JEE Main 2021 25th February Evening Shift Question Paper with Solutions

From energy conservation

mgh = ${1 \over 2}mv_0^2 + {1 \over 2}I{\omega ^2}$

mgh = ${1 \over 2}mv_0^2 + {1 \over 2} \times {1 \over 5}m{a^2} \times {{{v_0}^2} \over {{a^2}}}$

gh = ${1 \over 2}{v_0}^2 + {1 \over 5}{v_0}^2$

gh = ${7 \over {10}}{v_0}^2$

h = ${7 \over {10}}{{{v_0}^2} \over g}$

from triangle, $\sin \theta = {h \over l}$

then h = l sin$\theta$

$l\sin \theta = {7 \over {10}}{{{v_0}^2} \over g}$

$l = {7 \over {10}}{{{v_0}^2} \over {g\sin \theta }}$

$\because$ relation b/w kinetic energy & momentum is

$P = \sqrt {2mKE}$ ($\because$ KE = same)

$\Rightarrow$ ${{{p_1}} \over {{p_2}}} = \sqrt {{{{m_1}} \over {{m_2}}}}$

$\Rightarrow$ ${n \over 2} = \sqrt {{4 \over {16}}}$

$\Rightarrow$ $n = 1$

The point a covers 30$^\circ$ in 0.1 sec.

Means ${\pi \over 6}\buildrel {} \over \longrightarrow 0.1$ sec.

$1\buildrel {} \over \longrightarrow {{0.1} \over {{\pi \over 6}}}$

$2\pi \buildrel {} \over \longrightarrow {{0.1 \times 6} \over \pi } \times 2\pi$

$T = 1.2$ sec.

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

$\omega = {{2\pi } \over {1.2}}$

Restoration force $(F) = m{\omega ^2}A$

Then Restoration force per unit mass $\left( {{F \over m}} \right) = {\omega ^2}A$

$\left( {{F \over m}} \right) = {\left( {{{2\pi } \over {1.2}}} \right)^2} \times 0.36$

$\cong 9.87$ N

When they brought into contact & then separated by a distance = 0.5 m

Then charge distribution will be



The electrostatic force acting b/w the sphere is

${F_e} = {{k{q_1}{q_2}} \over {{r^2}}}$

$= {{9 \times {{10}^9} \times 1 \times {{10}^{ - 9}} \times 1 \times {{10}^{ - 9}}} \over {{{(0.5)}^2}}}$

$= {{900} \over {25}} \times {10^{ - 9}}$

$\Rightarrow$ ${F_e} = 36 \times {10^{ - 9}}N$

For particle (1)

$20 + h = 10t + {1 \over 2}g{t^2}$ ...... (i)

For particle (2)

$h = {1 \over 2}g{t^2}$ ..... (ii)

put equation (ii) in equation (i)

$20 + {1 \over 2}g{t^2} = 10t + {1 \over 2}g{t^2}$

t = 2 sec.

Put in equation (ii)

$h = {1 \over 2}g{t^2}$

$= {1 \over 2} \times 10 \times {2^2}$

h = 20 m

The height of the building $= 25 + 20$ = 45 m

Given, $\overrightarrow P \times \overrightarrow Q = \overrightarrow Q \times \overrightarrow P$

It is possible only when $\overrightarrow P \times \overrightarrow Q = \overrightarrow Q \times \overrightarrow P$ = 0

$\Rightarrow$ We know, $\overrightarrow P \times \overrightarrow Q$ = PQsin $\theta$

Only if $\overrightarrow P = 0$

or $\overrightarrow Q = 0$

or sin $\theta$ = 0 $\Rightarrow$ $\theta$ = 0 or 180^o

The angle b/w $\overrightarrow P$ & $\overrightarrow Q$ is $\theta$(0$^\circ$ < $\theta$ < 360$^\circ$)

So, $\theta$ = 180$^\circ$

(a) Rectifier : used to convert a a.c. voltage into d.c. voltage.

(b) Stabilizer : used for constant output voltage even when the input voltage or load current change

(c) Transformer : used either for stepping up or stepping down the a.c. voltage.

(d) Filter : used to remove any ripple in the rectified output voltage.

