MHT CET 2021 22nd September Evening Shift (Physics & Chemistry) Question Paper with Solutions

To answer this question, we need to understand the relationship between moment of inertia, angular momentum, and kinetic energy in the context of rotational motion. The conservation of angular momentum plays a crucial role in this problem.

First, we know that the angular momentum (L) is conserved when there is no external torque acting on the system. Angular momentum can be defined as the product of the moment of inertia (I) and the angular velocity (ω), given by the formula:

$L = I\omega$

Given in the question, when the child stretches his arms out, the moment of inertia of the system becomes double. If we let the initial moment of inertia be $I$, then the final moment of inertia after the child stretches his arms becomes $2I$.

Since angular momentum is conserved and initially, the system had a certain angular momentum $L = I\omega$, after the change, it still maintains the same angular momentum, now with the new moment of inertia, leading to a new angular velocity $\omega'$ such that:

$L = I\omega = 2I\omega'$

From this equation, we solve for $\omega'$ to find that:

$\omega' = \frac{\omega}{2}$

This indicates that the angular velocity is halved when the moment of inertia is doubled.

The kinetic energy (K) of a rotating body is given by:

$K = \frac{1}{2}I\omega^2$

Initially, the kinetic energy is:

$K = \frac{1}{2}I\omega^2$

After the moment of inertia is doubled and the angular velocity is halved:

$K' = \frac{1}{2}(2I)\left(\frac{\omega}{2}\right)^2$

$K' = \frac{1}{2} \cdot 2I \cdot \frac{\omega^2}{4}$

$K' = \frac{1}{2}I\omega^2 \cdot \frac{1}{2}$

$K' = \frac{1}{2}K$

Therefore, the kinetic energy of the system after the child stretches his arms out, resulting in the moment of inertia becoming double, is $\frac{K}{2}$.

The correct answer is Option A $\frac{K}{2}$.

To find the ratio of the speeds $V_A$ to $\mathrm{V}_{\mathrm{B}}$ of two particles $A$ and $B$ after falling through the same electric potential difference, we should consider the work done by the electric field on each particle due to the potential difference. This work done will convert into kinetic energy for each particle assuming they start from rest.

The work done on a charge by an electric potential difference $V$ is given by the formula:

where

• $W$ is the work done or energy gained,
• $q$ is the charge of the particle, and
• $V$ is the electric potential difference.

Since the initial kinetic energy of both particles is zero (as they are initially at rest), the entire work done on each will convert into its kinetic energy. The kinetic energy ($K$) of a particle is given by:

where

• $m$ is the mass of the particle, and
• $v$ is its speed.

Thus, for particle $A$ with charge $+q$,

And for particle $B$ with charge $+4q$,

Now, to find the ratio of their speeds, $V_A$ to $V_B$, we can set up a ratio using the above equations. Dividing the equation for particle $A$ by the equation for particle $B$,

Simplifying the equation gives,

Taking the square root of both sides,

This means the ratio of their speeds $V_A : V_B$ is $1:2$.

Therefore, the correct answer is:

Option A: $1: 2$.

Energy of the photon is given by

$\mathrm{E}=\mathrm{h} \nu$

The frequency of photon does not change hence the energy does not change.

$T_1=2 \pi \sqrt{\frac{m}{k_1}} \quad T_2=2 \pi \sqrt{\frac{m}{k_2}}$

When the two springs are connected in series, the effective spring constant is given by

$\frac{1}{\mathrm{k}}=\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2}$

When connected in series

$$\begin{aligned} \mathrm{T} & =2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \\ \therefore \mathrm{T}^2 & =4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}}=4 \pi^2 \mathrm{~m}\left[\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2}\right]=4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}_1}+4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}_2} \\ & =\mathrm{T}_1^2+\mathrm{T}_2^2 \\ \therefore \mathrm{T} & =\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2} \end{aligned}$$

Magnetic field inside the toroid is given by

$$\begin{aligned} & \mathrm{B}=\mu_0 \mathrm{nI} \\ \therefore & \frac{\mathrm{B}_2}{\mathrm{~B}_1}=\frac{I_2}{\mathrm{I}_1}=3 \end{aligned}$$

(If cross-sectional radius of the solenoid is changed there will be no effect on the magnetic field, assuming the number of turns remains same.)

