JEE Main 2023 31st January Evening Shift Question Paper with Solutions

Current sensitivity $=$ Voltage sensitivity $\times R$

Current sensitivity is made $1.5$ times.

$R$ also increase $1.5$ times.

Hence voltage sensitivity $=\frac{1.5 \times \text { current sensitivity }}{1.5 \times R}$

$=\text { no change }$

$\phi=\frac{h c}{\lambda}=\frac{1240}{350} \mathrm{eV}=3.54 \mathrm{eV}$

$\therefore$ Only metal B will emit photoelectron.

$\mathrm{E}_{H}=13.6$

$$\begin{aligned} & \mathrm{E}_{\mathrm{Li}^{2+}}=13.6 \frac{Z^{2}}{n^{2}}=13.6 \times \frac{9}{9}=13.6 \mathrm{eV} \\ & =136 \times 10^{-1} \mathrm{eV} \end{aligned}$$

Shift $=\left(d-\frac{d}{\mu}\right)=30 \mathrm{~cm}$

$$\begin{aligned} & \Rightarrow d\left[1-\frac{1}{\frac{5}{3}}\right]=30 \\ & \Rightarrow d=\frac{30 \times 5}{2}=75 \mathrm{~cm} \end{aligned}$$

A. Microwave → IV

B. UV rays → III

C. Infra-red → I

D. X-ray → II

$\vec{L}=\vec{r} \times \vec{p} \Rightarrow[\mathrm{L}]=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\right]\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\right]$

$=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]$

$$\begin{aligned} \vec{\tau}=\vec{r} \times \vec{F} \Rightarrow[\tau] & =\left[\mathrm{L}^{1}\right]\left[\mathrm{MLT}^{-2}\right] \\ & =\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right] \end{aligned}$$

Stress $\equiv$ Pressure $=\frac{F}{A} \Rightarrow[$ Stress $]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$

Pressure Gradient $=\frac{d P}{d x} \Rightarrow[$ Pressure Gradient $]$ $=\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]$

In steady state, capacitor behaves as an open circuit. Circuit is :



$\Rightarrow i_{A B}=\frac{6}{3}=2 \mathrm{~A}$

$i_{A D}=\frac{6}{12}=0.5 \mathrm{~A}$

$\Rightarrow V_{B}+2 \times 2-10 \times 0.5=V_{D}$

$\Rightarrow V_{B}-V_{D}=1$ volt

$\omega=100 \pi$

$$\begin{aligned} & \text { So } Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \\ & =\sqrt{80^{2}+(100-40)^{2}} \\ & =100 \Omega \\ & i_{0}=\frac{V_{0}}{Z}=\frac{2500}{100} \mathrm{~A}=25 \mathrm{~A} \end{aligned}$$

$\Delta Q=n C_{P} \Delta T=$

$n\left( {{f \over 2} + 1} \right)R\Delta T$

= $n\left( {{5 \over 2} + 1} \right)R\Delta T$

= $n\left( {{7 \over 2}} \right)R\Delta T$

Given, $\Delta Q$ = 735

$\Rightarrow$ $n\left( {{7 \over 2}} \right)R\Delta T$ = 735

$\Rightarrow$ $nR\Delta T = 735 \times {2 \over 7}$

Also, $\Delta$U = ${f \over 2}nR\Delta T$

= ${5 \over 2}nR\Delta T$

= ${5 \over 2} \times 735 \times {2 \over 7}$

= 525 J

For transmitting a signal, antenna should have a size comparable to the wavelength of the signal i.e., at least ${1 \over 4}$ in dimension, so that the antenna properly senses the time variation of the signal.

The amplitude of the carrier wave is varied in accordance with the message signal is termed as amplitude modulation.

So, statement I is correct but statement II is incorrect.

$\omega$ = 628 rad/s

$X_{L}=L \omega$

$$\begin{aligned} & =5 \mathrm{mH} \times 628 \\ & =3.14 \Omega \end{aligned}$$

$v=\sqrt{\frac{Y}{\rho}}=\sqrt{\frac{3.2 \times 10^{11}}{8 \times 10^{3}}}$

$$\begin{aligned} & =\sqrt {0.4 \times {{10}^8}} \\ & = \sqrt {40 \times {{10}^6}} \\ & =6.32 \times 10^{3} \mathrm{~m} / \mathrm{s} \end{aligned}$$

The weight of an object varies with altitude due to the change in gravitational force. The force of gravity decreases with the square of the distance from the center of the Earth.