If A = B = 0, then output y = 1

If A = B = 1, then output y = 1

Here R = radius of the earth

From energy conservation

${{ - G{m_e}m} \over R} + {1 \over 2}m{v_i}^2 = {{ - G{m_e}m} \over {11R}} + 0$

${1 \over 2}m{v_i}^2 = {{10} \over {11}}{{G{m_e}m} \over R}$

${v_i} = \sqrt {{{20} \over {11}}{{G{m_e}} \over R}}$

${v_i} = \sqrt {{{10} \over {11}}} {v_e}$

{$\because$ escape velocity ${v_e} = \sqrt {{{2G{m_e}} \over R}}$}

Then the value of x = 10

$\Delta$E = 10.2 eV

${{hc} \over \lambda }$ = 10.2 eV

$\lambda$ = ${{hc} \over {(10.2)e}}$

= ${{12400} \over {10.2}}\mathop A\limits^o$

= 121.56 nm

$\simeq$ 121.8 nm

From P-V diagram,

Given PV^1/2 = constant ...... (1)

We know that

PV = nRT

$P \propto \left( {{T \over V}} \right)$

Put in equation (1)

$\left( {{T \over V}} \right){(V)^{1/2}}$ = constant

$T \propto {V^{1/2}}$

$\Rightarrow$ ${{{T_2}} \over {{T_1}}} = \sqrt {{{{V_2}} \over {{V_1}}}}$

$\Rightarrow$ ${{{T_2}} \over {{T_1}}} = \sqrt {{{2{V_1}} \over {{V_1}}}}$ [As V₂ = 2V₁ ]

$\Rightarrow$ ${{{T_2}} \over {{T_1}}} = \sqrt 2$

$\because$ ${v_1}\cos \alpha = {v_2}\cos \beta$

${{{v_1}} \over {{v_2}}} = {{\cos \beta } \over {\cos \alpha }}$

Then the ratio of kinetic energies

${{{k_1}} \over {{k_2}}} = {{{1 \over 2}m{v_1}^2} \over {{1 \over 2}m{v_2}^2}} = {\left( {{{{v_1}} \over {{v_2}}}} \right)^2} = {\left( {{{\cos \beta } \over {\cos \alpha }}} \right)^2}$

$\Rightarrow$ ${{{k_1}} \over {{k_2}}} = {{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}$

The initial phase angle ${\phi _0} = \pi - {\pi \over 6} = {{5\pi } \over 6}$

T sin$\theta$ = ${{k{q^2}} \over {{r^2}}}$

T cos$\theta$ = mg

tan$\theta$ = ${{k{q^2}} \over {mg{r^2}}}$

q² = ${{\tan \theta mg{r^2}} \over k}$

$\because$ tan$\theta$ = ${{0.1} \over {0.5}} = {1 \over 5}$

${q^2} = {1 \over 5} \times {{10 \times {{10}^{ - 6}} \times 10 \times 0.2 \times 0.2} \over {9 \times {{10}^9}}}$

$q = {{2\sqrt 2 } \over 3} \times {10^{ - 8}}$

after comparison from the given equation a = 20

The current ${i_1} = \left( {{{{R_2}} \over {{R_1} + {R_2}}}} \right)i$

$= \left( {{2 \over {4 + 2}}} \right) \times 6$

${i_1} = 2A$

Speed of transverse wave is

$V = \sqrt {{T \over \mu }}$

$\ln V = {1 \over 2}\ln T - {1 \over 2}\ln \mu$

${{\Delta V} \over V} = {1 \over 2}{{\Delta T} \over T}$

$= {1 \over 2} \times 4$

$\Rightarrow$ ${{\Delta V} \over V} = 2\%$

The translational kinetic energy & rotational kinetic energy both obey Maxwell's distribution independent of each other.

T.K.E. of diatomic molecules = ${3 \over 2}kT$

R.K.E. of diatomic molecules = ${2 \over 2}kT$

So statement I is true but statement II is false.

In a ferromagnetic material, below the curie temperature a domain is defined as a macroscopic region with saturation magnetization.

Given mass of electron = m_e

Mass of proton = m_p

$\therefore$ given m_p = 1836 m_e

From de-Broglie wavelength

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

${{{\lambda _e}} \over {{\lambda _p}}} = {{{m_p}} \over {{m_e}}}$

$= {{1836{m_e}} \over {{m_e}}}$

${{{\lambda _e}} \over {{\lambda _p}}} = 1836$

Firstly, we know that the intensity (I) of a wave is defined as the power (P) per unit area (A). For a spherical wave emanating from a point source, the area of the sphere is $4\pi r^2$ where r is the distance from the source. So,

I = \frac{P}{4\pi r^2} \tag{1} ........(1)

This intensity can also be related to the electric field (E) in an electromagnetic wave using the equation :

I = \frac{1}{2} c \varepsilon_0 E^2 \tag{2} .........(2)

where $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=$ speed of light in vacuum and $\varepsilon_0$ is the permittivity of free space.

