MHT CET 2020 16th October Morning Shift (Physics & Chemistry) Question Paper with Solutions

The spacing of the interference fringes in Young's double slit experiment can be calculated using the formula:

$\Delta y = \frac{\lambda L}{d}$

where:

• $\Delta y$ is the fringe spacing, i.e., the distance between adjacent bright or dark fringes on the screen,
• $\lambda$ is the wavelength of the light used,
• $L$ is the distance from the slits to the screen, and
• $d$ is the separation between the two slits.

Given this formula, let's discuss how each option affects the fringe spacing:

Option A: Moving the screen away from the slits

If we increase $L$, the distance from the slits to the screen, according to the formula, $\Delta y$ would also increase, causing the fringes to be more widely spaced, not more closely spaced.

Option B: Using blue light instead of green light

The wavelength of light ($\lambda$) is a key factor in determining fringe spacing. Blue light has a shorter wavelength than green light. Decreasing $\lambda$ (by switching from green to blue light) would cause $\Delta y$ to decrease, according to the formula. This means the fringes will be more closely spaced, making this option a potentially correct answer.

Option C: Using red light instead of green light

Since red light has a longer wavelength than green light, switching to red light would increase $\lambda$, leading to an increase in $\Delta y$, according to the formula. This would make the fringes more widely spaced, not more closely spaced.

Option D: Reducing the separation between the slits

Reducing the separation between the slits ($d$) while keeping the wavelength and the screen distance constant would cause $\Delta y$ to increase, according to the formula, thereby spacing the fringes further apart, not closer together.

Based on this analysis, Option B: Using blue light instead of green light is the correct choice, as it would cause the observed fringes to be more closely spaced due to the shorter wavelength of blue light.

The velocity of an electron in the first orbit of a hydrogen atom according to Bohr's model can be derived from the quantization of angular momentum and the electrostatic force balance with centripetal force. The velocity $v_n$ of an electron in the nth orbit is given by:

$v_n = \frac{e^2}{2 \varepsilon_0 h} \cdot \frac{1}{n}$

For the ground state (first orbit, $n=1$), the velocity $v_1$ is:

$v_1 = \frac{e^2}{2 \varepsilon_0 h}$

To find the ratio of this velocity to the speed of light $c$, we divide the expression for $v_1$ by $c$:

$\frac{v_1}{c} = \frac{\frac{e^2}{2 \varepsilon_0 h}}{c}$

$= \frac{e^2}{2 \varepsilon_0 h c}$

Therefore, the correct option is:

Option C: $\frac{e^2}{2 \varepsilon_0 h c}$

When a coil of $n$ turns moves through a magnetic flux from $\phi_1$ to $\phi_2$, the emf induced in the coil is

$\theta=-\frac{n d \phi}{d t}=\frac{-n\left(\phi_2-\phi_1\right)}{t}=\frac{n\left(\phi_1-\phi_2\right)}{t} \quad \text{..... (i)}$

As, resistance $R$ of coil and $\frac{R}{2}$ are in series, so equivalent resistance is

$R_{\text {eq }}=R+\frac{R}{2}=\frac{3 R}{2} \quad \text{..... (ii)}$

The induced current in the circuit is,

$I=\frac{\theta}{R_{\text {eq }}}=\frac{n\left(\phi_1-\phi_2\right)}{t\left(\frac{3 R}{2}\right)}=\frac{2 n\left(\phi_1-\phi_2\right)}{3 R t}$

The situation is as shown below

The magnetic field at $P$ due to first magnet is

$B_1=\frac{\mu_0}{4 \pi} \times \frac{2 M_1}{d^3} \text { (along } M_1 \text { ) }$

and magnetic field at $P$ due to second magnet is

$B_2=\frac{\mu_0}{4 \pi} \times \frac{M_2}{d^3} \text { (opposite to } M_2 \text { ) }$

The resultant field at $P$,

$$\begin{aligned} B & =\sqrt{B_1^2+B_2^2}=\frac{\mu_0}{4 \pi d^3} \sqrt{4 M_1^2+M_2^2} \\ & =\frac{\mu_0}{4 \pi} \sqrt{5} \frac{M}{d^3} \quad\left(\because M_1=M_2=M\right) \end{aligned}$$

