JEE Main 2020 3rd September Morning Shift Question Paper with Solutions

Applying Angular momentum conservation

mv$l$ = $\left( {{{M{l^2}} \over 3} + m{l^2}} \right)$$\omega$

$\Rightarrow$ $\omega = {{1 \times 6 \times 1} \over {{2 \over 3} + 1}}$ = ${{18} \over 5}$

Now, using energy conservation

${1 \over 2}I{\omega ^2} = \left( {m + M} \right)gh$

$\Rightarrow$ ${1 \over 2}\left( {{{M{l^2}} \over 3} + m{l^2}} \right){\omega ^2} = \left( {m + M} \right)g{x_{COM}}\left( {1 - \cos \theta } \right)$

$\Rightarrow$ ${1 \over 2}\left( {{2 \over 3} + 1} \right){\left( {{{18} \over 5}} \right)^2} = \left( {m + M} \right)g \times {{M \times {l \over 2} + ml} \over {\left( {M + m} \right)}}\left( {1 - \cos \theta } \right)$

$\Rightarrow$ ${1 \over 2} \times {5 \over 3} \times {{18 \times 18} \over {25}} = 10 \times \left( {2 \times {1 \over 2} + 1} \right)\left( {1 - \cos \theta } \right)$

$\Rightarrow$ ${5 \over 6} \times {{18 \times 18} \over {25}} = 20\left( {1 - \cos \theta } \right)$

$\Rightarrow$ $\left( {1 - \cos \theta } \right) = {{18} \over 5} \times {3 \over {20}}$

$\Rightarrow$ ${\cos \theta = 1 - {{27} \over {50}}}$

$\Rightarrow$ ${\cos \theta = {{23} \over {50}}}$

$\Rightarrow$ $\theta$ $\simeq$ 63^o

V₀ = $\sqrt {{{GM} \over {{R_e}}}}$

Applying Conservation of Angular Momentum,

mVR_e = mV'R

$\Rightarrow$ m$\sqrt {{3 \over 2}} {V_0}{R_e}$ = mV'R

$\Rightarrow$ V' = $\sqrt {{3 \over 2}} {{{V_0}{R_e}} \over R}$

Applying Conservation of Energy,

${ - {{GMm} \over {{R_e}}}}$ + ${1 \over 2}m{V^2}$ = ${ - {{GMm} \over R_{max}}}$ + ${1 \over 2}mV{'^2}$

$\Rightarrow$ ${ - {{GMm} \over {{R_e}}}}$ + ${1 \over 2}m\left( {{3 \over 2}V_0^2} \right)$ = ${ - {{GMm} \over R_{max}}}$ + ${1 \over 2}m \times$${\left( {V'} \right)^2}$

$\Rightarrow$ ${ - {{GMm} \over {{R_e}}} + }$ ${1 \over 2}m \times {3 \over 2}{{GM} \over {{R_e}}}$ = $- {{GMm} \over R_{max}} + {1 \over 2}m \times {3 \over 2}{V_0}{{R_e^2} \over {{R^2}}}$

$\Rightarrow$ ${ - {{GMm} \over {{R_e}}} + }$ ${1 \over 2}m \times {3 \over 2}{{GM} \over {{R_e}}}$ = $- {{GMm} \over R_{max}} + {1 \over 2}m \times {3 \over 2}{{GM} \over {{R_e}}}{{R_e^2} \over {{R^2}}}$

$\Rightarrow$ $- {1 \over {{R_e}}} + {3 \over {4{R_e}}}$ = $- {1 \over R} + {{3{R_e}} \over {4{R^2}}}$

$\Rightarrow$ ${1 \over {4{R_e}}}$ = $- {1 \over R} + {{3{R_e}} \over {4{R^2}}}$

$\Rightarrow$ R = 3R_e

q₁ + q₂ = 12 - 3 = 9 $\mu$C .....(1)

and V₁ = V₂

$\Rightarrow$ ${{K{q_1}} \over {{{2R} \over 3}}} = {{K{q_2}} \over {{R \over 3}}}$

$\Rightarrow$ q₁ = 2q₂ .....(2)

