JEE Main 2024 29th January Evening Shift Question Paper with Solutions

$$\begin{aligned} & \overrightarrow{\mathrm{E}}=6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{A}}=30 \hat{\mathrm{i}} \\ & \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} \\ & \phi=(6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(30 \hat{\mathrm{i}}) \\ & \phi=6 \times 30=180 \end{aligned}$$

$$\begin{aligned} & \mathrm{PV}=\mathrm{nRT} \\ & \mathrm{PV}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}} \mathrm{RT} \\ & \mathrm{N}=\text { Total no. of molecules } \\ & \mathrm{P}=\frac{\mathrm{N}}{\mathrm{V}} \mathrm{kT} \\ & 1.38 \times 1.01 \times 10^5=2 \times 10^{25} \times 1.38 \times 10^{-23} \times \mathrm{T} \\ & 1.01 \times 10^5=2 \times 10^2 \times \mathrm{T} \\ & \mathrm{T}=\frac{1.01 \times 10^3}{2} \approx 500 \mathrm{~K} \end{aligned}$$

$$\begin{aligned} \frac{E}{B} & =C \\ \frac{E}{B} & =3 \times 10^8 \\ B & =\frac{E}{3 \times 10^8}=\frac{9.6}{3 \times 10^8} \\ B & =3.2 \times 10^{-8} T \\ \hat{B} & =\hat{v} \times \hat{E} \\ & =\hat{i} \times \hat{j}=\hat{k} \end{aligned}$$

So,

$\overrightarrow{\mathrm{B}}=3.2 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}$

$$\begin{aligned} \mathrm{Y} & =\mathrm{A} \cdot \mathrm{B}+\overline{\mathrm{A}} \cdot \mathrm{B} \\ & =(\mathrm{A}+\overline{\mathrm{A}}) \cdot \mathrm{B} \\ \mathrm{Y} & =1 \cdot \mathrm{B} \\ \mathrm{Y} & =\mathrm{B} \end{aligned}$$

According to Rutherford atomic model, most of mass of atom and all its positive charge is concentrated in tiny nucleus & electron revolve around it.

According to Thomson atomic model, atom is spherical cloud of positive charge with electron embedded in it.

Hence, Statement I is true but statement II false.

Given that

$$\begin{aligned} & \mathrm{m}=900 \mathrm{~gm}=\frac{900}{1000} \mathrm{~kg}=\frac{9}{10} \mathrm{~kg} \\ & \mathrm{r}=1 \mathrm{~m} \\ & \omega=\frac{2 \pi \mathrm{N}}{60}=\frac{2 \pi(10)}{60}=\frac{\pi}{3} \mathrm{rad} / \mathrm{sec} \\ & \mathrm{T}-\mathrm{mg}=\mathrm{mr} \omega^2 \\ & \mathrm{~T}=\mathrm{mg}+\mathrm{mr} \omega^2 \\ & =\frac{9}{10} \times 9.8+\frac{9}{10} \times 1\left(\frac{\pi}{3}\right)^2 \\ & =8.82+\frac{9}{10} \times \frac{\pi^2}{9} \\ & =8.82+0.98 \\ & =9.80 \mathrm{~N} \end{aligned}$$

$$\begin{aligned} & \mathrm{n}_1 \times \frac{\mathrm{hc}}{\lambda_1}=200 \\ & \mathrm{n}_2 \times \frac{\mathrm{hc}}{\lambda_2}=200 \\ & \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{\lambda_1}{\lambda_2}=\frac{300}{500} \\ & \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{3}{5} \end{aligned}$$

$$\begin{aligned} & \mathrm{m}=2=\frac{-v}{u} \\ & 2=\frac{-(15-u)}{-u} \\ & 2 u=15-u \\ & 3 u=15 \Rightarrow u=5 \mathrm{~cm} \\ & v=15-u=15-5=10 \mathrm{~cm} \\ & \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ & =\frac{1}{10}+\frac{1}{(-5)}=\frac{1-2}{10}=\frac{-1}{10} \\ & f=-10 \mathrm{~cm} \end{aligned}$$

$$\begin{aligned} & P_{\text {avg }}=V_{\text {rms }} I_{r m s} \cos (\Delta \phi) \\ & =\frac{100}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \cos \left(\frac{\pi}{3}\right) \\ & =\frac{10^4}{2} \times \frac{1}{2} \times 10^{-3} \\ & =\frac{10}{4}=2.5 \mathrm{~W} \end{aligned}$$

