AIPMT 1997 Physics Question Paper with Answer and Solution

(A) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:
$\vec{\tau} = \vec{r} \times \vec{F}$
Given $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$,we calculate the determinant:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix}$
Expanding the determinant:
$\vec{\tau} = \hat{i}(3 \times 5 - 1 \times 1) - \hat{j}(7 \times 5 - 1 \times (-3)) + \hat{k}(7 \times 1 - 3 \times (-3))$
$\vec{\tau} = \hat{i}(15 - 1) - \hat{j}(35 + 3) + \hat{k}(7 + 9)$
$\vec{\tau} = 14\hat{i} - 38\hat{j} + 16\hat{k}$

(B) Given: Initial velocity $u = 0 \, m/s$,final velocity $v = 144 \, km/h = 144 \times \frac{5}{18} \, m/s = 40 \, m/s$,and time $t = 20 \, s$.
Using the first equation of motion,$v = u + at$:
$40 = 0 + a \times 20 \Rightarrow a = 2 \, m/s^2$.
Now,using the second equation of motion,$s = ut + \frac{1}{2}at^2$:
$s = 0 \times 20 + \frac{1}{2} \times 2 \times (20)^2 = 400 \, m$.
Therefore,the distance covered is $400 \, m$.

(C) The velocity $v$ is the first derivative of position $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}(at^2 - bt^3) = 2at - 3bt^2$
The acceleration $a_{acc}$ is the derivative of velocity $v$ with respect to time $t$:
$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(2at - 3bt^2) = 2a - 6bt$
To find the time when acceleration is zero,set $a_{acc} = 0$:
$2a - 6bt = 0$
$6bt = 2a$
$t = \frac{2a}{6b} = \frac{a}{3b}$

(B) The initial kinetic energy of the ball is $E = \frac{1}{2}mv^2$,where $v$ is the initial velocity.
At the highest point of the trajectory,the vertical component of the velocity becomes zero,while the horizontal component remains constant.
The horizontal component of velocity is $v_x = v \cos \theta$.
At the highest point,the velocity of the ball is $v_h = v \cos \theta$.
The kinetic energy at the highest point is $E' = \frac{1}{2}m(v_h)^2 = \frac{1}{2}m(v \cos \theta)^2$.
$E' = (\frac{1}{2}mv^2) \cos^2 \theta = E \cos^2 \theta$.
Given $\theta = 45^\circ$,we have $\cos 45^\circ = \frac{1}{\sqrt{2}}$.
Therefore,$E' = E (\frac{1}{\sqrt{2}})^2 = E (\frac{1}{2}) = \frac{E}{2}$.

(C) The formula for work done is given by $W = Fs \cos \theta$,where $W$ is work,$F$ is force,$s$ is displacement,and $\theta$ is the angle between the force and the direction of motion.
Given: $W = 25 \, J$,$F = 5 \, N$,and $s = 10 \, m$.
Substituting the values into the formula:
$25 = 5 \times 10 \times \cos \theta$
$25 = 50 \times \cos \theta$
$\cos \theta = \frac{25}{50} = \frac{1}{2}$
Since $\cos 60^\circ = \frac{1}{2}$,the angle $\theta = 60^\circ$.

(C) The relationship between linear momentum $(p)$,mass $(m)$,and kinetic energy $(K.E.)$ is given by the formula: $p = \sqrt{2m(K.E.)}$.
Since the kinetic energy $(K.E.)$ is equal for both bodies,we have $p \propto \sqrt{m}$.
Therefore,the ratio of their linear momentums is $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$.
Given $m_1 = m$ and $m_2 = 4m$,we substitute these values:
$\frac{p_1}{p_2} = \sqrt{\frac{m}{4m}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(B) Initial velocity of the first ball,$u_1 = 36 \,km/h = 36 \times \frac{5}{18} = 10 \,m/s$.
Initial velocity of the second ball,$u_2 = 0 \,m/s$.
Masses are $m_1 = 2 \,kg$ and $m_2 = 3 \,kg$.
By the law of conservation of linear momentum,$m_1 u_1 + m_2 u_2 = (m_1 + m_2)V$,where $V$ is the common velocity after the collision.
$2 \times 10 + 3 \times 0 = (2 + 3)V \implies 20 = 5V \implies V = 4 \,m/s$.
Initial kinetic energy,$K_i = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} \times 2 \times (10)^2 + 0 = 100 \,J$.
Final kinetic energy,$K_f = \frac{1}{2} (m_1 + m_2) V^2 = \frac{1}{2} \times 5 \times (4)^2 = \frac{1}{2} \times 5 \times 16 = 40 \,J$.
Loss in kinetic energy,$\Delta K = K_i - K_f = 100 - 40 = 60 \,J$.

