {{R \over g}}$

$= 2\pi \sqrt {{{6400 \times {{10}^3}} \over {10}}}$

$\Rightarrow {T \over {60}}$ = 83.775 minutes

$\simeq$ 84 minuites

Let length of each wire is l and area A. When they are connected in parallel then their effective area 2A.

From formula we know,

${R_{eq}} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}$

$\Rightarrow {{\rho l} \over {2A}} = {{{\rho _1}{l \over A} \times {\rho _2}{l \over A}} \over {{\rho _1}{l \over A} + {\rho _2}{l \over A}}}$

$\Rightarrow {\rho \over 2} = {{{\rho _1} \times {\rho _2}} \over {{\rho _1} + {\rho _2}}}$

$\Rightarrow {\rho \over 2} = {{6 \times 3} \over {6 + 3}} = 2$

$\Rightarrow \rho = 4$

$\because$ $\widehat E \times \widehat B = \widehat C$

$\therefore$ $\widehat E \times \widehat B$ should point in the direction of propagation of wave (y-direction here)

$\therefore$ possible combinations are (E_x, B_z) or (E_z, B_x)

Force at point A,

mg + p₀A = pA

$\Rightarrow$ p = p₀ + ${{mg} \over A}$

Given, area of A = 0.4 m² = 0.4 $\times$ 10⁴ cm² and area of B = 1 cm²

applying continuity equation

AV₁ = av [V₁ = velocity at point A]

$\Rightarrow$ V₁ = ${a \over A}v$

As A >>> a so ${a \over A}$ very small

$\therefore$ V₁ <<< v

So we can neglect V₁ by assuming V₁ = 0

On applying Bernoulli's theorem at points A and B,

${p_0} + {{mg} \over A} + {{\rho V_1^2} \over 2} + \rho gh = {p_0} + {{\rho {v^2}} \over 2} + 0$

$\Rightarrow {p_0} + {{mg} \over A} + 0 + \rho gh = {p_0} + {{\rho {v^2}} \over 2}$

$\Rightarrow {{mg} \over A} + \rho gh = {{\rho {v^2}} \over 2}$

$\Rightarrow {{24 \times 10} \over {0.4}} + 1000 \times 10 \times 0.4 = {{1000 \times {v^2}} \over 2}$

$\Rightarrow$ v $\simeq$ 3 m/s

For translational equilibrium,

T + f = mg sin60$^\circ$ ..... (1)

For rotational equilibrium,

Torque about center of mass should be zero.

$\therefore$ $\tau$_com = 0

$\Rightarrow$ f $\times$ R $-$ T $\times$ R = 0

$\Rightarrow$ f = R ..... (2)

Putting value of (2) in (1),

f + f = mg sin60$^\circ$

$\Rightarrow$ f = ${{mg\sin 60^\circ } \over 2} = {{\sqrt 3 mg} \over 4}$ = 0.433 mg

$\therefore$ cylinder need minimum 0.433 mg friction force to stay in equilibrium.

But here maximum friction force possible,

f_max = $\mu$_sN

= $\mu$_smg cos60$^\circ$

= 0.4 mg $\times$ ${1 \over 2}$

= 0.2 mg

As maximum possible friction force is less than 0.433 mg. so cylinder will not stay in equilibrium. It will slide on the surface. So friction will be kinetic friction.

f_k = $\mu$_k N

=$\mu$_k mg cos60$^\circ$

= 0.4 mg $\times$ ${1 \over 2}$

= ${mg \over 5}$

Gravitational force line passes through the sun so torque about sun always zero for the planet.

$\therefore$ Angular momentum about sum is constant.

