MHT CET 2023 12th May Morning Shift (Physics & Chemistry) Question Paper with Solutions

The converging or diverging ability of a lens or mirror is referred to as its "focal power." Focal power, measured in diopters (D), is the reciprocal of the focal length (in meters). It indicates how strongly the lens or mirror converges (positive focal power) or diverges (negative focal power) light.

Therefore, the correct answer is :

Option A : focal power

$\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$

Dividing both the sides by $\mathrm{C}_{\mathrm{v}}$,

$$\begin{array}{lll} \therefore & \gamma-1=\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}} & \ldots .\left(\because \frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma\right) \\ \therefore & \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{R}}{\gamma-1} \end{array}$$

$$\begin{aligned} \mathrm{v} & =\omega \sqrt{\mathrm{a}^2-\mathrm{x}^2} \\ & \text { At } \mathrm{x}=\frac{\mathrm{a}}{2}, \\ \therefore \quad \mathrm{v} & =\omega \sqrt{\mathrm{a}^2-\frac{\mathrm{a}^2}{4}} \\ & =\omega \frac{\sqrt{3 \mathrm{a}}}{2} \\ & =\frac{2 \pi}{\mathrm{T}} \times \frac{\sqrt{3} \mathrm{a}}{2} \quad .....\left(\because \omega=\frac{2\pi}{T}\right)\\ \therefore \quad \mathrm{v} & =\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}} \end{aligned}$$

Dipole of moment p is at a distance x from N

At N, |E. F. due to dipole 1| = |E. F. due to dipole 2|

$$\begin{aligned} & \therefore \quad \frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 \mathrm{p}}{\mathrm{x}^3}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2(27 \mathrm{p})}{(24-\mathrm{x})^3} \\ & \therefore \quad \frac{1}{\mathrm{x}^3}=\frac{27}{(24-\mathrm{x})^3} \Rightarrow \mathrm{x}=6 \mathrm{~cm}. \end{aligned}$$

From figure,

$$\begin{aligned} & \tan \theta=\frac{\mathrm{h}}{\mathrm{d}} \\ \therefore \quad & \frac{\mathrm{v}^2}{\mathrm{rg}}=\frac{\mathrm{h}}{\mathrm{d}} \quad \ldots .\left(\because \tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}}\right) \\ \therefore \quad & \mathrm{r}=\frac{\mathrm{v}^2 \mathrm{~d}}{\mathrm{gh}} \end{aligned}$$

For monoatomic gas,

$\gamma_1=\frac{5}{3}$

For rigid diatomic gas,

$$\begin{aligned} \gamma_2 & =\frac{7}{5} \\ \therefore \quad \frac{\gamma_2}{\gamma_1} & =\frac{7}{5} \times \frac{3}{5}=\frac{21}{25} \end{aligned}$$

$l=300 \mathrm{~cm}=3 \mathrm{~m}$

Total resistance of wire,

$\mathrm{R}=3 \times 2=6 \Omega$

Since, the potentiometer is balanced. Voltage across wire segment $=1.5 \mathrm{~V}$

$$\begin{array}{ll} \therefore & \mathrm{IR}=1.5 \mathrm{~V} \\ \therefore & \mathrm{I}=\frac{1.5}{6}=250 \mathrm{~mA} \end{array}$$

Phase difference, $\phi=\frac{2 \pi}{\lambda} \Delta l$

For first point, $\phi_1=\frac{2 \pi}{\lambda}\left(\frac{\lambda}{4}\right)$

$$\begin{array}{l} \therefore \quad \phi_1 =\frac{\pi^{\mathrm{c}}}{2} \\ \mathrm{I}_1 =2 \mathrm{I}_0\left(1+\cos \phi_1\right) \\ \therefore \quad \mathrm{I}_1 =2 \mathrm{I}_0 \quad \ldots\left[\because \cos \phi_1=\cos (\pi / 2)=0\right] \end{array}$$

Similarly, for second point, $\phi_2=\frac{2 \pi^{\mathrm{c}}}{3}$

$$\begin{aligned} & \therefore \quad \mathrm{I}_2=2 \mathrm{I}_0\left(1+\cos \phi_2\right) \\ & \therefore \quad \mathrm{I}_2=2 \mathrm{I}_0\left(1-\frac{1}{2}\right) \quad \ldots\left[\because \cos \left(\frac{2 \pi}{3}\right)=\frac{-1}{2}\right] \\ & \therefore \quad \mathrm{I}_2=\mathrm{I}_0 \\ \end{aligned}$$