(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
From this formula,we can see that $v_e \propto \sqrt{\frac{M}{R}}$.
Let the initial mass be $M$ and radius be $R$. The initial escape velocity is $v_{e1} = 11.2 \, km/s$.
According to the problem,the new mass $M' = 2M$ and the new radius $R' = \frac{R}{2}$.
The new escape velocity $v_{e2}$ is given by:
$v_{e2} = \sqrt{\frac{2G(2M)}{R/2}} = \sqrt{4 \cdot \frac{2GM}{R}} = 2 \cdot \sqrt{\frac{2GM}{R}} = 2 \cdot v_{e1}$.
Substituting the value of $v_{e1}$:
$v_{e2} = 2 \cdot 11.2 \, km/s = 22.4 \, km/s$.

(B) In a thermodynamic process,the work done by a gas is equal to the area under the $PV$ curve with respect to the volume axis.
For a given expansion from volume ${V_1}$ to ${V_2}$,the pressure $P$ remains constant in an isobaric process,whereas it decreases in both isothermal and adiabatic processes.
Since the isobaric process maintains the highest pressure throughout the expansion,the area under the $PV$ curve is the largest.
Therefore,the work done follows the order: ${W_{adiabatic}} < {W_{isothermal}} < {W_{isobaric}}$.
Thus,the work done is greatest for an isobaric expansion.

(C) The length of the string is $l = 10 \; m$. The number of segments is $p = 5$. The wave velocity is $v = 20 \; m/s$.
For a string vibrating in $p$ segments,the length $l$ is related to the wavelength $\lambda$ by the formula $l = p \cdot \frac{\lambda}{2}$.
Substituting the values,we get $10 = 5 \cdot \frac{\lambda}{2}$,which simplifies to $\lambda = \frac{10 \cdot 2}{5} = 4 \; m$.
The frequency $f$ is given by the relation $f = \frac{v}{\lambda}$.
Substituting the values,$f = \frac{20 \; m/s}{4 \; m} = 5 \; Hz$.

(B) The fundamental frequency of an open tube of length $L$ is given by $f_0 = \frac{v}{2L}$,where $v$ is the speed of sound in air.
When the tube is dipped vertically into water such that half of its length is inside the water,the tube effectively becomes a closed pipe (closed at one end by the water surface) with a new length $L' = L/2$.
The fundamental frequency of a closed pipe of length $L'$ is given by $f' = \frac{v}{4L'}$.
Substituting $L' = L/2$ into the formula,we get $f' = \frac{v}{4(L/2)} = \frac{v}{2L}$.
Comparing this with the original frequency,we find $f' = f_0$.

(B) couple is defined as a pair of two equal,parallel,and opposite forces acting at different points on a rigid body. Since the net force acting on the body is zero $(F_{net} = F - F = 0)$,there is no linear acceleration or linear motion. However,because the forces act at different points,they create a net torque about any point,which results in purely rotational motion.

(A) Given: Force $F = 6 \; N$,mass $m = 1 \; kg$,initial velocity $u = 0 \; m/s$,and final velocity $v = 30 \; m/s$.
According to Newton's second law of motion,the acceleration $a$ is given by $a = \frac{F}{m} = \frac{6 \; N}{1 \; kg} = 6 \; m/s^2$.
Using the first equation of motion,$v = u + at$,we can find the time $t$:
$30 = 0 + 6 \times t$
$t = \frac{30}{6} = 5 \; s$.
Therefore,the force acts on the body for $5 \; seconds$.