$\overrightarrow L$ = Constant

we know, L = M${v_ \bot }$ r

Now,

$dA = {1 \over 2}$ $\times$ base $\times$ height

$= {1 \over 2} \times r \times {v_ \bot }dt$

$\Rightarrow {{dA} \over {dt}} = {{r{v_ \bot }} \over 2} = {r \over 2} \times {L \over {Mr}} = {L \over {2M}}$

Given :

X_L = 10$\Omega$

X_C = 4$\Omega$

R = 6$\Omega$

$\therefore$ Power factor = cos$\theta$ = ${R \over Z}$

$= {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }}$

$= {6 \over {\sqrt {{6^2} + {{(10 - 4)}^2}} }}$

$= {6 \over {6\sqrt 2 }} = {1 \over {\sqrt 2 }}$

Given, radius of sphere = (7.50 $\pm$ 0.85) cm

$\therefore$ r = 7.50 and dr = 0.85

We know, volume of a sphere $v = {4 \over 3}\pi {r^3}$

taking log both sides, we get

$\ln v = \ln {{4\pi } \over 3} + 3\ln r$

Differentiating both sides,

${{dv} \over v} = 0 + 3{{dr} \over r}$

$\therefore$ Fractional error in volume ${{dv} \over v} = 3{{dr} \over r}$

$\therefore$ % error in volume,

${{dv} \over v} \times 100 = 3{{dr} \over r} \times 100$

$= 3 \times {{0.85} \over {7.50}} \times 100$

= 34%

For common emitter configuration

$\beta$ = ${{\Delta {I_C}} \over {\Delta {I_B}}} = {{\left( {4 - 2} \right)\,mA} \over {\left( {20 - 10} \right)\,\mu A}}$

= ${2 \over {10}} \times {{{{10}^{ - 3}}} \over {{{10}^{ - 6}}}}$ = 200

Before collision



After collision,



Applying momentum conservation,

m₁v₁ = m₂v $-$ m₁v ..... (1)

As collision is elastic,

$\therefore$ e = 1

$\Rightarrow$ ${{v - ( - v)} \over {0 - {v_1}}} = 1$

$\Rightarrow$ v₁ = 2v ..... (2)

Put value of v₁ in equation (1),

m₁ (2v) = (m₂ $-$ m₁)v

$\Rightarrow$ 2m₁ = m₂ $-$ m₁

$\Rightarrow$ 3m₁ = m₂ $\Rightarrow$ ${{{m_2}} \over {{m_1}}} = {3 \over 1}$

for gas 1, S₁ = ${f \over 2}{n_1}R$

for gas 2, S₂ = ${f \over 2}{n_2}R$

after removal of piston,

S = ${f \over 2}({n_1} + {n_2})R = {S_1} + {S_2}$

The slope of the given v versus x graph is $m = - {{{v_0}} \over {{x_0}}}$ and intercept is c = + v₀. Hence, v varies with x as

$v = - \left( {{{{v_0}} \over {{x_0}}}} \right)x + {v_0}$ ...... (1)

where v₀ and x₀ are constants of motion. Differentiating with respect to time t, we have

${{dv} \over {dt}} = - \left( {{{{v_0}} \over {{x_0}}}} \right){{dx} \over {dt}}$

or $a = - \left( {{{{v_0}} \over {{x_0}}}} \right)v$ ........ (2)

Using Eq. (1) in Eq. (2), we get

$a = - \left( {{{{v_0}} \over {{x_0}}}} \right)\left( { - {{{v_0}} \over {{x_0}}}x + {v_0}} \right)$

$a = {\left( {{{{v_0}} \over {{x_0}}}} \right)^2}x - {{v_0^2} \over {{x_0}}}$

Thus the graph of a versus x is a straight line having a positive slope = ${\left( {{{{v_0}} \over {{x_0}}}} \right)^2}$ and negative intercept = $- {{v_0^2} \over {{x_0}}}$. Hence the correct choice is (a).

Given potential field U(r) = ${{ - C} \over r}$

$\because$ $F = {{dU} \over {dr}} = {C \over {{r^2}}}$

& ${F_c} = {{m{v^2}} \over r}$

$\therefore$ ${{m{v^2}} \over r} = {C \over {{r^2}}}$

$\Rightarrow r = {C \over {m{v^2}}}$

$\Rightarrow r$ $\propto$ ${1 \over {{v^2}}}$

$\therefore$ the graph between r & v will be hyperbolic.

(a) Electric monopoles exist while magnetic monopoles do not exist.

(b) Magnetic field lines at the ends and outside of solenoid cannot be confined. (c) Magnetic field lines are confined within a toroid.

