JEE Main 2021 25th July Morning Shift Question Paper with Solutions

(a) $\overrightarrow C = \overrightarrow A + \overrightarrow B$

Option (iv)

(b) $\overrightarrow A = \overrightarrow B + \overrightarrow C = \overrightarrow C + \overrightarrow B$

Option (iii)

(c) $\overrightarrow B = \overrightarrow A + \overrightarrow C$

Option (i)

(d) $\overrightarrow A + \overrightarrow B + \overrightarrow C = 0$

Option (ii)

$\lambda = {h \over p} = {h \over {\sqrt {2mE} }} \propto {1 \over {\sqrt m }}$

m_$\alpha$ > m_p > m_e

So, $\lambda$_e > $\lambda$_p > $\lambda$_$\alpha$

${I_{\max }} = {V \over R} = {{20V} \over {10K\Omega }} = 2$ mA

For LR - decay circuit

$I = {I_{\max }}{e^{ - Rt/L}}$

$I = 2mA{e^{{{ - 10 \times {{10}^3} \times 1 \times {{10}^{ - 6}}} \over {10 \times {{10}^{ - 3}}}}}}$

I = 2mA e^$-$1

I = 2 $\times$ 0.37 mA

$I = {{74} \over {100}}$ mA

x = 74

Net force on free charged particle

$F = {{k{q^2}} \over {{{(d + x)}^2}}} - {{k{q^2}} \over {{{(d - x)}^2}}}$

$F = - k{q^2}\left[ {{{4dx} \over {{{({d^2} - {x^2})}^2}}}} \right]$

$a = - {{4k{q^2}d} \over m}\left( {{x \over {{d^4}}}} \right)$

$a = - \left( {{{4k{q^2}} \over {m{d^3}}}} \right)x$

So, angular frequency

$\omega = \sqrt {{{4k{q^2}} \over {m{d^3}}}}$

$\omega = \sqrt {{{4 \times 9 \times {{10}^9} \times 10} \over {1 \times {{10}^{ - 6}} \times {1^3}}}}$

$\omega = 6 \times {10^8}$ rad/sec

$\omega = 6000 \times {10^5}$ rad/sec

$\overrightarrow r = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$

$\overrightarrow v = 20\alpha t\widehat i + 5\beta \widehat j$

$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v )$

$= m[10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j] \times [20\alpha t\widehat i + 5\beta \widehat j]$

$\overrightarrow L = m[50\alpha \beta {t^2}\widehat k - 10\alpha \beta ({t^2} - 5t)\widehat k]$

At t = 0, $\overrightarrow L = \overrightarrow 0$

$50\alpha \beta {t^2} - 100\alpha \beta ({t^2} - 5t) = 0$

t $-$ 2 (t $-$ 5) = 0

t = 10 sec

$\phi = \overrightarrow B .\overrightarrow S$

$\phi = {4 \over \pi } \times {10^{ - 3}}\left( {1 - {t \over {100}}} \right).\pi {R^2}$

$\phi = 4 \times {10^{ - 3}} \times {(1)^2}\left( {1 - {t \over {100}}} \right)$

$\varepsilon = {{ - d\phi } \over {dt}}$

$\varepsilon = {{ - d} \over {dt}}\left( {4 \times {{10}^{ - 3}}\left( {1 - {t \over {100}}} \right)} \right)$

$\varepsilon = 4 \times {10^{ - 3}}\left( {{1 \over {100}}} \right) = 4 \times {10^{ - 5}}V$

When B = 0

$1 - {t \over {100}} = 0$

t = 100 sec

Heat $= {{{\varepsilon ^2}} \over R}t$

Heat $= {{{{(4 \times {{10}^{ - 5}})}^2}} \over {2 \times {{10}^{ - 6}}}} \times 100$ J

Heat $= {{16 \times {{10}^{ - 10}} \times 100} \over {2 \times {{10}^{ - 6}}}}$ J