(A) According to Stefan-Boltzmann Law,the rate of energy emission (power) $P$ from a black body is directly proportional to the fourth power of its absolute temperature $T$.
Mathematically,$P \propto T^{4}$.
Given the temperature $T = 500 \; K$,the rate of energy emission is proportional to $(500)^{4}$.

(B) Let the two rectangular simple harmonic motions be represented as:
$x = a \sin(\omega t)$
$y = a \sin(\omega t + \frac{\pi}{2})$
Since $\sin(\omega t + \frac{\pi}{2}) = \cos(\omega t)$,we have:
$y = a \cos(\omega t)$
Squaring and adding both equations:
$x^2 + y^2 = a^2 \sin^2(\omega t) + a^2 \cos^2(\omega t)$
$x^2 + y^2 = a^2 (\sin^2(\omega t) + \cos^2(\omega t))$
$x^2 + y^2 = a^2$
This is the equation of a circle with radius $a$ centered at the origin. Therefore,the resultant motion is circular.

(D) The position of the centre of mass of a system of particles is defined by the formula $\vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$.
From this formula,it is clear that the centre of mass depends on the masses $(m_i)$ and the positions $(\vec{r}_i)$ of the particles.
The relative distance between particles is a function of their positions.
However,the centre of mass is a geometric property of the distribution of mass and does not depend on the external or internal forces acting on the particles.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
This implies $T \propto \sqrt{l}$.
Taking the natural logarithm on both sides: $\ln T = \ln(2\pi) + \frac{1}{2} \ln l - \frac{1}{2} \ln g$.
Differentiating both sides, we get the fractional change: $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l}$.
Given that the length is increased by $2\%$, we have $\frac{\Delta l}{l} = 0.02$.
Substituting this value: $\frac{\Delta T}{T} = \frac{1}{2} \times 0.02 = 0.01$.
Therefore, the time period increases by $1\%$.

(B) vector quantity is a physical quantity that has both magnitude and direction.
Distance,heat,and energy are scalar quantities because they only have magnitude.
Angular momentum is defined as the cross product of position vector $\vec{r}$ and linear momentum $\vec{p}$,given by $\vec{L} = \vec{r} \times \vec{p}$.
Since it is defined by a cross product and has a specific direction (determined by the right-hand rule),angular momentum is a vector quantity.

(B) When a shell explodes in flight,the explosion is caused by internal forces.
According to the law of conservation of linear momentum,if the net external force acting on a system is zero,the total linear momentum of the system remains conserved.
In this case,the gravitational force is an external force,but during the short interval of the explosion,the internal forces are much larger than the external forces,making the impulse due to external forces negligible.
Therefore,the total linear momentum of the shell is conserved.
Kinetic energy is generally not conserved in an explosion because internal chemical energy is converted into kinetic energy,leading to an increase in the total kinetic energy of the fragments.

(C) The temperatures of the reservoirs are given as $T_{1} = 100^{\circ}C$ and $T_{2} = -23^{\circ}C$.
First,convert these temperatures to the Kelvin scale:
$T_{1} = 100 + 273 = 373\,K$
$T_{2} = -23 + 273 = 250\,K$
The efficiency $\eta$ of a Carnot engine is given by the formula:
$\eta = 1 - \frac{T_{2}}{T_{1}}$
Substituting the values:
$\eta = \frac{T_{1} - T_{2}}{T_{1}} = \frac{373 - 250}{373}$

(C) Let $T_A$ and $T_B$ be the time periods of planet $A$ and $B$ around the sun respectively. Given that $T_A = 8 T_B$.
According to Kepler's third law of planetary motion,the square of the time period is proportional to the cube of the semi-major axis (distance from the sun),i.e.,$T^2 \propto R^3$.
Therefore,$\left(\frac{T_A}{T_B}\right)^2 = \left(\frac{R_A}{R_B}\right)^3$.
Substituting the given values:
$\left(\frac{8 T_B}{T_B}\right)^2 = \left(\frac{R_A}{R_B}\right)^3$
$8^2 = \left(\frac{R_A}{R_B}\right)^3$
$64 = \left(\frac{R_A}{R_B}\right)^3$
Taking the cube root on both sides:
$\frac{R_A}{R_B} = (64)^{1/3} = 4$.
Thus,the distance of planet $A$ from the sun is $4$ times the distance of planet $B$ from the sun.

