MHT CET 2021 24th September Morning Shift (Physics & Chemistry) Question Paper with Solutions

In a parallel LCR circuit, the total impedance is maximum at the resonant frequency ${𝑓}_{𝑟}$:

$𝑍={𝑍}_{\text{max}}$

Since the r.m.s. current supplied to the circuit is $𝐼=\frac{𝑉}{𝑍}$, the current reaches a minimum at resonance. Away from resonance, the impedance decreases because the inductive branch offers low reactance at low frequencies and the capacitive branch offers low reactance at high frequencies, resulting in a large current.

Thus, the curve of current versus frequency exhibits a minimum at ${𝑓}_{𝑟}$, which is represented by graph (Q).

Hence, the correct option is (A).

The correct answer is Option C - shape of the loop. Here's why:

The torque (τ) acting on a current-carrying loop in a uniform magnetic field is given by:

$τ = NIAB sinθ$

Where:

• N is the number of turns in the loop
• I is the current flowing through the loop
• A is the area of the loop
• B is the strength of the magnetic field
• θ is the angle between the magnetic field and the normal to the loop

From the formula, we can see that the torque is directly proportional to:

• N (number of turns): More turns mean more current loops, leading to a larger torque.
• I (current): Higher current results in stronger magnetic moments and thus greater torque.
• A (area): A larger loop area creates a larger magnetic dipole moment, leading to a larger torque.
• B (magnetic field strength): A stronger magnetic field exerts a greater force on the current, resulting in more torque.

However, the formula does not explicitly include the shape of the loop. The shape affects the loop's magnetic moment, but ultimately, the torque depends on the area of the loop, not its specific shape.

Let's consider an example. Imagine two loops: one square and one circular, both with the same area. Even though they have different shapes, they will experience the same torque in the same magnetic field because their areas are equal.

Therefore, the torque acting on a current-carrying loop in a uniform magnetic field does not depend on the shape of the loop.

To determine the heat supplied at constant volume for a monoatomic gas, let's start by defining some fundamental relationships and properties of monoatomic gases.

The ratio of specific heats $\gamma$ for a monoatomic gas is given as:

$\gamma = \frac{C_p}{C_v} = \frac{5}{3}$

Where:

• $C_p$: Specific heat at constant pressure
• $C_v$: Specific heat at constant volume

The work done at constant pressure, $\mathrm{W}$, for a rise in temperature $\Delta T$ is defined as:

$W = P \Delta V$

Using the ideal gas law, $PV = nRT$, we know that at constant pressure, $V \propto T$, so:

$P \Delta V = n R \Delta T$

Thus, the work done at constant pressure can be written as:

$W = n R \Delta T$

The heat supplied at constant pressure $Q_p$ is given by:

$Q_p = n C_p \Delta T$

Since the work done, $W$, at constant pressure is equivalent to $n R \Delta T$, it can also be related to $Q_p$ using the specific heat at constant pressure:

$Q_p = W \cdot \frac{C_p}{R}$

The specific heat at constant volume, $Q_v$, is given by:

$Q_v = n C_v \Delta T$

Since $C_p / C_v = \gamma$:

$C_p = \gamma C_v$

Therefore, we can substitute to get:

$C_v = \frac{C_p}{\gamma}$

Substitute $C_v$ into the expression for the heat supplied at constant volume:

$Q_v = n \left(\frac{C_p}{\gamma}\right) \Delta T$

Since $n R \Delta T = W$, and using the relationship between $C_p$ and $R$ for a monoatomic gas:

$C_p = \frac{\gamma R}{\gamma - 1}$

For a monoatomic gas, $\gamma = \frac{5}{3}$:

$C_p = \frac{\frac{5}{3} R}{\frac{5}{3} - 1} = \frac{\frac{5}{3} R}{\frac{2}{3}} = \frac{5}{2} R$

Therefore, $C_v$ is:

$C_v = \frac{C_p}{\gamma} = \frac{\frac{5}{2} R}{\frac{5}{3}} = \frac{3R}{2}$

Thus:

$Q_v = n C_v \Delta T = n \frac{3R}{2} \Delta T$

Since $W = n R \Delta T$, we can write:

$Q_v = \frac{3}{2} \cdot W$

Hence, the heat supplied at constant volume for the same rise in temperature of the gas is:

Option D: $\frac{3 W}{2}$

The weight of the disc is balanced by the force due to the surface tension and the upthrust of water.

The component of surface tension in vertically upward direction is $T \cos \theta$ and the force acting due to it is $2 \pi \mathrm{T} \cos \theta$.

