MHT CET 2021 21st September Evening Shift (Physics & Chemistry) Question Paper with Solutions

The charge stored by the charged condenser is

$\mathrm{Q}=\mathrm{C}_1 \mathrm{~V}_1$

When the uncharged condenser is connected in parallel, the effective capacitance becomes $\left(C_1+C_2\right)$

Hence the potential $V_2=\frac{Q}{C_1+C_2}=\frac{C_1 V_1}{C_1+C_2}$

In a series LCR circuit, the voltage across the inductor ${𝑉}_{𝐿}$ leads the current by ${90}^{\circ }$, whereas the voltage across the capacitor ${𝑉}_{𝐶}$ lags the current by ${90}^{\circ }$.

The phase difference between ${𝑉}_{𝐿}$ and ${𝑉}_{𝐶}$ is therefore:

$\Delta 𝜙={90}^{\circ }-(-{90}^{\circ })={180}^{\circ }$

At resonance, ${𝑋}_{𝐿}={𝑋}_{𝐶}$, which gives ${𝑉}_{𝐿}={𝑉}_{𝐶}$. Since the two voltages have equal magnitude and are ${180}^{\circ }$ out of phase, they cancel each other.

Hence, the correct option is (D).

The dark band between the $4^{\text {th }}$ and $5^{\text {th }}$ bright band in the $5^{\text {th }}$ dark band. For $5^{\text {th }}$ dark band, the path difference is

$\left(5-\frac{1}{2}\right) \lambda=4.5 \lambda=4.5 \times 6000=2.7 \times 10^{-4} \mathrm{~cm}$

Pressure $\mathrm{P}=\frac{1}{3} \rho \mathrm{c}^2 \quad \therefore \mathrm{c}=\sqrt{\frac{3 \mathrm{p}}{\rho}}$

$\therefore \quad \frac{C_1}{C_2}=\sqrt{\frac{\rho_2}{\rho_1}}=\sqrt{\frac{16}{1}}=4$

Let the capacitance of capacitors are $\mathrm{C}$ initially.

So, afterward $\mathrm{C}_1=\mathrm{kC}$ and $\mathrm{C}_2=\mathrm{C}$

p.d. across capacitor $\mathrm{C}_2$ is

$V_2=\frac{C_2 V}{C_1+C_2}=\frac{k C V}{C+k C}=\frac{k V}{k+1}$

Inductive reactance = R = 2$\pi$fL

If L and f both are doubled, the inductive reactance will become 4R.

$$\begin{aligned} & 2 \times\left[\frac{1}{2 \ell_1} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}\right]=\frac{v}{4 \ell_2} \\ & \quad \frac{1}{0.5} \sqrt{\frac{50}{\mathrm{~m}}}=\frac{320}{4 \times 0.8} \\ & \therefore \quad \mathrm{m}=0.02 \mathrm{~kg} / \mathrm{m} \\ & \therefore \quad \text { Total mass of the string } \\ & \quad=0.02 \times 0.5 \mathrm{~kg}=10 \mathrm{gm}\end{aligned}$$

$\tan\theta=\frac{1}{20}=\frac{h}{1}$

$\therefore \mathrm{h=\frac{1}{20}m=0.05m=5cm}$

To address this question, we need to understand the relationship between kinetic energy and momentum. The kinetic energy (KE) of a body is given by the formula:

$KE = \frac{1}{2}mv^2$

where m is mass and v is velocity. The momentum (p) of a body, on the other hand, is given by

$p = mv$

Given that the bodies P, Q, and R have masses m, 2m, and 3m respectively, and all have the same kinetic energy, let us find a relationship between their momenta.

Since their kinetic energies are the same, we can write

$\frac{1}{2}mv_p^2 = \frac{1}{2}(2m)v_q^2 = \frac{1}{2}(3m)v_r^2$

where v_p, v_q, and v_r are the velocities of P, Q, and R, respectively. Simplifying this,

$mv_p^2 = 2mv_q^2 = 3mv_r^2$

We can remove the factor of m from each side, as it's common, which simplifies to:

$v_p^2 = 2v_q^2 = 3v_r^2$

Let's express the velocities of Q and R in terms of P's velocity to see the relation clearly:

$v_q^2 = \frac{v_p^2}{2} \quad \text{and} \quad v_r^2 = \frac{v_p^2}{3}$

Now, we find the square roots of both sides to get the actual velocities:

$v_q = v_p / \sqrt{2} \quad \text{and} \quad v_r = v_p / \sqrt{3}$

Since the momentum $p = mv$, we can substitute these velocity relationships into the equation for momentum:

$p_p = mv_p$

$p_q = 2m(v_p / \sqrt{2}) = \sqrt{2}mv_p$

$p_r = 3m(v_p / \sqrt{3}) = \sqrt{3}mv_p$

Comparing the momenta:

$p_p$ refers to the momentum of P with a factor of $mv_p$.

