AIEEE 2005 Physics Question Paper with Answer and Solution

75 QuestionsEnglishWith Solutions

PhysicsQ175 of 75 questions

Page 1 of 1 · English

1
PhysicsMediumMCQAIEEE · 2005
$A$ particle moves towards east with velocity $5\, m/s$. After $10\, s$,its direction changes towards north with the same velocity. The average acceleration of the particle is
A
Zero
B
$\frac{1}{\sqrt{2}}\,m/s^2$ in North-West direction
C
$\frac{1}{\sqrt{2}}\,m/s^2$ in North-East direction
D
$\frac{1}{\sqrt{2}}\,m/s^2$ in South-West direction

Solution

(B) Initial velocity $\vec{v}_i = 5\hat{i}\,m/s$.
Final velocity $\vec{v}_f = 5\hat{j}\,m/s$.
Change in velocity $\Delta \vec{v} = \vec{v}_f - \vec{v}_i = 5\hat{j} - 5\hat{i}$.
Magnitude of change in velocity $|\Delta \vec{v}| = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\,m/s$.
The direction of $\Delta \vec{v}$ is North-West (since it points in the direction of $-\hat{i} + \hat{j}$).
Average acceleration $\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\,m/s^2$.
Thus,the average acceleration is $\frac{1}{\sqrt{2}}\,m/s^2$ in the North-West direction.
2
PhysicsMediumMCQAIEEE · 2005
Out of the following pairs,which one does not have identical dimensions?
A
Moment of inertia and moment of force
B
Work and torque
C
Angular momentum and Planck's constant
D
Impulse and momentum

Solution

(A) To determine which pair does not have identical dimensions,we analyze the dimensional formulas of each:
$1$. Moment of inertia $(I = MR^2)$ has dimensions $[ML^2T^0]$. Moment of force (Torque,$\tau = r \times F$) has dimensions $[ML^2T^{-2}]$. These are not identical.
$2$. Work $(W = F \cdot d)$ has dimensions $[ML^2T^{-2}]$. Torque $(\tau = r \times F)$ has dimensions $[ML^2T^{-2}]$. These are identical.
$3$. Angular momentum $(L = mvr)$ has dimensions $[ML^2T^{-1}]$. Planck's constant $(h = E/f)$ has dimensions $[ML^2T^{-1}]$. These are identical.
$4$. Impulse $(J = F \Delta t)$ has dimensions $[MLT^{-1}]$. Momentum $(p = mv)$ has dimensions $[MLT^{-1}]$. These are identical.
Therefore,the pair that does not have identical dimensions is Moment of inertia and moment of force.
3
PhysicsDifficultMCQAIEEE · 2005
$A$ bullet fired into a fixed target loses half of its velocity after penetrating $3\,cm$. How much further will it penetrate before coming to rest,assuming that it faces constant resistance to motion? (in $cm$)
A
$1.5$
B
$1$
C
$3$
D
$2$

Solution

(B) Let the initial velocity of the bullet be $u$.
After penetrating $3\,cm$,its velocity becomes $u/2$.
Using the equation of motion $v^2 = u^2 - 2as$:
$(u/2)^2 = u^2 - 2a(3)$
$u^2/4 = u^2 - 6a$
$6a = 3u^2/4$
$a = u^2/8$
Let the bullet penetrate an additional distance $x$ before coming to rest.
For this motion,initial velocity is $u/2$,final velocity is $0$,and acceleration is $-a = -u^2/8$.
Using $v^2 = u^2 - 2as$:
$0^2 = (u/2)^2 - 2(u^2/8)x$
$0 = u^2/4 - (u^2/4)x$
$u^2/4 = (u^2/4)x$
$x = 1\,cm$.
Solution diagram
4
PhysicsDifficultMCQAIEEE · 2005
The relation between time $t$ and distance $x$ is given by $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. The retardation of the particle is:
A
$2\alpha v^3$
B
$2\beta v^3$
C
$2\alpha \beta v^3$
D
$2\beta^2 v^3$

