MHT CET 2012 Physics Question Paper with Answer and Solution

(B) Impulse is defined as the change in momentum of the object.
Let the initial velocity be $u = 30 \,ms^{-1}$ and the final velocity be $v = -20 \,ms^{-1}$ (since it rebounds in the opposite direction).
The mass of the ball is $m = 0.1 \,kg$.
Impulse $J = \Delta p = m(v - u)$.
Taking the direction of the initial velocity as positive:
$J = m(v_{final} - v_{initial}) = 0.1 \times (-20 - 30) = 0.1 \times (-50) = -5 \,N-s$.
The magnitude of the impulse exerted by the wall on the ball is $|J| = 5 \,N-s$.

(D) When an elevator moves vertically upward with an acceleration $a$,the effective weight of the passenger increases.
According to Newton's second law of motion,the forces acting on the passenger are the normal force $R$ (exerted by the floor) acting upwards and the gravitational force $m g$ acting downwards.
The equation of motion is $R - m g = m a$.
Therefore,the normal force exerted by the floor on the passenger is $R = m g + m a$.

(B) Let the mass of the fireman be $m$ and his acceleration be $a$. The weight of the fireman is $mg$.
The breaking strength of the rope is given as $\frac{2}{3}mg$.
When the fireman slides down with acceleration $a$,the tension $T$ in the rope is given by $T = m(g - a)$.
For the rope not to break,the tension $T$ must be less than or equal to the breaking strength.
To find the minimum acceleration $a$,we set the tension equal to the breaking strength:
$m(g - a) = \frac{2}{3}mg$
Dividing both sides by $m$:
$g - a = \frac{2}{3}g$
$a = g - \frac{2}{3}g = \frac{g}{3}$
Therefore,the minimum acceleration is $\frac{g}{3}$.

(A) When the car moves with an acceleration $a$ in the forward direction,a pseudo force $F_p = ma$ acts on the plumb bob in the backward direction in the non-inertial frame of the car.
Let $m$ be the mass of the bob. The forces acting on the bob are:
$1$. Weight $mg$ acting vertically downwards.
$2$. Pseudo force $ma$ acting horizontally backwards.
$3$. Tension $T$ in the string.
In the equilibrium position relative to the car,the net force on the bob is zero. Resolving the forces,we have:
$T \sin \theta = ma$
$T \cos \theta = mg$
Dividing the two equations,we get:
$\frac{T \sin \theta}{T \cos \theta} = \frac{ma}{mg}$
$\tan \theta = \frac{a}{g}$
Therefore,the angle made by the string with the vertical is $\theta = \tan ^{-1}\left(\frac{a}{g}\right)$.

(B) The system consists of two blocks $A$ and $B$ connected by a string. Since the table is frictionless and the system is pulled with an acceleration $a = 2 \,ms^{-2}$, we can analyze the motion of block $A$ separately.
Block $A$ is being pulled by the tension $T$ in the string. According to Newton's second law of motion, the force acting on block $A$ is given by:
$T = M_A \times a$
Given:
$M_A = 20 \,kg$
$a = 2 \,ms^{-2}$
Substituting the values:
$T = 20 \,kg \times 2 \,ms^{-2} = 40 \,N$
Therefore, the tension in the string is $40 \,N$.

(A) The plane rises $5$ in $12$,which means the height is $5$ and the length of the incline is $12$.
Therefore,$\sin \theta = \frac{5}{12}$.
Since the body just slides down,the angle of inclination $\theta$ is equal to the angle of repose.
The coefficient of friction $\mu$ is given by $\mu = \tan \theta$.
We know that $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (\frac{5}{12})^2} = \sqrt{1 - \frac{25}{144}} = \sqrt{\frac{119}{144}} = \frac{\sqrt{119}}{12}$.
Thus,$\mu = \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{5/12}{\sqrt{119}/12} = \frac{5}{\sqrt{119}}$.
Calculating the value,$\sqrt{119} \approx 10.9087$.
$\mu = \frac{5}{10.9087} \approx 0.4583 \approx 0.46$.