$\therefore \mathrm{B}_2=3 \mathrm{B}_1=3 \times 0.2=0.6 \mathrm{~T}$

For isothermal process, $\mathrm{P}_1 \mathrm{V}_1=\mathrm{P}_2 \mathrm{V}_2$

$\mathrm{P}_2=\mathrm{P}_1-\frac{\mathrm{P}_1}{10}=\frac{9}{10} \mathrm{P}_1$

$$\begin{aligned} \therefore \mathrm{P}_1 \mathrm{V}_1 & =\frac{9}{10} \mathrm{P}_1 \mathrm{V}_2 \\ \therefore \mathrm{V}_2 & =\frac{10}{9} \mathrm{V}_1 \\ & =1.11 \mathrm{V}_1 \\ & =\mathrm{V}_1+0.11 \mathrm{V}_1 \end{aligned}$$

The critical angle, $\theta$, is the angle of incidence in medium $x$ beyond which light is totally internally reflected within medium $x$ and cannot pass into medium $Y$. This situation occurs when the angle of refraction in medium $Y$ is $90^\circ$.

According to Snell's Law, the relationship between the angles of incidence and refraction and the indices of refraction of the two media is given by:

$n_{x} \sin{\theta} = n_{Y} \sin{90^\circ}$

Where $n_{x}$ and $n_{Y}$ are the indices of refraction of medium $x$ and medium $Y$ respectively. Since $\sin{90^\circ} = 1$, we can simplify this to:

$n_{x} \sin{\theta} = n_{Y}$

The speed of light in any medium is related to the speed of light in a vacuum ($c$) and the index of refraction of that medium ($n$) by the equation $V = \frac{c}{n}$.

Therefore, we can express the indices of refraction in terms of the speeds of light in the respective media:

$\frac{c}{V_{x}} \sin{\theta} = \frac{c}{V_{Y}}$

Rearranging it to solve for $V_{Y}$:

$V_{Y} = \frac{V_{x}}{\sin{\theta}}$

Thus, the correct answer is:

Option D

$\frac{\mathrm{V}_{\mathrm{x}}}{\sin \theta}$

To find the value of $n$ given the two rings of radius 'R' and 'nR' and knowing that the ratio of their moments of inertia is $1:8$, we first establish the formula for the moment of inertia of a ring about an axis passing through its centre and perpendicular to the plane of the ring.

The moment of inertia ($I$) of a ring about an axis passing through its centre and perpendicular to the plane of the ring is given by the formula:

$I = mR^2$

where $m$ is the mass of the ring and $R$ is its radius.

Given the mass per unit length $\lambda$, the mass of the first ring with radius $R$ can be expressed as:

$m_1 = \lambda \cdot 2\pi R$

The moment of inertia of the first ring, therefore, is:

$I_1 = (\lambda \cdot 2\pi R) \cdot R^2 = 2\pi\lambda R^3$

The second ring has a radius of $nR$. Therefore, its mass $m_2$ is:

$m_2 = \lambda \cdot 2\pi(nR)$

And the moment of inertia of the second ring is:

$I_2 = (\lambda \cdot 2\pi(nR)) \cdot (nR)^2 = 2\pi\lambda n^3 R^3$

It's given that the ratio of their moments of inertia is $1:8$, so:

$\frac{I_1}{I_2} = \frac{1}{8}$

Substituting the expressions for $I_1$ and $I_2$, we get:

$\frac{2\pi\lambda R^3}{2\pi\lambda n^3 R^3} = \frac{1}{8}$

Simplifying this expression, we see that $\pi$, $\lambda$, and $R^3$ cancel out, leaving us with:

$\frac{1}{n^3} = \frac{1}{8}$

Solving for $n$, we find:

$n^3 = 8$

Taking the cube root of both sides gives us:

$n = 2$

Therefore, the value of $n$ is $2$, which corresponds to Option A.