Solution

(A) Given the relation: $t = \alpha x^2 + \beta x$
Differentiating with respect to $x$:
$\frac{dt}{dx} = 2\alpha x + \beta$
Since velocity $v = \frac{dx}{dt}$,we have:
$v = \frac{1}{2\alpha x + \beta} \implies 2\alpha x + \beta = \frac{1}{v}$
Acceleration $a$ is given by $a = v \frac{dv}{dx}$.
First,find $\frac{dv}{dx}$ by differentiating $v = (2\alpha x + \beta)^{-1}$ with respect to $x$:
$\frac{dv}{dx} = -1(2\alpha x + \beta)^{-2} \cdot (2\alpha) = -\frac{2\alpha}{(2\alpha x + \beta)^2}$
Substitute $(2\alpha x + \beta) = \frac{1}{v}$ into the expression:
$\frac{dv}{dx} = -2\alpha \cdot v^2$
Now,calculate acceleration:
$a = v \cdot (-2\alpha v^2) = -2\alpha v^3$
Retardation is the negative of acceleration:
$\text{Retardation} = -a = 2\alpha v^3$
5
PhysicsDifficultMCQAIEEE · 2005
$A$ car,starting from rest,accelerates at the rate $f$ through a distance $S$,then continues at constant speed for time $t$,and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is $15S$,then:
A
$S = \frac{1}{2}f{t^2}$
B
$S = \frac{1}{4}f{t^2}$
C
$S = \frac{1}{72}f{t^2}$
D
$S = \frac{1}{6}f{t^2}$

Solution

(C) Let the car start from point $A$ from rest and move up to point $B$ with acceleration $f$ over distance $S$.
The velocity of the car at point $B$ is $v = \sqrt{2fS}$ (using $v^2 = u^2 + 2as$).
The car moves distance $BC = x$ with this constant velocity $v$ in time $t$,so $x = vt = \sqrt{2fS} \cdot t$ ... $(i)$.
At point $C$,the car begins to decelerate at rate $a' = \frac{f}{2}$ until it comes to rest at point $D$. Let the distance $CD = y$.
Using $v^2 = u^2 - 2a'y$,where final velocity is $0$: $0 = v^2 - 2(\frac{f}{2})y \implies y = \frac{v^2}{f} = \frac{2fS}{f} = 2S$ ... (ii).
The total distance $AD = AB + BC + CD = S + x + 2S = 15S$.
$3S + x = 15S \implies x = 12S$.
Substituting $x = 12S$ into equation $(i)$: $12S = \sqrt{2fS} \cdot t$.
Squaring both sides: $144S^2 = 2fS \cdot t^2$.
Dividing by $2S$: $72S = f \cdot t^2 \implies S = \frac{1}{72}ft^2$.
6
PhysicsDifficultMCQAIEEE · 2005
$A$ parachutist after bailing out falls $50\, m$ without friction. When the parachute opens,it decelerates at $2\, m/s^2$. He reaches the ground with a speed of $3\, m/s$. At what height did he bail out (in $, m$)?
A
$293$
B
$111$
C
$91$
D
$182$

Solution

(A) $1$. Free fall phase (from point $A$ to $B$): The parachutist falls freely under gravity for a distance $s_1 = 50\, m$. Initial velocity $u_1 = 0$,acceleration $a_1 = 9.8\, m/s^2$.
The velocity $v$ at point $B$ is given by $v^2 = u_1^2 + 2a_1s_1 = 0 + 2 \times 9.8 \times 50 = 980$.
So,$v = \sqrt{980}\, m/s$.
$2$. Deceleration phase (from point $B$ to $C$): The parachute opens,and the parachutist decelerates at $a_2 = -2\, m/s^2$. The final velocity at the ground is $v_f = 3\, m/s$. Let the distance covered be $h$.
Using $v_f^2 = v^2 + 2a_2h$:
$(3)^2 = 980 + 2(-2)h$
$9 = 980 - 4h$
$4h = 980 - 9 = 971$
$h = 971 / 4 = 242.75\, m$.
$3$. Total height: The total height from which he bailed out is $H = s_1 + h = 50 + 242.75 = 292.75\, m \approx 293\, m$.
Solution diagram
7
PhysicsDifficultMCQAIEEE · 2005
For a given velocity,a projectile has the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then:
A
${t_1}{t_2} \propto {R^2}$
B
${t_1}{t_2} \propto R$
C
${t_1}{t_2} \propto \frac{1}{R}$
D
${t_1}{t_2} \propto \frac{1}{R^2}$