Given, $F_1=F_2=1 \mathrm{~N}, m=2 \mathrm{~kg}$ and $\theta=60^{\circ}$

Resultant force,

$$\begin{aligned} F & =\sqrt{F_1^2+F_2^2+2 F_1 F_2 \cos \theta} \\ & =\sqrt{1^2+1^2+2 \cos 60^{\circ}} \\ & =\sqrt{2+2(0.5)}=\sqrt{3} \mathrm{~N} \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & m a=\sqrt{3} \\ \text { or } & a=\frac{\sqrt{3}}{m}=\frac{\sqrt{3}}{2}=\sqrt{\frac{3}{4}}=\sqrt{0.75} \mathrm{~ms}^{-2} \end{array}$$

Let's start by understanding the basic relationships in play when a force is applied to rods within the elastic limit. The extension ($\Delta L$) seen in the rod when a tensile force is applied can be described by Hooke's law for elastic deformation, which is given by:

$\Delta L = \frac{FL}{AE}$

where:

• $F$ is the force applied,
• $L$ is the original length of the rod,
• $A$ is the cross-sectional area of the rod,
• $E$ is the Young's modulus of the material (a property of the material).

Given that both rods are of the same material, the value of $E$ is constant for both. Additionally, since both rods are subjected to the same tensile force ($T$), $F$ is constant as well. The question tells us that the volume of both rods is the same, and the diameter of the first rod is half that of the second rod. Let's denote the diameter of the first rod as $D$ and that of the second rod as $2D$.

The cross-sectional area of a rod with a circular cross-section is given by $A = \pi \left(\frac{D}{2}\right)^2$. So, for the first rod, the area $A_1$ is:

$A_1 = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4}$

And for the second rod, the area $A_2$ is:

$A_2 = \pi \left(\frac{2D}{2}\right)^2 = \pi D^2$

Since the volume of both rods is the same and they are made up of the same material, their volume is $V = AL$, indicating that their length and area are inversely proportional. If the diameter of the second rod is twice that of the first, its cross-sectional area is quadruple the first rod's area, which implies that the length of the first rod is four times that of the second to maintain the same volume for both rods.

Therefore, the length of the first rod $L_1 = 4L_2$, where $L_2$ is the length of the second rod.

Substituting these values back into the extension formula and comparing the extensions of the first and second rods, we get:

$\Delta L_1 = \frac{T \times 4L_2}{\frac{\pi D^2}{4}E} = \frac{16TL_2}{\pi D^2E}$

$\Delta L_2 = \frac{T \times L_2}{\pi D^2E}$

Comparing $\Delta L_1$ to $\Delta L_2$, we find:

$\frac{\Delta L_1}{\Delta L_2} = \frac{\frac{16TL_2}{\pi D^2E}}{\frac{TL_2}{\pi D^2E}} = 16$

Therefore, the correct ratio of extension of the first rod to the second rod is $16:1$, which corresponds to Option C.

At extreme position, the potential energy of simple pendulum,

$P E=\frac{1}{2} m A^2 \omega^2 \quad \text{.... (i)}$

where, $\omega=$ angular frequency $=\frac{2 \pi}{T}$

For simple pendulum, $T=2 \pi \sqrt{\frac{L}{g}}$

$\therefore \quad \omega=\frac{2 \pi}{2 \pi} \sqrt{\frac{g}{L}}=\sqrt{\frac{g}{L}} \quad\left[\because T=\frac{2 \pi}{\omega}\right]$

On putting value of $\omega$ in Eq. (i), we get

$\mathrm{PE}=\frac{1}{2} m A^2\left(\sqrt{\frac{g}{L}}\right)^2=\frac{m g A^2}{2 L}$

The magnifying power at a telescope is given by

$m=-\frac{f_o}{f_e}$

Therefore, the magnifying power will be high when objective lens has large focal length and eye-piece has small focal length.

At maximum height, the velocity of body becomes zero. Let maximum height be $h$.