Now puttion the value of (2) in (1), we get

2q₂ + q₂ = 9

$\Rightarrow$ 3q₂ = 9

$\Rightarrow$ q₂ = 3 $\mu$C

$\Rightarrow$ q₁ = 6 $\mu$C

Q₂ = CV = 8 $\times$ 20 = 160 $\mu$C

Q₁ = 750 - 160 = 590 $\mu$C

Given, $\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i$ T

$\therefore$ E₀ = CB₀

= 3 × 10⁸ × 3 × 10^–8 = 9 V/m

We know, $\left( {\overrightarrow E \times \overrightarrow B } \right)||\overrightarrow C$

And here $\widehat B = \widehat i\& \widehat C = - \widehat j$

$\therefore$ $\widehat E = - \widehat k$

So, $\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$ V/m

$\phi$ = BAcos$\theta$

$\Rightarrow$ $\phi$ = BAcos($\omega t$)

${{d\phi } \over {dt}} = - BA\omega \sin \left( {\omega t} \right)$

We know, $\varepsilon$ = $- {{d\phi } \over {dt}}$ = $BA\omega \sin \left( {\omega t} \right)$

Power (P) = ${{{\varepsilon ^2}} \over R} = {{{B^2}{A^2}{\omega ^2}{{\sin }^2}\left( {\omega t} \right)} \over R}$

$\therefore$ P_av = ${{{B^2}{A^2}{\omega ^2}} \over 2R}$

= ${{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}} \over {2R}}$ [As A = $\pi$ab]

Capillary rise

h = ${{2T\cos \theta } \over {\rho gr}}$

$\Rightarrow$ T = ${{\rho grh} \over {2\cos \theta }}$

= ${{\left( {900} \right)\left( {10} \right)\left( {15 \times {{10}^{ - 5}}} \right)\left( {15 \times {{10}^{ - 2}}} \right)} \over 2}$

= 1012.5 $\times$ 10^–4

= 101.25 × 10^–3 = 101.25 mN/m

$\simeq$ 101.00 mN/m

K₁ = ${{hc} \over {500}} - {\phi _0}$

K₂ = ${{hc} \over {200}} - {\phi _0}$

$\because$ K₂ = 3K₁

$\Rightarrow$ $3\left[ {{{hc} \over {500}} - {\phi _0}} \right] = \left[ {{{hc} \over {200}} - {\phi _0}} \right]$

$\Rightarrow$ $\phi$₀ = 0.61 eV

1.5 – 0.5 – R $\times$ 10^–2 = 0

$\Rightarrow$ R = 100 $\Omega$

Internal resitance of battery

$\int {dr} = \int\limits_a^b {{{\rho dr} \over {2\pi rl}}}$

$\Rightarrow$ r = ${{\rho \over {2\pi l}}\ln {b \over a}}$

Power in external resistance P_R = ${\left( {{ \in \over {r + R}}} \right)^2}R$

Maximum Power in external resistance is generated when it is equal to internal resistance of battery. So P_R is maximum when r = R.

$\therefore$ R = ${{\rho \over {2\pi l}}\ln {b \over a}}$

Initial velocity, v = $\sqrt {2gh}$

= $\sqrt {2 \times 10 \times 20}$

= 20 m/s

Now work done by the machine,

W_F = $\Delta$k

$\Rightarrow$ F.d = $\Delta$k

$\Rightarrow$ F = ${{\Delta k} \over d}$

= ${{{1 \over 2} \times 0.15 \times 400 - 0} \over {0.2}}$

= 150 N

$\Delta$V = V$\gamma$$\Delta$T

$\therefore$ V₁$\gamma$₁ = V₂$\gamma$₂

$\Rightarrow$ 500 $\times$ 6 $\times$ 10^-6 = V_m $\times$ 1.5 $\times$ 10^-4

$\Rightarrow$ V_m = ${{500 \times 6 \times {{10}^{ - 6}}} \over {1.5 \times {{10}^{ - 4}}}}$