From equations (1) and (2), we can solve for E and square root it to find the peak value of the electric field $E_{\text{peak}}$ :

$E_{\text{peak}} = \sqrt{\frac{2P}{c \varepsilon_0 4\pi r^2}} = \sqrt{\frac{2P \mu_0 c}{4\pi r^2}}$

Since $\varepsilon_0=\frac{1}{\mu_0 c^2}$

Given the efficiency of the bulb is 10%, the actual power radiated is $0.10 \times 8\, \text{W} = 0.8\, \text{W}$.

So, substituting $P = 0.8\, \text{W}$, $r = 10\, \text{m}$, $c = 3 \times 10^8\, \text{m/s}$, and $\mu_0 = 4\pi \times 10^{-7}\, \text{T m/A}$, we have :

$E_{\text{peak}} = \sqrt{\frac{2 \times 0.8 \times 4\pi \times 10^{-7} \times 3 \times 10^8}{4\pi \times 100}} = \frac{x}{10} \sqrt{\frac{\mu_0 c}{\pi}}$

Comparing with the expression in the question, we find that $x = 2$.

Therefore, the answer is $x = 2$.

Given

e = electronic charge

c = speed of light in free space

h = Planck's constant

We know, E = ${{hc} \over \lambda }$

and $F = {1 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {{d^2}}}$ $\Rightarrow$ ${{{q^2}} \over {4\pi {\varepsilon _0}}} = F{d^2}$

${1 \over {4\pi {\varepsilon _0}}}{{{e^2}} \over {hc}}$

= ${{F{d^2}} \over {E\lambda }}$

= ${{Fd.d} \over {E\lambda }}$

= ${d \over \lambda }$

$=$ dimensionless

$= \left[ {{M^0}{L^0}{T^0}} \right]$

$\because$ $n = {w \over {{Q_{in}}}} = {1 \over 4}$

${1 \over 4} = 1 - {{{T_1}} \over {{T_2}}}$

${{{T_1}} \over {{T_2}}} = {3 \over 4}$

When the temperature of the sink is reduced by 52K then its efficiency is doubled.

${1 \over 2} = 1 - {{\left( {{T_1} - 52} \right)} \over {{T_2}}}$

${{{T_1} - 52} \over {{T_2}}} = {1 \over 2}$

${{{T_1}} \over {{T_2}}}{{ - 52} \over {{T_2}}} = {1 \over 2}$

${3 \over 4} - {{52} \over {{T_2}}} = {1 \over 2}$

${{52} \over {{T_2}}} = {1 \over 4}$

${T_2} = 208$ K

Flux through cube, $\phi = {q \over {8{\varepsilon _0}}}$

Flux through surfaces ABEH, ADGH, ABCD will be zero.

$\phi$(EFGH) = $\phi$(DCFG) = $\phi$(EBCF) = ${1 \over 3}\left( {{q \over {8{\varepsilon _0}}}} \right)$

= ${q \over {24{\varepsilon _0}}}$

Given frequency of massage signal = f_m

Frequency of carrier signal = f_c

The wavelength of the corresponding signal in air is $\Rightarrow$ $\lambda = {c \over {{f_c}}}$

Since $\phi$ remain same, circuit is in resonance

$\therefore$ ${I_{rms}} = {{{v_{rms}}} \over z}$ = ${{{v_{rms}}} \over R}$

$= {{220} \over {110}}$

$\Rightarrow$ ${I_{rms}} = 2A$

From the photoelectric effect equation

${{hc} \over \lambda } = \phi + e{v_s}$

so, $e{v_{{s_1}}} = {{hc} \over {{\lambda _1}}} - \phi$ .....(i)

$e{v_{{s_2}}} = {{hc} \over {{\lambda _2}}} - \phi$ ......(ii)

Subtract equation (i) from equation (ii)

$e{v_{{s_1}}} - e{v_{{s_2}}} = {{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _2}}}$

${v_{{s_1}}} - {v_{{s_2}}} = {{hc} \over e}\left( {{1 \over {{\lambda _1}}} - {1 \over {{\lambda _2}}}} \right)$

$(0.710 - 1.43) = 1240\left( {{1 \over {491}} - {1 \over {{\lambda _2}}}} \right)$

${{ - 0.72} \over {1240}} = {1 \over {491}} - {1 \over {{\lambda _2}}}$

${1 \over {{\lambda _2}}} = {1 \over {491}} + {{0.72} \over {1240}}$

${1 \over {{\lambda _2}}} = 0.00203 + 0.00058$

${1 \over {{\lambda _2}}} = 0.00261$

${\lambda _2} = 383.14$

${\lambda _2} \simeq 382$ nm

In n-type semiconductor pentavalent impurity is added. Each pentavalent impurity donates a free electron. So the Fermi-level of n-type semiconductor will go upward.