The amount of heat radiated by an object is given by the formula:

$Q = E \times A \times t$

where:

• $Q$ is the amount of heat radiated,
• $E$ is the emissive power (in $kcal \ s^{-1} \ m^{-2}$),
• $A$ is the surface area (in $m^2$), and
• $t$ is the time in seconds.

Given that:

• Emissive power, $E = 0.7 \ kcal \ s^{-1} \ m^{-2}$
• Area, $A = 0.04 \ m^2$
• Time, $t = 20 \ s$

Substituting the given values into the formula gives:

$Q = 0.7 \times 0.04 \times 20$

$Q = 0.56 \ kcal$

Hence, the amount of heat radiated in 20 seconds is $0.56 \ k \mathrm{cal}$, which corresponds to Option D.

To find the magnitude of the induced current in the circuit at $t = 0.2$ second, we will use Faraday's law of electromagnetic induction. According to Faraday's law, the induced electromotive force (EMF) in a circuit is equal to the negative rate of change of magnetic flux through the circuit. Mathematically, this is expressed as:

Given that the magnetic flux, $\phi=5t^2-6t+9$, we first need to find its derivative with respect to time $(t)$ to get the EMF:

Now, substituting $t = 0.2$ seconds into the derivative, we get the EMF at $t = 0.2$:

The magnitude of EMF is 4 V (we consider magnitude so we'll not worry about the negative sign which indicates direction).

Using Ohm's Law, we find the induced current, $I$, in a circuit with the resistance $R$, as:

Given that the resistance $R = 20 \, \Omega$, we substitute the EMF value to find the induced current:

Therefore, the magnitude of the induced current at $t = 0.2$ second is 0.2 A, which corresponds to Option C.

The correct statement regarding semiconductors is that their valence band and conduction band are separated by a small energy gap. Therefore, the correct choice is:

Option B: is separated by small energy gap.

In semiconductors, the energy gap between the valence band (where the electrons occupy at lower energy levels) and the conduction band (where electrons can move freely, allowing the material to conduct electricity) is relatively small, typically less than 3 eV. This small energy gap allows electrons to be easily excited from the valence band to the conduction band with an increase in temperature or upon exposure to light, making semiconductors capable of conducting electricity under certain conditions.

Option A is incorrect because a large energy gap is characteristic of insulators, not semiconductors. Insulators have energy gaps usually greater than 3 eV, which prevents electrons from easily moving to the conduction band, thereby inhibiting electrical conductivity.

Option C is incorrect because it doesn't accurately describe the state of the bands in semiconductors. The valence band in semiconductors is typically full, and the conduction band can have electrons, albeit significantly less than metals.

Option D is incorrect because it describes a characteristic of metals or some special kinds of materials like semimetals or graphene, where valence and conduction bands overlap, allowing for free movement of electrons and therefore good electrical conductivity. This is not the case for semiconductors.

The intensity is given by

$\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}$

Maximum intensity $\mathrm{K}=4 \mathrm{I}_0$ when $\phi=0$ when path difference is $\frac{\lambda}{3}, \phi=\frac{2 \pi}{3}$

$\therefore \quad \mathrm{I}=\mathrm{k} \cos ^2 \frac{2 \pi}{3}=\mathrm{k}\left(-\frac{1}{2}\right)^2=\frac{\mathrm{k}}{4}$

$$\begin{aligned} & \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T} \\ & =\mathrm{n} \frac{\mathrm{R}}{\gamma-1} \Delta \mathrm{T} \\ & =\frac{\mathrm{P} \Delta \mathrm{V}}{\gamma-1}=\frac{\mathrm{PV}}{\gamma-1} \\ & \because \Delta \mathrm{V}=2 \mathrm{V}-\mathrm{V} \\ & \end{aligned}$$

The correct option for the velocity acquired by an electron when it travels a distance '$L$' in a uniform electric field of strength '$E$' is found by considering the work done by the electric field on the electron and equating it to the kinetic energy gained by the electron.