Hence, $\frac{\mathrm{I}_1+\mathrm{I}_2}{\mathrm{I}_0}=\frac{2 \mathrm{I}_0+\mathrm{I}_0}{\mathrm{I}_0}=3$

In the given circuit, $C_3$ and $C_4$ are in series and $\mathrm{C}_3=\mathrm{C}_4=8 \mu \mathrm{F}$

$\therefore \quad \frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}_3}+\frac{1}{\mathrm{C}_4}$

$$\begin{aligned} \therefore \quad \mathrm{C}_{\mathrm{S}} & =\frac{\mathrm{C}_3^2}{2 \mathrm{C}_3} \\ \mathrm{C}_{\mathrm{S}} & =\frac{\mathrm{C}_3}{2} \\ \therefore \quad \mathrm{C}_{\mathrm{S}} & =4 \mu \mathrm{F} \quad \text{.... (i)} \end{aligned}$$

$\mathrm{C}_5$ and $\mathrm{C}_6$ are in parallel and $\mathrm{C}_5=\mathrm{C}_6=4 ~\mu \mathrm{F}$

$\mathrm{C}_{\mathrm{P}}=\mathrm{C}_5+\mathrm{C}_6$

$\therefore \quad \mathrm{C}_{\mathrm{P}}=8 \mu \mathrm{F}$

$\therefore \quad$ Equivalent circuit is as shown in figure.

Now, $\mathrm{C}_2$ and $\mathrm{C}_{\mathrm{P}}$ are in series and their combination in parallel with $\mathrm{C}_{\mathrm{S}}$

$$\begin{aligned} \therefore \quad \mathrm{C}_{\mathrm{E}} & =\frac{\mathrm{C}_2 \mathrm{C}_{\mathrm{p}}}{\mathrm{C}_2 \mathrm{C}_{\mathrm{p}}}+\mathrm{C}_{\mathrm{s}} \\ & \mathrm{C}_{\mathrm{E}}=\frac{(8)(8)}{16}+4 \\ \therefore \quad & \mathrm{C}_{\mathrm{E}}=8 \mu \mathrm{F} \end{aligned}$$

Now, $\mathrm{C}_1$ and $\mathrm{C}_{\mathrm{E}}$ are in series,

$$\begin{aligned} & \therefore \quad C=\frac{C_1 C_E}{C_1+C_E} \\ & \therefore \quad C=\frac{(8)(8)}{16} \\ & \therefore \quad C=4 \mu F \end{aligned}$$

$$\begin{aligned} & \therefore \quad \mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}} \cdot(\mathrm{A}+\mathrm{B}) \\ & \therefore \quad \mathrm{Y}=(\overline{\mathrm{A}}+\overline{\mathrm{B}}) \cdot(\mathrm{A}+\mathrm{B}) \quad \ldots(\because \overline{\mathrm{A} \cdot \mathrm{B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}) \\ & \therefore \quad \mathrm{Y}=\overline{\mathrm{A}} \cdot \mathrm{B}+\overline{\mathrm{B}} \cdot \mathrm{A} \\ & \therefore \quad \mathrm{Y}=\mathrm{A} \cdot \overline{\mathrm{B}}+\mathrm{B} \cdot \overline{\mathrm{A}} \\ \end{aligned}$$

This represents XOR gate.

Force per unit length exerted by B on A, $\mathrm{F}_1=\frac{\mu_0(\mathrm{I})(\mathrm{I})}{2 \pi \mathrm{d}}=\frac{\mu_0 \mathrm{I}^2}{2 \pi \mathrm{d}}$ (outside the plane of paper)

Force per unit length exerted by $\mathrm{C}$ on $\mathrm{A}$,

$\mathrm{F}_2=\frac{\mu_0(\mathrm{I})(2 \mathrm{I})}{2 \pi(2 \mathrm{~d})}=\frac{\mu_0 \mathrm{I}^2}{2 \pi \mathrm{d}}$ (Inside the plane of paper)

$\therefore \quad F_1=-F_2$

$$\begin{aligned} & l_2=2 l_1, \mathrm{R}_2=2 \mathrm{R}_1, \mathrm{~T}_1=\mathrm{T}_2 \\ & \mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\ & \text { Where, } \mathrm{m}=\text { mass per unit length }=\frac{\left(\pi \mathrm{R}^2 l\right) \rho}{l} \\ & \therefore \quad \mathrm{m} \propto \mathrm{R}^2 \\ & \therefore \quad \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{l_1}{l_2} \times \frac{\mathrm{R}_1}{\mathrm{R}_2} \\ & \therefore \quad \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{l_1}{2 l_1} \times \frac{\mathrm{R}_1}{2 \mathrm{R}_1} \\ & \therefore \quad \mathrm{n}_2=\frac{\mathrm{n}_1}{4}=\frac{\mathrm{n}}{4} \end{aligned}$$