The upthurst is equal to the weight of the water displaced (W).

$\therefore$ Weight of the disc $=2 \pi \mathrm{r} \mathrm{T} \cos \theta+\mathrm{W}$

$$\begin{aligned} & \frac{Q}{A t}=\frac{k \Delta \theta}{d} \\ & \therefore 10=k \times \frac{9}{1.8 \times 10^{-2}} \\ & \therefore \quad k=\frac{18 \times 10^{-2}}{9}=2 \times 10^{-2} \mathrm{kcal} / \mathrm{mS}^{\circ} \mathrm{C} \end{aligned}$$

Photoelectric current is proportional to the intensity of the incident light.

When string is horizontal its speed is given by

$V=\sqrt{3 \mathrm{gr}}$

Centripetal acceleration $=\frac{\mathrm{V}^2}{\mathrm{r}}=3 \mathrm{~g}$

When length decreases, the frequency of wire will increase. If $\mathrm{n}$ is the frequency of the tuning fork, then we have fork, then we have

$\mathrm{n} \times 25=(\mathrm{n}+6) \times 24$

$\therefore \mathrm{n}=144 \mathrm{~Hz}$

To determine the temperature at which the average translational kinetic energy of a molecule in a gas becomes equal to the kinetic energy of an electron accelerated from rest through a potential difference of 'V' volts, we start by comparing their respective kinetic energies.

The kinetic energy of an electron accelerated through a potential difference V is given by:

$E_{\text{electron}} = eV$

where:

• $e$ is the electronic charge.
• $V$ is the potential difference in volts.

For a molecule in an ideal gas, the average translational kinetic energy is given by:

$E_{\text{translational}} = \frac{3}{2} k_B T$

where:

• $k_B$ is the Boltzmann constant.
• $T$ is the absolute temperature in Kelvin.

Now, we equate the kinetic energy of the electron to the average translational kinetic energy of a molecule in the gas:

$eV = \frac{3}{2} k_B T$

Solving for temperature T, we get:

$T = \frac{2 eV}{3 k_B}$

For a gas with number of molecules $N$ and using the relationship between Boltzmann constant $k_B$ and gas constant $R$ (i.e., $k_B = \frac{R}{N}$), we substitute this value into the equation:

$T = \frac{2 eV}{3} \cdot \frac{N}{R}$

Simplifying, we obtain:

$T = \frac{2 eVN}{3R}$

Thus, the correct option is:

Option A

$\frac{2 \mathrm{eVN}}{3 \mathrm{R}}$

The internal energy of an ideal gas is given by the following formula:

$U = \frac{f}{2} nRT$

where:

• $U$ is the internal energy
• $f$ is the degree of freedom
• $n$ is the number of moles
• $R$ is the ideal gas constant
• $T$ is the temperature

We are given that the internal energy of $n_1$ moles of hydrogen at temperature $T$ is equal to the internal energy of $n_2$ moles of helium at temperature $2T$. Therefore, we can write the following equation:

$\frac{5}{2}n_1RT = \frac{3}{2}n_2R(2T)$

Simplifying the equation, we get:

$5n_1 = 6n_2$

Therefore, the ratio $n_1: n_2$ is $\boxed{6:5}$.

Induced emf $\mathrm{e}=\frac{-\mathrm{N}\left(\phi_2-\phi_1\right)}{\mathrm{t}}=\frac{-\mathrm{NBA}\left(\cos \theta_2-\cos \theta_1\right)}{\mathrm{t}}$

Assuming that the coil was initially placed perpendicular to the magnetic flux $\theta_1=0^{\circ}$ and $\theta_2=90^{\circ}$.

$$\begin{aligned} e & =\frac{-\mathrm{NBA}\left(\cos 90^{\circ}-\cos 0^{\circ}\right)}{t} \\ & =\frac{-\mathrm{NBA}(0-1)}{\mathrm{t}}=\frac{\mathrm{NBA}}{\mathrm{t}} \\ & =\frac{600 \times 5 \times 10^{-5} \times 0.06}{0.2}=9 \times 10^{-3} \mathrm{~V} \end{aligned}$$

The wavelength corresponding to $2.5 \mathrm{~eV}$ is given by

$$\begin{aligned} & \text { Energy }(\mathrm{eV})=\frac{12400}{\lambda(\mathop A\limits^o)} \\ & \therefore \lambda=\frac{12400}{2.5}=5000 \mathop A\limits^o \end{aligned}$$

To get detected, the signal should have energy greater than $2.5 \mathrm{~eV}$ or wavelength less than $5000 \mathop A\limits^o$.