Solution

(B) For a given velocity $u$,the range $R$ is the same for two angles of projection,$\theta$ and $(90^\circ - \theta)$.
The time of flight for angle $\theta$ is $t_1 = \frac{2u \sin \theta}{g}$.
The time of flight for angle $(90^\circ - \theta)$ is $t_2 = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos \theta}{g}$.
Multiplying the two times of flight:
$t_1 t_2 = \left( \frac{2u \sin \theta}{g} \right) \left( \frac{2u \cos \theta}{g} \right) = \frac{4u^2 \sin \theta \cos \theta}{g^2}$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$t_1 t_2 = \frac{2u^2 (2 \sin \theta \cos \theta)}{g^2} = \frac{2(u^2 \sin 2\theta)}{g^2}$.
Since the range $R = \frac{u^2 \sin 2\theta}{g}$,we can substitute this into the equation:
$t_1 t_2 = \frac{2R}{g}$.
Since $g$ is constant,we conclude that $t_1 t_2 \propto R$.
8
PhysicsMediumMCQAIEEE · 2005
$A$ particle of mass $0.3 \, kg$ is subjected to a force $F = -kx$ with $k = 15 \, N/m$. What will be its initial acceleration if it is released from a point $20 \, cm$ away from the origin? (in $m/s^2$)
A
$5$
B
$10$
C
$3$
D
$15$

Solution

(B) Given: Mass $m = 0.3 \, kg$,force constant $k = 15 \, N/m$,and displacement $x = 20 \, cm = 0.2 \, m$.
According to Hooke's Law,the magnitude of the force is $F = kx$.
Substituting the values: $F = 15 \times 0.2 = 3 \, N$.
Using Newton's Second Law,$F = ma$,the acceleration $a$ is given by $a = \frac{F}{m}$.
Substituting the values: $a = \frac{3}{0.3} = 10 \, m/s^2$.
Therefore,the initial acceleration is $10 \, m/s^2$.
9
PhysicsMediumMCQAIEEE · 2005
$A$ block is kept on a frictionless inclined surface with an angle of inclination '$\alpha$'. The incline is given a horizontal acceleration '$a$' to keep the block stationary relative to the incline. Then '$a$' is equal to:
Question diagram
A
$g$
B
$g \tan \alpha$
C
$g / \tan \alpha$
D
$g \csc \alpha$

Solution

(B) Let the mass of the block be $m$. To keep the block stationary relative to the inclined surface,we analyze the forces in the frame of the incline.
$1$. The forces acting on the block are its weight $(mg)$ acting downwards,the normal force $(N)$ perpendicular to the surface,and the pseudo force $(ma)$ acting horizontally in the direction opposite to the acceleration of the incline.
$2$. Resolving the forces along the inclined surface:
- The component of the gravitational force down the incline is $mg \sin \alpha$.
- The component of the pseudo force up the incline is $ma \cos \alpha$.
$3$. For the block to remain stationary relative to the incline,these two forces must balance each other:
$ma \cos \alpha = mg \sin \alpha$
$4$. Dividing both sides by $m \cos \alpha$:
$a = g \frac{\sin \alpha}{\cos \alpha}$
$a = g \tan \alpha$
Solution diagram
10
PhysicsMediumMCQAIEEE · 2005
Consider a car moving on a straight road with a speed of $100\, m/s$. The distance at which the car can be stopped is ........ $m$. $[\mu_k = 0.5, g = 10\, m/s^2]$
A
$100$
B
$400$
C
$800$
D
$1000$

