JEE Main 2023 30th January Evening Shift Question Paper with Solutions

Intensity at 5 m $= {5 \over {4\pi \times {5^2}}}\left( {{W \over {{m^2}}}} \right)$

$= {1 \over {20\pi }}\left( {{W \over {{m^2}}}} \right)$

Intensity due to electric field $= {1 \over {40\pi }}\left( {{W \over {{m^2}}}} \right)$

$= \left( {{W \over {40\pi }}} \right)$

K.E. per molecule $= \left( {{f \over 2}KT} \right)$

${{\mathrm{average{{(K.E)}_{hydrogen}}}} \over {\mathrm{average{{(K.E)}_{oxygen}}}}} = {{{f_{\mathrm{hydrogen}}}} \over {{f_{\mathrm{oxygen}}}}} = 1$

$P = \sqrt {2\,eVm}$

$\lambda = \left( {{h \over {{P_1}}}} \right)$ ..... (i)

${{3\lambda } \over 2} = {h \over {{P_2}}}$ ..... (ii)

Dividing (i) by (ii)

$\Rightarrow {2 \over 3} = \left( {{{{P_2}} \over {{P_1}}}} \right) = \sqrt {{{{v_2}} \over {{v_1}}}}$

$\Rightarrow {4 \over 9} = \left( {{{{v_2}} \over {{v_1}}}} \right)$

${{{v_1}} \over {{v_2}}} = \left( {{9 \over 4}} \right)$

Force between two current carrying wire

$= {{{\mu _0}{I_1}{I_2}} \over {2\pi d}} \times L$

Here, ${I_1}$ & ${I_2}$ are equal

$F = {{{\mu _0}{I^2}} \over {2\pi d}} \times L$

$F \propto {I^2}$

${{{F_I}} \over {{F_{2I}}}} = {{{I^2}} \over {4{I^2}}} = {1 \over 4}$

$P = Fv$

$m{{vdv} \over {dt}} = P$

$m\int_0^v {vdv = \int_0^t {Pdt} }$

${{m{v^2}} \over 2} = Pt$

$v = \sqrt {{{2P} \over m}} {t^{1/2}}$

$\int_0^s {dx = \sqrt {{{2P} \over m}} \int_0^t {{t^{1/2}}dt} }$

$s = {2 \over 3}\sqrt {{{2P} \over m}} {t^{3/2}}$

or $s = {2 \over 3}\sqrt {{{2P} \over 2}} \times {4^{3/2}}$

$= {{16} \over 3}\sqrt P ~m$

So, $\alpha = 4$

${U_P} = - {{GMm} \over {2R}}$

${U_S} = - {{GMm} \over R}$

$\Rightarrow$ Energy conservation

${1 \over 2}m{v^2} - {{GMm} \over R} = - {{GMm} \over {2R}}$

${v^2} = {{GM} \over R}$

$v = \sqrt {{{GM} \over R}} = \sqrt {gR}$

The circuit is a Wheatstone bridge, so

${{{{6 \times 3} \over {6 + 3}}} \over {{{x \times 1} \over {x + 1}}}} = {{1 + 2} \over x}$

$\Rightarrow {{2(x + 1)} \over x} = {3 \over x}$

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

So $n = 2$

Average speed

$= {{\mathrm{Total\,dis\tan ce}} \over {\mathrm{Total\,time}}}$

$= {8 \over {{4 \over 3} + {4 \over 5}}}$

$= 3.75$ km/h

$\Rightarrow {\lambda _{eff}} = {\lambda _1} + {\lambda _2}$

$\Rightarrow {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over {{{({t_{1/2}})}_1}}} + {{\ln 2} \over {{{({t_{1/2}})}_2}}}$

$\Rightarrow {t_{1/2}} = {{{{({t_{1/2}})}_1} \times {{({t_{1/2}})}_2}} \over {{{({t_{1/2}})}_1} + {{({t_{1/2}})}_2}}} = {{300 \times 30} \over {300 + 30}}s = {{300} \over {11}}s$