(B) The unit of magnetic field strength or magnetic induction $(B)$ in the $SI$ system is the Tesla $(T)$. One Tesla is defined as one Weber per square meter $(1 \ T = 1 \ Wb/m^2)$. Therefore,Tesla is the unit for measuring magnetic induction.

(D) In the given circuit,the battery is connected to two branches in parallel. One branch contains a single $3\,\Omega$ resistor,and the other branch contains two $3\,\Omega$ resistors in series.
$1$. The resistance of the first branch is $R_1 = 3\,\Omega$.
$2$. The resistance of the second branch is $R_2 = 3\,\Omega + 3\,\Omega = 6\,\Omega$.
$3$. These two branches are in parallel,so the equivalent resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$
$\Rightarrow R_{eq} = 2\,\Omega$.
$4$. Using Ohm's law,the total current $I$ supplied by the $2\,V$ battery is:
$I = \frac{V}{R_{eq}} = \frac{2\,V}{2\,\Omega} = 1\,A$.

(A) Kirchhoff's first law,also known as the Kirchhoff's Current Law $(KCL)$,states that the algebraic sum of currents meeting at a junction in an electric circuit is zero,i.e.,$\Sigma i = 0$.
This law implies that the total charge entering a junction must equal the total charge leaving the junction in the same time interval.
Since electric charge cannot be created or destroyed at a junction,this law is a direct consequence of the law of conservation of charge.

(C) The resistance $R$ is defined by the relation $R = V/I$. In an $I-V$ graph,the slope of the curve is given by $dI/dV$,which is equal to $1/R$.
For negative resistance,the slope $dI/dV$ must be negative.
Looking at the graph,in the portion $CD$,as the voltage $V$ increases,the current $I$ decreases.
Therefore,the slope $dI/dV$ is negative in the region $CD$,which corresponds to negative resistance.

(A) The unit of energy is kilowatt hour $(kWh)$.
$1 \; kWh = 1 \; kW \times 1 \; h$
Since $1 \; kW = 1000 \; W$ and $1 \; h = 3600 \; s$,
$1 \; kWh = 1000 \; W \times 3600 \; s = 3,600,000 \; W \cdot s$.
Since $1 \; W \cdot s = 1 \; J$,we have $1 \; kWh = 3.6 \times 10^6 \; J$.
This can also be written as $36 \times 10^5 \; J$.

(A) In a series circuit,the current $I$ flowing through both bulbs is the same.
The power dissipated in a resistor is given by the formula $P = I^2 R$.
Since $I$ is constant for both bulbs,the power dissipated is directly proportional to the resistance,i.e.,$P \propto R$.
Therefore,the ratio of the powers dissipated is equal to the ratio of their resistances:
$\frac{P_1}{P_2} = \frac{R_1}{R_2} = \frac{1}{2}$.

(A) The resistance $R$ of the bulb is calculated using its rated power $P_R$ and rated voltage $V_R$:
$R = \frac{V_R^2}{P_R} = \frac{200^2}{100} = \frac{40000}{100} = 400\,\Omega$.
When connected to a supply voltage $V = 160\, V$,the power consumed $P$ is given by:
$P = \frac{V^2}{R} = \frac{160^2}{400} = \frac{25600}{400} = 64\, W$.
Alternatively,using the ratio formula:
$P_{consumed} = \left( \frac{V}{V_R} \right)^2 \times P_R = \left( \frac{160}{200} \right)^2 \times 100 = (0.8)^2 \times 100 = 0.64 \times 100 = 64\, W$.