$p_q$ has a factor of $\sqrt{2}mv_p$, which indicates that the momentum of Q is greater than that of P because $\sqrt{2} > 1$.

$p_r$ has a factor of $\sqrt{3}mv_p$, which indicates that the momentum of R is greater than both P and Q since $\sqrt{3} > \sqrt{2}$.

Therefore, the body with the greatest momentum is body R, making the correct option:

Option B: R

$q = 4 \times {10^{ - 7}}$

$q = Ne$

$N = {q \over e} = {{4 \times {{10}^{ - 7}}} \over {1.6 \times {{10}^{ - 19}}}} = 2.5 \times {10^{12}}$

Currents in $\mathrm{A}$ and $\mathrm{B}$ are in the same direction. Hence force $\mathrm{F}_1$ exerted by $\mathrm{B}$ on $\mathrm{A}$ will attractive (towards B). Current in $\mathrm{A}$ and $\mathrm{C}$ are in opposite directions. Hence force $\mathrm{F}_2$, exentos by $\mathrm{C}$ on $\mathrm{A}$ will be repulsive (away from $\mathrm{C}$ ). Thus $\mathrm{F}_1$ and $\mathrm{F}_2$ are opposite in direction.

$$\begin{aligned} & F_1=\frac{\mu_0}{2 \pi} \cdot \frac{I_1 I_2}{x} \cdot L=\frac{\mu_0}{2 \pi} \cdot \frac{I^2}{x} \cdot L \\ & F_2=\frac{\mu_0}{2 \pi} \cdot \frac{2 I^2}{2 x} \cdot L=\frac{\mu_0}{2 \pi} \cdot \frac{I^2}{x} \cdot L \end{aligned}$$

$\therefore F_1$ and $F_2$ have some magnitude.

$\therefore \mathrm{F}_1=-\mathrm{F}_2$

For a lens $\frac{1}{f}=\left(\mu^{\prime}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

where $\mu^{\prime}=\frac{\mu_{\mathrm{g}}}{\mu_{\ell}}=1$

$\therefore \frac{1}{f}=0 \text { or } f=\infty$

Given the instantaneous value of an alternating current as $\mathrm{i}=50 \sin (100 \pi \mathrm{t})$, and we're asked to find at what time interval it achieves a value of $25 \mathrm{A}$. Using the given information ($\sin 30^{\circ}=0.5$), we set up the equation:

$25 = 50 \sin(100 \pi t)$

To find the value of $t$, we divide both sides by 50:

$\frac{25}{50} = \sin(100 \pi t)$

$0.5 = \sin(100 \pi t)$

Given $\sin 30^{\circ}=0.5$, we have:

$\sin(100 \pi t) = \sin 30^{\circ}$

Therefore,

$100 \pi t = 30^{\circ}$

To convert degrees into radians (since we're dealing with radians in the sine function), remember that $180^{\circ} = \pi$ radians. Thus, $30^{\circ}$ is $\frac{\pi}{6}$ radians:

$100 \pi t = \frac{\pi}{6}$

Cancelling $\pi$ from both sides and solving for $t$ yields:

$100t = \frac{1}{6}$

$t = \frac{1}{600} \mathrm{~S}$

Therefore, the correct answer is:

Option D: $\frac{1}{600} \mathrm{~S}$

Let $\mathrm{R}$ be the thermal resistance of each rod when connected in series, the total resistance will be

$\mathrm{R}_{\mathrm{s}}=2 \mathrm{R}$

Whey they are connected in parallel, the effective resistance will be

$\mathrm{R}_{\mathrm{p}}=\frac{\mathrm{R}}{2}$

$\therefore \frac{\mathrm{R}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{s}}}=\frac{1}{4}$