$\Rightarrow \alpha = 300$

$\omega = \sqrt {{K \over m}} \Rightarrow \omega \propto {1 \over {\sqrt m }}$

${{{\omega _2}} \over {{\omega _1}}} = \sqrt {{{{m_1}} \over {{m_2}}}} = \sqrt {{1 \over 2}}$

$\eta = 1 - {{{T_{\mathrm{cold}}}} \over {{T_{\mathrm{hot}}}}}$

${T_{\mathrm{cold}}} = \mathrm{OK} \Rightarrow \eta = 1$ (max)

A is correct.

R is also correct and explains A.

Theoretical attenuation $\to$ (IV)

Transducer $\to$ (III)

Demodulation $\to$ (II)

Repeater $\to$ (I)

$4{v^2} = 50 - {x^2}$

or $v = {1 \over 2}\sqrt {50 - {x^2}}$

Comparing the above equation with $v = \omega \sqrt {{A^2} - {x^2}}$

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

& $A = \sqrt {50}$

so ${{2\pi } \over T} = {1 \over 2}$

$\Rightarrow T = 4\pi \sec$

$= 4 \times {{22} \over 7}\sec$

$T = {{88} \over 7}\sec$

so $x = 88$

${E_I} \ne 0$ (inside region)

${E_{II}} = 0$ (conducting region)

${E_{III}} \ne 0$

$= {{KQ} \over {{r^2}}}\,\,\,(r > b)$

Momentum of bullets per unit time

$= {{180 \times {{20} \over {1000}} \times 100} \over {60}}$ kg m/s$^2$

= 6 N

$\Rightarrow$ Force on gun = 6 N

We cannot calculate recoil velocity with the given data.

If we consider recoil velocity at $t = 1$ s, then

${V_{\mathrm{recoil}}} = u + at$

$= 0 + {6 \over {10}} \times 1$ = 0.6 m/s

$({\mu _1} - 1){A_1} = ({\mu _2} - 1){A_2}$

$\Rightarrow (1.54 - 1)6 = (1.72 - 1){A_2}$

${A_2} = \left( {{{0.54} \over {0.72}} \times 6} \right) = {{18} \over 4} = \left( {{9 \over 2}} \right) = 4.5^\circ$

Equivalent circuit can be drawn as

$\therefore$ $\frac{1}{R_{AB}}=\frac{7}{12}+\frac{1}{4}+\frac{2}{3}$

$R_{AB}=\frac{2}{3}~\Omega$

$I' = I{\cos ^2}\left( {{{k\Delta x} \over 2}} \right)$

so ${I_1} = 4{I_0}{\cos ^2}\left( {{{2\pi } \over {2\lambda }} \times {\lambda \over 4}} \right)$

${I_1} = 2{I_0}$

& ${I_2} = 4{I_0}{\cos ^2}\left( {{{2\pi } \over {2\lambda }} \times {\lambda \over 3}} \right)$

${I_2} = {I_0}$

So ${{{I_1} + {I_2}} \over {{I_0}}} = 3$

Drawing the FBD of the point where $F$ is applied

$T' = \sqrt 3 g$

$\Rightarrow T\cos 30^\circ = \sqrt 3 g$

$\Rightarrow T = 2g = 20$ N

${B_{net}} = {{{\mu _0}i} \over {4\pi r}}(\sin \alpha + \sin \beta ) \times 3$

$= {{{\mu _0} \times 2} \over {4\pi \times (2 \times {{10}^{ - 2}})}} \times \left( {{{\sqrt 3 } \over 2} + {{\sqrt 3 } \over 2}} \right) \times 3$

$= {10^{ - 7}} \times {10^2}(3\sqrt 3 )$

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

Torque $\to$ kg m$^2$ s$^{-2}$ (IV)

Energy density $\to$ kg m$^{-1}$ s$^{-2}$ (I)

Pressure gradient $\to$ kg m$^{-2}$ s$^{-2}$ (III)

Impulse $\to$ kg m s$^{-1}$ (II)

$R = {R_0}{A^{{1 \over 3}}}$, using this

$\rho = {M \over {{4 \over 3}\pi {R^3}}} = {{A{m_P}} \over {{4 \over 3}\pi R_0^3A}} = {{{m_P}} \over {{4 \over 3}\pi R_0^3}}$

$\rho$ is independent of mass number.

