MHT CET 2023 13th May Evening Shift (Physics & Chemistry) Question Paper with Solutions

Both diffraction and photoelectric effect can be explained by the help of wave nature and particle nature of light.

Refraction and reflection is explained by the help of wave nature only.

Let, $f_1$ and $f_2$ be the frequency of the two turning forks, given

$f_1-f_2=5$ ..... (i)

Length of wire in sonometer is

$L_1=0.96 \mathrm{~m} \text { and } L_2=0.97 \mathrm{~m}$

Frequency of vibration of the wire is given by

$f=\frac{1}{2 l} \sqrt{T / m}$

According to the question,

$$\begin{aligned} & \frac{f_1}{f_2}=\frac{l_2}{l_1}=\frac{0.97}{0.96}>1 \\ \therefore \quad & f_1>f_2 \end{aligned}$$

Now, from Eq. (i)

$$\begin{array}{ll} \Rightarrow & f_1-f_2=5 \\ \Rightarrow & \frac{0.97}{0.96} f_2-f_2=5 \\ \Rightarrow & {\left[\frac{0.97}{0.96}-1\right] f_2=5} \\ \Rightarrow & {\left[\frac{0.97-0.96}{0.96}\right] f_2=5} \\ \Rightarrow & 0.01 f_2=5 \times 0.96 \\ \Rightarrow & 0.01 f_2=4.8 \\ \Rightarrow & f_2=480 \mathrm{~Hz} \\ \Rightarrow & f_1=5+f_2=485 \mathrm{~Hz} \end{array}$$

The ball executes free fall until it hits the upper surface of glycerine.

$\therefore$ Velocity of ball when it reaches the upper surface of glycerine is given as,

$$\begin{aligned} v & =\sqrt{2 g h} \\ & =\sqrt{2 \times 9.8 \times 1.5} \\ \Rightarrow \quad v^2 & =29.4 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$$

Using the formula, $v^2-u^2=2 a s$ inside the glycerine.

We have,

$0-29.4=-2 \times(2.66) \times h$

($\because h=$ distance travelled inside glycerine)

$\Rightarrow \quad h=5.5 \mathrm{~m}$

$\therefore$ The height of glycerine in jar will be $5.5 \mathrm{~m}$.

Magnetic field on the axis of a circular coil,

$B_A=\frac{\mu_0 i R^2}{2\left(R^2+Z^2\right)^{3 / 2}}$

Magnetic field at the centre of the circular coil,

$B_C=\frac{\mu_0 i}{2 R}$

According to question,

$$\begin{aligned} B_A & =\frac{\mu_0 i R^2}{2\left(R^2+3 R^2\right)^{3 / 2}} \quad(\because Z=\sqrt{3} R) \\ \Rightarrow \quad B_A & =\frac{\mu_0 i R^2}{2 \times\left(4 R^2\right)^{3 / 2}} \\ \Rightarrow \quad \frac{\mu_0 i R^2}{2 \times 8 \times R^3} & =\frac{\mu_0 i R^2}{8 \times 2 R^3} \\ & =\frac{\mu_0 i}{8 \times 2 R} \end{aligned}$$

Thus, the ratio, $B_A: B_C=1: 8$

Since, the particle experiences a downward acceleration due to gravity at peak (highest point), and traces a circular arc, it has no component of acceleration in $x$-direction.

Let, radius of circle be $R$.

Horizontal velocity of the particle of height point.

$v_x=u\cos \theta \quad \left(\because a_x=0\right)$

$\therefore a=\frac{v_x^2}{R}$ (for circular motion centripetal acceleration $=\frac{v^2}{R}$)

$\Rightarrow R=\frac{v_x^2}{a}=\frac{u^2 \cos ^2 \theta}{g}\quad (\because a=g)$

Pressure $p$ of the gas is given by,

$p=\frac{1}{3} \frac{m N}{V} v_{\mathrm{rms}}^2$

Where, $m$ is mass, $N$ is the number of moles, $V$ is the volume and $v_{\mathrm{rms}}$ is the root mean square velocity of the particles of the gas.