(D) The magnetic field $B$ at a distance $r$ from a long straight current-carrying wire is given by the formula $B = \frac{\mu_0}{4\pi} \frac{2i}{r}$.
Here,$\mu_0$ is the permeability of free space,$i$ is the current flowing through the wire,and $r$ is the perpendicular distance from the wire.
As per the formula,the magnetic field $B$ depends only on the current $i$ and the distance $r$ from the wire.
It is independent of the radius or diameter of the wire.
Since the current $i$ remains the same and the distance $r$ is considered far away (implying the same observation point),the magnetic field strength remains unchanged.

(C) According to the Lorentz force formula,$\vec{F} = q(\vec{v} \times \vec{B})$.
Here,the velocity $\vec{v}$ is directed towards the east,and the magnetic field $\vec{B}$ is directed vertically upwards.
Using the right-hand rule for the cross product $\vec{v} \times \vec{B}$,the force $\vec{F}$ acts in the horizontal plane towards the north.
Since the magnetic force is always perpendicular to the velocity of the particle,it does no work on the particle $(W = \vec{F} \cdot \vec{d} = 0)$.
Consequently,the kinetic energy and the speed of the particle remain constant.
$A$ particle moving perpendicular to a uniform magnetic field follows a circular path with a constant speed.

(A) The force per unit length between two parallel current-carrying wires is given by the formula: $F/L = \frac{\mu_0}{4\pi} \frac{2i_1 i_2}{d}$.
Given: $\mu_0/4\pi = 10^{-7}\,T\cdot m/A$,$i_1 = i_2 = 10\,A$,and $d = 10\,cm = 0.1\,m$.
Substituting the values: $F/L = 10^{-7} \times \frac{2 \times 10 \times 10}{0.1} = 10^{-7} \times \frac{200}{0.1} = 10^{-7} \times 2000 = 2 \times 10^{-4}\,N/m$.
Since the currents are in the same direction,the force between the wires is attractive.

(C) The earth's magnetic field lines are perpendicular to the surface at the poles and parallel to the surface at the equator. Charged particles moving towards the equator experience a magnetic force $F = q(v \times B)$ that deflects them. To overcome this magnetic influence and reach the equatorial region,the charged particles must possess higher kinetic energy compared to those reaching the polar regions.

(A) The transformation ratio of a transformer is given by the formula: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.
Given:
Primary turns $(N_p)$ = $500$
Secondary turns $(N_s)$ = $5000$
Primary voltage $(V_p)$ = $20\, V$
Substituting the values:
$\frac{V_s}{20} = \frac{5000}{500}$
$\frac{V_s}{20} = 10$
$V_s = 200\, V$.
In a transformer,the frequency of the output voltage remains the same as the input frequency because the magnetic flux changes at the same rate as the input current. Therefore,the frequency is $50\, Hz$.

(D) The instantaneous power in an $ac$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\phi$ is the phase difference between voltage and current.
Since the power depends on the power factor $\cos \phi$,the power dissipated in the circuit depends on the phase difference between $V$ and $I$.
Therefore,the correct option is $D$.

(A) The kinetic energy $(K)$ gained by an electron accelerated through a potential difference $(V)$ is given by the formula $K = e \cdot V$.
Here,the charge of an electron $e = 1.602 \times 10^{-19} \, C$ and the potential difference $V = 100 \, V$.
Substituting these values into the formula:
$K = (1.602 \times 10^{-19} \, C) \times (100 \, V)$
$K = 1.602 \times 10^{-17} \, J$.
Therefore,the correct option is $A$.

(C) According to the experimental observations of the photoelectric effect,the photoelectric current is directly proportional to the intensity of the incident light,provided the frequency of the light is above the threshold frequency.
$1$. The frequency of light determines the kinetic energy of the emitted photoelectrons,not the number of electrons (current).
$2$. The photocurrent saturates at a certain voltage,so it is not directly proportional to the applied voltage for all values.
$3$. The stopping potential depends on the frequency of the incident light,not its intensity.
Therefore,the correct statement is that the photocurrent increases with increasing intensity of light.