Solution

(D) Given: Initial velocity $u = 100\, m/s$,Final velocity $v = 0\, m/s$,Coefficient of kinetic friction $\mu_k = 0.5$,Acceleration due to gravity $g = 10\, m/s^2$.
Using the work-energy theorem,the work done by the frictional force is equal to the change in kinetic energy:
$W = \Delta K$
$-f_k \cdot s = 0 - \frac{1}{2} m u^2$
$-(\mu_k m g) s = -\frac{1}{2} m u^2$
$s = \frac{u^2}{2 \mu_k g}$
Substituting the values:
$s = \frac{(100)^2}{2 \times 0.5 \times 10} = \frac{10000}{10} = 1000\, m$.
11
PhysicsDifficultMCQAIEEE · 2005
The time taken by an object to slide down a $45^{\circ}$ rough inclined plane is $n$ times the time it takes to slide down a perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the inclined plane is:
A
$\left( 1 - \frac{1}{n^2} \right)$
B
$1 + \frac{1}{n^2}$
C
$\sqrt{1 - \frac{1}{n^2}}$
D
$\sqrt{\frac{1}{1 - n^2}}$

Solution

(A) For a smooth inclined plane,the acceleration is $a_s = g \sin \theta$. The time taken to cover distance $s$ is $t_s = \sqrt{\frac{2s}{g \sin \theta}}$.
For a rough inclined plane,the acceleration is $a_r = g(\sin \theta - \mu \cos \theta)$. The time taken is $t_r = \sqrt{\frac{2s}{g(\sin \theta - \mu \cos \theta)}}$.
Given $t_r = n t_s$,so $t_r^2 = n^2 t_s^2$.
$\frac{2s}{g(\sin \theta - \mu \cos \theta)} = n^2 \frac{2s}{g \sin \theta}$.
$\sin \theta - \mu \cos \theta = \frac{\sin \theta}{n^2}$.
$\mu \cos \theta = \sin \theta \left( 1 - \frac{1}{n^2} \right)$.
$\mu = \tan \theta \left( 1 - \frac{1}{n^2} \right)$.
Since $\theta = 45^{\circ}$,$\tan 45^{\circ} = 1$,so $\mu = 1 - \frac{1}{n^2}$.
12
PhysicsDifficultMCQAIEEE · 2005
The upper half of an inclined plane of inclination $\theta$ is perfectly smooth,while the lower half is rough. $A$ body starting from rest at the top comes back to rest at the bottom. The coefficient of friction $\mu$ for the lower half is given by:
A
$\mu = \sin \theta$
B
$\mu = \cot \theta$
C
$\mu = 2 \cos \theta$
D
$\mu = 2 \tan \theta$

Solution

(D) Let the total length of the inclined plane be $l$. The upper half has length $l/2$ and is smooth,while the lower half has length $l/2$ and is rough with coefficient of friction $\mu$.
For the upper half (smooth):
The acceleration is $a_1 = g \sin \theta$. Starting from rest $(u=0)$,the velocity $v$ at the midpoint is given by:
$v^2 = u^2 + 2 a_1 (l/2) = 0 + 2(g \sin \theta)(l/2) = gl \sin \theta$.
For the lower half (rough):
The acceleration is $a_2 = g(\sin \theta - \mu \cos \theta)$. The body starts with velocity $v$ and comes to rest $(v_f = 0)$ at the bottom after covering distance $l/2$:
$v_f^2 = v^2 + 2 a_2 (l/2) = 0$.
Substituting $v^2 = gl \sin \theta$ and $a_2 = g(\sin \theta - \mu \cos \theta)$:
$gl \sin \theta + 2[g(\sin \theta - \mu \cos \theta)](l/2) = 0$.
$gl \sin \theta + gl(\sin \theta - \mu \cos \theta) = 0$.
$2 \sin \theta - \mu \cos \theta = 0$.
$\mu \cos \theta = 2 \sin \theta$.
$\mu = 2 \tan \theta$.
Solution diagram
13
PhysicsMediumMCQAIEEE · 2005
$A$ spherical ball of mass $20 \, kg$ is stationary at the top of a hill of height $100 \, m$. It slides down a smooth surface to the ground,then climbs up another hill of height $30 \, m$ and finally slides down to a horizontal base at a height of $20 \, m$ above the ground. The velocity attained by the ball is ............... $m/s$.
A
$10$
B
$10\sqrt{30}$
C
$40$
D
$20$