The work done by the electric field on the electron is given by the product of the force ($F$) exerted by the electric field on the electron, and the distance ($L$) over which this force acts. The force exerted by the electric field on a charge is given by $F = qE$, where $q$ is the charge of the electron, and $E$ is the strength of the electric field. Therefore, the work done ($W$) by the electric field can be expressed as:

$W = F \cdot L = qEL$

According to the principle of conservation of energy, the work done by the electric field is converted into kinetic energy ($K$) of the electron. The kinetic energy of an object with mass $m$ and velocity $v$ is given by $K = \frac{1}{2}mv^2$. Therefore, we equate the work done to the kinetic energy to find the velocity:

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

Solving this equation for $v$ to find the velocity of the electron after traversing the distance $L$ yields:

$v = \sqrt{\frac{2qEL}{m}}$

Hence, the correct answer is

Option C: $\sqrt{\frac{2qEL}{m}}$

For first resonance $L_1=\frac{\lambda}{4}$

For second resonance $\mathrm{L}_2=\frac{3 \lambda}{4}$

$$\begin{aligned} \therefore \quad & \mathrm{L}_2-\mathrm{L}_1=\frac{\lambda}{2} \text { or } \lambda=2\left(\mathrm{~L}_2-\mathrm{L}_1\right) \\ & \mathrm{V}=\mathrm{n} \lambda=2 \mathrm{n}\left(\mathrm{L}_2-\mathrm{L}_1\right) \end{aligned}$$

The weight of the needle is balanced by the force due to surface tension

$$\begin{aligned} \mathrm{W} & =2 \mathrm{TL} \\ & =2 \times 70 \times 7 \\ & =980 \text { dyne } \\ & =1 \mathrm{~g} \mathrm{~wt} \end{aligned}$$

The thermal resistance $R$ of a material is given by the ratio of the temperature difference ($\Delta T$) across the material to the rate of heat flow ($Q$) through the material. Mathematically, this can be expressed as:

$R = \frac{\Delta T}{Q}$

For this particular problem, we are given:

• Temperature difference, $\Delta T = 28^{\circ} \mathrm{C}$
• Rate of heat flow, $Q = 1400 \mathrm{~cal} \mathrm{~s}^{-1}$

Using the formula for thermal resistance, we can calculate $R$ as follows:

$R = \frac{28^{\circ} \mathrm{C}}{1400 \mathrm{~cal} \mathrm{~s}^{-1}}$

$R = 0.02 \frac{{ }^{\circ} \mathrm{C} \mathrm{s}}{\mathrm{cal}}$

Therefore, the thermal resistance of the copper rod is $0.02 \frac{{ }^{\circ} \mathrm{C} \mathrm{s}}{\mathrm{cal}}$. The correct answer is Option B.

To find the ratio of the kinetic energy (K.E.) to the total energy (E) of an electron in the $n^{\text {th }}$ Bohr orbit, let's first recall the expressions for kinetic energy, potential energy, and total energy in a Bohr orbit.

The kinetic energy of an electron in the $n^{\text {th }}$ orbit is given by:

$K.E. = \frac{k e^2}{2r_n}$

where $k$ is Coulomb's constant, $e$ is the charge of an electron, and $r_n$ is the radius of the $n^{\text {th }}$ orbit.