Moment of inertia $\mathrm{I}=2 \mathrm{~m}\left(\frac{\mathrm{d}}{2}\right)^2=\frac{\mathrm{md}^2}{2}$

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

$$\begin{aligned} & \therefore \omega^2=\frac{2 k}{I}=2 \mathrm{k} \cdot \frac{2}{\mathrm{md}^2}=\frac{4 \mathrm{k}}{\mathrm{md}^2} \\ & \therefore \omega=\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \end{aligned}$$

Force on each conductor is given by

$\mathrm{F}=\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{I}_1 \mathrm{I}_2}{\mathrm{~d}} \cdot \ell$

The force will be attractive. If the direction of current is reversed in one conductor the force will become repulsive.

$\therefore F^{\prime}=-\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{3 d} \cdot \ell=-\frac{2}{3} F$

In the part of an a.c. circuit

$$\begin{aligned} & V_A-V_B=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}+\mathrm{IR} \\ & 0.5=8 \mathrm{~L}+0.5 \times 0.2 \\ & \therefore 8 \mathrm{~L}=0.4 \\ & \therefore \quad \mathrm{L}=\frac{0.4}{8}=0.05 \mathrm{~H} \end{aligned}$$

In series combination, the effective spring constant

$\mathrm{k}_1=\frac{\mathrm{k}}{2}$

In the parallel combination, the effective spring constant is

$$\begin{aligned} k_2 & =2 k \\ f_1 & =\frac{1}{2 \pi} \sqrt{\frac{k_1}{m}}=\frac{1}{2 \pi} \sqrt{\frac{k}{2 m}} \\ f_2 & =\frac{1}{2 \pi} \sqrt{\frac{k_2}{m}}=\frac{1}{2 \pi} \sqrt{\frac{2 k}{m}} \\ \therefore \frac{f_1}{f_2} & =\sqrt{\frac{1}{4}}=\frac{1}{2} \end{aligned}$$

Maximum wavelength of Balmer series is given by

$\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 \mathrm{R}}{36}\quad \text{.... (1)}$

Maximum wavelength is given by

$\frac{1}{\lambda^{\prime}}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{\infty}\right)=\frac{\mathrm{R}}{4} \quad \text{.... (2)}$

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

$\frac{\lambda}{\lambda^{\prime}}=\frac{9}{5}$

In the first case &#x27; in math mode at position 7: \theta̲ is the critica…" style="color:#cc0000">\theta$ is the critical angle</p> <p>Hence\sin \theta=\frac{1}{\mu}$</p> <p>In the second case$\frac{\sin \theta}{\sin r}=\mu \quad \therefore \sin r=\frac{\sin \theta}{\mu}$</p> <p>$\therefore \sin r=\frac{1}{\mu^2} \quad \therefore r=\sin ^{-1}\left(\frac{1}{\mu^2}\right)$$

Since $\mathrm{L}=3 \frac{\lambda}{2}$, it is third harmonic or second overtone.

For isothermal expansion we have

$$\begin{aligned} & P_1 V_1=P_2 V_2 \\ \therefore \quad & P_2=P_1 \frac{V_1}{V_2}=P_1 \times \frac{1}{2}=\frac{P}{2} \\ \therefore & P_i=\frac{P}{2} \end{aligned}$$

For adiabatic process :

$$\begin{aligned} & P_1 V_1^\gamma=P_2 V_2^\gamma \\ \therefore \quad & P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma=P_1\left(\frac{1}{2}\right)^\gamma=\frac{P}{2^\gamma} \\ \therefore \quad & P_a=\frac{P}{2^\gamma} \\ \therefore \quad & \frac{P_i}{P_a}=\frac{2^\gamma}{2}=2^{\gamma-1} \end{aligned}$$

For adiabatic expansion we have

$$\begin{aligned} & \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\ & \therefore \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}=\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^{\gamma-1} \quad[\because \mathrm{v} \propto \mathrm{L}] \\ & =\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^{\frac{5}{3}-1} \quad \text{[For monoatocomic gas}~ \gamma=\frac{5}{3}]\\ & =\left(\frac{L_1}{L_2}\right)^{\frac{2}{3}} \\ \end{aligned}$$

$$\begin{aligned} & \text { Acceleration } \mathrm{a}=\omega^2 \mathrm{x} \\ & \therefore \quad \omega^2=\frac{\mathrm{a}}{\mathrm{x}}=\frac{0.5}{0.02}=25 \\ & \therefore \quad \omega=5 \mathrm{rad} / \mathrm{s} \\ & \quad \mathrm{V}_{\max }=\mathrm{A} \omega^2=0.1 \times 5=0.5 \mathrm{~m} / \mathrm{s} \end{aligned}$$