(d) Magnetic field lines inside a bar magnet are parallel.

(e) For perfectly diamagnetic material $\chi$ = -1.

Resolving power (RP) $\propto$ ${1 \over \lambda }$

We know, de-Broglie wavelength

$\lambda = {h \over {mv}}$

$\therefore$ RP $\propto$ ${mv \over {h}}$

$\therefore$ ${{R{P_e}} \over {R{P_p}}} = {{{m_e}} \over {{m_p}}} = 1837$

As galvanometer shows zero deflection so it act's as balanced wheatstone bridge.

$\therefore$ ${{12} \over 6} = {x \over {72 - x}}$

$\Rightarrow$ 14 $-$ 2x = x

$\Rightarrow$ 3x = 144

$\Rightarrow$ x = 48 cm

If spring compressed by x,

then work done by spring = 0 $-$ ${1 \over 2}$ $\times$ 4 $\times$ 10²

Applying work energy theorem,

$-$${1 \over 2}$ kx² = $-$${1 \over 2}$ $\times$ 4 $\times$ 10²

$\Rightarrow$ 100x² = 4 $\times$ 10²

$\Rightarrow$ x = 2

$\therefore$ Final length of the spring = 8 $-$ 2 = 6 m

for adiabatic expansion :

PV^$\gamma$ = const.

$\Rightarrow$ ln P + $\gamma$ln v = const.

$\Rightarrow$ differentiating both sides;

${{dp} \over p} + \gamma {{dv} \over v} = 0$

$\Rightarrow {{dp} \over p} = - \gamma {{dv} \over V}$

Positron emission or Beta plus decay is a subtype of radioactive decay called Beta decay, in which a proton inside a nucleus is converted into a neutron while releasing a positron and an electron neutrino.

So, decay of a proton to neutron is possible only inside the nucleus. Free proton cannot decay to neutron as mass of proton is less compared to neutron so to decay into higher mass, proton need extra energy which free proton can’t get, only proton inside nucleus can get.

$|\overrightarrow u | = |\overrightarrow v |$ .... (1)

$\overrightarrow u = u\cos 45\widehat i + u\sin 45\widehat j$ ..... (2)

$\overrightarrow v = v\cos 45\widehat i - v\sin 45\widehat j$ ..... (3)

$|\Delta \overrightarrow P | = |m(\overrightarrow v - \overrightarrow u )|$ .... (4)

$\Delta P = 2mu\sin 45^\circ$

$= 2 \times 5 \times {10^{ - 3}} \times 5\sqrt 2 \times {1 \over {\sqrt 2 }}$

= 50 $\times$ 10^$-$3

= 5 $\times$ 10^$-$2

${F_{total}} = {{k{q_1}{q_2}} \over {r_1^2}} + {{k{q_1}{q_3}} \over {r_2^2}} + {{k{q_1}{q_4}} \over {r_3^2}} + .....$

$= 9 \times {10^9} \times {10^{ - 6}}\left[ {1 + {{\left( {{1 \over 2}} \right)}^2} + {{\left( {{1 \over {{2^2}}}} \right)}^2} + {{\left( {{1 \over {{2^3}}}} \right)}^2} + {{\left( {{1 \over {{2^\infty }}}} \right)}^2}} \right]$

$= 9 \times {10^9} \times {10^{ - 6}}\left[ {{1 \over {1 - {1 \over 4}}}} \right]$

$\because$ $\left[ {{S_\infty } = {a \over {1 - r}}} \right]$ for G.P.

$= 9 \times {10^3} \times {4 \over 3}$ = 12 $\times$ 10³ N

Bohr’s theory is applicable for unielectronic species only Li⁺ has two electrons.

Bohr’s theory could not explain the splitting of spectral lines in the presence of external magnetic field (Zeeman effect)

Statement I – false

Statement II – true

Both C₂H₅OH and AgCN can generate nucleophile.

KCN generates nitriles on substitution reactions with haloalkanes where AgCN generates isonitriles on substitution reactions with haloalkanes. Because KCN is ionic and has ‘C’ nucleophilic centre whereas AgCN is covalent and has ‘N’ nucleophilic centre.