And In p-type semiconductor trivalent impurity is added. Each trivalent impurity creates a hole in the valence band. So the Fermi-level of p-type semiconductor will go downward.

Yes, you're absolutely correct. This equation you provided is for the angular size of the central maximum (or central diffraction disk) in a circular aperture diffraction pattern (like a pinhole) :

$\sin \theta = \frac{1.22\lambda}{D}$

where :

• θ is the angle subtended by the radius of the central maximum at the pinhole,


• λ is the wavelength of the light, and


• D is the diameter of the aperture or pinhole.

If D (the diameter of the pinhole) is increased, then sinθ (and hence θ itself, for small θ) will decrease, implying that the size of the central maximum or diffraction disk will decrease. This is because less diffraction (bending of light) occurs when the pinhole is larger.

At the same time, increasing the size of the pinhole allows more light to pass through, which increases the intensity (brightness) of the light in the diffraction pattern.

So, increasing the diameter of the pinhole decreases the size of the diffraction pattern but increases its intensity, which confirms Option C.

Due to parallel combination K_eff = 2k + 2k = 4k

$\because$ $T = 2\pi \sqrt {{m \over {{k_{eff}}}}}$

$= 2\pi \sqrt {{m \over {4k}}}$

$T = \pi \sqrt {{m \over k}}$

Given wavelength of an x-ray beam = 10$\mathop A\limits^o$

$\because$ E = ${{hc} \over \lambda } = m{c^2}$

$\Rightarrow$ m = ${h \over {c\lambda }}$

The mass of a fictitious particle having the same energy as that of the x-ray photons = ${x \over 3}$h kg

$\therefore$ ${x \over 3}h = {h \over {c\lambda }}$

$x = {3 \over {c\lambda }}$

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

x = 10

The gas performs isothermal irreversible work (W).

where, $\Delta$U = 0 (change in internal energy)

From, 1st law of thermodynamics,

$\Rightarrow$ $\Delta$U = $\Delta$Q + W

$\Rightarrow$ 0 = $\Delta$Q + W

$\Rightarrow$ $\Delta$Q = $-$W

Now, $W = - {p_{ext}}({V_2} - {V_1})$

$= - {p_{ext}}\left( {{{nRT} \over {{p_2}}} - {{nRT} \over {{p_1}}}} \right) = - {p_{ext}} \times nRT\left( {{1 \over {{p_2}}} - {1 \over {{p_1}}}} \right)$

Given, p_ext = 4.3 MPa, p₁ = 2.1 MPa, p₂ = 1.3 MPa, n = 5 mol, T = 293 K and R = 8.314 J mol^$-$1 K^$-$1

$= - 4.3 \times 5 \times 8.314 \times 293\left( {{1 \over {1.3}} - {1 \over {2.1}}} \right)$

= $-$ 15347.70 J mol^$-$1

= $-$ 15.347 kJ mol^$-$1 $\simeq$ $-$ 15 kJ mol^$-$1

$\Rightarrow$ $\Delta$Q = 15 kJ mol^$-$1

1. Viscosity of hydrophilic sol > viscosity of H₂O

2. Hydrophilic sol is more stable so can't be easily coagulated.

3. Hydrophilic sols are reversible sols.

4. No electrolytes are required to stabilise hydrophilic sol.

Composition of German Silver $\to$ 56% Cu + 24% Zn + 20% Ni

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

$\alpha - sulphur\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{{ < 369K}}^{{ > 369K}}} \beta - sulphur$

at room temperature $\alpha$-sulphur (Rhombic) is most stable form.

Acidic strength


Since carboxylic acids are more acidic than phenols.

–I and – R effect increase the acidic strength where as +I and +R effect decrease the acidic strength of carboxylic acids.

Statement I is true. Dimethyl glyoxime (DMG) is a chelating reagent commonly used for the identification of Ni²⁺ ions in aqueous solution. In the presence of ammonium hydroxide (NH₄OH), Ni²⁺ reacts with dimethyl glyoxime to form a deep red-colored complex, which can be used to confirm the presence of Ni²⁺ ions.

Statement II is also true. Dimethyl glyoxime is a neutral bidentate ligand, which means it has two donor atoms that can coordinate with a metal ion, and its overall charge is neutral.