Let Frequency of $1^{\text {st }}$ tuning fork be $=\mathrm{n}_1$

$\therefore \quad$ frequency of $41^{\text {st }}$ tuning fork $=\mathrm{n}_{41}$

Now,

$$\begin{array}{ll} & \mathrm{n}_{41}=\mathrm{n}_1+(41-1) \times 5 \\ & \text { But, } \mathrm{n}_{41}=2 \mathrm{n}_1 \\ \therefore \quad & 2 \mathrm{n}_1=\mathrm{n}_1+200 \\ \therefore \quad & \mathrm{n}_1=200 \mathrm{~Hz} \\ \therefore \quad & \mathrm{n}_{41}=400 \mathrm{~Hz} \end{array}$$

$$\begin{aligned} & \mathrm{m}=\mathrm{nI} A \\ & \mathrm{~m}=\mathrm{nI}\left(\pi \mathrm{r}^2\right) \end{aligned}$$

$\mathrm{L}_1=\mathrm{L}_2=\mathrm{L}=60 \mathrm{~mH}$

When two inductors are connected in parallel, their equivalent inductance is given by,

$$\begin{aligned} & \frac{1}{\mathrm{~L}_{\mathrm{eq}}}=\frac{1}{\mathrm{~L}_1}+\frac{1}{\mathrm{~L}_2} \\ & \therefore \quad \mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}}{2}=30 \mathrm{~mH} \\ & \mathrm{u}_{\mathrm{B}}=\frac{1}{2} \mathrm{~L}_{\mathrm{eq}} \mathrm{I}^2 \\ & \therefore \quad \mathrm{u}_{\mathrm{B}}=\frac{1}{2} \times 30 \times 10^{-3} \times 2.2 \times 2.2 \\ & \therefore \quad \mathrm{u}_{\mathrm{B}}=0.0726 \mathrm{~J} \\ \end{aligned}$$

Momentum of particle moving towards east

$\overrightarrow{\mathrm{p}_1}=\mathrm{mv \hat {i }}$

Momentum of particle moving towards North

$\overrightarrow{\mathrm{p}_2}=\mathrm{mv} \hat{\mathrm{j}}$

Momentum after collision,

$\vec{p}=2 m\left(v_x \hat{i}+v_y \hat{j}\right)$

Applying momentum conservation,

$$\begin{array}{ll} & \overrightarrow{p_1}+\overrightarrow{p_2}=\vec{p} \\ & m v \hat{i}+m v \hat{j}=2 m\left(v_x \hat{i}+v_y \hat{j}\right) \\ \therefore \quad & 2 m v_x=m v \\ & v_x=\frac{v}{2} \end{array}$$

Similarly, $\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{v}}{2}$

$v_R=\sqrt{v_x^2+v_y^2}=\sqrt{\left(\frac{v}{2}\right)^2+\left(\frac{v}{2}\right)^2}=\frac{v}{\sqrt{2}}$

On the right hand side of the block, springs are connected in parallel

$\therefore$ Their effective spring constant is given by

$$\begin{aligned} & \mathrm{K}_1=\mathrm{K}+2 \mathrm{~K} \\ & \mathrm{~K}_1=3 \mathrm{~K} \end{aligned}$$

On the left hand side of the block, springs are connected in series.

$\therefore$ Their effective spring constant is given by,

$$\begin{aligned} & \frac{1}{\mathrm{~K}_2}=\frac{1}{2 \mathrm{~K}}+\frac{1}{2 \mathrm{~K}} \\ \therefore \quad & \mathrm{~K}_2=\mathrm{K} \end{aligned}$$

$\therefore \quad$ Effective spring constant of the system is given by,

$$\begin{aligned} & \quad \mathrm{K}_E=3 \mathrm{~K}+\mathrm{K}=4 \mathrm{~K} \\ & \therefore \quad \omega=\sqrt{\frac{\mathrm{K}_E}{M}}=\sqrt{\frac{4 K}{M}} \\ & \therefore \quad \mathrm{f}=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{4 K}{M}} \end{aligned}$$

$\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$

At a height '$h$' from earth's surface,

$$\begin{aligned} \mathrm{xT} & =2 \pi \sqrt{\frac{l}{\mathrm{~g}_{\mathrm{h}}}} \\ \therefore \quad \mathrm{x} & =\sqrt{\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}} \quad \text{..... (i)} \end{aligned}$$