(D) The penetrating power of radiation depends on its energy and mass. $\alpha -$ particles are heavy and have low penetrating power. $\beta -$ rays are lighter and have more penetrating power than $\alpha -$ particles. $\gamma -$ rays are high-energy electromagnetic waves with no mass and no charge,giving them the highest penetrating power among the options provided. Therefore,$\gamma -$ rays are the most penetrating.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = \frac{-13.6 \ eV}{n^2}$.
For the first excited state,the quantum number is $n = 2$.
Substituting $n = 2$ into the formula:
$E_2 = \frac{-13.6 \ eV}{2^2} = \frac{-13.6 \ eV}{4} = -3.4 \ eV$.

(B) The radius of an electron orbit in a hydrogen atom is given by the formula $r_n = n^2 a_0$,where $n$ is the principal quantum number and $a_0$ is the Bohr radius.
For the ground state,$n = 1$.
For the first excited state,$n = 2$.
Substituting $n = 2$ into the formula,we get $r_2 = (2)^2 a_0 = 4 a_0$.
Therefore,the radius of the first excited state is $4$ times the Bohr radius.

(B) The correct answer is $(b)$. $A$ neutron moderator is a medium that reduces the speed of fast neutrons,thereby turning them into thermal neutrons capable of sustaining a nuclear chain reaction involving uranium$-235$.
Heavy water $(D_2O)$ serves as an effective neutron moderator in a nuclear reactor. It slows down fast-moving neutrons through elastic collisions,which increases the probability of these neutrons causing further fission in uranium nuclei.
Explanation:
In a fission reaction,neutrons are released with high kinetic energy. These fast neutrons are less likely to cause further fission. By using a moderator like heavy water,the neutrons lose kinetic energy and become 'thermal' or 'slow' neutrons,which are much more efficient at sustaining the chain reaction.

(C) To obtain a $P$-type semiconductor,the intrinsic semiconductor (like Germanium or Silicon) must be doped with a trivalent impurity atom.
Among the given options,Arsenic $(As)$,Antimony $(Sb)$,and Phosphorus $(P)$ are pentavalent elements (Group $15$),which are used to create $N$-type semiconductors.
Indium $(In)$ is a trivalent element (Group $13$),which creates a deficiency of electrons (holes) when added to Germanium,thus forming a $P$-type semiconductor.
Therefore,the correct option is $C$.

(B) The given truth table is:
$A=0, B=0 \implies Y=1$
$A=1, B=0 \implies Y=0$
$A=0, B=1 \implies Y=0$
$A=1, B=1 \implies Y=0$
For a $NOR$ gate,the output is defined by the Boolean expression $Y = \overline{A + B}$.
Calculating for each case:
$1$. For $A=0, B=0$: $Y = \overline{0+0} = \overline{0} = 1$.
$2$. For $A=1, B=0$: $Y = \overline{1+0} = \overline{1} = 0$.
$3$. For $A=0, B=1$: $Y = \overline{0+1} = \overline{1} = 0$.
$4$. For $A=1, B=1$: $Y = \overline{1+1} = \overline{1} = 0$.
This matches the given truth table. Therefore,the correct option is $B$.

(B) The maximum current $I$ that the diode can handle is determined by its maximum power rating $P$ and its constant voltage drop $V_d$.
Given $P = 100\; mW = 100 \times 10^{-3}\; W$ and $V_d = 0.5\; V$.
The maximum current is $I = \frac{P}{V_d} = \frac{100 \times 10^{-3}}{0.5} = 0.2\; A$.
Applying Kirchhoff's Voltage Law to the circuit,the voltage across the resistor $R$ is $V_R = V_{source} - V_d = 1.5\; V - 0.5\; V = 1.0\; V$.
Using Ohm's law,$V_R = I \times R$,so $R = \frac{V_R}{I} = \frac{1.0}{0.2} = 5\; \Omega$.

(A) The speed of light in free space is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
The speed of light in a medium is given by $v = \frac{1}{\sqrt{\mu \varepsilon}}$.
The refractive index $n$ of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium:
$n = \frac{c}{v} = \frac{1/\sqrt{\mu_0 \varepsilon_0}}{1/\sqrt{\mu \varepsilon}} = \sqrt{\frac{\mu \varepsilon}{\mu_0 \varepsilon_0}}$.