(B) Let the length of the incline be $L$ and the angle of inclination be $\theta = 45^{\circ}$.
For a smooth incline,the acceleration is $a_1 = g \sin \theta$. The time taken is $t_1 = \sqrt{\frac{2L}{g \sin \theta}}$.
For a rough incline,the acceleration is $a_2 = g(\sin \theta - \mu \cos \theta)$. The time taken is $t_2 = \sqrt{\frac{2L}{g(\sin \theta - \mu \cos \theta)}}$.
Given that $t_2 = n t_1$,we have $\frac{t_2}{t_1} = n$.
Squaring both sides,$\frac{t_2^2}{t_1^2} = n^2$,which implies $\frac{\sin \theta}{\sin \theta - \mu \cos \theta} = n^2$.
Rearranging the terms: $\sin \theta = n^2 \sin \theta - n^2 \mu \cos \theta$.
$n^2 \mu \cos \theta = (n^2 - 1) \sin \theta$.
$\mu = \frac{n^2 - 1}{n^2} \tan \theta$.
Since $\theta = 45^{\circ}$,$\tan 45^{\circ} = 1$.
Therefore,$\mu = 1 - \frac{1}{n^2}$.

(C) When a body is projected up an inclined plane,both the component of gravity acting down the plane and the frictional force act in the direction opposite to the motion.
Retardation $a = g \sin \theta + \mu g \cos \theta = g(\sin \theta + \mu \cos \theta)$.
Given $\theta = 45^{\circ}$ and $\mu = 0.5$.
Substituting the values:
$a = g(\sin 45^{\circ} + 0.5 \cos 45^{\circ})$
$a = g\left(\frac{1}{\sqrt{2}} + 0.5 \times \frac{1}{\sqrt{2}}\right)$
$a = \frac{g}{\sqrt{2}}(1 + 0.5)$
$a = \frac{1.5 g}{\sqrt{2}} = \frac{3 g}{2 \sqrt{2}}$.

(A) Let $\alpha$ be the angle of inclination of the plane.
Case $1$: Force $F$ acting along the plane.
For the weight to be just supported,$F = W \sin \alpha - f_s$,where $f_s$ is the static friction. At the point of impending motion,$f_s = \mu R = \mu W \cos \alpha = W \tan \theta \cos \alpha$.
Thus,$F = W \sin \alpha - W \tan \theta \cos \alpha = W \frac{\sin(\alpha - \theta)}{\cos \theta}$.
Case $2$: Force $F$ acting horizontally.
For the weight to be just supported,$F \cos \alpha = W \sin \alpha + f_s$. At the point of impending motion,$f_s = \mu R = \mu (W \cos \alpha + F \sin \alpha) = \tan \theta (W \cos \alpha + F \sin \alpha)$.
Solving for $F$,we get $F = W \tan(\alpha + \theta)$.
Since the problem states that the weight can be supported by the same force $F$ in both cases,we equate the two expressions for $F$:
$W \frac{\sin(\alpha - \theta)}{\cos \theta} = W \tan(\alpha + \theta)$.
This implies $\sin(\alpha - \theta) = \sin(\alpha + \theta) \cdot \frac{\cos \theta}{\cos(\alpha + \theta)}$.
Solving this trigonometric equation leads to the condition where $F/W = \tan \theta$.

(D) Given: Mass $m = 4 \times 10^{7} \,kg$, Force $F = 5 \times 10^{4} \,N$, Initial velocity $u = 0$, Distance $s = 4 \,m$.
Using Newton's second law, $F = ma$:
$5 \times 10^{4} = 4 \times 10^{7} \times a$
$a = \frac{5 \times 10^{4}}{4 \times 10^{7}} = 1.25 \times 10^{-3} \,ms^{-2}$.
Using the kinematic equation $v^{2} = u^{2} + 2as$:
$v^{2} = 0^{2} + 2 \times (1.25 \times 10^{-3}) \times 4$
$v^{2} = 2 \times 1.25 \times 4 \times 10^{-3} = 10 \times 10^{-3} = 10^{-2} \,m^{2}s^{-2}$.
$v = \sqrt{10^{-2}} = 0.1 \,ms^{-1}$.

(C) When a sudden jerk is given to string $C$,an impulsive force is applied to it.
Because the mass has inertia,it resists the sudden change in motion.
This impulsive tension develops in string $C$ first,exceeding its breaking strength.
Since the impulse takes finite time to propagate through the mass to string $A$,string $C$ breaks before the tension in string $A$ can increase significantly.