The potential energy (P.E.) of an electron in that orbit (due to the electrostatic force of attraction between the positively charged nucleus and the negatively charged electron) is:

$P.E. = - \frac{k e^2}{r_n}$

This expression is negative because the potential energy in an electrostatically bound system is considered zero at an infinite distance and negative within the system due to the attraction.

The total energy (E) of an electron in the $n^{\text {th }}$ Bohr orbit is the sum of its kinetic energy and potential energy:

$E = K.E. + P.E. = \frac{k e^2}{2r_n} - \frac{k e^2}{r_n}$

$E = - \frac{k e^2}{2r_n}$

Notice that the total energy is $- \frac{1}{2}$ times the kinetic energy and also half of the potential energy in magnitude but with a negative sign, indicating that the electron is bound to the nucleus.

To find the ratio of kinetic energy to total energy:

$\frac{K.E.}{E} = \frac{\frac{k e^2}{2r_n}}{-\frac{k e^2}{2r_n}}$

This simplifies to:

$\frac{K.E.}{E} = -1$

Therefore, the correct answer is Option B), which represents the ratio of the kinetic energy of an electron to the total energy of it in the $n^{\text {th }}$ Bohr orbit as $1: -1$.

We have $\frac{1}{2} m v_{\max }^2=\frac{h c}{\lambda}-\phi=\frac{h c-\phi \lambda}{\lambda}$

$$\begin{array}{ll} \therefore \quad \mathrm{V}_{\max }^2=\frac{2(\mathrm{hc}-\phi \lambda)}{\mathrm{m} \lambda} \\ \therefore \quad\mathrm{V}_{\max }=\sqrt{\frac{2(\mathrm{hc}-\phi \lambda)}{\mathrm{m} \lambda}} \end{array}$$

When two spherical conductors are charged to the same potential, the relationship between the potential ($V$), surface charge density ($\sigma$), and radius ($R$) of the conductor can be utilized to find the ratio of the surface charge densities $\sigma_1$ and $\sigma_2$. The electric potential on the surface of a charged spherical conductor is given by:

$V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{R}$

Where:

• $V$ is the potential,
• $Q$ is the total charge,
• $R$ is the radius of the sphere,
• $\epsilon_0$ is the permittivity of free space.

The surface charge density ($\sigma$) is defined as the charge per unit area, so:

$\sigma = \frac{Q}{4\pi R^2}$

Combining these expressions, we can express the potential in terms of the surface charge density as follows:

$V = \frac{\sigma R}{\epsilon_0}$

Given that the two spheres are charged to the same potential ($V_1 = V_2 = V$), and the radii are $R_1 = 4\, \mathrm{cm}$ and $R_2 = 5\, \mathrm{cm}$, we can set up the equality:

$\frac{\sigma_1 R_1}{\epsilon_0} = \frac{\sigma_2 R_2}{\epsilon_0}$

Upon simplifying, $\epsilon_0$ cancels out on both sides, allowing us to solve for the ratio of $\sigma_1$ to $\sigma_2$ directly:

$\sigma_1 R_1 = \sigma_2 R_2$

Thus,

$\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}$

Plugging in the given radii values:

$\frac{\sigma_1}{\sigma_2} = \frac{5}{4}$

This corresponds to the ratio of surface charge densities:

$\sigma_1 : \sigma_2 = 5 : 4$

Therefore, the correct answer is:

Option A $5: 4$

The frequency of a wave on a string depends on the tension in the string and the linear mass density (mass per unit length) of the string. The formula for the frequency, when we consider the tension $T$ and the linear mass density $\mu$, is given by:

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

where:

• $f$ is the frequency of the wave,


• $L$ is the length of the vibrating segment of the string,


• $T$ is the tension in the string,


• $\mu$ is the linear mass density of the string ($\mu = \frac{m}{L}$, with $m$ being the mass and $L$ the length).