Applying principle of conservation of energy, we get

$$\begin{aligned} & \frac{1}{2} m v^2-\frac{G M m}{R}=\frac{1}{2} m(0)^2-\frac{G M m}{(R+h)} \\ & \text { [where, } m=\text { mass of body } \\ & \text { and } M=\text { mass of earth.] } \\ & \Rightarrow \frac{1}{2} m v^2=G M m\left(\frac{1}{R}-\frac{1}{R+h}\right) \\ & \Rightarrow \quad \frac{h}{R(R+h)}=\frac{v^2}{2 G M} \\ & \Rightarrow \quad \frac{h}{R+h}=\frac{v^2R}{2GM}=\frac{v^2R}{2gR^2} \quad (\because GM=gR^2) \\ & \begin{aligned} & \Rightarrow \quad \frac{R+h}{h} =\frac{2 g R}{v^2} \\ & \Rightarrow \quad \frac{R}{h}+1 =\frac{2 g R}{v^2} \\ & \Rightarrow \quad \frac{R}{h} =\frac{2 g R}{v^2}-1=\frac{2 g R-v^2}{v^2} \\ \text{or} \quad & h =\frac{v^2 R}{2 g R-v^2} \end{aligned} \end{aligned}$$

The force on a charge $q$ due to electric field is

$F_E=q E \quad \text{..... (i)}$

and due to magnetic field is

$\mathrm{F}_B=q(\mathrm{v} \times \mathrm{B})$

When charge particle moves in combined field,

Resultant force,

$$\begin{aligned} \mathbf{F} & =\mathbf{F}_E+\mathbf{F}_B \\ & =q \mathbf{E}+q(\mathbf{v} \times \mathbf{B}) \end{aligned}$$

To determine the number of beats produced per second by two waves, we need to calculate the difference in their frequencies. The beat frequency is given by the absolute difference between the frequencies of the two waves. In the given equations, $Y_1=0.25 \sin 316 t$ and $Y_2=0.25 \sin 310 t$, the numerical coefficients in front of $t$ represent the angular frequency $\omega$ of each wave, since the general form of a wave equation is $y = A \sin(\omega t + \phi)$, where $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is the time, and $\phi$ is the phase angle.

The angular frequency $\omega$ is related to the frequency $f$ of the wave by the equation $\omega = 2\pi f$. Thus, to find the frequencies ($f_1$ and $f_2$) of the waves $Y_1$ and $Y_2$, we have:

• For $Y_1$: $316 = 2\pi f_1 \Rightarrow f_1 = \frac{316}{2\pi} = \frac{158}{\pi}$ Hz.
• For $Y_2$: $310 = 2\pi f_2 \Rightarrow f_2 = \frac{310}{2\pi} = \frac{155}{\pi}$ Hz.

The beat frequency $f_{\text{beat}}$ is the difference between $f_1$ and $f_2$:

$f_{\text{beat}} = |f_1 - f_2| = \left| \frac{158}{\pi} - \frac{155}{\pi} \right| = \frac{|158 - 155|}{\pi} = \frac{3}{\pi} \, \text{Hz}$

Thus, the number of beats produced per second is $\frac{3}{\pi}$, which corresponds to Option C.

Given, magnetic field intensity,

$H=2000 \mathrm{Am}^{-1} \text {, }$

Magnetic flux, $\phi=6 \times 10^{-4} \mathrm{~Wb}$,

Cross-sectional area, $A=3 \mathrm{~cm}^2=3 \times 10^{-4} \mathrm{~m}^2$

As, we know, $\phi=B A$ and $B=\mu H$

$$\begin{aligned} \Rightarrow \quad \mu & \frac{B}{H}=\frac{\phi}{A H}=\frac{6 \times 10^{-4}}{3 \times 10^{-4} \times 2000} \\ & =10^{-3} \mathrm{~Wb} \mathrm{~A}^{-1} \mathrm{~m}^{-1} \end{aligned}$$

The path followed by ray of light is shown in figure

As, the ray emerges normally from other surface. So, angle of emergence is zero.

The angle of deviation i.e., between incident and emergent ray is

$\delta=(i+e)-A=i-A \quad \text{.... (i)} ~(\because e=0)$

Also, for minimum deviation, $\delta=(\mu-1) A\quad \text{..... (ii)}$

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

$$\begin{aligned} i-A & =(\mu-1) A \\ \Rightarrow \quad i & =(\mu-1) A+A=\mu A \end{aligned}$$

The electric field intensity at a distance $r$ from the conducting sphere is

$E=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \quad \text{..... (i)}$

The surface charge density of sphere is

$\sigma=\frac{Q}{4 \pi R^2} \quad \text{..... (ii)}$

Dividing Eqs. (ii) by (i), we get

$$\begin{aligned} & \frac{\sigma}{E}=\frac{Q}{4 \pi R^2} \times \frac{4 \pi \varepsilon_0 r^2}{Q}=\varepsilon_0\left(\frac{r}{R}\right)^2 \\ \text{or} \quad & \sigma=\varepsilon_0 E\left(\frac{r}{R}\right)^2 \end{aligned}$$