In parallel combination, the resistance becomes one-fourth and hence time will also become one-fourth. $\quad \therefore$ Time taken $\frac{8}{4}=2 \mathrm{~s}$

$$\begin{aligned} & A=\pi r^2, \quad \frac{A_2}{A_1}=\frac{1}{9} \\ & \therefore \frac{r_2^2}{r_1^2}=\frac{1}{3} \\ & h \propto \frac{1}{r} \quad \therefore \frac{h_2}{h_1}=\frac{r_1}{r_2}=3 \\ & \therefore h_2=3 h_1\end{aligned}$$

Magnetic field inside the solenoid is given by

$$\begin{aligned} B=\frac{\mu_0 \mathrm{NI}}{\mathrm{L}} \\ \therefore \frac{\phi}{\mathrm{A}}=\frac{\mu_0 \mathrm{NI}}{\mathrm{L}} & \\ \therefore \text { Magnetic moment, NIA } & =\frac{\phi \mathrm{L}}{\mu_0} \\ & =\frac{1.5 \times 10^{-6} \times 0.6}{4 \times 3.14 \times 10^{-7}} \\ & =0.75 \mathrm{Am}^2 \end{aligned}$$

$$\begin{aligned} & (\text { K.E. })_1=h v_1-h v_0 \quad \text{.....(1)}\\ & (\text { K.E. })_2=h v_2-h v_0 \quad \text{.....(2)} \end{aligned}$$

Dividing Eq.(1) by Eq.(2) :

$\frac{\left(\mathrm{K} . \mathrm{E}_1\right.}{(\mathrm{K} . \mathrm{E})_2}=\frac{1}{\mathrm{k}}=\frac{v_1-v_0}{v_2-v_0}$

Solving for $v_0$ we get :

$v_0=\frac{k v_1-v_2}{k-1}$

Given a hollow metal sphere with a radius $r$, if it is charged, the electric field $E$ outside the sphere behaves just as if all the charge were concentrated at the center.

From the properties of spherical conductors, the potential $V$ at a distance $r$ from the center due to a point charge $Q$ is given by :

$V(r) = \frac{kQ}{r}$

where $k$ is Coulomb's constant.

Now, let the potential at the surface of the sphere be $V_1$ and the potential at a distance $3r$ be $V_2$.

Given :

$V_2 - V_1 = V$

$\frac{kQ}{3r} - \frac{kQ}{r} = V$

$\frac{kQ(1 - 3)}{3r} = V$

$\frac{-2kQ}{3r} = V$

$Q = \frac{-3rV}{2k}$ ... (i)

The electric field $E$ at a distance $3r$ from the center of the sphere is given by:

$E = \frac{kQ}{(3r)^2}$

Using equation (i),

$E = \frac{k \times (-3rV/2k)}{(3r)^2}$

$E = \frac{-3rV}{2 \times 9r^2}$

$E = \frac{-V}{6r}$

The two equations can be written as

$\mathrm{Y}_1=2 \sin 2 \pi\left(\frac{4 \mathrm{t}}{0.2}-\frac{4 \mathrm{x}}{2}\right)=2 \sin 2 \pi\left(\frac{\mathrm{t}}{0.05}-\frac{\mathrm{x}}{0.5}\right)$

and $\mathrm{Y}_2=4 \sin 2 \pi\left(\frac{4 \mathrm{t}}{0.16}-\frac{4 \mathrm{x}}{1.6}\right)=2 \sin 2 \pi\left(\frac{\mathrm{t}}{0.04}-\frac{\mathrm{x}}{0.4}\right)$

Comparing with standard equation

$\mathrm{Y}=\mathrm{A} \sin 2 \pi\left(\frac{\mathrm{t}}{\mathrm{T}}-\frac{\mathrm{x}}{\lambda}\right)$

We get, for the first wave,

$\mathrm{T}=0.05 \mathrm{~s} \text { and } \lambda=0.5 \mathrm{~m}$

For the second wave,

$\mathrm{T}=0.04 \mathrm{~s} \text { and } \lambda=0.4 \mathrm{~m}$

Hence their periods (hence frequencies) are not same. Their wavelength are also not same.