$$\begin{aligned} & x=t^3-6 t^2+20 t+15 \\ & \frac{d x}{d t}=v=3 t^2-12 t+20 \\ & \frac{d v}{d t}=a=6 t-12 \\ & \text { When } a=0 \\ & 6 t-12=0 ; t=2 \mathrm{sec} \\ & \text { At } t=2 \text { sec } \\ & v=3(2)^2-12(2)+20 \\ & v=8 \mathrm{~m} / \mathrm{s} \end{aligned}$$

$$\begin{aligned} & \mathrm{T}^2 \propto \mathrm{r}^3 \\ & \frac{\mathrm{T}_1^2}{\mathrm{r}_1^3}=\frac{\mathrm{T}_2^2}{\mathrm{r}_2^3} \\ & \frac{(200)^2}{\mathrm{r}^3}=\frac{\mathrm{T}_2^2}{\left(\frac{\mathrm{r}}{4}\right)^3} \\ & \frac{200 \times 200}{4 \times 4 \times 4}=\mathrm{T}_2^2 \\ & \mathrm{~T}_2=\frac{200}{4 \times 2} \\ & \mathrm{~T}_2=25 \text { days } \end{aligned}$$

$$\begin{aligned} & Y=\frac{\text { stress }}{\text { strain }} \\ & Y=\frac{\frac{\mathrm{F}}{\frac{\pi \mathrm{r}^2}{\ell}}}{\frac{\ell}{\mathrm{L}}} \end{aligned}$$

$\mathrm{F}=\mathrm{Y} \pi \mathrm{r}^2 \times \frac{\ell}{\mathrm{L}} \quad \text{.... (i)}$

$$\begin{aligned} & \mathrm{Y}=\frac{\frac{\mathrm{F} / 2}{\pi \mathrm{r}^2 / 4}}{\frac{\Delta \ell}{\mathrm{L}}} \\ & \mathrm{F}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \times 2 \times \frac{\pi \mathrm{r}^2}{4} \end{aligned}$$

From (i)

$$\begin{aligned} & \mathrm{Y} \pi \mathrm{r}^2 \frac{\ell}{\mathrm{L}}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \frac{\pi \mathrm{r}^2}{2} \\ & \Delta \ell=2 \ell \end{aligned}$$

$\mathrm{f}_{\mathrm{eq}}=\frac{\mathrm{n}_1 \mathrm{f}_1+\mathrm{n}_2 \mathrm{f}_2}{\mathrm{n}_1+\mathrm{n}_2}$

For diatomic gas $\mathrm{f}_{\mathrm{eq}}=5$

$$\begin{aligned} & 5=\frac{(\mathrm{N})(6)+(2)(3)}{\mathrm{N}+2} \\ & 5 \mathrm{~N}+10=6 \mathrm{~N}+6 \\ & \mathrm{~N}=4 \end{aligned}$$

$\ell=10 \mathrm{~m}$,

Initial energy $=\mathrm{mg} \ell$

$$\begin{aligned} & \text { So, } \frac{9}{10} \mathrm{mg} \ell=\frac{1}{2} \mathrm{mv}^2 \\ & \Rightarrow \frac{9}{10} \times 10 \times 10=\frac{1}{2} \mathrm{v}^2 \\ & \mathrm{v}^2=180 \\ & \mathrm{v}=\sqrt{180}=6 \sqrt{5} \mathrm{~m} / \mathrm{s} \end{aligned}$$

Volume constant

$$\begin{aligned} & \frac{4}{3} \pi R^3=27 \times \frac{4}{3} \times \pi r^3 \\ & R^3=27 r^3 \\ & R=3 r \\ & r=\frac{R}{3} \\ & r^2=\frac{R^2}{9} \end{aligned}$$

$$\begin{aligned} & \text { Work done }=T . \Delta A \\ & =27 T\left(4 \pi r^2\right)-T 4 \pi R^2 \\ & =27 T 4 \pi \frac{R^2}{9}-4 \pi R^2 T \\ & =8 \pi R^2 T \end{aligned}$$

$$\begin{aligned} & \mathrm{Q}=\frac{\mathrm{a}^4 \mathrm{~b}^3}{\mathrm{c}^2} \\ & \frac{\Delta \mathrm{Q}}{\mathrm{Q}}=4 \frac{\Delta \mathrm{a}}{\mathrm{a}}+3 \frac{\Delta \mathrm{b}}{\mathrm{b}}+2 \frac{\Delta \mathrm{c}}{\mathrm{c}} \end{aligned}$$