(B) The change in momentum $\Delta p$ is given by the final momentum minus the initial momentum.
Initial momentum $p_i = m \times v = 2 \,kg \times 100 \,ms^{-1} = 200 \,kg \cdot ms^{-1}$.
Since the body rebounds with the same speed in the opposite direction,the final momentum $p_f = m \times (-v) = 2 \,kg \times (-100 \,ms^{-1}) = -200 \,kg \cdot ms^{-1}$.
The change in momentum $\Delta p = p_f - p_i = -200 - 200 = -400 \,kg \cdot ms^{-1}$.
The magnitude of the change in momentum is $|\Delta p| = 400 \,kg \cdot ms^{-1}$.
The force exerted on the wall is given by Newton's Second Law: $F = \frac{|\Delta p|}{\Delta t}$.
Given $\Delta t = 1/50 \,s$,we have $F = \frac{400}{1/50} = 400 \times 50 = 20,000 \,N = 2 \times 10^{4} \,N$.

(C) According to Stokes' Law,the viscous force $F$ acting on a spherical body of radius $r$ moving with terminal velocity $v$ is given by $F = 6 \pi \eta r v$.
For a body falling in a viscous fluid,the terminal velocity $v$ is proportional to the square of the radius,i.e.,$v \propto r^2$.
Substituting this into the force equation: $F \propto r \cdot r^2 = r^3$.
Since the volume $V$ of a sphere is proportional to the cube of its radius $(V \propto r^3)$,we have $F \propto V$.
Therefore,if the volume of the ball is doubled $(V' = 2V)$,the viscous force acting on it will also be doubled $(F' = 2F)$.

(D) The kinematic equation $x=ut+\frac{1}{2}at^2$ describes the motion of an object under constant acceleration.
This equation is derived from the definitions of velocity and acceleration using calculus or graphical methods.
It is a kinematic relationship and does not directly follow from Newton's laws of motion,which relate force,mass,and acceleration $(F=ma)$.
Therefore,the correct option is $D$.

(B) Inductive reactance $(X_L)$ is defined as the opposition offered by an inductor to the flow of alternating current.
It is given by the formula $X_L = \omega L = 2\pi f L$.
Since it represents the opposition to current flow in an $AC$ circuit, it is analogous to resistance.
Therefore, the $SI$ unit of inductive reactance is the same as that of resistance, which is the ohm $(\Omega)$.

(B) At resonance, the inductive reactance $(X_L)$ is equal to the capacitive reactance $(X_C)$, i.e., $X_L = X_C$.
Therefore, the net impedance $(Z)$ of the $L-C-R$ circuit is equal to the resistance $(R)$, i.e., $Z = R$.
The power factor $(\cos \phi)$ is defined as the ratio of resistance to impedance: $\cos \phi = \frac{R}{Z}$.
Substituting $Z = R$, we get $\cos \phi = \frac{R}{R} = 1$.
Thus, at resonance, the $L-C-R$ circuit behaves as a purely resistive circuit, and the power factor is $1$.

(A) The resonant frequency of an $L-C-R$ circuit is given by the formula:
$v_{0} = \frac{1}{2 \pi \sqrt{L C}}$
This implies that $v_{0} \propto \frac{1}{\sqrt{L C}}$.
If the inductance $L$ is doubled $(L' = 2L)$ and the capacitance $C$ is doubled $(C' = 2C)$,the new resonant frequency $v_{0}'$ will be:
$v_{0}' = \frac{1}{2 \pi \sqrt{(2L)(2C)}} = \frac{1}{2 \pi \sqrt{4LC}} = \frac{1}{2 \pi \cdot 2 \sqrt{LC}}$
$v_{0}' = \frac{1}{2} \left( \frac{1}{2 \pi \sqrt{LC}} \right) = \frac{1}{2} v_{0}$
Therefore,the resonant frequency will decrease to one-half of the original value.

(A) In an $L-C$ circuit,resonance occurs when the inductive reactance $(X_L)$ equals the capacitive reactance $(X_C)$.
$X_L = X_C$
$\omega L = \frac{1}{\omega C}$
$\omega^2 = \frac{1}{LC}$
$\omega = \frac{1}{\sqrt{LC}}$
Since $\omega = 2 \pi f$,the resonant frequency $f$ is given by:
$f = \frac{1}{2 \pi \sqrt{LC}}$

(B) The brightness of the lamp depends on the current flowing through the circuit.
In an $RC$ series circuit,the capacitive reactance is given by $X_{C} = \frac{1}{\omega C}$.
When the capacitance $C$ is reduced,the capacitive reactance $X_{C}$ increases.
The total impedance of the circuit is $Z = \sqrt{R^2 + X_{C}^2}$.
As $X_{C}$ increases,the total impedance $Z$ of the circuit increases.
According to Ohm's law for $AC$ circuits,the current $I = \frac{V}{Z}$.
Since $Z$ increases,the current $I$ flowing through the lamp decreases.
Therefore,the lamp shines less brightly.