For first wave velocity $=\frac{\lambda}{\mathrm{T}}=\frac{0.5}{0.05}=10 \mathrm{~m} / \mathrm{s}$

For second wave velocity $=\frac{0.4}{0.04}=10 \mathrm{~m} / \mathrm{s}$

Hence velocity in same,

pressure in side the first bubble $=P+\frac{4 T}{r_1}$

Pressure inside the second bubble $=P+\frac{4 T}{r_2}$

using the formula PV =n R $\theta \quad \theta=$ absolute temp.

we have : $\left(P+\frac{4 T}{r_1}\right) \cdot \frac{4 \pi}{3} r_1^3=n_1 R^{\prime} \theta \quad$ ( $R^{\prime}$ is molar gas constant)

$\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}\right) \cdot \frac{4 \pi}{3} \mathrm{r}_2^3=\mathrm{n}_2 \mathrm{R}^{\prime} \theta$

and $\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \cdot \frac{4 \pi}{3} \mathrm{R}^3=\left(\mathrm{n}_1+\mathrm{n}_2\right) \mathrm{R}^{\prime} \theta$

$$\begin{aligned} & \therefore\left(P+\frac{4 T}{R}\right) \cdot \frac{4 \pi}{3} R^3=\left(\frac{P+4 T}{r_1}\right) \cdot \frac{4 \pi}{3} r_1^3+\left(P+\frac{4 T}{r_2}\right) \cdot \frac{4 \pi r^3}{3} \\ & \therefore\left(p+\frac{4 T}{R}\right) R^3=\left(P+\frac{4 T}{r_1}\right) r_1^3+\left(p+\frac{4 T}{r_2}\right) r_2^3 \end{aligned}$$

on solving : $T=\frac{p\left(R^3-r_1^3-r_2^3\right)}{4\left(r_1^2+r_2^2-R^2\right)}$

Rise in a capillary tube is given by

$$\begin{aligned} \mathrm{h} & =\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}} \\ \therefore \mathrm{h} & \propto \frac{1}{\mathrm{~g}} \end{aligned}$$

Smaller the value of $g$, greater will be $h$. Out of the given options, value $g$ will be minimum for option (A).

The magnetic moment $μ$ associated with the angular momentum $L$ of a revolving electron is given by :

$\mu = \frac{-e}{2m} L$

The negative sign indicates the direction of the magnetic moment is opposite to the direction of angular momentum. However, for the sake of magnitude, we'll focus on :

$\mu = \frac{eL}{2m}$

So, the correct answer is Option C : $\frac{eL}{2m}$.

Kinetic energy $k=\frac{1}{2} I \omega^2$

$$\begin{aligned} \therefore \frac{\mathrm{K}_2}{\mathrm{~K}_1} & =\frac{\mathrm{I}_2 \omega_2^2}{\mathrm{I}_1 \omega_1^2} \quad \therefore \frac{1}{4}=\frac{\mathrm{I}_2}{\mathrm{I}_1} \cdot 4 \quad \therefore \frac{\mathrm{I}_2}{\mathrm{I}_1}=\frac{1}{16} \\ \frac{\mathrm{L}_2}{\mathrm{~L}_1} & =\frac{\mathrm{I}_2 \omega_2}{\mathrm{I}_1 \omega_1}=\frac{1}{16} \times 2=\frac{1}{8} \\ \therefore \mathrm{L}_2 & =\frac{\mathrm{L}_1}{8} \end{aligned}$$

Fringe width $\mathrm{X}=\frac{\lambda \mathrm{D}_1}{\mathrm{~d}_1}=\frac{\lambda \mathrm{D}_2}{\mathrm{~d}_2}$

$$\begin{aligned} & \therefore \quad \frac{d_2}{d_1}=\frac{D_2}{D_1} \quad \therefore 2=\frac{D_2}{D_1} \\ & \therefore \quad D_2=2 D_1=2 \times 75=150 \mathrm{~cm} \end{aligned}$$

The kinetic energy per mole E = $\frac{3}{2}$RT

$\therefore \mathrm{R=\frac{2E}{3T}}$

The moment of inertia of the upper ring about its diameter is given by

$\mathrm{I}_1=\frac{\mathrm{MR}^2}{2}$

The moment of inertia of the two lower rings about a tangent in their plane is given by