$\therefore$ A is false

$v = {v_0} - \mu gt$

$\Rightarrow v = 18 - 0.3 \times 10 \times 2 = 12$ m/s

$\Rightarrow$ Kinetic energy $= {1 \over 2}m{v^2} + {1 \over 2}{{m{v^2}} \over 2}$

$= {3 \over 4}m{v^2} = {3 \over 4} \times 0.5 \times 144\,\mathrm{J} = 54\,\mathrm{J}$

${I_{rms}} = {{{V_{rms}}} \over z} = {{200\sqrt 2 } \over {\sqrt {{{100}^2} + {{(200 - 100)}^2}} }}$

$= {{200\sqrt 2 } \over {100\sqrt 2 }}$

= 2 A

Let the correct temperature be X$^\circ$C

$\Rightarrow {{X - 0} \over {100 - 0}} = {{41 - 5} \over {95 - 5}} \Rightarrow X = 40$

$\Rightarrow$ Temperature is 273 + 40 K = 313 K

$Y = \overline {\left( {\overline A + AB} \right)\left( {\overline B + AB} \right)} = \left( {A + B} \right)\left( {\overline A + \overline B } \right)$

$= A\overline B + B\overline A$ (XOR gate)

$\because$ $\Delta l = {{Fl(4)} \over {Y\pi {d^2}}}$

${{\Delta {l_1}} \over {\Delta {l_2}}} = {{\Delta {l_1}} \over {d_1^2}} \times {{d_2^2} \over {{l_2}}}$

${{0.2} \over {\Delta {l_2}}} = {1 \over 2} \times {(2.4)^2}$

$\Delta {l_2} = {{2 \times 0.2} \over {{{(2.4)}^2}}}$

$= 6.9 \times {10^{ - 2}}$ mm

$\phi=B.A$

$\phi=\mathrm{BNA}\cos\omega t$

So, $Emf = {{ - d\phi } \over {dt}} = NBA\omega \sin \omega t$

So maximum value of emf is

${E_{\max }} = NBA\omega$

$= 100 \times 3 \times 14 \times {10^{ - 2}} \times {{360 \times 2\pi } \over {60}} = 1584$

Maximum electrons accommodated in $\mathrm{(n = 4)}$ is $\mathrm{2n^2=32}$ electrons

The correct order of bond strength is

$\mathrm{H_2O > H_2S > H_2Se > H_2Te}$

Boric acid is solid because of strong hydrogen bond in it.

The peptide has seven amino acid units therefore it has six peptide bonds.

A clean water would have BOD value less than 5 ppm.

$\therefore$ Order of $\mathrm{pK_a}$ will be = b > d > a > c

Note :

Acid Strength : -

Acids : Acids are those molecules which donates H⁺ ion in aqueous solution (water) and increases concentration of H⁺ ion in water.

$H - A(aq)$ $\rightleftharpoons$ ${H^ + }(aq) + {A^ - }(aq)$

Dissociation constant of this reaction is K_a. It is called ionisation constant.

$\to$ So we can say if a acid form more H⁺ ion after adding it in H₂O then that acid is called strong acid. If acid produce less H⁺ ion in H₂O then that acid is weak acid.

$\to$ Acid Strength $\propto$ Stability of conjugate base.

$\to$ Acid Strength $\propto$ K_a

$\to$ Acid Strength $\propto {1 \over {p{K_a}}}$

Remember :

$p{K_a} = - \log {K_a}$

$= \log {({K_a})^{ - 1}} = \log {1 \over {{K_a}}}$