Now, $\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}$

$\therefore \quad \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{4 \mathrm{R}^2}\quad$ .....($\because$ h = R)

$\therefore \quad \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{4}\quad$ .... (ii)

$\therefore \quad$ From equations (i) and (ii),

$x=\sqrt{\frac{g}{g / 4}}=\sqrt{4}=2$

$E=e A \sigma T^4$

But for sphere, $A=4 \pi R^2$

$$\begin{aligned} \therefore \quad & E=e\left(4 \pi R^2\right) \sigma T^4 \\ \therefore \quad & \frac{E_1}{E_2}=\frac{R_1^2 T_1^4}{R_2^2 T_2^4} \\ \therefore \quad & \frac{E}{E_2}=\frac{R^2 T^4}{\left(\frac{R}{3}\right)^2(3 T)^4} \\ \therefore \quad & E_2=9 E \end{aligned}$$

$$\begin{array}{ll} &\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2 \\ \because & \mathrm{R}_1=\frac{\mathrm{x}}{\mathrm{KA}}, \mathrm{R}_2=\frac{4 \mathrm{x}}{2 \mathrm{KA}} \\ \therefore & \mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{x}}{\mathrm{KA}}+\frac{2 \mathrm{x}}{\mathrm{KA}}=\frac{3 \mathrm{x}}{\mathrm{KA}} \end{array}$$

Rate of heat transfer of composite slab is given by,

$\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{R}_{\mathrm{eq}}}=\frac{\mathrm{KA}\left(\mathrm{T}_2-\mathrm{T}_1\right)}{3 \mathrm{x}}$

$\therefore \quad \mathrm{f}=\frac{1}{3}$

$y_1=8 \sin (20 x-3 t)$

Substituting $x=5 \mathrm{~cm}$ and $\mathrm{t}=2 \mathrm{~s}$,

$\mathrm{y}_1=8 \sin (40)$

Similarly, $\mathrm{y}_2=8 \sin (45)$

$\therefore$ phase difference $=45-40=5$ radian

The net magnetic field is acting in the direction of $\mathrm{GF}$ as shown in figures.

Resolving $\overrightarrow{\mathrm{B}}$ into its components, amongst the components, only $\vec{B} \sin \theta$ exerts force on side $\mathrm{EF}$ of current carrying loop.

$\therefore \quad \mathrm{F}_{\mathrm{EF}}=\mathrm{I} \times \mathrm{d}(\mathrm{EF}) \times \mathrm{B} \sin \theta$

From figure (a), $\sin \theta=\frac{12}{13}$

$$\begin{array}{ll} \therefore & \mathrm{F}_{\mathrm{EF}}=2 \times 0.05 \times 0.75 \times \frac{12}{13} \\ \therefore & \mathrm{F}_{\mathrm{EF}}=\frac{9}{130} \mathrm{~N} \\ \therefore & \mathrm{x}=9 \end{array}$$

$$\begin{array}{ll} & \text { Fringe width, } W=\frac{\lambda D}{d} \\ & \text { For constant } W \text { and } \lambda \\ & D \propto d \\ \therefore \quad & \frac{D_1}{D_2}=\frac{d_1}{d_2}=\frac{2}{3} \end{array}$$

$$\begin{aligned} & \lambda_1=\lambda,\left(\mathrm{V}_0\right)_1=\mathrm{V} \\ & \lambda_2=3 \lambda,\left(\mathrm{V}_0\right)_2=\frac{\mathrm{V}}{6} \end{aligned}$$

Photo electric equation is given by

$\mathrm{eV}_0=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)$ ..... (i)

In first case,

$\mathrm{eV}=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)$

For second case,

$\frac{\mathrm{eV}}{6}=\mathrm{hc}\left(\frac{1}{3 \lambda}-\frac{1}{\lambda_0}\right)$ ..... (ii)

Dividing equation (i) by equation (ii),

$$\begin{aligned} 6 & =\frac{\frac{1}{\lambda}-\frac{1}{\lambda_0}}{\frac{1}{3 \lambda}-\frac{1}{\lambda_0}}=\frac{3 \lambda_0-\lambda}{\lambda_0-3 \lambda} \\ \therefore \quad \lambda_0 & =5 \lambda \end{aligned}$$

According to Rydberg's formula,

$\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$ ...... (i)

When electron jumps from $2^{\text {nd }}$ exited state to first exited state,

$\mathrm{n}_2=3, \mathrm{n}_1=2, \lambda=\lambda_0$, we get

$\frac{1}{\lambda_0}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$

When electron jumps from $3^{\text {rd }}$ exited state to $2^{\text {nd }}$ orbit,