Heat = 0.08 J

Heat = 80 mJ

$\omega = \sqrt {{{{k_{eq}}} \over \mu }}$

$\mu$ = reduced mass

springs are in series connection

${k_{eq}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}$

${k_{eq}} = {{k \times 4k} \over {5k}} = {{4k} \over 5}$

${k_{eq}} = {{4 \times 20} \over 5}$ N/m = 16 N/m

$\mu = {{{m_1}{m_2}} \over {{m_1} + {m_2}}} = {{0.2 \times 0.8} \over {0.2 + 0.8}} = 0.16$ kg

$\omega = \sqrt {{{16} \over {0.16}}} = \sqrt {100} = 10$

Max. voltage that can be measured by this potentiometer will be equal to potential drop across AB

R_AB = 10 $\times$ 0.1 $\times$ 100 = 100 ohm.

$\therefore$ ${V_{AB}} = {6 \over {20 + 100}} \times 100 = 6 \times {{100} \over {120}} = 5$ V

$\lambda = {h \over p}$

both the particles will move with momentum same in magnitude & opposite in direction.

So De-Broglie wavelength of both will be same i.e. ratio 1 : 1

Zero error occurs when the screw gauge does not read zero when its jaws are closed. This error needs to be accounted for in the final measurement. The least count (LC) of a screw gauge, calculated as the pitch (p) divided by the number of divisions (N) on the circular scale, helps us determine the value of this error.

The least count of both screw gauges is LC = $p / N=0.1 / 100=0.001 \mathrm{~cm}$.

For screw gauge A, the zero error is calculated by adding the main scale reading (MSR) at zero and the circular scale reading (CSR) at zero, multiplied by the least count. Screw gauge A's MSR at zero is 0 and its CSR at zero is 5. Therefore, its zero error is ( 0 + 5 $\times$ 0.001 = 0.005 ) cm.

For screw gauge B, the zero error involves a negative main scale reading and a circular scale reading. The MSR at zero is -1 (since the main scale is 0.1 cm per division, this corresponds to -0.1 cm) and the CSR at zero is 92. Thus, the zero error for B is ( (-1 $\times$ 0.1) + 92 $\times$ 0.001 = -0.008 ) cm.

To find the difference between the final circular scale readings of screw gauges A and B, we look at the difference in their zero errors. The absolute difference is ( |0.005 - (-0.008)| = |0.005 + 0.008| = 0.013 ) cm. This value represents the absolute difference in the final readings of the two screw gauges, accounting for their individual zero errors.

However, to express this difference in terms of the circular scale divisions (since the least count is 0.001 cm per division), we multiply this difference by 1000 (since 1 cm = 1000 mm and each division represents 0.001 cm) :

$\text{Difference in Divisions} = 0.013 \text{ cm} \times 1000$
$= 13 \text{ divisions}$

Reflected wave will have direction opposite to incident wave.

In series combination $\Delta$l = l₁ + l₂

$Y = {{F/A} \over {\Delta l/l}} \Rightarrow \Delta l = {{Fl} \over {AY}}$

$\Rightarrow \Delta l \propto {l \over Y}$

Equivalent length of rod after joining is = 2l

As, lengths are same and force is also same in series

$\Delta l = \Delta {l_1} + \Delta {l_2}$

${{{l_{eq}}} \over {{Y_{eq}}}} = {l \over {{Y_1}}} + {l \over {{Y_2}}} \Rightarrow {{2l} \over Y} = {l \over {{Y_1}}} + {l \over {{Y_2}}}$

$\therefore$ $Y = {{2{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}$

$\sin {\theta _C} = {1 \over \mu } = {1 \over {2{\mu _2}}} < \sin {\theta _C}$

sin$\theta$ > sin$\theta$_C

$\theta$ > $\theta$_C

Total internal reflection will happen.

p_i = p_f

2 $\times$ 4 = 2 $\times$ 1 + m₂ $\times$ v₂

m₂v₂ = 6 ..... (i)

by coefficient of restitution

$1 = {{{v_2} - 1} \over 4} \Rightarrow {v_2} = 5$ m/s

by (i)

m₂ $\times$ 5 = 6

m₂ = 1.2 kg

${v_{cm}} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}$

${v_{cm}} = {{2 \times 1 + 1.2 \times 5} \over {2 + 1.2}} = {8 \over {3.2}} = {{25} \over {10}}$

x = 25

PV^r = const.