(D) In a purely inductive circuit,the voltage $V$ leads the current $i$ by a phase angle of $\pi / 2$ radians $(90^{\circ})$.
Conversely,this means the current $i$ lags behind the voltage $V$ by $\pi / 2$ radians.
This is represented by the phasor diagram where $V$ is along the positive x-axis and $i$ is along the negative y-axis.

(A) The alternating current is represented by the equation $I = I_0 \sin(\omega t)$.
To find the average value over a complete cycle of period $T$,we integrate the current over the interval $[0, T]$ and divide by the total time $T$.
$I_{avg} = \frac{1}{T} \int_{0}^{T} I_0 \sin(\omega t) dt$.
Since $\omega = \frac{2\pi}{T}$,the integral of $\sin(\omega t)$ over a full period is zero because the positive area of the sine wave cancels out the negative area.
Therefore,the average value of alternating current over a complete cycle is $0$.

(B) The instantaneous current is given by the equation $I = I_{0} \sin (\omega t + \phi)$,where $I_{0}$ is the peak current.
Comparing this with the given equation $I = 5 \sin (\omega t + \phi)$,we find the peak current $I_{0} = 5 \text{ A}$.
The root mean square $(rms)$ value of an alternating current is related to the peak current by the formula $I_{rms} = \frac{I_{0}}{\sqrt{2}}$.
Substituting the value of $I_{0}$,we get $I_{rms} = \frac{5}{\sqrt{2}} \text{ A}$.

(D) An alternating quantity varies periodically with time. One complete set of positive and negative values of an alternating quantity is defined as a $cycle$. It represents the complete variation of the quantity before it starts repeating itself.

(B) The root mean square $(I_{\text{rms}})$ value of an alternating current is defined as the square root of the mean of the squares of the instantaneous current over one complete cycle.
For a sinusoidal alternating current given by $I = I_{0} \sin(\omega t)$,the $I_{\text{rms}}$ value is calculated as:
$I_{\text{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} I^{2} dt} = \frac{I_{0}}{\sqrt{2}}$
where $I_{0}$ is the peak value (amplitude) of the current.

(C) The average power consumed in an $AC$ circuit is given by the formula $P = V_{rms} I_{rms} \cos \phi$, where $\cos \phi$ is the power factor.
In a purely inductive circuit, the phase difference $\phi$ between voltage and current is $90^{\circ}$.
Since $\cos 90^{\circ} = 0$, the power consumed becomes $P = V_{rms} I_{rms} \times 0 = 0$.
Therefore, if the inductance is very high (acting as a pure inductor) and the resistance is negligible, the power consumed in the $AC$ circuit is zero.

(D) An alternating voltage is defined as a voltage that changes its magnitude and direction periodically with time. The standard mathematical expression for an alternating voltage is $V(t) = V_m \sin(\omega t)$,where $V_m$ is the peak voltage and $\omega$ is the angular frequency. Since this expression follows a sine function,the alternating voltage varies sinusoidally with time.

(B) The standard frequency of alternating current $(AC)$ mains supplied in India is $50 \text{ Hz}$.
Frequency is defined as the number of cycles per second,which is equivalent to $50 \text{ cycles/second}$ or $50 \text{ s}^{-1}$.

(A) choke coil is an inductor with high inductance and low resistance. It is used to control the current in $AC$ circuits without dissipating significant power as heat. In $DC$ circuits,an inductor acts as a simple wire with negligible resistance,so it cannot be used to control current effectively. Therefore,a choke coil is used as a resistance (impedance) in $AC$ circuits.

(C) In $AC$ circuits,a resistor dissipates energy in the form of heat due to the $I^2R$ loss,regardless of the phase difference.
However,an ideal choke coil is an inductor with negligible resistance.
In an ideal inductor,the phase difference between voltage and current is $90^{\circ}$,which makes the power factor $\cos(90^{\circ}) = 0$.
Therefore,the average power consumed by an ideal choke coil is $P = V_{rms} I_{rms} \cos(90^{\circ}) = 0$.
Thus,a choke coil controls the current in an $AC$ circuit without wasting electrical energy as heat.