Let $x$ be the distance from the centre of earth, where gravitational force is zero and $m_0$ be the point mass as shown below

Using Newton's law of gravitation,

Force between earth and point mass,

$F_1=\frac{G \times 81 m \times m_0}{x^2} \quad \text{..... (i)}$

and force between moon and point mass,

$F_2=\frac{G \times m \times m_0}{(R-x)^2} \quad \text{.... (ii)}$

According to the question,

$$\begin{array}{ll} & F_1=F_2 \\ \Rightarrow & \frac{G \times 81 m \times m_0}{x^2}=\frac{F_2}{(R-x)^2} \\ \Rightarrow & \frac{81}{x^2}=\frac{1}{(R-x)^2} \\ \Rightarrow & \frac{9}{x}=\frac{1}{R-x} \\ \Rightarrow \quad & x=\frac{9}{10} R \end{array}$$

Let's consider the two vectors to be $\mathbf{A}$ and $\mathbf{B}$, each with magnitude $a$. According to the question, the magnitude of the resultant vector is equal to the magnitude of either of the two vectors, i.e., resultant vector $R = a$.

To calculate the angle between the two vectors, we can use the cosine rule for vectors, which gives us the magnitude of the resultant vector $R$ in terms of the magnitudes of $\mathbf{A}$ and $\mathbf{B}$ and the angle $\theta$ between them:

$R^2 = A^2 + B^2 + 2AB\cos\theta$

Given that $A = B = a$ and $R = a$, our equation becomes:

$a^2 = a^2 + a^2 + 2a \cdot a \cdot \cos\theta$

Simplifying it, we get:

$a^2 = 2a^2 + 2a^2\cos\theta$

$0 = a^2 + 2a^2\cos\theta$

$-1 = 2\cos\theta$

Solving for $\cos\theta$:

$\cos\theta = -\frac{1}{2}$

The angle $\theta$ whose cosine is $-\frac{1}{2}$ is $\cos^{-1}(-0.5)$. Therefore, the correct answer is:

Option A: $\cos^{-1}(-0.5)$

The height of water column in a capillary tube is given by

$h=\frac{2 \pi \cos \theta}{\rho r g}$

$\Rightarrow \quad h \propto \frac{1}{r} \quad \text{.... (i)}$

Also, mass of liquid in a column of radius $r$ and height $h$, is

$$\begin{aligned} & m=\left(\pi r^2 h\right) \rho \\ & m \propto r^2 h \quad \text{.... (ii)} \end{aligned}$$

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

$$\begin{aligned} & m \propto r^2 \times \frac{1}{r} \\ & m \propto r \\ \Rightarrow \quad & \frac{m^{\prime}}{m}=\frac{\frac{r}{4}}{r} \Rightarrow m^{\prime}=\frac{m}{4} \end{aligned}$$

Given, force, $F=A \sin (C t)+B \cos (D x)$

Here, $(C t)$ and $(D x)$ are dimensionless as they represents angles.

$\therefore \quad [C]=\frac{1}{[t]}=\left[\mathrm{T}^{-1}\right]$

and $\quad [D]=\frac{1}{[x]}=\left[\mathrm{L}^{-1}\right]$

Dimensions of $\frac{[C]}{[D]}=\frac{\left[\mathrm{T}^{-1}\right]}{\left[\mathrm{L}^{-1}\right]}=\left[\mathrm{LT}^{-1}\right]$

This is same as the dimensions of velocity.

In Bohr's theory of the hydrogen atom, the force acting on the electrons is the electrical force between the positively charged nucleus (proton) and the negatively charged electron. This force is given by Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The relevant formula from Coulomb's law is

$F = k \frac{e^2}{r^2}$

where $F$ is the force between the electron and the nucleus, $k$ is Coulomb's constant, $e$ is the charge of the electron (the same magnitude for the proton), and $r$ is the distance between the nucleus and the electron.

In Bohr's model, the radius of the $n$th orbit (where $n$ is the principal quantum number) is directly proportional to $n^2$. This relation is given by the formula for the radius of the $n$th orbit as,

$r_n = n^2 \cdot a_0$

where $a_0$ is the Bohr radius, the radius of the electron's orbit in its ground state (when $n=1$).