$\frac{\Delta \mathrm{Q}}{\mathrm{Q}} \times 100=4\left(\frac{\Delta \mathrm{a}}{\mathrm{a}} \times 100\right)+3\left(\frac{\Delta \mathrm{b}}{\mathrm{b}} \times 100\right)+2\left(\frac{\Delta \mathrm{c}}{\mathrm{c}} \times 100\right)$

$$\begin{aligned} \% \text { error in } \mathrm{Q} & =4 \times 3 \%+3 \times 4 \%+2 \times 5 \% \\ & =12 \%+12 \%+10 \% \\ & =34 \% \end{aligned}$$

$$\begin{aligned} & \Delta x=\frac{7 \lambda}{4} \\ & \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=\frac{2 \pi}{\lambda} \times \frac{7 \lambda}{4}=\frac{7 \pi}{2} \\ & \mathrm{I}=\mathrm{I}_{\max } \cos ^2\left(\frac{\phi}{2}\right) \\ & \frac{\mathrm{I}}{\mathrm{I}_{\max }}=\cos ^2\left(\frac{\phi}{2}\right)=\cos ^2\left(\frac{7 \pi}{2 \times 2}\right)=\cos ^2\left(\frac{7 \pi}{4}\right) \\ & =\cos ^2\left(2 \pi-\frac{\pi}{4}\right) \\ & =\cos ^2 \frac{\pi}{4} \\ & =\frac{1}{2} \\ & \end{aligned}$$

$\mathrm{R}_{\mathrm{eq}}=2 \Omega+2 \Omega+1 \Omega=5 \Omega$

$\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{10}{5}=2 \mathrm{~A}$

Current in resistance

$$\begin{aligned} R_3 & =2 \times\left(\frac{4}{4+4}\right) \\ & =2 \times \frac{4}{8} \\ & =1 \mathrm{~A} \end{aligned}$$

Apply energy conservation between A & B

$$\begin{aligned} & \frac{1}{2} \mathrm{mV}_{\mathrm{L}}^2=\frac{1}{2} \mathrm{mV}_{\mathrm{H}}^2+\mathrm{mg}(2 \mathrm{~L}) \\ & \because \mathrm{V}_{\mathrm{L}}=\sqrt{5 \mathrm{gL}} \end{aligned}$$

So, $\mathrm{V}_{\mathrm{H}}=\sqrt{\mathrm{gL}}$

$\frac{(\mathrm{K} . \mathrm{E})_{\mathrm{A}}}{(\mathrm{K} . \mathrm{E})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}(\sqrt{5 \mathrm{gL}})^2}{\frac{1}{2} \mathrm{~m}(\sqrt{\mathrm{gL}})^2}=\frac{5}{1}$

$$\begin{aligned} & \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\mathrm{p}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}} \\ & \mathrm{R} \propto \sqrt{\mathrm{m}} \\ & \mathrm{m} \propto \mathrm{R}^2 \\ & \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^2 \end{aligned}$$

\begin{aligned} & \left|\vec{a}_c\right|=\left|\vec{a}_t\right| \\ & \frac{v^2}{r}=\frac{d v}{d t} \\ & \Rightarrow \int_\limits4^v \frac{d v}{v^2}=\int_\limits0^t \frac{d t}{r} \\ & \Rightarrow\left[\frac{-1}{v}\right]_4^v=\frac{t}{r} \\ & \Rightarrow \frac{-1}{v}+\frac{1}{4}=2 t \end{aligned}

$$\begin{aligned} & \Rightarrow \mathrm{v}=\frac{4}{1-8 \mathrm{t}}=\frac{\mathrm{ds}}{\mathrm{dt}} \\ & 4 \int_0^{\mathrm{t}} \frac{\mathrm{dt}}{1-8 \mathrm{t}}=\int_0^{\mathrm{s}} \mathrm{ds} \\ & (\mathrm{r}=0.5 \mathrm{~m} \\ & \mathrm{s}=2 \pi \mathrm{r}=\pi) \\ & 4 \times \frac{[\ell \mathrm{n}(1-8 \mathrm{t})]_0^{\mathrm{t}}}{-8}=\pi \\ & \ell \mathrm{n}(1-8 \mathrm{t})=-2 \pi \\ & 1-8 \mathrm{t}=\mathrm{e}^{-2 \pi} \\ & \mathrm{t}=\left(1-\mathrm{e}^{-2 \pi}\right) \frac{1}{8} \mathrm{~s} \end{aligned}$$