TV^{r $-$ 1} = const.

$T{(l)^{{5 \over 3} - 1}}$ = const.

${{{T_1}} \over {{T_2}}} = {\left( {{{{l_2}} \over {{l_1}}}} \right)^{{2 \over 3}}}$

${\left( {{{\Delta Q} \over {\Delta t}}} \right)_A} = {\left( {{{\Delta Q} \over {\Delta t}}} \right)_B}$

$m{S_A}{\left( {{{\Delta T} \over {\Delta t}}} \right)_A} = m{S_B}{\left( {{{\Delta T} \over {\Delta t}}} \right)_B}$

$\Rightarrow$ ${{{S_A}} \over {{S_B}}} = {{{{\left( {{{\Delta T} \over {\Delta t}}} \right)}_B}} \over {{{\left( {{{\Delta T} \over {\Delta t}}} \right)}_B}}} = {{90/6} \over {120/3}} = {{15} \over {40}} = {3 \over 8}$

${{3{N_0}} \over 4} = {N_0}{e^{ - \lambda {t_1}}}$

${{{N_0}} \over 2} = {N_0}{e^{ - \lambda {t_2}}}$

$\ln (3/4) = - \lambda {t_1}$ ..... (i)

$\ln (1/2) = - \lambda {t_2}$ ..... (i)

$\ln (3/4) - \ln (1/2) = \lambda ({t_2} - {t_1})$ ....(i)

$\Delta t = {{\ln (3 /2)} \over \lambda }$

Angular momentum conservation equation ${v_0}{x_2} = {v_1}{x_1}$

${v_1} = {{{v_0}{x_2}} \over {{x_1}}}$

Bandwidth = 2 $\times$ f_m

= 2 $\times$ 10⁵ HZ = 200 KHZ

Applying work energy theorem :

w_g + w_T = $\Delta$K

$-$mgl(1 $-$ cos60$^\circ$) = ${1 \over 2}$mv² $-$ ${1 \over 2}$mu²

v² = u² $-$ 2gl(1 $-$ cos60$^\circ$)

v² = 9 $-$ 2 $\times$ 10 $\times$ 0.5$\left( {{1 \over 2}} \right)$

v² = 4

v = 2 m/s

Power, $P = {{{V^2}} \over {{R_B}}}$

${R_B} = {{{V^2}} \over P} = {{100 \times 100} \over {200}}$

${R_B} = 50\Omega$



To produce same power, same voltage (i.e. 100 V) should be across the bulb.

Hence, R = R_B

R = 50 $\Omega$

I_z = I_x + I_y (using perpendicular axis theorem) & I = mk² (K : radius of gyration)

so, mK_z² = mK_x² + mK_y²

K_z² = K_x² + K_y²

so radius of gyration about axes x, y & z won't be same hence assertion A is not correct reason R is correct statement (property of a rigid body)

Taking an element of width dx at a distance x(x < d/2) from left plate

$dc = {{({\varepsilon _0} + kx)A} \over {dx}}$

Capacitance of half of the capacitor

${1 \over C} = \int\limits_0^{d/2} {{1 \over {dc}} = {1 \over A}\int\limits_0^{d/2} {{{dx} \over {{\varepsilon _0} + kx}}} }$

${1 \over C} = {1 \over {kA}}\ln \left( {{{{\varepsilon _0} + kd/2} \over {{\varepsilon _0}}}} \right)$

Capacitance of second half will be same

${C_{eq}} = {C \over 2} = {{kA} \over {2\ln \left( {{{2{\varepsilon _0} + kd} \over {2{\varepsilon _0}}}} \right)}}$

In 4 sec. 1^st drop will travel

$\Rightarrow {1 \over 2}$ $\times$ (9.8) $\times$ (4)² = 78.4 m

$\therefore$ 2^nd drop would have travelled

$\Rightarrow$ 78.4 $-$ 34.3 = 44.1 m.