$\Rightarrow$ V_m = 20 cc

$\tan 30 = {x \over d}$

$d = {x \over {\tan 30}}$

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

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

For one part of the wire with N turns,

$B = {{{\mu _0}IN} \over {4\pi d}}(\sin {\theta _1} + \sin {\theta _2})$

For 6 identical parts of the wire,

${B_{net}} = 6B$

$= {{6{\mu _0}IN} \over {4\pi d}}(\sin 30 + \sin 30)$

$= {{{\mu _0}I} \over \pi }\left( {{{6 \times 50} \over {4 \times 5\sqrt 3 \times {{10}^{ - 2}}}}} \right)\left( {2 \times {1 \over 2}} \right)$

$= 500\sqrt 3 \left( {{{{\mu _0}I} \over \pi }} \right)$

${P_{in}} = {P_0} + {{4T} \over {{R_1}}}$

$\Rightarrow 1.01 = 1 + {{4T} \over {{R_1}}}$

$\Rightarrow {{4T} \over {{R_1}}} = 0.01$

$1.02 = 1 + {{4T} \over {{R_2}}}$

$\Rightarrow {{4T} \over {{R_2}}} = 0.02$

$\therefore {{{R_2}} \over {{R_1}}} = {1 \over 2}$

$\Rightarrow {R_1} = 2{R_2}$

${{{V_1}} \over {{V_2}}} = {{R_1^3} \over {R_2^3}} = {{8R_2^3} \over {R_2^3}} = {8 \over 1}$

$\beta$ = ${{\lambda D} \over d}$

and $\theta$ = ${\beta \over D}$

$\Rightarrow$ $\theta$ = ${\lambda \over d}$

= ${{500 \times {{10}^{ - 9}}} \over {0.05 \times {{10}^{ - 3}}}}$

= 0.01 rad

= 0.57^o

tan r = ${{15} \over {30}}$ = ${1 \over 2}$

sin r = ${1 \over {\sqrt 5 }}$

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

$\Rightarrow$ ${1 \over {\sqrt 2 }} = \mu \left( {{1 \over {\sqrt 5 }}} \right)$

$\Rightarrow$ $\mu$ = $\sqrt {{5 \over 2}}$ = 1.58

$\mu = {N \over {100}}$ = 1.58

$\Rightarrow$ N = 158

Using a screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured as:

Measurement = (Main scale reading) + (Circular scale reading × Least count)

where the least count is calculated as the pitch of the screw gauge divided by the number of divisions on the circular scale:

Least count = (Pitch of screw gauge) / (Number of circular scale divisions)

Least count = ${{0.1} \over {50}}$ = 0.002 cm

Now if we multiply division of circular scale with least count then we get 0^th digit of fraction part even.

Here only option B has 0^th digit of fraction part even.

${I_1}{\omega _1} = {I_2}{\omega _2}$

$\Rightarrow$ $\left( {{{M{R^2}} \over 2} + m{R^2}} \right){\omega _1} = {{M{R^2}} \over 2}{\omega _2}$

$\Rightarrow$ $\left( {1 + {{2m{R^2}} \over {M{R^2}}}} \right){\omega _1} = {\omega _2}$

$\Rightarrow$ $\left( {1 + {{2 \times 80} \over {200}}} \right){\omega _1} = {\omega _2}$

$\Rightarrow$ ${\omega _2} = \left( {1.8} \right){\omega _1}$

$\Rightarrow$ 2$\pi$f₂ = 2$\pi$f₁ $\times$ 1.8

$\Rightarrow$ f₂ = 5 $\times$ 1.8 = 9

Bursting of helium ballon is irreversible adiabatic because Energy can not be restored.

Degree of freedom (f) = 3 + 3 = 6 and n = 1

U = ${f \over 2}nRT$

= ${6 \over 2} \times 1 \times RT$

= 3RT

T₁ = 2g

T₂ = 8g

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

$\therefore$ V $\propto$ $\sqrt T$

Also V = f$\lambda$

$\therefore$ V₁ = f₁$\lambda$₁

and V₂ = f₂$\lambda$₂

We know frequency of sources are same.