$\therefore \quad p_1=p_0=\frac{1}{3} \frac{m_1 N}{V_1}\left(v_{\mathrm{rms}}^2\right)_1$ ..... (i)

If $m_2=\frac{m_1}{2}$ and $\left(v_{\mathrm{rms}}\right)_2=2\left(v_{\mathrm{rms}}\right)_1$

For simplicity write $v_2=2 v_1$

Hence, $p_2=\frac{1}{3} \frac{m_2 N}{V} v_2^2$ ..... (ii)

Divide Eq. (i) by Eq. (ii)

$$\begin{aligned} & \frac{p_0}{p_2}=\frac{\frac{1}{3} \frac{m_1 N}{V} v_1^2}{\frac{1}{3} \frac{m_2 N}{V} v_2^2} \\ & \Rightarrow \quad \frac{p_0}{p_2}=\frac{m_1 v_1^2}{\frac{m_1}{2}\left(2 v_1\right)^2} \Rightarrow \frac{p_0}{p_2}=\frac{1}{2} \\ & \Rightarrow \quad p_2=2 p_0 \\ \end{aligned}$$

$\therefore$ Resulting pressure is $2 p_0$

The magnetic field at the centre of a current carrying loop is

$B=\frac{\mu_0}{4 \pi}\left(\frac{2 \pi I}{a}\right)=\frac{\mu_0 l}{2 a}$ ..... (i)

and magnetic moment $M$ is

$M=I\left(\pi a^2\right)$ .... (ii)

$\text { Thus, } \quad X=\frac{B}{M}=\frac{\mu_0 I}{2 a} \times \frac{I}{I \pi a^2}=\frac{\mu_0}{2 \pi a^3}$ .... (iii)

Now, according to question, when both the current and radius are doubled, then the ratio will be

$\Rightarrow \frac{\mu_0}{2 \pi(2 a)^3}=\frac{\mu_0}{8\left(2 \pi a^3\right)}$

From Eq. (iii), $\frac{\mu_0}{2 \pi a^3}=X$

So, $\frac{\mu_0}{8\left(2 \pi a^3\right)}=\frac{X}{8}$

Distance travelled by body $B, s_B=v t$

Distance travelled by $A, s_A=u t+\frac{1}{2} a t^2=a t^2$

When they meet, $s_A=s_B$

$$\begin{aligned} \Rightarrow & \quad v t =\frac{1}{2} a t^2 \\ \Rightarrow & \quad t =\frac{2 v}{a} \end{aligned}$$

We know that, efficiency of full wave rectifier $=0.8$

efficiency of half wave rectifier $=0.4$

According to question, $0.8=x, 0.4=y$

$\Rightarrow x=2 y$

Alternate solution We know that, efficiency of full wave rectifier is twice the efficiency of half wave rectifier. Therefore, $x=2 y$.

Time period for the force $F_1=T_1$

Time period for the force $F_2=T_2$

If both forces acts simultaneously then the resultant force is $F$,

$F=F_1+F_2$

Let, time period for the force $F$ be $T$

Now, $F=F_1+F_2$

$$\begin{aligned} & m \omega^2 y=m \omega_1^2 y+m \omega_2^2 y \quad\left(\because F=m \omega^2 y\right) \\ & \omega^2=\omega_1^2+\omega_2^2 \end{aligned}$$

$$\begin{aligned} & \Rightarrow \quad\left(\frac{2 \pi}{T}\right)^2=\left(\frac{2 \pi}{T_1}\right)^2+\left(\frac{2 \pi}{T_2}\right)^2 \\ & \Rightarrow \quad \frac{1}{T^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2} \\ & \Rightarrow \quad \frac{1}{T^2}=\frac{T_2^2+T_1^2}{T_1^2 T_2^2} \\ & \Rightarrow \quad T^2=\frac{T_1^2 T_2^2}{T_1^2+T_2^2} \Rightarrow T=\sqrt{\frac{T_1^2 T_2^2}{T_1^2+T_2^2}} \\ & \end{aligned}$$

Initial intensity of bright fringes, $(\sqrt{I}+\sqrt{I})^2$

Final intensity of bright fringes after covering one slit with glass sheet $=\left(\sqrt{I}+\sqrt{\frac{I}{2}}\right)^2$

Clearly, the intensity of bright fringes has decreased. For dark fringes initial intensity $=(\sqrt{I}-\sqrt{I})^2$

But, new intensity $=\left(\sqrt{I}-\frac{\sqrt{I}}{2}\right)^2 \neq 0$

Hence, intensity of dark fringes increases.

Convection is a method of heat transfer, in which the molecules/particle travel from one place to another to transmit heat energy. Heating of copper and iron rod is an example of conduction and heat transfer from sun is due to radiation.