Time for 2^nd drop

$\Rightarrow {1 \over 2}$(9.8)t² = 44.1

t = 3 sec

$\therefore$ each drop have time gap of 1 sec

$\therefore$ 1 drop per sec

Impulse = change in momentum

Ball (a) $\left| {\overrightarrow {\Delta p} } \right|$ = 2mu = J₁

Ball (b) $\left| {\overrightarrow {\Delta p} } \right|$ = 2mu cos45$^\circ$ = J₂

${{{J_1}} \over {{J_2}}} = {1 \over {\cos 45^\circ }} = \sqrt 2$

A = $\lambda$N

$\lambda = {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over {3 \times 24 \times 60 \times 60}}$sec^$-$1 = 2.67 $\times$ 10^$-$6 sec^$-$1

N = Number of atoms in 2 mg Au

$= {{2 \times {{10}^{ - 3}}} \over {198}} \times 6 \times {10^{23}}$ = 6.06 $\times$ 10¹⁵

$A = \lambda N = 1.618 \times {10^{13}} = 16.18 \times {10^{12}}$ dps

Fringe Width, $\beta = {{\lambda D} \over d}$

${\beta _{\max }} \Rightarrow {d_{\min }}$ and ${\beta _{\min }} \Rightarrow {d_{\max }}$

$d = {d_0} + {a_0}\sin \omega t$

${d_{\max }} = {d_0} + {a_0}$ and ${d_{\min }} = {d_0} - {a_0}$

$\therefore$ ${\beta _{\min }} = {{\lambda D} \over {{d_0} + {a_0}}}$ and

$\therefore$ ${\beta _{\max }} = {{\lambda D} \over {{d_0} - {a_0}}}$

${\beta _{\max }} - {\beta _{\min }} = {{\lambda D} \over {{d_0} - {a_0}}} - {{\lambda D} \over {{d_0} + {a_0}}} = {{2\lambda D{a_0}} \over {d_0^2 - a_0^2}}$

C_P $-$ C_V = R for ideal gas and gas behaves as ideal gas at high temperature, so T_P > T_Q

$\Delta {H_{sub}} = \Delta {H_{fus.}} + \Delta {H_{vap.}}$

= 2.8 + 98.2

= 101 kJ/mol

Gabriel phthalimide synthesis

In bauxite impurities of Fe₂O₃, TiO₂ and SiO₂ are present, Fe₂O₃ and TiO₂ are basic oxides therefore does not reacts with or dissolve in NaOH whereas SiO₂ is acidic oxide it gets dissolve in NaOH, hence leach out

$Si{O_2} + 2NaOH \to N{a_2}Si{O_3}(aq.) + {H_2}O$

Numbers of $\sigma$ bonds = 10

Bromine water gives tribromo products, other gives monobromo products in which para is major product.

$K = {{[C]_{eq}^2} \over {{{[A]}_{eq}}{{[B]}_{eq}}}} = {{{{(1 + 2x)}^2}} \over {(1 - x)(1 - x)}}$

$100 = {\left( {{{1 + 2x} \over {1 - x}}} \right)^2}$

$\left( {{{1 + 2x} \over {1 - x}}} \right) = 10$

$x = {3 \over 4}$

$[C]{e_{q.}} = 1 + 2x$

$= 1 + 2\left( {{3 \over 4}} \right)$

= 2.5 M

= 25 $\times$ 10^-1 M

1. ${[\mathop {Fe}\limits^{ + 3} {({H_2}O)_6}]^{3 + }}$

$F{e^{3 + }}:[Ar]3{d^5}$

Hybridisation : $s{p^3}{d^2}$

Magnetic nature : Paramagnetic (so this complex response to external magnetic field)

2. ${[Ni{(CN)_4}]^{2 - }}$

$N{i^{2 + }}:[Ar]3{d^8}$

Hybridisation : dsp²

Magnetic nature : diamagnetic

3. ${[Co{(CN)_6}]^{3 - }}$

$C{o^{3 + }}:[Ar]3{d^6}$

Hybridisation : ${d^2}s{p^3}$

Magnetic nature : diamagnetic

4. $[Ni{(CO)_4}]$

$Ni:[Ar]3{d^8}4{s^2}$

Hybridisation : $s{p^3}$

Magnetic nature : diamagnetic

Statement - I is incorrect

Be(OH)₂ dissolve in alkali due to it's amphoteric nature.