$\therefore$ f₁ = f₂

So $\sqrt {{{{T_1}} \over {{T_2}}}} = {{{\lambda _1}} \over {{\lambda _2}}}$

$\Rightarrow$ $\sqrt {{{2g} \over {8g}}} = {6 \over {{\lambda _2}}}$

$\Rightarrow$ $\lambda$₂ = 12 cm

Given I = $M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)$

M = $\rho$.V = $\rho$$\pi$R²L

$\Rightarrow$ R² = ${M \over {\rho \pi L}}$

$\therefore$ I = $M\left( {{M \over {4\rho \pi L}} + {{{L^2}} \over {12}}} \right)$

$\Rightarrow$ ${{dI} \over {dL}}$ = $M\left( {{M \over {4\rho \pi }}\left( { - {1 \over {{L^2}}}} \right) - {{2L} \over {12}}} \right)$

For minimum I, ${{dI} \over {dL}} = 0$

$\therefore$ $M\left( {{M \over {4\rho \pi }}\left( { - {1 \over {{L^2}}}} \right) - {{2L} \over {12}}} \right)$ = 0

$\Rightarrow$ ${{{M^2}} \over {4\rho \pi {L^2}}} = {{2LM} \over {12}}$

$\Rightarrow$ ${M \over {4\rho \pi {L^2}}} = {L \over 6}$

$\Rightarrow$ ${{\rho \pi {R^2}L} \over {4\rho \pi {L^2}}} = {L \over 6}$

$\Rightarrow$ ${{{R^2}} \over {{L^2}}} = {2 \over 3}$

$\Rightarrow$ ${R \over L} = \sqrt {{2 \over 3}}$

$\Rightarrow$ ${L \over R} = \sqrt {{3 \over 2}}$

f = 750 Hz, V_rms = 20 V,

R = 100 $\Omega$, L = 0.1803 H,

C = 10$\mu$ F, S = 2 J/°C

|Z| = $\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}$

= $\sqrt {{R^2} + {{\left( {\omega L - {1 \over {\omega C}}} \right)}^2}}$

= $\sqrt {{R^2} + {{\left( {2\pi fL - {1 \over {2\pi fC}}} \right)}^2}}$

= $\sqrt {{{(100)}^2} + {{\left( {2 \times 3.14 \times 750 \times 0.1803 - {1 \over {2 \times 3.14 \times 750 \times {{10}^{ - 5}}}}} \right)}^2}}$

= 834 $\Omega$

In AC, power (P) = i_rmsV_rms cos $\phi$

and i_rms = ${{{V_{rms}}} \over {\left| Z \right|}}$

Power factor (cos $\phi$) = ${R \over {\left| Z \right|}}$

$\therefore$ P = ${{{V_{rms}}} \over {\left| Z \right|}}.{V_{rms}}.{R \over {\left| Z \right|}}$

= ${\left( {{{{V_{rms}}} \over {\left| Z \right|}}} \right)^2}R$

= ${\left( {{{20} \over {834}}} \right)^2} \times 100$

= 0.0575 J/S

Also, H = Pt = S$\Delta$$\theta$

$\Rightarrow$ t = ${{2\left( {10} \right)} \over {0.0575}}$ = 348 sec

First order decay

N(t) = N₀e^-$\lambda$t

Given ${{N\left( t \right)} \over {{N_0}}} = {9 \over {16}} =$ e^-$\lambda$t

N(t/2) = N₀e^{-$\lambda$(t/2)}

${{N\left( {t/2} \right)} \over {{N_0}}} = \sqrt {{e^{ - \lambda t}}}$ = $\sqrt {{9 \over {16}}}$ = ${3 \over 4}$

$\Rightarrow$ $N\left( {t/2} \right) = {3 \over 4}{N_0}$

Given,

${\overrightarrow V }$ = $\left( {2\widehat i + 3\widehat j + 4\widehat k} \right)$ ms^–1

${\overrightarrow B }$ = $\left( {5\widehat i + 3\widehat j - 6\widehat k} \right)$× 10^–3 T

q = 1 $\mu$C

$\overrightarrow F = q\left( {\overrightarrow V \times \overrightarrow B } \right)$

= {10^{ - 6}} \times {10^{ - 3}} \times \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 3 & 4 \cr 5 & 3 & { - 6} \cr } } \right|

= (${ - 30\widehat i + 32\widehat j - 9\widehat k}$) $\times$ 10^-9

Volume strength of H₂O₂ at 1 atm 273 kelvin
= M × 11.2 = 8.9 × 11.2 = 99.68 $\simeq$ 100

Mole fraction of glucose in aqueous solution = 0.1

Let total mole is 1 mol then mole of glucose will be 0.1 and mole of water will be 0.9.