Total magnetic flux linked

$$\begin{aligned} \phi & =B A \\ \therefore \text { induced emf, }|e| & =\frac{\Delta \phi}{\Delta t}=\frac{\Delta B A}{\Delta t}=B \cdot \frac{\Delta A}{\Delta T} \\ & =B \cdot \frac{\pi R^2}{\left(\frac{2 \pi}{\omega}\right)}=\frac{B \omega R^2}{2} \end{aligned}$$

Acceleration due to gravity of planet $A$ is given below.

$g_A=\frac{G M_A}{R_A^2}$ ..... (i)

$\because$ We know that, (Mass $=$ Density $\times$ Volume) $(M=D \times V)$

So, from Eq. (i)

$$\begin{aligned} g_A & =\frac{G D_A \times \frac{4}{3} \pi R_A^3}{R_A^2}=G D_A \times \frac{4}{3} \pi R_A \\ g_A & =\frac{4}{3} \pi G R_A D_A \quad \text{.... (ii)} \end{aligned}$$

Similarly, Acceleration due to gravity of planet $B$ is

$g_B=\frac{G M_B}{R_B^2}$ .... (iii)

$\text { and } \quad g_B=\frac{4}{3} \pi G D_B R_B$ .... (iv)

We have to find the ratio of $g_A: g_B$

So, according to question, we have

$$\begin{aligned} & \frac{R_A}{R_B}=\frac{x}{y} \quad \text{.... (v)}\\ \text{and} \quad & \frac{D_A}{D_B}=\frac{m}{n} \quad \text{.... (vi)} \end{aligned}$$

Now, taking the ratio of Eqs. (ii) and (iv), we get

$\frac{g_A}{g_B}=\frac{R_A D_A}{R_B D_B}$

Put the value from Eqs. (v) and (vi) in this equation, we get

$\frac{g_A}{g_B}=\frac{x \cdot m}{y \cdot n}$

or $\quad g_A / g_B=m x / n y$

$\because$ Work done in rotating dipole by $\theta$ angle from equilibrium,

$$\begin{aligned} & W=p E[1-\cos \theta] \quad \text{..... (i)}\\ & \Rightarrow \quad W=p E\left[1-\cos 180^{\circ}\right] \\ & \Rightarrow \quad W=p E[1+1]=2 p E \\ & \text { or } \quad p E=\frac{W}{2} \quad \text{..... (ii)}\\ & \text { and } \quad W^{\prime}=p E\left(1-\cos 60^{\circ}\right] \\ & \Rightarrow \quad W^{\prime}=p E\left[1-\frac{1}{2}\right]=\frac{p E}{2} \\ & \Rightarrow \quad W^{\prime}=\frac{p E}{2} \quad \text{..... (iii)}\\ \end{aligned}$$

From Eq. (ii) put the value of $p E$ in Eq. (iii), we get

$\Rightarrow \quad W^{\prime}=\frac{W}{2 \times 2}=\frac{W}{4} \Rightarrow W^{\prime}=\frac{W}{4}$

Total length of wire $=10 \mathrm{~m}$

The equivalent resistance of the circuit, $R_{\mathrm{eq}}=10+20 =30 \Omega$

The current flowing through the wire, $i=V / R$

$=\frac{3 V}{30 \Omega}=0.1 \mathrm{~A}$

$\therefore$ Potential gradient along the wire,

$$\begin{aligned} & =\frac{\text { potential difference across } 20 \Omega}{\text { length of } 20 \Omega \text { wire }} \\ & =\frac{0.1 \times 20 \Omega}{10 \mathrm{~m}}=0.20 \mathrm{~V} / \mathrm{m} \end{aligned}$$

$\because$ We know that, Transformation ratio is

$k=\frac{V_2}{V_1}$ .... (i)

and Turn ratio is

$\alpha=\frac{1}{k}=\frac{V_1}{V_2}=\frac{N_1}{N_2}$ ..... (ii)

Here, $N_1=$ Primary winding turns

$N_2=$ Secondary winding turns

$V_1=$ Primary winding voltage

$V_2=$ Secondary winding voltage

$$\begin{array}{rlrl} \text { Given : } N_1 & =1000 \\ N_2 & =3000 \text { and } V_1=80 \mathrm{~V} \\ V_2 & =? \\ \therefore \quad \frac{V_2}{V_1} & =\frac{N_2}{N_1} \\ \Rightarrow \quad & V_2 =\frac{N_2}{N_1} \cdot V_1 \quad \text{..... (iii)} \end{array}$$

put the values of $N_1, N_2$ and $V_1$ in above equation, we get

$V_2=\frac{3000}{1000} \times 80=240 \mathrm{~V}$

Now, potential difference across each turn

$=\frac{240}{3000}=0.08 \mathrm{~V}$

The electric field due to a single charge has point symmetry, or single charged forms spherical equipotential surface.