(B) $AC$ measuring instruments,such as $AC$ ammeters and voltmeters,are designed to measure the root mean square $(rms)$ value of the alternating current or voltage. This is because the heating effect of current,which is the principle behind most analog measuring instruments,is proportional to the square of the current,making the $rms$ value the most physically significant quantity for power calculations.

(D) The power factor of an $AC$ circuit is given by $\cos \phi = \frac{R}{Z}$,where $R$ is the resistance and $Z$ is the impedance.
When the power factor is unity,$\cos \phi = 1$,which implies $\phi = 0^\circ$.
Since $\cos \phi = \frac{R}{Z} = 1$,we get $R = Z$.
This condition occurs at resonance in an $LCR$ circuit,where the inductive reactance $X_L$ and capacitive reactance $X_C$ cancel each other out $(X_L = X_C)$.
Consequently,the circuit behaves as a purely resistive circuit.

(B) The quality factor ($Q$-factor) of a resonant circuit is defined as the ratio of resonant frequency to the bandwidth of the circuit.
Mathematically,$Q = \frac{\omega_{0} L}{R}$.
At resonance,the angular frequency is given by $\omega_{0} = \frac{1}{\sqrt{LC}}$.
Substituting the value of $\omega_{0}$ into the $Q$-factor formula:
$Q = \frac{1}{\sqrt{LC}} \cdot \frac{L}{R} = \frac{1}{R} \cdot \frac{\sqrt{L}}{\sqrt{C}} = \frac{1}{R} \sqrt{\frac{L}{C}}$.

(D) Mutual inductance is the phenomenon where a change in current in one coil induces an electromotive force $(EMF)$ in a neighboring coil. $A$ transformer consists of two coils,the primary and the secondary,wound on a common core. When an alternating current flows through the primary coil,it creates a changing magnetic flux that links with the secondary coil,inducing an $EMF$ in it. Therefore,the transformer operates on the principle of mutual inductance.

(B) transformer is an electrical device that operates on the principle of mutual induction. It is used to change the magnitude of alternating voltage and current. Specifically,it converts high alternating voltage at low current into low alternating voltage at high current (step-down) or vice-versa (step-up),while keeping the frequency constant.

(B) transformer operates on the principle of mutual induction.
Mutual induction requires a changing magnetic flux,which is produced by a time-varying current.
Since $AC$ (Alternating Current) changes its magnitude and direction periodically,it produces a changing magnetic flux in the primary coil,which induces an $EMF$ in the secondary coil.
$DC$ (Direct Current) is constant and does not produce a changing magnetic flux; therefore,a transformer cannot work on $DC$.

(B) step-down transformer is a device used to decrease the voltage of an alternating current. According to the transformer equation,$\frac{V_s}{V_p} = \frac{N_s}{N_p}$,where $V_s$ and $V_p$ are the voltages of the secondary and primary coils,and $N_s$ and $N_p$ are the number of turns in the secondary and primary coils,respectively. For a step-down transformer,the output voltage $V_s$ is less than the input voltage $V_p$. Therefore,the number of turns in the primary coil $(N_p)$ must be greater than the number of turns in the secondary coil $(N_s)$.

(B) The energy stored in an inductor is given by the formula $U = \frac{1}{2} L i^2$.
When a coil is connected to a battery of $EMF$ $E$ and resistance $R$,the steady-state current $i$ is given by $i = \frac{E}{R}$.
Substituting the values: $E = 10 \ V$,$L = 4 \ H$,and $R = 5 \ \Omega$.
First,calculate the steady-state current: $i = \frac{10 \ V}{5 \ \Omega} = 2 \ A$.
Now,calculate the stored energy: $U = \frac{1}{2} \times 4 \ H \times (2 \ A)^2$.
$U = 2 \times 4 \ J = 8 \ J$.

(C) An inductor stores energy in the form of a magnetic field created by the current flowing through it.
The energy $U$ stored in an inductor is given by the formula $U = \frac{1}{2} L I^{2}$,where $L$ is the self-inductance of the inductor and $I$ is the current flowing through it.
Therefore,the correct option is $C$.

(A) Lenz's law states that the direction of the induced current is such that it opposes the change in magnetic flux that produced it. If the induced current were to assist the change,it would violate the law of conservation of energy,as it would create energy out of nothing. Therefore,Lenz's law is a direct consequence of the law of conservation of energy.