Solution

(C) Since the surface is smooth,mechanical energy is conserved throughout the motion.
Let the initial height be $h_1 = 100 \, m$ and the final height be $h_2 = 20 \, m$.
The initial velocity $u = 0 \, m/s$.
According to the law of conservation of energy:
$PE_{initial} + KE_{initial} = PE_{final} + KE_{final}$
$mgh_1 + 0 = mgh_2 + \frac{1}{2}mv^2$
$mg(h_1 - h_2) = \frac{1}{2}mv^2$
$v = \sqrt{2g(h_1 - h_2)}$
Substituting the values $(g = 10 \, m/s^2)$:
$v = \sqrt{2 \times 10 \times (100 - 20)}$
$v = \sqrt{20 \times 80}$
$v = \sqrt{1600}$
$v = 40 \, m/s$.
Solution diagram
14
PhysicsMediumMCQAIEEE · 2005
$A$ block of mass $M$ moving on a frictionless horizontal surface collides with a spring of spring constant $K$ and compresses it by a length $L$. The maximum momentum of the block during the collision process is
Question diagram
A
Zero
B
$\frac{M L^2}{K}$
C
$\sqrt{MK} L$
D
$\frac{K L^2}{2M}$

Solution

(C) When a block of mass $M$ moving with velocity $v$ collides with a spring,its kinetic energy is converted into the elastic potential energy of the spring at the point of maximum compression $L$.
By the law of conservation of energy:
$\frac{1}{2} M v^2 = \frac{1}{2} K L^2$
Solving for $v$:
$v^2 = \frac{K}{M} L^2 \implies v = L \sqrt{\frac{K}{M}}$
The momentum $P$ of the block is given by $P = Mv$.
Substituting the value of $v$:
$P = M \left( L \sqrt{\frac{K}{M}} \right) = \sqrt{M^2 \cdot \frac{K}{M}} L = \sqrt{MK} L$
Thus,the maximum momentum of the block is $\sqrt{MK} L$.
15
PhysicsDifficultMCQAIEEE · 2005
$A$ mass $m$ moves with a velocity $v$ and collides inelastically with another identical mass at rest. After the collision,the $1^{st}$ mass moves with velocity $\frac{v}{\sqrt{3}}$ in a direction perpendicular to the initial direction of motion. Find the speed of the $2^{nd}$ mass after the collision.
Question diagram
A
$\frac{2}{\sqrt{3}}v$
B
$\frac{v}{\sqrt{3}}$
C
$v$
D
$\sqrt{3}v$