Carbonyl carbon is lost as $CO_3^{2 - }$.

(a) Mercury $\to$ Distillation refining

(b) Copper $\to$ Electrolytic refining

(c) Silicon $\to$ Zone refining

(d) Nickel $\to$ Vapour phase refining

In pure water



$\therefore$ K_sp = (s)² = (8 $\times$ 10^-4)² = 64 $\times$ 10^-8

In H₂SO₄ solution,



As S₁ < < 0.01 so, S₁ + 0.01 $\simeq$ 0.01

$\therefore$ K_sp = [Cd⁺²] [So$_4^{ - 2}$]

$\Rightarrow$ 64 $\times$ 10^-8 = S₁ $\times$ 0.01

$\Rightarrow$ S₁ = 64 $\times$ 10^-6

CuSO₄.5H₂O $\Rightarrow$ [Cu(H₂O)₄]SO₄.H₂O

In CuSO₄.5H₂O, Cu is co-ordinated with 4 water molecule and two more oxygen atom from sulphate ion and fifth water molecule is hydrogen bonded.

So secondary valency = 4

No. of hydrogen bond per molecule = 1

Non-biodegradable wastes are generated by thermal power which produce fly ash.

Nowadays most of the detergents available are biodegradable. The bacteria responsible for degrading biodegradable detergent feed on it and grow rapidly. While growing, they may use up all the oxygen dissolved in water which cause Eutrophication.

During nitration of aniline, meta-nitroaniline is also formed as product due to formation of NH₃⁺ group which is meta directing group under strongly acidic medium. The percentage of p, m and o product is 51%, 47% and 2%, respectively.

Oxides & Hydroxide sol are positive so TiO₂ is positive sol.

Sulphide sol are negatively charge so CdS is negatively charge sol.

$\Delta$T_b = Boiling point of the solution $-$ Boiling point of the pure solvent

= 100.52 $-$ 100

= 0.52

$\therefore$ $\Delta$T_b = K_b (iM)

$\Rightarrow$ 0.52 = i $\times$ 0.52 $\times$ 2

$\Rightarrow$ i = ${1 \over 2}$

We know,

i = 1 + $\left( {{1 \over n} - 1} \right)\beta$

here, $\beta$ = degree of dimerization

n = number of particle associated

n = 2 for dimerization

$\therefore$ ${1 \over 2} = 1 + \left( {{1 \over 2} - 1} \right)\beta$

$\Rightarrow \beta = 1$

$\therefore$ % association = 100

Given t_1/2 = 1 min

$\therefore$ $K = {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over 1}$

From formula we know,

$Kt = \ln \left( {{{100} \over {100 - x}}} \right)$

$\Rightarrow {{\ln 2} \over 1} = {1 \over t}\ln \left( {{{100} \over {0.1}}} \right)$

$\Rightarrow 0.3 = {1 \over t}\ln \left( {{{10}^3}} \right)$

$\Rightarrow 0.3 = {1 \over t} \times 3$

$\Rightarrow t = 10$ min

From the given value of HCl, it is clear that most appropriate volume of HCl used is 5 ml because it occurs most number of times.

Na₂CO₃ + 2 HCl $\to$ 2 NaCl + CO₂ + H₂O

n factor of Na₂CO₃ = 2

n factor of HCl = 1

equivalent of Na₂CO₃ = equivalent of HCl

$\Rightarrow$ ${{10} \over {1000}} \times 2 \times M = {5 \over {1000}} \times 1 \times 0.2$

$\Rightarrow M = {1 \over {20}}M$

= 0.05 M

= 50 $\times$ 10^$-$3 M

= 50 mM

In basic medium, oxidising action of H₂O₂.

Mn²⁺ + H₂O₂ $\to$ Mn⁺⁴ + 2OH^–

In basic medium, reducing action of H₂O₂.

I₂ + H₂O₂ + 2OH^– $\to$ 2I^– + 2H₂O + O₂

In acidic medium, oxidising action of H₂O₂.

PbS(s) + 4H₂O₂(aq) $\to$ PbSO₄(s) + 4H₂O(l)

Covalent or network solid are insulator (except graphite) and have very high melting point.