(A) The principle of electromagnetic induction,specifically Faraday's law,states that a change in magnetic flux linked with a coil induces an electromotive force $(emf)$ in it.
This principle is the fundamental working mechanism of an electric generator,which converts mechanical energy into electrical energy by rotating a coil within a magnetic field.
While motors and galvanometers also involve magnetic fields,their primary operating principle is the magnetic effect of current (Lorentz force),not electromagnetic induction.

(A) The back emf $e$ in a $DC$ motor is given by the relation $e \propto \omega$,where $\omega$ is the angular velocity of the motor armature.
Since the back emf is directly proportional to the angular velocity,it reaches its maximum value when the motor attains its maximum speed.

(D) The induced emf $(e)$ in a coil due to self-induction is given by the formula: $e = L \frac{di}{dt}$.
Here, the induced emf $e = 2 \,V$ and the rate of change of current $\frac{di}{dt} = 4 \,A s^{-1}$.
Substituting these values into the formula:
$2 = L \times 4$
$L = \frac{2}{4} \,H$
$L = 0.5 \,H$.
Therefore, the self-inductance of the coil is $0.5 \,H$.

(C) The mutual inductance $M$ between two coils is related to their self-inductances $L_{1}$ and $L_{2}$ by the formula $M = k\sqrt{L_{1} L_{2}}$,where $k$ is the coefficient of coupling.
Given that the flux in one coil is completely linked with the other,the coupling is perfect,meaning $k = 1$.
Therefore,the expression simplifies to $M = \sqrt{L_{1} L_{2}}$.

(A) The induced $emf$ $(e)$ in a coil is given by the formula: $e = L \frac{di}{dt}$.
Here,$L_1 = 8 \ mH$ and $L_2 = 2 \ mH$.
The rate of change of current,$\frac{di}{dt}$,is the same for both coils.
Therefore,the ratio of the induced $emf$s is:
$\frac{e_1}{e_2} = \frac{L_1 (di/dt)}{L_2 (di/dt)} = \frac{L_1}{L_2}$.
Substituting the given values:
$\frac{e_1}{e_2} = \frac{8 \ mH}{2 \ mH} = \frac{4}{1}$.
Thus,the ratio is $4: 1$.

(B) The electromagnetic spectrum is categorized based on wavelength ranges.
- The wavelength range of $100 \text{ Å}$ to $400 \text{ Å}$ falls within the ultraviolet $(UV)$ region.
- $X$-rays typically range from $0.01 \text{ Å}$ to $100 \text{ Å}$.
- The visible region ranges from approximately $4000 \text{ Å}$ to $7000 \text{ Å}$.
- The infrared region starts above $7000 \text{ Å}$.
Therefore,the correct classification for the range $100 \text{ Å}$ to $400 \text{ Å}$ is the $UV$ region.

(C) Electromagnetic waves are waves that are created as a result of vibrations between an electric field and a magnetic field. They do not require a medium to travel.
Light rays,$X$-rays,and $Gamma$ rays are all parts of the electromagnetic spectrum.
Alpha rays consist of high-energy helium nuclei ($He^{2+}$ particles),which are charged particles with mass. Therefore,alpha rays are particle radiation,not electromagnetic waves.

(C) The existence of electromagnetic waves was first experimentally confirmed by the German physicist Heinrich Hertz in $1887$. He used a spark-gap transmitter to produce and detect electromagnetic waves,thereby validating Maxwell's theoretical predictions.

(B) Electromagnetic waves travel in free space or vacuum with the velocity of light,which is approximately $3 \times 10^{8} \ m/s$.

(A) The relationship between the speed of light $(c)$, frequency $(\nu)$, and wavelength $(\lambda)$ is given by the formula: $c = \nu \lambda$.
Given:
Wavelength $\lambda = 10 \, m$.
Speed of light $c = 3 \times 10^{8} \, m/s$.
Rearranging the formula to solve for frequency: $\nu = \frac{c}{\lambda}$.
Substituting the values: $\nu = \frac{3 \times 10^{8} \, m/s}{10 \, m} = 3 \times 10^{7} \, Hz$.

(C) According to electromagnetic theory,light is an electromagnetic wave. It consists of time-varying electric and magnetic fields that oscillate in space and time. These electric and magnetic field vectors are mutually perpendicular to each other and also perpendicular to the direction of propagation of the wave.