$\mathrm{I}_2=\mathrm{I}_3=\frac{3}{2} \mathrm{MR}^2$

Total moment of inertia

$\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3$

$=\frac{\mathrm{MR}^2}{2}+\frac{3}{2} \mathrm{MR}^2+\frac{3}{2} \mathrm{MR}^2=\frac{7}{2} \mathrm{MR}^2$

The standard equation of a wave can be written as

$\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})$

Speed of wave $V=\frac{\omega}{k}=\frac{60}{3}=20 \mathrm{~m} / \mathrm{s}$

$$\begin{aligned} & \text { Also, } \mathrm{V}=\sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\ & \therefore \quad \mathrm{m}=\frac{\mathrm{T}}{\mathrm{V}^2}=\frac{0.4}{400}=10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \end{aligned}$$

$$\begin{aligned} & \mu_{\mathrm{g}}=1.5=\frac{3}{2} \quad \mu_w=1.33=\frac{4}{3} \\ & \therefore \quad{ }_w \mu_g=\frac{\mu_g}{\mu_w}=\frac{3}{2} \times \frac{3}{4}=\frac{9}{8} \\ & \sin C=\frac{1}{{ }_w \mu_g}=\frac{8}{9} \\ & \therefore \quad C=\sin ^{-1}\left(\frac{8}{9}\right)\end{aligned}$$

Change in momentum in each collision is [mv $-$ ($-$ mv)] = 2mv

Hence change in momentum in 5 collisions is 5 $\times$ 2 mV = 10 mV

Maximum potential energy is attained at the highest point which gets converted into kinetic energy at the lowest point.

$$\begin{aligned} & \quad \mathrm{h}=2-0.75=1.25 \mathrm{~m} \\ & \quad \frac{1}{2} \mathrm{mv}^2=\mathrm{mgh} \\ & \therefore \quad v^2=2 \mathrm{gh}=2 \times 10 \times 1.25=25 \\ & \therefore \quad \mathrm{v}=5 \mathrm{~m} / \mathrm{s} \end{aligned}$$

The diode is forward biased. The potential difference between P and Q is $(3-(-5))=8$ V.

$\mathrm{I=\frac{V}{R}=\frac{8}{2\times10^3}=4\times10^{-3}A}$

De-Broglie wavelength is given by

$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}}$

If $\lambda_1$ and $\lambda_2$ are de-Broglie wavelengths of proton and alpha particle then

$\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{m_2 q_2}{m_1 q_1}}=\sqrt{4 \times 2}=\sqrt{8}=2 \sqrt{2}$

In a step up transformer, the current in the secondary is less than the current in the primary.

It is a balanced Wheatstone bridge. Hence no current flows through the galvanometer and it can be removed from the circuit. $5 \Omega$ and $1 \Omega$ resistances are in series. Their equivalent resistance is $5+1=6 \Omega$. $12.5 \Omega$ and $2.5 \Omega$ resistances are also in series. Their equivalent resitance is $15 \Omega$. $6 \Omega$ and $15 \Omega$ resistances are in parallel. If $I_1$ and $I_2$ are the currents in them then

$$\begin{array}{ll} 6 \mathrm{I}_1=15 \mathrm{I}_2 & \therefore \mathrm{I}_2=0.4 \mathrm{I}_1 \\ \mathrm{I}_1+\mathrm{I}_2=2.1 \mathrm{~A} & \therefore \mathrm{I}_1=\frac{2.1}{1.4}=1.5 \mathrm{~A} \end{array}$$

Centripetal acceleration $a=\frac{v^2}{r}$

For hydrogen atom, $V \propto \frac{1}{\mathrm{n}}$ and $\mathrm{r} \propto \mathrm{n}^2$

$\therefore a \propto \frac{1}{n^4} \quad \therefore \frac{a_2}{a_5}=\left(\frac{5}{3}\right)^4=\frac{625}{81}$

$\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{r}^2} \quad$ where $\mathrm{r}$ is the distance from the centre of the earth

In the first case, $r_1=R+R=2 R$

In the second case, $r_2=R+\frac{R}{2}=\frac{3}{2} R$

$$\begin{aligned} \therefore \frac{g_2}{g_1} & =\left(\frac{r_1}{r_2}\right)^2=\frac{4 \times 4}{9}=\frac{16}{9} \\ g_2 & =\frac{16}{9} g_1=\frac{16}{9} \times \frac{9}{4}=\frac{4}{9} g \end{aligned}$$