When the radius of the string is doubled, its cross-sectional area becomes four times larger (since the area of a circle is $A = \pi r^2$, doubling $r$ increases $A$ by a factor of 4). This implies that the mass of a given length of the string also becomes four times larger, and therefore, the linear mass density ($\mu$) increases by a factor of 4, because the length of the string we are examining hasn't changed.

Now, taking this into account, the new linear mass density is $4\mu$, and substituting this into the frequency formula gives us:

$f_{new} = \frac{1}{2L} \sqrt{\frac{T}{4\mu}} = \frac{1}{2L} \cdot \frac{1}{2} \sqrt{\frac{T}{\mu}} = \frac{1}{2} f$

This shows that the frequency of the wave, and therefore the number of waves produced per second (since frequency and the number of waves per second are equivalent), will be halved if the radius of the string is doubled while keeping the tension the same. The correct answer is thus:

Option B $\frac{\mathrm{n}}{2}$

The ball is moving with constant velocity. Hence the net force acting on it is zero.

The weight of the ball $=\mathrm{W}=\frac{4}{3} \pi \mathrm{r}^3 \rho_{\mathrm{b}} \mathrm{g}$ (in downward direction)

The viscous force $\mathrm{F}_{\mathrm{v}}$ is in downward direction

The buoyant force $\mathrm{F}_{\mathrm{B}}=\frac{4}{3} \pi \mathrm{r}^3 \rho_{\ell} \mathrm{g} \quad$ (in upward direction)

$$\begin{aligned} & \therefore \mathrm{F}_{\mathrm{v}}+\mathrm{W}=\mathrm{F}_{\mathrm{B}} \\ & \therefore \mathrm{F}_{\mathrm{v}}=\mathrm{F}_{\mathrm{B}}-\mathrm{W} \\ & =\frac{4}{3} \pi r^3 g\left(\rho_{\ell}-\rho_{\mathrm{b}}\right)=\frac{4}{3} \pi \mathrm{r}^3 \mathrm{~g} \times 3 \rho_{\mathrm{b}} \quad\left(\because \rho_l=4 \rho_{\mathrm{b}}\right) \\ & \therefore \frac{F_v}{W}=3 \\ \end{aligned}$$

Let's analyze the relationship between the energy (E) and angular momentum (L) of an electron in the $n^{\text{th}}$ Bohr orbit according to Bohr's theory.

The energy of an electron in the $n^{\text{th}}$ Bohr orbit is given by:

$E = -\frac{k e^2}{2r_n}$

where $k$ is a constant, $e$ is the electronic charge, and $r_n$ is the radius of the $n^{\text{th}}$ orbit.

The angular momentum ($L$) of the electron in the $n^{\text{th}}$ Bohr orbit is quantized and given by:

$L = n\hbar$

where $\hbar$ is the reduced Planck's constant, and $n$ is the principal quantum number.

The radius of the $n^{\text{th}}$ orbit ($r_n$) according to Bohr's model is given by:

$r_n = n^2 \frac{\hbar^2}{k' m_e e^2}$

where $m_e$ is the mass of the electron and $k'$ is another constant.

By substituting the expression for $r_n$ in the equation for energy $E$, we get:

$E = -\frac{k e^2}{2 \left(n^2 \frac{\hbar^2}{k' m_e e^2}\right)}$

$E \propto -\frac{1}{n^2}$

Now, since $L = n\hbar$, it implies that $n \propto L$. Thus, the relationship between energy ($E$) and angular momentum ($L$) can be rewritten in terms of $L$:

$E \propto -\frac{1}{L^2}$

Therefore, among the provided options, the correct one is:

Option C: $\mathrm{E} \propto \mathrm{L}^{-2}$

$$\begin{aligned} \mathrm{u} & =15 \mathrm{~m} / \mathrm{s}, \mathrm{a}=-0.3 \mathrm{~m} / \mathrm{s}^2, \mathrm{t}=60 \mathrm{~s} \\ \mathrm{~s} & =\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \\ & =15 \times 60+\frac{1}{2} \times(-0.3) \times(60)^2 \\ & =900-540 \\ & =360 \mathrm{~m} \end{aligned}$$