So, $\alpha=8$

$$\begin{aligned} & \mathrm{B}_{\mathrm{H}}=0.60 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^2 \\ & \text { Induced emf e}=\mathrm{B}_{\mathrm{H}} \mathrm{v} \ell \\ &=0.60 \times 10^{-4} \times 10 \times 5 \\ &=3 \times 10^{-3} \mathrm{~V} \end{aligned}$$

For $\mathrm{n}^{\text {th }}$ minima

$\mathrm{b} \sin \theta=\mathrm{n} \lambda$

($\lambda$ is small so $\sin \theta$ is small, hence $\sin \theta \simeq \tan \theta$)

$\mathrm{btan} \theta=\mathrm{n} \lambda$

$\mathrm{b} \frac{\mathrm{y}}{\mathrm{D}}=\mathrm{n} \lambda$

$\Rightarrow \mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{b}}\left(\text { Position of } \mathrm{n}^{\mathrm{th}}\right. \text { minima) }$

$$\begin{aligned} & \mathrm{B} \rightarrow 1^{\text {st }} \text { minima, } \mathrm{A} \rightarrow 3^{\mathrm{rd}} \text { minima } \\ & \mathrm{y}_3=\frac{3 \lambda \mathrm{D}}{\mathrm{b}}, \mathrm{y}_1=\frac{\lambda \mathrm{D}}{\mathrm{b}} \\ & \Delta \mathrm{y}=\mathrm{y}_3-\mathrm{y}_1=\frac{2 \lambda \mathrm{D}}{\mathrm{b}} \\ & 3 \times 10^{-3}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{\mathrm{~b}} \\ & \mathrm{~b}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{3 \times 10^{-3}} \\ & \mathrm{~b}=2 \times 10^{-4} \mathrm{~m} \\ & x=2 \end{aligned}$$

At steady state, capacitor behaves as an open circuit and current flows in circuit as shown in the diagram.

$$\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=9 \Omega \\ & \mathrm{i}=\frac{9 \mathrm{~V}}{9 \Omega}=1 \mathrm{~A} \\ & \Delta \mathrm{V}_{6 \Omega}=1 \times 6=6 \mathrm{~V} \\ & \mathrm{~V}_{\mathrm{A}}=3 \mathrm{~V} \end{aligned}$$

So, potential difference across 6$\mu$F is 6 V.

$$\begin{aligned} \text { Hence } \mathrm{Q} & =\mathrm{C} \Delta \mathrm{V} \\ & =6 \times 6 \times 10^{-6} \mathrm{C} \\ & =36 \mu \mathrm{C} \end{aligned}$$

For minimum potential difference electron has to make transition from $n=3$ to $n=2$ state but first electron has to reach to $\mathrm{n}=3$ state from ground state. So, energy of bombarding electron should be equal to energy difference of $\mathrm{n=3}$ and $\mathrm{n=1}$ state.

$$\begin{aligned} & \Delta \mathrm{E}=13.6\left[1-\frac{1}{3^2}\right] \mathrm{e}=\mathrm{eV} \\ & \frac{13.6 \times 8}{9}=\mathrm{V} \\ & \mathrm{V}=12.09 \mathrm{~V} \approx 12.1 \mathrm{~V} \end{aligned}$$

So, $\alpha=121$

$y-x-4=0$

$d_1$ is perpendicular distance of given line from origin.

$\mathrm{d}_1=\left|\frac{-4}{\sqrt{1^2+1^2}}\right| \Rightarrow 2 \sqrt{2} \mathrm{~m}$

So

$$\begin{aligned} |\overrightarrow{\mathrm{L}}|=\mathrm{mvd}_1 & =5 \times 3 \sqrt{2} \times 2 \sqrt{2} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s} \\ & =60 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s} \end{aligned}$$

From phasor diagram particle has to move from $\mathrm{P}$ to $\mathrm{Q}$ in a circle of radius equal to amplitude of SHM.