(B) When light travels from one medium to another,its frequency $(n)$ remains constant because it depends on the source of the light.
When light enters a medium with refractive index $\mu > 1$,its velocity $(v')$ decreases and is given by $v' = \frac{v}{\mu}$.
Since the relation between velocity,frequency,and wavelength is $v = n\lambda$,and $n$ is constant,the new wavelength $\lambda'$ is given by $\lambda' = \frac{v'}{n} = \frac{v/\mu}{n} = \frac{\lambda}{\mu}$.
Therefore,the new frequency,wavelength,and velocity are $n, \frac{\lambda}{\mu}, \frac{v}{\mu}$ respectively.

(B) According to the lens maker's formula,the focal length $f$ is related to the refractive index $\mu$ as $f \propto \frac{1}{\mu - 1}$.
From Cauchy's dispersion formula,the refractive index $\mu$ is inversely proportional to the wavelength $\lambda$ (i.e.,$\mu \propto \frac{1}{\lambda}$).
Since the wavelength of violet light is the shortest $({\lambda_V} < {\lambda_G} < {\lambda_R})$,the refractive index for violet light is the highest $({\mu_V} > {\mu_G} > {\mu_R})$.
Consequently,the focal length is inversely related to the refractive index,meaning the focal length for violet light is the shortest.
Therefore,the correct relationship is ${f_V} < {f_G} < {f_R}$.

(B) For an astronomical telescope in normal adjustment,the length of the tube is given by $L = f_o + f_e = 44\, cm$.
The angular magnification is given by $|m| = \frac{f_o}{f_e} = 10$,which implies $f_o = 10 f_e$.
Substituting $f_o = 10 f_e$ into the length equation: $10 f_e + f_e = 44$.
$11 f_e = 44$,so $f_e = 4\, cm$.
Therefore,the focal length of the objective is $f_o = 10 \times 4 = 40\, cm$.

(A) The magnetic moment $M$ of a bar magnet is given by the product of its pole strength $m$ and its magnetic length $2l$,so $M = m \times 2l$.
When the magnet is cut into two equal parts perpendicular to its length,the pole strength $m$ of each part remains the same,but the length of each part becomes half,i.e.,$l' = l$.
Therefore,the new magnetic moment $M'$ of each part is $M' = m \times l = (m \times 2l) / 2 = M / 2$.
Thus,$M' = 0.5 M$.

(C) The resistance $R$ of a wire is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
For the three wires,the resistances are:
$R_1 = \rho \frac{l}{A}$
$R_2 = \rho \frac{2l}{A/2} = 4 \rho \frac{l}{A} = 4R_1$
$R_3 = \rho \frac{l/2}{2A} = \frac{1}{4} \rho \frac{l}{A} = 0.25R_1$
Comparing the resistances,$R_3 < R_1 < R_2$.
Therefore,the resistance is minimum for the wire with length $l/2$ and cross-sectional area $2A$.

(D) The current gain $\alpha$ is defined as the ratio of collector current $(I_{C})$ to emitter current $(I_{E})$: $\alpha = \frac{I_{C}}{I_{E}}$.
The current gain $\beta$ is defined as the ratio of collector current $(I_{C})$ to base current $(I_{B})$: $\beta = \frac{I_{C}}{I_{B}}$.
From Kirchhoff's current law for a transistor,the emitter current is the sum of base and collector currents: $I_{E} = I_{B} + I_{C}$.
Substituting $I_{E}$ in the expression for $\alpha$: $\alpha = \frac{I_{C}}{I_{B} + I_{C}}$.
Dividing the numerator and denominator by $I_{C}$: $\alpha = \frac{1}{\frac{I_{B}}{I_{C}} + 1}$.
Since $\frac{1}{\beta} = \frac{I_{B}}{I_{C}}$,we substitute this into the equation: $\alpha = \frac{1}{\frac{1}{\beta} + 1} = \frac{1}{\frac{1+\beta}{\beta}} = \frac{\beta}{1+\beta}$.
Therefore,the correct relation is $\alpha = \frac{\beta}{1+\beta}$.