Distance from traffic light $=400-360=40 \mathrm{~m}$

To find the magnetic field produced at the center of the circular path by a revolving charge, we can use the expression derived from Biot-Savart law for a moving charge. The magnetic field $B$ at the center of a circular loop carrying a steady current $I$ is given by:

$B = \frac{\mu_0 I}{2R}$

However, in this scenario, we don't have a steady current but a charged particle moving in a circular path. The equivalent "current" in this situation is the charge passing a point per unit time. Since the particle completes one revolution per second (1 r.p.s), it's equivalent to having a current where the charge $Q$ passes the point of observation once per second. Therefore, the equivalent current $I$ is the charge divided by the time period $T$ of one revolution, which is 1 second here:

$I = \frac{Q}{T} = \frac{100e}{1 \, \mathrm{s}} = \frac{100 \times 1.6 \times 10^{-19} \, \mathrm{C}}{1 \, \mathrm{s}} = 1.6 \times 10^{-17} \, \mathrm{A}$

Given the radius $R = 0.8 \, \mathrm{m}$ and substituting the values into the equation for $B$:

$B = \frac{\mu_0 I}{2R} = \frac{\mu_0 \times 1.6 \times 10^{-17}}{2 \times 0.8}$

Simplifying this:

$B = \frac{\mu_0 \times 1.6 \times 10^{-17}}{1.6} = \mu_0 \times 10^{-17}$

Therefore, the magnetic field produced at the center of the circle in SI unit is:

$10^{-17} \mu_0$

Which corresponds to Option A.

The charge oscillates according to the equation

$$\begin{aligned} & q=q_0 \cos \omega t \\ & i=-\frac{d q}{d t}=\omega q_0 \sin \omega t=i_0 \sin \omega t \end{aligned}$$

where $i_0=\omega q_0, \omega=\frac{1}{\sqrt{\mathrm{LC}}}, q_0=C V$

$\therefore \quad i_0=\frac{1}{\sqrt{\mathrm{LC}}} \cdot \mathrm{CV}=\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}$

The field produced by charge on each plate is $\frac{E}{2}$.

Hence force on each plate is given by

$\mathrm{F}=\mathrm{Q} \times$ (Field produced by the other plate)

$=\frac{\mathrm{QE}}{2}$

(I) NAND gate gives output ' 0 ' when inputs are 1, 1

(II) NOR gate gives output ' 1 ' when inputs are 0, 0

(III) AND gate gives output ' 0 ' when inputs are 0, 1

(IV) EX-OR gate gives output ' 1 ' when inputs are 1, 0

$\therefore$ Ans. II and IV

The gravitational acceleration at a distance 'r' from the center of the Earth, where 'r' is greater than the radius of the Earth (R), is given by the formula:

$g' = \frac{GM}{r^2}$

Here,

• $g'$ is the gravitational acceleration at distance 'r' from the center of the Earth,
• $G$ is the universal gravitational constant,
• $M$ is the mass of the Earth,
• $r$ is the distance from the center of the Earth to the point where the gravitational acceleration is being calculated.

In this case, we're looking at a height 'R' above the Earth's surface, so the distance 'r' from the center of the Earth becomes $r = R + R = 2R$. Substituting this into our formula gives us:

$g' = \frac{GM}{(2R)^2}$

$g' = \frac{GM}{4R^2}$

Now, the acceleration due to gravity on the surface of the Earth (g) is given by:

$g = \frac{GM}{R^2}$

Comparing the two expressions, we can now express $g'$ in terms of $g$:

$g' = \frac{g}{4}$

Thus, the correct answer is Option C, which states that the gravitational acceleration at a height 'R' above the Earth's surface is $\frac{g}{4}$. This result demonstrates how gravitational acceleration decreases as one moves away from the surface of the Earth.