For output Y to be '1', the output of the OR gate should be '0'. This is possible only when both the inputs are zero.

$$\begin{aligned} & Q_1=C V_1, Q_2=C V_2, Q=Q_1+Q_2=2 C V \\ & \therefore C V_1+C V_2=2 C V \\ & \therefore \quad V=\frac{V_1+V_2}{2} \end{aligned}$$

$$\begin{aligned} &\text { Decrease in energy } =\frac{1}{2} C V_1^2+\frac{1}{2} C V_2^2-\frac{1}{2}(2 C)\left(\frac{V_1+V_2}{2}\right)^2 \\ & =\frac{1}{2} C\left[V_1^2+V_2^2-2\left(\frac{V_1+V_2}{2}\right)^2\right] \\ & =\frac{1}{2} C\left[V_1^2+V_2^2-\frac{1}{2}\left(V_1+V_2\right)^2\right] \\ & =\frac{1}{4} C\left[2 V_1^2+2 V_2^2-\left(V_1+V_2\right)^2\right] \\ & =\frac{1}{4} C\left(V_1-V_2\right)^2 \end{aligned}$$

Reynold number is given by

$$\begin{aligned} & R_n=\frac{v_c \rho d}{\eta} \\ & \therefore \quad v_c=\frac{R_n \eta}{\rho d} \\ & R_n=2500, \eta=10^{-3} \mathrm{Ns} / \mathrm{m}^2, R_n=2500, \rho=10^3 \mathrm{~kg} / \mathrm{m}^3 \\ & d=2 r=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m} \end{aligned}$$

Substituting these values and calculating we get critical velocity $\mathrm{v}_{\mathrm{c}}=0.125 \mathrm{~m} / \mathrm{s}$

To determine the current in the primary winding, we can use the principle of conservation of energy, which states that the power input to the transformer should be equal to the power output (ignoring losses for simplicity).

The primary voltage is given as $220 \mathrm{~V}$, and the current in the secondary is given as $2 \mathrm{~A}$. The ratio of the primary winding to the secondary winding is given as $1:20$.

The formula for the ratio of the voltage in a transformer is:

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

Given the winding ratio $\frac{N_p}{N_s} = \frac{1}{20}$, we can calculate the voltage in the secondary winding:

$\frac{V_p}{V_s} = \frac{1}{20} \implies V_s = 220 \mathrm{~V} \times 20 = 4400 \mathrm{~V}$

Now, using the power equation for the transformer:

$P_{\text{primary}} = P_{\text{secondary}}$

Since power $P$ is given by $P = VI$ (where $V$ is voltage and $I$ is current), we can write:

$V_p \times I_p = V_s \times I_s$

Substituting the given values:

$220 \mathrm{~V} \times I_p = 4400 \mathrm{~V} \times 2 \mathrm{~A}$

Solving for $I_p$:

$I_p = \frac{4400 \mathrm{~V} \times 2 \mathrm{~A}}{220 \mathrm{~V}} = 40 \mathrm{~A}$

Thus, the current in the primary winding is $40 \mathrm{~A}$. Therefore, the correct option is:

Option C

To determine the magnetic energy stored in the medium surrounding the coil, we need to use the relationship between the magnetic flux, current, number of turns in the coil, and the energy stored in the magnetic field.

Magnetic energy stored in an inductor (coil) is given by the formula:

$\text{Magnetic Energy} = \frac{1}{2} L I^2$

where:

$L$ is the inductance of the coil.

$I$ is the current through the coil.

According to the definition of inductance:

$L = \frac{N \phi}{I}$

where:

$N$ is the number of turns in the coil.

$\phi$ is the magnetic flux through the coil.