$$\begin{aligned} & \cos \phi=\frac{\frac{\sqrt{3} \mathrm{~A}}{2}}{\mathrm{~A}}=\frac{\sqrt{3}}{2} \\ & \phi=\frac{\pi}{6} \end{aligned}$$

Now, $\frac{\pi}{6}=\omega \mathrm{t}$

$$\begin{aligned} & \frac{\pi}{6}=\frac{2 \pi}{T} t \\ & \frac{\pi}{6}=\frac{2 \pi}{6 \pi} t \end{aligned}$$

$\mathrm{t}=\frac{\pi}{2}$

So, $x=2$

Given, $\mathrm{i}_1=0.3 \mathrm{~A}, \mathrm{i}_1+\mathrm{i}_2+\mathrm{i}_3=0.9 \mathrm{~A}$

So, $\mathrm{V}_{\mathrm{AB}}=\mathrm{i}_1 \times 20 \Omega=20 \times 0.3 \mathrm{~V}=6 \mathrm{~V}$

$$\begin{aligned} & \mathrm{i}_2=\frac{6 \mathrm{~V}}{15 \Omega}=\frac{2}{5} \mathrm{~A} \\ & \mathrm{i}_1+\mathrm{i}_2+\mathrm{i}_3=\frac{9}{10} \mathrm{~A} \\ & \frac{3}{10}+\frac{2}{5}+\mathrm{i}_3=\frac{9}{10} \\ & \frac{7}{10}+\mathrm{i}_3=\frac{9}{10} \\ & \mathrm{i}_3=0.2 \mathrm{~A} \\ & \mathrm{So}, \mathrm{i}_3 \times \mathrm{R}_1=6 \mathrm{~V} \\ & (0.2) \mathrm{R}_1=6 \\ & \mathrm{R}_1=\frac{6}{0.2}=30 \Omega \end{aligned}$$

$$\begin{aligned} \mathrm{q} & =4 \mu \mathrm{C}, \overrightarrow{\mathrm{v}}=4 \times 10^6 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s} \\ \overrightarrow{\mathrm{B}} & =2 \hat{\mathrm{k} T} \\ \overrightarrow{\mathrm{F}} & =\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \\ & =4 \times 10^{-6}\left(4 \times 10^6 \hat{\mathrm{j}} \times 2 \hat{\mathrm{k}}\right) \\ & =4 \times 10^{-6} \times 8 \times 10^6 \hat{\mathrm{i}} \\ \overrightarrow{\mathrm{F}} & =32 \hat{\mathrm{i}} \mathrm{N} \\ \mathrm{x} & =32 \end{aligned}$$

$\mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell}$

$$\begin{aligned} & \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} \\ & \mathrm{V}=\mathrm{A} \ell \Rightarrow \ell=\frac{\mathrm{V}}{\mathrm{A}} \\ & \Delta \ell=\frac{\mathrm{FV}}{\mathrm{A}^2 \mathrm{Y}} \end{aligned}$$

$Y ~\& V$ is same for both the wires

$$\begin{aligned} & \Delta \ell \propto \frac{\mathrm{F}}{\mathrm{A}^2} \\ & \frac{\Delta \ell_1}{\Delta \ell_2}=\frac{\mathrm{F}_1}{\mathrm{~A}_1^2} \times \frac{\mathrm{A}_2^2}{\mathrm{~F}_2} \\ & \Delta \ell_1=\Delta \ell_2 \\ & \mathrm{~F}_1 \mathrm{~A}_2^2=\mathrm{F}_2 \mathrm{~A}_1^2 \\ & \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\mathrm{A}_1^2}{\mathrm{~A}_2^2}=\left(\frac{4}{1}\right)^2=16 \end{aligned}$$

A - II, B - IV, C - III, D - I

Fact based.

A-II, B-III, C-I, D-IV

Fact based.

Nessler's Reagent Reaction :

$\underset{\text { (Nessler's Reagent) }}{2 \mathrm{~K}_2 \mathrm{HgI}_4}+\mathrm{NH}_3+3 \mathrm{KOH} \rightarrow \underset{\substack{\text { (lodine of Millon's base } \\ \text { (Brown precipitate }}}{\mathrm{HgO} . \mathrm{Hg}\left(\mathrm{NH}_2\right) \mathrm{I}}+7 \mathrm{KI}+2 \mathrm{H}_2 \mathrm{O}$

Statement- 1 is false because chlorine has most negative electron gain enthalpy in its group.