The gravitational force (weight) at a height h from the surface of the Earth is given by:

$W' = W \left(\frac{R}{{R + h}}\right)^2$

where:

• W' is the weight at height h,
• W is the weight at the Earth's surface,
• R is the radius of the Earth,
• h is the height above the Earth's surface.

In this problem, the height h is given as nine times the radius of the Earth (h = 9R). Substituting h = 9R into the equation gives:

$W' = W \left(\frac{R}{{R + 9R}}\right)^2 = W \left(\frac{1}{10}\right)^2$

Simplifying this gives:

$W' = \frac{W}{100}$

Therefore, the weight of the body at that height will be 1/100th of its weight on the surface of the Earth.

The correct answer is Option C: $\frac{W}{100}$.

$H \propto i^{2}$ for $t=$ constant

$\Rightarrow$ $\frac{H}{H^{\prime}}=\left(\frac{4}{16}\right)^{2}$

$\Rightarrow$ $H^{\prime}=16 H$

$\Delta \ell=\frac{F \ell}{S Y}$

$F$ is same for both wire and $\Delta \ell$ is also same

$$\begin{aligned} & \frac{\Delta \ell}{F}=\frac{\ell}{S Y} \\ & \Rightarrow \frac{\ell_{A}}{S_{A} Y_{A}}=\frac{\ell_{B}}{S_{B} Y_{B}} \\ & \Rightarrow \frac{5}{2.5 \times Y_{A}}=\frac{6}{3 \times Y_{B}} \\ & \Rightarrow \frac{Y_{A}}{Y_{B}}=1 \end{aligned}$$

We know, $h^{\prime}=e^{2} h$

$h^{\prime}=(0.5)^{2} \times 20 m=5 m$

$$\begin{aligned} & \mathrm{\omega}=\frac{2 \pi}{\mathrm{T}}=\frac{2 \pi}{4}=\frac{\pi}{2} \mathrm{rad} / \mathrm{s} \\ & \theta=\mathrm{\omega t} \\ & \theta=\frac{\pi}{2} \times 3 \\ & \theta=\frac{3 \pi}{2} \mathrm{rad} \end{aligned}$$



$$\begin{aligned} & r=10 \mathrm{~m} \\ & T=4 \mathrm{sec} \\ & d=\sqrt{2}(10) \mathrm{m} \end{aligned}$$

Given, $y_{1}=10 \sin \left(\omega t+\frac{\pi}{3}\right) \mathrm{cm}$ and

$y_{2}=5(\sin \omega t+\sqrt{3} \cos \omega t)$

$=10 \sin \left(\omega t+\frac{\pi}{3}\right)$

Thus the phase difference between the waves is 0 .

$\text { so } A=A_{1}+A_{2}=20 \mathrm{~cm}$

Charge on $\mathrm{C}_{1}=\mathrm{KCV}$

And charge on $\mathrm{C}_{2}=\mathrm{CV}$

When they are connected in parallel charge will be equally divided so charge on one capacitor is

$q=\frac{K+1}{2} \mathrm{CV}$

So, $V=\frac{q}{K C}=\frac{K+1}{2 K}=55 \mathrm{~V}$

Since range is same.

$$\begin{aligned} & \Rightarrow \theta_{1}+\theta_{2}=90^{\circ} \\ & \Rightarrow \theta_{2}=30^{\circ} \\ & \Rightarrow \left(H_{\max }\right)_{1}+\left(H_{\max }\right)_{2}=\frac{u^{2} \sin ^{2} \theta_{1}}{2 g}+\frac{u^{2} \sin ^{2} \theta_{2}}{2 g} \\ & =\frac{40^{2}}{20}\left(\frac{1}{4}+\frac{3}{4}\right)=80 \mathrm{~m} \end{aligned}$$

$V=\frac{\sum K Q_{i}}{r_{i}}$

Here, $Q_{i}$ and $r_{i}$ are positive.

$\therefore V > 0$

The correct statement is:

(A) Net potential of the system cannot be zero at a point but net electric field can be zero at that point.

Explanation:

In a group of positive charges, the net potential at a point is the sum of the potentials due to each individual charge. The potential due to a point charge is given by the Coulomb's law, which is non-zero except at the location of the charge itself. Therefore, the net potential due to a group of positive charges can never be zero at a point.