The work done in moving a charge on this spherical surface is always zero.

First overtone for open pipe, $V_{01}=\frac{2 \mathrm{v}}{2 I_0}$

First overtone for close pipe, $V_{C 1}=\frac{3 V}{4 I_C}$

According to question, $V_{01}=V_{C 1}$

$$\begin{array}{ll} \Rightarrow & \frac{2 \mathrm{~V}}{2 I_0}=\frac{3}{4} \frac{\mathrm{V}}{I_C} \\ \Rightarrow & \frac{I_C}{I_0}=\frac{3}{4} \end{array}$$

Let the length of the rod $A B$ be $l$ and mass of the rod be $m$.

Moment of inertia of the uniform rod about an axis passing through one of its end $V$ given as,

$\mathrm{MI}=\frac{m l^2}{3}$

Rotational kinetic energy, $(\mathrm{KE})_{\mathrm{rod}}=\frac{1}{2} l \omega^2$

Initially the rod is held, vertical, so potential energy of rod is given by $(\mathrm{PE})_i=m g \frac{I}{2}\quad \left(\because \frac{l}{2}=\mathrm{COM}\right)$

Initial kinetic energy, $(\mathrm{KE})_i=0$

The rod is allowed to fall, when it just touches the ground its potential energy becomes zero, i.e., $(\mathrm{PE})_f=0$

Let, angular velocity of rod at that instant be $\omega$, then kinetic energy at that instant is

$(\mathrm{KE})_f=\frac{1}{2} / \omega^2$

From conservation of energy,

$(\mathrm{KE})_i+(\mathrm{PE})_i=(\mathrm{KE})_f+(\mathrm{PE})_f$

$$\begin{array}{rlrl} \Rightarrow & 0+m g \frac{l}{2} =\frac{1}{2} l \omega^2+0 \\ \Rightarrow & m g l =l \omega^2 \\ \Rightarrow & m g l =\frac{m l^2}{3} \omega^2 \\ \Rightarrow & \omega =\sqrt{\frac{3 g}{l}} \end{array}$$

$\because$ We know that, time required to empy the tank is

$t=\frac{A}{A_0} \sqrt{\frac{2 H}{g}}$

and $t \propto \sqrt{H}$

$\therefore \quad \frac{t_2}{t_1}=\sqrt{\frac{H_2}{H_1}}$ .... (i)

and we know that (given data)

$$\begin{aligned} & H_1=h \text { and } t_1=t \\ & H_2=4 h \text { and } t_2=? \end{aligned}$$

Put $H_1$ and $H_2$ value in Eq. (i), we get

$$\begin{aligned} \frac{t_2}{t_1} & =\sqrt{\frac{4 h}{h}}=\sqrt{4}=2 \\ \Rightarrow \quad \frac{t_2}{t} & =2 \Rightarrow t_1=2 t \end{aligned}$$

Doping increases the charge concentration of an intrinsic semiconductor. This increases conductivity, and so resistivity decreases.

Work done in blowing a soap bubble, $W=$ Surface Tension $\times$ change in area

$\Rightarrow \quad T \times \Delta A$

As, we know that,

$V=\frac{4}{3} \pi r^3$

and $A=4 \pi r^2$

$\Rightarrow \quad A \propto(V)^{\frac{2}{3}}$

or $\quad \Delta A \propto(\Delta V)^{\frac{2}{3}}$

According to the question,

$$\begin{aligned} & \frac{W_1}{W_2}=\frac{\Delta A_1}{\Delta A_2}=\left(\frac{\Delta V_1}{\Delta V_2}\right)^{2 / 3} \\ \Rightarrow \quad W_2 & =W_1\left(\frac{2 V}{V}\right)^{2 / 3}=W \cdot(2)^{\frac{2}{3}}=(4)^{\frac{1}{3}} \cdot W \end{aligned}$$

Given, path difference $(\Delta x)=3 \mu \mathrm{m}=3 \times 10^{-6} \mathrm{~m}$

Wavelength of length, $\lambda=5000 \mathop A\limits^o=5 \times 10^{-7} \mathrm{~m}$

angular position is given by

$\theta=\frac{\Delta x}{\lambda}=\frac{3 \times 10^{-6}}{5 \times 10^{-7}}=6$

Since, $n=$ integer (6) is even, therefore, we will have 6th order maxima.