Now, substituting the expression for $L$ in the energy formula:

$\text{Magnetic Energy} = \frac{1}{2} \left( \frac{N \phi}{I} \right) I^2$

Simplifying this expression:

$\text{Magnetic Energy} = \frac{1}{2} N \phi I$

Therefore, the magnetic energy stored in the medium surrounding the coil is:

$\frac{N \phi I}{2}$

The correct option is:

Option B: $\frac{\mathrm{N} \phi \mathrm{I}}{2}$

Potential energy is a half of the total energy

$$\begin{aligned} & \therefore \frac{1}{2} \mathrm{~m} \omega^2 x^2=\frac{1}{2}\left[\frac{1}{2} m \omega^2 A^2\right] \\ & \therefore x^2=\frac{A^2}{2} \text { or } x=\frac{A}{\sqrt{2}} \end{aligned}$$

Maximum force $F_m=\omega^2 \mathrm{~A}$

Force at a distance $\mathrm{x}$ is $\mathrm{F}^{\prime}=\omega^2 \mathrm{x}$

$$\begin{aligned} & \therefore \frac{F^{\prime}}{F_m}=\frac{x}{A}=\frac{1}{\sqrt{2}} \\ & \therefore F^{\prime}=\frac{F_m}{\sqrt{2}}=\frac{50}{\sqrt{2}}=25 \sqrt{2} \end{aligned}$$

$$\begin{aligned} & \mathrm{IA}=\mathrm{I}+4 \mathrm{I}+2 \sqrt{I} \cdot \sqrt{4 I} \cdot \cos \frac{\pi}{2} \\ & =5 I \\ & \mathrm{I}_8=I+4 I+2 \sqrt{I} \cdot \sqrt{4 I} \cdot \cos \pi \\ & =I \\ & \therefore I_A-I_B=5 I-I=4 I \end{aligned}$$

$V=\sqrt{\frac{k}{\rho}}$

The electric potential decreases in the direction of the field. Hence field is maximum at B.

Here's how to solve this problem:

When the lift is stationary, the spring balance reads the actual weight of the bag, which is $49 \mathrm{~N}$. This means the mass of the bag is:

$m = \frac{W}{g} = \frac{49 \mathrm{~N}}{9.8 \mathrm{~m/s^2}} = 5 \mathrm{~kg}$

Now, when the lift accelerates downwards, the apparent weight of the bag will change. Here's why:

The net force acting on the bag is the difference between its weight (mg) and the upward force exerted by the spring balance (let's call it T). This net force is what causes the bag to accelerate downwards with the lift.

Applying Newton's second law (F = ma):

$mg - T = ma$

We can rearrange this to solve for the reading of the spring balance (T):

$T = mg - ma = m(g - a)$

Plugging in the values:

$T = 5 \mathrm{~kg} (9.8 \mathrm{~m/s^2} - 5 \mathrm{~m/s^2}) = 24 \mathrm{~N}$

Therefore, the reading of the spring balance will be 24 N when the lift accelerates downwards.

So the correct answer is Option C.

$$\begin{aligned} & B+C=420 \\ & A+B=220 \mathrm{~cm} \quad \therefore C=420-220=200 \mathrm{~cm} \\ & B+C=320 \mathrm{~cm} \quad \therefore B=320-200=120 \mathrm{~cm} \\ & A=220-B=100 \mathrm{~cm} \\ & \therefore A: B: C:: 1: 1.2: 2 \end{aligned}$$

Maximum height $\mathrm{h}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}$

For the first stone $\theta=90^{\circ}, \sin 90^{\circ}=1$

$\therefore \mathrm{h}_1=\frac{\mathrm{u}^2}{2 \mathrm{~g}}=\frac{\mathrm{v}^2}{2 \mathrm{~g}}$

For the second stone $h_2=\frac{v^2 \sin ^2 30^{\circ}}{2 g}=\lambda$

$=\frac{\mathrm{v}^2}{2 \mathrm{~g}} \times \frac{1}{4}$

The masses are same. Hence ratio of potential energies

$\frac{\mathrm{p}_1}{\mathrm{p}_2}=4$

$$\begin{aligned} & \left(\frac{R_1}{R_2}\right)^3=\left(\frac{T_1}{T_2}\right)^2=(8)^2=64 \\ \therefore & \frac{R_1}{R_2}=(64)^{\frac{1}{3}}=4 \\ \therefore & R_1=4 R_2 \\ \therefore & R_1-R_2=4 R_2-R_2=3 R_2 \end{aligned}$$

To determine the difference between the wavelengths of the sound waves produced by the two tuning forks, we can use the formula for the wavelength $\lambda$ of a wave:

$\lambda = \frac{v}{f}$

where $\lambda$ is the wavelength, $v$ is the velocity of sound, and $f$ is the frequency.

Given:

Wavelength for frequency $320 \mathrm{~Hz}$:

$\lambda_1 = \frac{320 \mathrm{~ms}^{-1}}{320 \mathrm{~Hz}} = 1 \mathrm{~m}$