Statement - II is correct

Solubility of alkaline earth metal hydroxide in water increases down the group due to rapid decreases in lattice energy as compared to hydration energy.

${C_4}{H_{10}} + {{13} \over 2}{O_2}\buildrel {} \over \longrightarrow 4C{O_2} + 5{H_2}O$

Moles of ${H_2}O = {{72} \over {18}} = 4$

Moles of C₄H₁₀ used $= {1 \over 5} \times 4$

Wight of C₄H₁₀ used $= {4 \over 5} \times 58$

= 46.4 gm = 464 $\times$ 10^-1 g

milli-equivalents of Fe²⁺ = milli-equivalents of K₂Cr₂O₇

M $\times$ 10 $\times$ 1 = 0.02 $\times$ 15 $\times$ 6

$\Rightarrow$ M = 0.18 = 18 $\times$ 10^$-$2 M

For Zero order reactions

rate = K [Reactant]⁰

$\Rightarrow$ r = k


t_1/2 = ${{{{[A]}_0}} \over {2K}}$ for Zero order


For First order reaction $\to$

r = K [Concentration]


Reactant concentration after time t $\to$

C_t = C₀ e^-kt

(a) $\mathop C\limits^{ + 6} r{O_3}$ : Since there are three oxygen atoms each with an oxidation number of -2, the total oxidation number from the oxygen atoms is -6. To make the compound neutral, Chromium (Cr) must therefore have an oxidation number of +6.

(b) $\mathop {F{e_2}}\limits^{ + 3} {O_3}$ : Here, there are three oxygen atoms, contributing a total oxidation number of -6. To balance this, the total oxidation number for the two iron (Fe) atoms must be +6, meaning each Fe atom has an oxidation number of +3.

(c) $\mathop {Mn}\limits^{ + 4} {O_2}$ : With two oxygen atoms, the total oxidation number is -4. Thus, the manganese (Mn) atom must have an oxidation number of +4 to balance this.

(d) $\mathop {{V_2}}\limits^{ + 5} {O_5}$ : In this case, there are five oxygen atoms for a total oxidation number of -10. To balance this, the total oxidation number for the two vanadium (V) atoms must be +10, meaning each V atom has an oxidation number of +5.

(e) $\mathop {C{u_2}}\limits^{ + 1} O$ : Here, there's one oxygen atom with an oxidation number of -2. To balance this, the total oxidation number for the two copper (Cu) atoms must be +2, meaning each Cu atom has an oxidation number of +1.

So, the order of the oxidation numbers from highest to lowest is :

(a) CrO$_3$ > (d) V$_2$O$_5$ > (c) MnO$_2$ > (b) Fe$_2$O$_3$ > (e) Cu$_2$O.

Therefore, the answer is Option C : (a) > (d) > (c) > (b) > (e).

CO₂, CCl₄ and Cl₂C = CCl₂ are used as dry cleaning agents for clothes.

H₂O₂ is used as bleaching agent in laundry.

From Henry's law

P_gas = K_H.X_gas

0.835 = 1.67 $\times$ 10³ $\times$ ${{n(C{O_2})} \over {{{0.9 \times 1000} \over {18}}}}$

$\Rightarrow$ n(CO₂) = 0.025

Millimoles of CO₂ = 0.025 $\times$ 1000 = 25

Mole of AgCl precipitated is equal to mole of Cl^- present in ionization sphere.



Since none of Cl^- is present in the co-ordination sphere. Therefore answer is zero.

Forms spherical micelles with CH₃(CH₂)₁₆ group pointing towards the centre of sphere.