Substituting the expression for $r_n$ into the formula for $F$, we get

$F = k \frac{e^2}{(n^2 a_0)^2} = k \frac{e^2}{n^4 a_0^2}$

This shows that the force $F$ is inversely proportional to $n^4$, indicating that as the principal quantum number $n$ increases, the force acting on the electrons decreases. Thus, the force acting on the electrons in a hydrogen atom, according to Bohr's theory, is related to the principal quantum number $n$ as $n^{-4}$.

Therefore, the correct option is:

Option A: $n^{-4}$

Amplitude Modulation (AM) is a modulation technique used in electronic communication, most commonly for transmitting information via a radio carrier wave. In AM, the amplitude (the strength or intensity) of the carrier wave is varied in proportion to that of the message signal, such as an audio signal. This variation in amplitude represents the information or data that is being transmitted.

Given the options:

Option A suggests that only the frequency of the carrier wave changes according to the information signal, which is not true for amplitude modulation. In frequency modulation (FM), it is the frequency that changes according to the information signal, not the amplitude.

Option B correctly describes amplitude modulation. In AM, the amplitude of the carrier wave changes in accordance with the amplitude of the information signal, while the frequency of the carrier wave remains constant.

Option C suggests that both amplitude and frequency of the carrier wave change according to the information signal. This description does not fit AM but rather describes more complex modulation schemes where both amplitude and frequency can be varied, such as amplitude and frequency modulation techniques used together or other forms of digital modulation.

Option D suggests that in amplitude modulation, both the amplitude and frequency of the carrier wave do not change according to the information signal. This option is incorrect because, in AM, it is specifically the amplitude of the carrier wave that changes to encode information.

Therefore, the correct answer is:

Option B: amplitude of carrier wave changes according to information signal

The fundamental frequency of a vibrating string can be expressed as:

$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

where:

• $f$ is the frequency of vibration,
• $L$ is the length of the string,
• $T$ is the tension in the string, and
• $\mu$ is the linear mass density of the string (mass per unit length).

From the given question, we have two strings with lengths $l$ and $2l$ vibrating at fundamental frequencies $N$ Hz and $1.5N$ Hz, respectively. Let's denote the tension in the shorter string as $T_1$ and in the longer string as $T_2$. Since the strings are identical, their linear mass densities ($\mu$) are the same.

The frequency of the shorter string ($l$) is given as:

$N = \frac{1}{2l} \sqrt{\frac{T_1}{\mu}}$

The frequency of the longer string ($2l$) is given as:

$1.5 N = \frac{1}{2(2l)} \sqrt{\frac{T_2}{\mu}} = \frac{1}{4l} \sqrt{\frac{T_2}{\mu}}$

We can manipulate these equations to solve for the ratio of tensions $T_1$ and $T_2$. First, rearrange the formula of the frequency for $T_1$ and $T_2$:

For the small string:

$T_1 = \mu N^2 (2l)^2 = 4l^2\mu N^2$

For the large string:

$T_2 = \mu (1.5N)^2 (4l)^2 =16l^2 \mu (1.5N)^2 = 16l^2 \mu (2.25N^2) = 36l^2\mu N^2$

The ratio of tensions $T_1 : T_2$ can be found as:

$\frac{T_1}{T_2} = \frac{4l^2\mu N^2}{36l^2\mu N^2} = \frac{4}{36} = \frac{1}{9}$

Therefore, the ratio of tensions for smaller length to larger length is $1:9$, which corresponds to Option B.

The motion of metal sphere is shown as

where, $F_B=$ Buoyant force and $F_V=$ viscous force. At terminal velocity, the net force on the sphere is zero.

$$\begin{aligned} W & =F_B+F_V \\ \Rightarrow \quad M g & =d_2 V g+F_V \\ & =d_2 \frac{M}{d_1} g+F_V \\ \therefore \quad F_V & =M g\left(1-\frac{d_2}{d_1}\right) \end{aligned}$$

To find the ratio of velocities of the emitted photoelectrons when the energy of the incident photon is $3 W$ and then $5 W$, where $W$ is the work function of the metal, we can use the photoelectric equation. The kinetic energy (KE) of the emitted photoelectron can be described by the equation:

$KE = h\nu - W$

where $h\nu$ is the energy of the incident photon and $W$ is the work function of the metal.