$\mathrm{n}_2=4, \mathrm{n}_1=2$, we get

$\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4^2}-\frac{1}{2^2}\right)$

$$\begin{aligned} \therefore \quad \frac{\lambda}{\lambda_0} & =\frac{\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}{\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)} \\ & =\frac{5}{36} \times \frac{16}{3}=\frac{20}{27} \\ \therefore \quad \lambda & =\frac{20}{27} \lambda_0 \\ \Rightarrow \mathrm{x} & =27 \end{aligned}$$

Using Parallel axis theorem, $\mathrm{I}_2=\mathrm{I}_1+\mathrm{Mh}^2$

For a disc, $I_1=\frac{1}{2} M R^2$ and given that, $h=\frac{2 R}{3}$

$$\begin{aligned} & \therefore \quad \mathrm{I}_2=\frac{1}{2} \mathrm{MR}^2+\mathrm{M}\left(\frac{4 \mathrm{R}^2}{9}\right) \\ & \therefore \quad \mathrm{I}_2=\frac{1}{2} \mathrm{MR}^2\left(1+\frac{8}{9}\right)=\mathrm{I}_1 \times \frac{17}{9} \\ & \therefore \quad \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{9}{17} \\ \end{aligned}$$

$$\begin{aligned} & \mathrm{T}=\frac{2 \pi}{\omega} \\ & \text { Hence, } \mathrm{T}_1=\mathrm{T}_2 \\ & \Rightarrow \omega_1=\omega_2 \\ & \therefore \quad \frac{\omega_1}{\omega_2}=\frac{1}{1} \end{aligned}$$

$B_1=\frac{\mu_0 I}{2 R}, B_2=\frac{\mu_0(2 I)}{2 R}$

Resultant magnetic field, $B=\sqrt{B_1+B_2}$

$$\begin{array}{ll} \therefore & B=\sqrt{\left(\frac{\mu_0 I}{2 R}\right)^2+\left(\frac{\mu_0(2 I)}{2 R}\right)^2} \\ \therefore & B=\frac{\mu_0 I}{2 R} \sqrt{1^2+2^2} \\ \therefore & B=\frac{\sqrt{5} \mu_0 I}{2 R} \end{array}$$

Applying KVL in the given circuit from point A to B,

$$\begin{array}{ll} & \mathrm{V}_{\mathrm{AB}}-\mathrm{IR}-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}-12=0 \\ & \mathrm{~V}_{\mathrm{AB}}-(2)(12)-6(-1)-12=0 \\ & \mathrm{~V}_{\mathrm{AB}}=24+12-6 \\ \therefore \quad & \mathrm{V}_{\mathrm{AB}}=30 \mathrm{~V} \end{array}$$

According to Brewster's law,

$$\begin{aligned} & \tan \theta_{\mathrm{B}}=\mathrm{n} \\ \therefore \quad & \tan 60^{\circ}=\mathrm{n} \end{aligned}$$

$$\begin{aligned} &\begin{array}{ll} \therefore & \mathrm{n}=\sqrt{3} \\ & \text { Now, } \frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\mathrm{n} \\ \therefore & \sin \mathrm{r}=\frac{\sin \mathrm{i}}{\mathrm{n}} \\ \therefore & \sin \mathrm{r}=\frac{\sin 45^{\circ}}{\sqrt{3}} \\ \therefore & \sin \mathrm{r}=\frac{1}{\sqrt{6}} \quad \ldots .\left(\because \sin 45^{\circ}=\frac{1}{\sqrt{2}}\right) \\ \therefore & \quad \mathrm{r}=\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right) \end{array}\\ \end{aligned}$$

Given that, $V_2=\frac{V_1}{4}, \frac{C_p}{C_v}=\gamma=1.5$

As the process is sudden, it is an adiabatic expansion,

$$\begin{aligned} & \therefore \quad \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\ & \therefore \quad \mathrm{T}_2=\mathrm{T}_1\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1} \\ & =\mathrm{T}_1(4)^{\gamma-1} \\ & =\mathrm{T}_1 \times(4)^{0.5} \\ & =2 \mathrm{~T}_1 \\ & \mathrm{~T}_2-\mathrm{T}_1=\mathrm{T}_1 \\ & \mathrm{~T}_2-\mathrm{T}_1=273 \mathrm{~K} \quad\left(\because \mathrm{T}_1=\text { Normal temperature }\right) \\ \end{aligned}$$