So mass % of water = ${{0.9 \times 18} \over {0.1 \times 180 + 0.9 \times 18}}$ $\times$ 100

= 47.37 $\simeq$ 47

Glycerol is separated in soap industries by distillation under reduced pressure.

Thermal power plants lead to acid rain.

Au + 4H⁺ + NO₃^- + 4Cl^- $\to$ AuCl₄^- + NO + 2H₂O

3Pt + 16H⁺ + 4NO₃^- + 18Cl^- $\to$ 3PtCl₆^2- + 4NO + 8H₂O

For Unnilennium

IUPAC symbol – Une

Atomic No. – 109

Note :

(1) $\,\,\,\,$ Bond strength $\propto$ Bond order

(2) $\,\,\,\,$ Bond length $\propto$ ${1 \over {Bond\,\,order}}$

(3) $\,$ Bond order $= {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Molecular orbital configuration of NO (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ *$

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Similarly Molecular orbital configuration of NO⁺ (14 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,$

$\therefore\,\,\,\,$ N_b = 10

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Similarly Molecular orbital configuration of NO²⁺ (13 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^1}}\,$

$\therefore\,\,\,\,$ N_b = 9

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {9 - 4} \right]$ = 2.5

Molecular orbital configuration of NO^- (16 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^1}^ *$

$\therefore\,\,\,\,$ N_b = 10

N_a = 6

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right]$ = 2

As Bond strength $\propto$ Bond order

$\therefore$ NO^– will have minimum bond strength.

Zero order reaction has complex mechanism.
Zero order reaction is a multistep reaction.

Diameter of dispersed particles should not be much smaller than wavelength of light used.

(1) On increasing temperature solubility of gases decreases.

(2) 55.5 molal solution of $\gamma$ = 55.5 mol of

P_$\gamma$ = K_HX_$\gamma$

= 2 $\times$ 10^-5 $\times$ ${{55.5} \over {55.5 + {{1000} \over {18}}}}$

= 10^–5 K bar

= 10^–2 bar

(3) P_$\delta$ = K_HX_$\delta$

= 0.5 $\times$ ${{55.5} \over {55.5 + {{1000} \over {18}}}}$

= 250 bar

(4) With temperature, the value of K_H (Henry’s constant) increases and solubility of gas in liquid decreases.

Pyrophosphoric acid

Number of P - OH bonds = 4

P = O bonds = 2

P – O – P bonds = 1

Nitro and diazo compounds do not give Kjeldahl estimation of Nitrogen.

[Ti(H₂O)₆]³⁺

Ti = [Ar]3d²4s²

Ti⁺³ = [Ar]3d¹


For octahedral

Crystal field stabilization energy (CFSE)

= (-0.4$\Delta$₀) $\times$ n_t2g + (+0.6$\Delta$₀) $\times$ n_eg + np

n_t2g = number of electrons in t_2g orbital

n_eg = number of electrons in e_g orbital

n = number of extra pairs

p = Pairing energy

Here n_t2g = 1

n_eg = 0

n = 0 ( For weak field ligands n always zero. Here H₂O is weak field ligand)

$\therefore$ Crystal field stabilization energy (CFSE)

= (-0.4$\Delta$₀) $\times$ 1 + (+0.6$\Delta$₀) $\times$ 0 + 0$\times$P

= -0.4$\Delta$₀ = –0.4 × 20300 = –8120 cm^–1

CFSE (in kJ) = ${{8120} \over {83.7}}$ = 97 kJ/mol