On the other hand, the net electric field at a point is the vector sum of the electric fields due to each individual charge. If the charges are arranged in such a way that their electric fields cancel out at a particular point, then the net electric field at that point can be zero, even though the charges are present. This can happen, for example, in a symmetrical arrangement of charges.

So, statement (A) is the correct statement.

Let $\gamma$ be the ratio of $\frac{C_{p}}{C_{v}}$

Then for adiabatic process

$$\begin{aligned} & P V^{\gamma}=\text { Constant } \\ & \Rightarrow \frac{P_{i}}{P_{f}}=\left(\frac{V_{f}}{V_{i}}\right)^{\gamma} \\ & \Rightarrow \frac{81}{16}=\left(\frac{27}{8}\right)^{\gamma} \\ & \Rightarrow \gamma=\frac{4}{3} \end{aligned}$$

$I=(2 \pi r) n$

$$\begin{aligned} & r \propto\left(\frac{I}{n}\right) \\ & B=n\left(\frac{\mu_{0} i}{2 r}\right) \propto\left(\frac{\mu_{0} i}{2 L}\right) n^{2} \\ & \frac{B_{1}}{B_{2}}=\left(\frac{N^{2}}{n^{2}}\right) \end{aligned}$$

$r \propto \frac{n^{2}}{Z}$

$$\begin{aligned} & \frac{r_{2 { nd }}}{r_{3 \mathrm{rd}}}=\left(\frac{n_{2}}{n_{3}}\right)^{2} \\ & \Rightarrow \frac{R}{r_{3 r d}}=\left(\frac{2}{3}\right)^{2} \\ & \Rightarrow r_{3 \mathrm{rd}}=\frac{9}{4} R \\ & =2.25 R \end{aligned}$$

$$\begin{aligned} & T \sin \theta=\frac{m v^{2}}{l \sin \theta} \\ & \cos \theta=\frac{m g}{T} ........(1) \\ & \sin^2 \theta=\frac{m v^{2}}{T l} ........(2) \end{aligned}$$

From (1) and (2),

$1=\left(\frac{m g}{T}\right)^{2}+\frac{m v^{2}}{T l}$

$\Rightarrow$ $1=\left(\frac{10}{400}\right)^{2}+\frac{v^{2}}{400}$

$\Rightarrow$ $v^{2}=399.78$

$\Rightarrow$ $v=20 \mathrm{~m} / \mathrm{s}$

$a=-\mu g$

$$\begin{aligned} & \because v=u+a t \\ & 0=20+(-\mu \times 10) \times 5 \\ & 50 \mu=20 \\ & \mu=\frac{2}{5}=0.4 \end{aligned}$$

The amount of heat energy required to raise the temperature of a substance can be calculated as:

Q = m $\times$ c $\times$ ΔT

where Q is the heat energy required, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.

The time required to heat a substance can be calculated as :

t = ${Q \over P}$

where t is the time required, and P is the power of the heating device.

The actual power output of the heating device can be calculated as:

P_actual = P_input $\times$ efficiency

where P_input is the input power to the device and efficiency is the fraction of input power that is actually converted to useful power output.

Substituting the given values:

Q = 2 kg $\times$ 4200 J/kg/K $\times$ (60 - 10) = 2 kg $\times$ 4200 J/kg/K $\times$ 50 K = 4200 $\times$ 50 $\times$ 2 J = 420,000 J

P_input = 2000 W = 2000 J/s

P_actual = 2000 $\times$ 0.7 = 1400 J/s

t = ${Q \over {{P_{actual}}}}$ = ${{420,000} \over {1400}}$ J/s = 300 s

So, the time required to heat 2 kg of water from 10°C to 60°C is approximately 300 s.

$\omega_{1}=\frac{\lambda_{1} D}{d}$ and $\omega_{2}=\frac{\lambda_{2} D}{d}$

$\omega_{1}=16 \mathrm{~mm}$ and $\omega_{2}=12 \mathrm{~mm}$

So $\operatorname{LCM}\left(\omega_{1}, \omega_{2}\right)=48 \mathrm{~mm}$

So at $48 \mathrm{~mm}$ distance both bright fringes will be found.