Given, angle of incidence = angle of emergence $=\frac{3}{4} \mathrm{~A}$

Here, $A=$ angle of prism

$\because$ prism is equilateral, $A=60^{\circ}$

$\therefore \quad i=e=60^{\circ} \times \frac{3}{4}=45^{\circ}$

From prism formula,

Angle of deviation, $\delta=i+e-A$

$=45+45-60=30^{\circ}$

Time period of oscillation when mass $M$ hangs by spring of spring constant $k$ is given by,

$T=2 \pi \sqrt{M / k}$

From Hooke's law, $k x=m g$

$\Rightarrow \quad \frac{1}{k}=\frac{x}{m g}$

When mass $M_1$ is added to the system

$T^{\prime}=2 \pi \sqrt{\frac{\left(M_1+M\right)}{k}}=2 \pi \sqrt{\frac{x\left(M_1+M\right)}{m g}}$

For an adiabatic process, there is no transfer of heat.

Since, $\Delta Q=0$

From first law of thermodynamics,

$$\begin{aligned} \Delta Q & =\Delta U+\Delta W \\ \text{or} \quad \Delta W & =-\Delta U \end{aligned}$$

Hence, all work done is utilised to change in internal energy.

From Einstein's photoelectric equation,

$$\begin{array}{rlrl} & \mathrm{KE}_{\max } & =h v-h v_0 \\ \text { or } & e V & =h v-h v_0 \\ \Rightarrow \quad & \quad e V & =\frac{h c}{\lambda}-\frac{h c}{\lambda_0} \end{array}$$

Here, $V$ is stopping potential and $\lambda$ is the incident wavelength, whereas $\lambda_0$ its threshold wavelength.

For case $1 e(4.8)=\frac{h c}{\lambda}-\frac{h c}{\lambda_0}$ ..... (i)

For case $2 e(1.6)=\frac{h c}{2 \lambda}-\frac{h c}{\lambda_0}$ ..... (ii)

From Eqs. (i) and (ii), we get

$$\begin{array}{ll} & e \times(3.2)=\frac{h c}{2 \lambda} \\ \Rightarrow \quad & e(1.6)=\frac{h c}{4 \lambda} \quad \text{.... (iii)} \end{array}$$

Putting the Eq. (iii) in Eq. (ii), we get

$$\begin{aligned} \Rightarrow \quad & \frac{h c}{4 \lambda}=\frac{h c}{2 \lambda}-\frac{h c}{\lambda_0} \\ \Rightarrow \quad & \frac{h c}{\lambda_0}=\frac{h c}{2 \lambda}-\frac{h c}{4 \lambda} \\ \Rightarrow \quad & \frac{h c}{\lambda_0}=\frac{h c}{4 \lambda} \\ \Rightarrow \quad & \lambda_0=4 \lambda \end{aligned}$$

When lift is stationary, reading of spring balance

$=49 \mathrm{~N}$

i.e., weight of bag, $=49 \mathrm{~N}$

$\Rightarrow \quad m=\frac{49}{g}$

When lift moves downward, $a=5 \mathrm{~m} / \mathrm{s}^2$

Weight (reading on spring balance) $=(g-a) \mathrm{m}$

$=(9.8-5) \times \frac{49}{9.8}=24 \mathrm{~N}$

The four identical uniform solid spheres are shown below. The radius of each sphere is $R$.

Lets take centre of the system at $O$.

Then, distance from the centre for $r_B=r_C$ and $r_A=r_D$

$\therefore$ Moment of inertial, $M_0 I=M r^2$, for each spheres will be $I_A, I_B, I_C$ and $I_D$.

$$\begin{array}{lll} \therefore & I_A=I_D \text { and } I_B=I_C \\ \text { Since, } & r_B=r_C < r_A=r_D \\ \text { Therefore, } & I_A=I_D > I_B=I_C \end{array}$$