For two different incident photon energies $3W$ and $5W$, let's denote the kinetic energies of the emitted photoelectrons as $KE_1$ and $KE_2$, respectively:

$KE_1 = 3W - W = 2W$

$KE_2 = 5W - W = 4W$

The kinetic energy is also given by the equation $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the photoelectron and $v$ is its velocity. Thus, for the two cases above, we have:

$\frac{1}{2}mv_1^2 = 2W$

$\frac{1}{2}mv_2^2 = 4W$

Dividing the second equation by the first gives:

$\frac{\frac{1}{2}mv_2^2}{\frac{1}{2}mv_1^2} = \frac{4W}{2W}$

Simplifying this, we get:

$\frac{v_2^2}{v_1^2} = 2$

Taking the square root of both sides to find the ratio of the velocities $v_2:v_1$, we find:

$\frac{v_2}{v_1} = \sqrt{2}$

Therefore, the correct answer is $1: \sqrt{2}$, which corresponds to Option C.

The conductivity of a semiconductor is given by

$\sigma=n_e e \mu_e+n_h e \mu_h$

where, $n_e$ & $n_h=$ number of electrons and holes per unit volume

and $\mu_e$ & $\mu_h=$ mobility of electrons and holes.

When impurity atoms are added in a semiconductor, then the concentration of holes and electrons increases. So, the conductivity increases and hence the resistivity $\left(\rho=\frac{1}{\sigma}\right)$ of semiconductor decreases.

To find the peak value of the output current in a common emitter amplifier, first, we need to understand how the input signal voltage is related to the base current, and how the base current is amplified to find the collector current (output current). The given parameters are:

• Input resistance, $R_{in} = 1000 \Omega$
• Peak value of input signal voltage, $V_{in(peak)} = 5 \mathrm{~mV}$
• Current gain, $\beta = 60$

The base current ($I_B$) can be calculated using Ohm's law. The input voltage is related to the base current through the input resistance as follows:

$I_{B(peak)} = \frac{V_{in(peak)}}{R_{in}}$

Plugging in the values:

$I_{B(peak)} = \frac{5 \mathrm{~mV}}{1000 \Omega} = \frac{5 \times 10^{-3} \mathrm{~V}}{1000 \Omega} = 5 \times 10^{-6} \mathrm{~A}$

Now, to find the peak value of the output current ($I_C(peak)$), we use the current gain ($\beta$). The collector current is related to the base current as follows:

$I_C = \beta \times I_B$

Thus, for the peak values:

$I_{C(peak)} = \beta \times I_{B(peak)}$

$I_{C(peak)} = 60 \times 5 \times 10^{-6} \mathrm{~A} = 3 \times 10^{-4} \mathrm{~A}$

Therefore, the correct answer is:

Option A: $3 \times 10^{-4} \mathrm{~A}$

Given, $G=20 \Omega$ and $I_g=1 \mathrm{~mA}=1 \times 10^{-3} \mathrm{~A}$

The equivalent resistance of 4 resistors of $16 \Omega$ each, connected in parallel is

$R_{\text {eq }}=\frac{R}{n}=\frac{16}{4}=4 \Omega$

The circuit diagram of an ammeter is

So, the maximum current that can be measured,

$I=\frac{I_g G+I_g R_{\mathrm{eq}}}{R_{\mathrm{eq}}}=\frac{I_g\left(G+R_{\mathrm{eq}}\right)}{R_{\mathrm{eq}}}$

$$\begin{aligned} & =\frac{1 \times 10^{-3}(20+4)}{4} \\ & =6 \times 10^{-3} \mathrm{~A} \text { or } 6 \mathrm{~mA} \end{aligned}$$

$$\begin{aligned} & \text {As, we know, } C_p=C_v+R \\ & \Rightarrow \quad \frac{C_p}{C_V}=1+\frac{R}{C_v} \\ \end{aligned}$$

Here, $\frac{R}{C_v}=0.4$ and also, $\frac{C_p}{C_v}=1+\frac{2}{f}$

$$\begin{aligned} \Rightarrow \quad 1+\frac{2}{f} & =1+0.4 \\ \text{or}\quad \frac{2}{f} & =0.4 \\ \text{or} \quad f & =\frac{2}{0.4}=5\end{aligned}$$

As, degree of freedom is 5, so the gas is made up of molecules which are rigid diatomic (no vibrations).