$m=\rho \pi R^{2} t$

$$\begin{aligned} & \text { so } R^{2}=\frac{m}{\rho \pi t} \\ & I=\frac{m R^{2}}{4}=\frac{m^{2}}{4 \rho \pi t} \end{aligned}$$

So $\frac{I_{1}}{I_{2}}=\frac{\rho_{2} t_{2}}{\rho_{1} t_{1}}=\frac{5}{3} \times \frac{0.5}{1}=\frac{5}{6}$

So $x=5$

Only primary amines react with PhSO₂Cl to produce a compounds which are soluble in KOH.

Option (B) and (D) are primary amines but the given compound is also optically active.

Hence the correct answer is (B).

(A) Elevation in boiling point temperature of water will be higher for 0.1 M NaCl as compared to 0.1 M urea.

(B) Azeotropic mixtures boil without change in their composition

(C) Osmosis always takes place from hypotonic (low concentration of solute) solution to hypertonic (high concentration of solute) solution.

(E) When KI solution is added to AgNO₃ solution, positively charged solution is formed due to adsorption of Ag⁺ ions from dispersion medium.

AgI/ Ag⁺ (positively charged)

(D) Let the mass of H₂SO₄ (32%) is 100.

$\therefore \text { Weight of } \mathrm{H}_2 \mathrm{SO}_4=32$

Moles of $\mathrm{H}_2 \mathrm{SO}_4=\frac{32}{98}$

Now, $4.09=\frac{32}{98 \times \mathrm{V}} \Rightarrow \mathrm{V}=79 \mathrm{ml}$

Density $=\frac{100}{79}=1.265$

Hence, correct answer is (C) B and D only.

(A) n = 3; l = 0; m = 0 ; 3s orbital

(B) n = 4; l = 0; m = 0 ; 4s orbital

(C) n = 3; l = 1; m = 0 ; 3p orbital

(D) n = 3; l = 2; m = 0 ; 3d orbital

As per Hund’s ruleo of energy is given by (n + l) value. If value of (n + l) remains same then energy is given by n only.

Formation of tetraacetate with AcO₂ means compound A has four –OH linkage.

Reduction of A with HI gives Isopentane i.e. molecule contains five carbon atoms.

Normally rain water has a pH of 5.6 due to presence of H⁺ ions formed by the reaction of rain water with carbon dioxide present in the atmosphere.

$$\begin{aligned} & \mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{CO}_2(\mathrm{~g}) \leftrightharpoons \mathrm{H}_2 \mathrm{CO}_3(\mathrm{aq}) \\ & \mathrm{H}_2 \mathrm{CO}_3(\mathrm{aq}) \leftrightharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{HCO}_3^{-}(\mathrm{aq}) \end{aligned}$$

Extent of back bonding, reduces down the group leading to more Lewis acidic strength.

Veronal is neurological medicine, Prontosil is antibiotic, rest all are disinfectants.

Calcium plays important role in neuromuscular function, interneuronal transmission, cell membrane etc.

Option A: Cast iron is obtained by melting pig iron (crude iron) with scrap iron and coke (a fuel made from coal) using a hot air blast. The resulting molten metal is poured into molds to form cast iron products.

Option B: Wrought Iron or Malleable iron is produced from cast iron by heating it in a reverberatory furnace lined with Haematite, which removes impurities through oxidation. The resulting product is malleable iron, which is a type of iron that can be shaped and molded.

Option C: van Arkel Method for Refining Zirconium or Titanium: This method is very useful for removing all the oxygen and nitrogen present in the form of impurity in certain metals like $\mathrm{Zr}$ and Ti. The crude metal is heated in an evacuated vessel with iodine. The metal iodide being more covalent, volatilises:

$\mathrm{Zr}+2 \mathrm{I}_2 \rightarrow \mathrm{ZrI}_4$

The metal iodide is decomposed on a tungsten filament, electrically heated to about $1800 \mathrm{~K}$. The pure metal deposits on the filament.

$\mathrm{ZrI}_4 \rightarrow \mathrm{Zr}+2 \mathrm{I}_2$

Option D: Zone refining is a method used to purify certain metals such as boron, indium, germanium, silicon, and galium. It involves heating the metal to a liquid state and then slowly moving it along a column to allow impurities to migrate to one end of the column, where they can be removed.

In dumas method for the estimation of N₂, the sample is heated with copper oxide and the gas evolved is passed over copper gauze.

Limiting reagent is carbon. One mole carbon produces one mole CO. Hence, volume at STP is 227 $\times$ 10^-1 litre.