As AB is a binary electrolyte,

$\therefore$ AB $\rightleftharpoons$ A⁺ + B^$-$, n = 2

$i = 1 + \alpha (n - 1) = 1 + {{75} \over {100}}(2 - 1) = 1.75$

Given, $\Delta$T_b = 2.5 K

K_b = 0.52 K kg mol^$-$1

$\therefore$ $\Delta$T_b = K_b $\times$ m $\times$ i

$\Rightarrow m = {{\Delta {T_b}} \over {{K_b} \times i}} = {{2.5} \over {0.52 \times 1.75}}$

$= 2.74 \simeq 3$ mol/kg

Energy of EMR = IE of the metal (A)

$= hv = {{hc} \over \lambda }$ atom^$-$1 $= {{hc} \over \lambda } \times {N_A}$ mol^$-$1

$= {{(6.63 \times {{10}^{ - 34}}) \times (3 \times {{10}^8}) \times (6.02 \times {{10}^{23}})} \over {(663 \times {{10}^{ - 9}})}}$ J mol^$-$1 [$\because$ $\lambda$ = 663 nm = 663 $\times$ 10^$-$9 m]

= 180600 J mol^$-$1 = 180.6 kJ mol^$-$1 $\simeq$ 181 kJ mol^$-$1

Cell-I $(HN{O_3} \to NO)$

$3Cu + 2NO_3^ - + 8{H^ + } \to 3C{u^{2 + }} + 2NO + 4{H_2}O$

${Q_1} = {{{{[C{u^{2 + }}]}^3} \times {{({p_{NO}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^8}}}$

$\because$ $E_1^o = 0.96 - ( - 0.34) = 1.3\,V$

${E_1} = 1.3 - {{0.059} \over 6}\log {Q_1}$

Cell-II $(HN{O_3} \to N{O_2})$

$Cu + 2NO_3^ - + 4{H^ + } \to C{u^{2 + }} + 2N{O_2} + 2{H_2}O$

${Q_2} = {{[C{u^2}] \times {{({p_{N{O_2}}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^4}}}$

$\because$ $E_2^o = 0.79 - ( - 0.34)\,V = 1.13\,V$

${E_2} = 1.13 - {{0.059} \over 2}\log {Q_2}$

Now, ${E_1} = {E_2}$

$1.3 - {{0.059} \over 6}\log {Q_1} = 1.13 - {{0.059} \over 2}\log {Q_2}$

$0.17 = {{0.059} \over 6}[\log {Q_1} - 3\log {Q_2} - = {{0.059} \over 6}\log {{{Q_1}} \over {{Q_2}}}$

$= {{0.059} \over 6}\log {{{{[C{u^{2 + }}]}^3} \times {{({p_{NO}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^8}}} \times {{{{[NO_3^ - ]}^6} \times {{[{H^ + }]}^{12}}} \over {{{[C{u^{2 + }}]}^3} \times {{({p_{N{O_2}}})}^6}}}$

$= {{0.059} \over 6}\log {{{{[{H^ + }]}^4} \times {{[NO_3^ - ]}^4}} \over {{{({p_{N{O_2}}})}^4}}}$ [$\because$ ${p_{NO}} = {p_{N{O_2}}}$]

$= {{0.059} \over 6}\log {{{{[HN{O_3}]}^4}} \over {{{({p_{N{O_2}}})}^4}}}$

Now, ${p_{N{O_2}}} \equiv [HN{O_3}]$

So, $0.17 = {{0.059} \over 6}\log {[HN{O_3}]^8}$

$= {{0.059} \over 6} \times 8\log [HN{O_3}]$

$\log [HN{O_3}] = 2.16$

$[HN{O_3}] = {10^{2.16}}M = {10^x}M$

$\therefore$ $x = 2.16$

$\Rightarrow 2x = 2 \times 2.16 = 4.32 \simeq 4$

CO₂ + H₂O $\to$ H₂CO₃

C + H₂O(steam) $\to$ CO + H₂

CH₄ + H₂O $\to$ CO + 3H₂

C₃H₈ + H₂O $\to$ 3CO + H₂

(a)   XeF₄  is sp³d² hybridised with 4 bond pairs and 1 lone pair and structure is square planar. Here all the bond lengths are equal.



(b) $\,\,\,\,$ BF$_4^ -$, 4 bond pair present so angle is 109^o 28^' and sp³ hybridised. So structure is regular tetrahedral. Here all the bond lengths are equal.



(c) SiF₄ is sp³ hybridisation and regular tetrahedral geometry. Here all the bond lengths are equal.



(d)   SF₄ is sp³d hybridised with 4 bond pairs and 1 lone pair and its expected trigonal bipyramidal geometry gets distorted due to presence of a lone pair of electrons and it becomes distorted tetrahedral or see-saw with the bond angles equal to < 120^o and 179^o instead of the expected angles of 120^o and 180^o respectively. Here axial and equitorial both bonds are presents. And we know axial bonds are longer and weaker.