Solution

(A) Let mass $A$ move with velocity $v$ and collide inelastically with mass $B$,which is initially at rest.
According to the problem,mass $A$ moves in a perpendicular direction after the collision with velocity $\frac{v}{\sqrt{3}}$. Let mass $B$ move at an angle $\theta$ with the horizontal with velocity $V$.
Initial horizontal momentum of the system (before collision) $= mv$.
Final horizontal momentum of the system (after collision) $= m \left( \frac{v}{\sqrt{3}} \right) \cos(90^{\circ}) + mV \cos \theta = mV \cos \theta$.
From the conservation of horizontal linear momentum: $mv = mV \cos \theta \implies v = V \cos \theta$ $...(i)$.
Initial vertical momentum of the system (before collision) $= 0$.
Final vertical momentum of the system (after collision) $= m \left( \frac{v}{\sqrt{3}} \right) - mV \sin \theta$.
From the conservation of vertical linear momentum: $m \left( \frac{v}{\sqrt{3}} \right) - mV \sin \theta = 0 \implies \frac{v}{\sqrt{3}} = V \sin \theta$ $...(ii)$.
Squaring and adding $(i)$ and $(ii)$:
$v^2 + \frac{v^2}{3} = V^2 (\cos^2 \theta + \sin^2 \theta)$.
$\frac{4v^2}{3} = V^2$.
$V = \frac{2}{\sqrt{3}}v$.
Solution diagram
16
PhysicsEasyMCQAIEEE · 2005
An elastic material of Young's modulus $Y$ is subjected to a stress $S$. The elastic energy stored per unit volume of the material is
A
$\frac{2Y}{S^2}$
B
$\frac{S^2}{2Y}$
C
$\frac{S}{2Y}$
D
$\frac{S^2}{Y}$

Solution

(B) The elastic potential energy stored per unit volume $(u)$ in a material is given by the formula:
$u = \frac{1}{2} \times \text{stress} \times \text{strain}$
We know that Young's modulus $(Y)$ is defined as the ratio of stress $(S)$ to strain $(\epsilon)$:
$Y = \frac{S}{\epsilon} \implies \epsilon = \frac{S}{Y}$
Substituting the expression for strain into the energy formula:
$u = \frac{1}{2} \times S \times \left( \frac{S}{Y} \right)$
$u = \frac{S^2}{2Y}$
Thus,the correct option is $B$.
17
PhysicsMediumMCQAIEEE · 2005
In a capillary tube experiment,a vertical $20 \,cm$ long capillary tube is dipped in water. The water rises up to a height of $8 \,cm$ due to capillary action. If this experiment is conducted in a freely falling elevator,the length of the water column becomes ....... $cm$.
A
$10$
B
$30$
C
$20$
D
$0$

Solution

(C) The height of the water column in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
In a freely falling elevator,the effective acceleration due to gravity $g_{eff}$ becomes $0$ because the elevator is in a state of weightlessness.
As $g_{eff} \to 0$,the height of the water column $h$ tends to infinity $(h \propto 1/g_{eff})$.
However,the water column cannot exceed the physical length of the capillary tube.
Therefore,the water will rise until it fills the entire length of the capillary tube,which is $20 \,cm$.
18
PhysicsMediumMCQAIEEE · 2005
$A$ gaseous mixture consists of $16\,g$ of helium and $16\,g$ of oxygen. The ratio $\frac{C_P}{C_V}$ of the mixture is
A
$1.4$
B
$1.54$
C
$1.59$
D
$1.62$

Solution

(D) Number of moles of Helium $(n_1)$ = $\frac{16}{4} = 4 \, \text{mol}$.
Number of moles of Oxygen $(n_2)$ = $\frac{16}{32} = 0.5 \, \text{mol}$.
For Helium (monatomic),$\gamma_1 = \frac{5}{3}$ and degrees of freedom $f_1 = 3$.
For Oxygen (diatomic),$\gamma_2 = \frac{7}{5}$ and degrees of freedom $f_2 = 5$.
The equivalent $\gamma$ for the mixture is given by $\frac{n_1 + n_2}{\gamma - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1}$.
Substituting the values: $\frac{4 + 0.5}{\gamma - 1} = \frac{4}{\frac{5}{3} - 1} + \frac{0.5}{\frac{7}{5} - 1}$.
$\frac{4.5}{\gamma - 1} = \frac{4}{2/3} + \frac{0.5}{2/5} = 6 + 1.25 = 7.25$.
$\gamma - 1 = \frac{4.5}{7.25} \approx 0.62$.
$\gamma \approx 1.62$.
19
PhysicsEasyMCQAIEEE · 2005
Which of the following statements is incorrect regarding the first law of thermodynamics?
A
It introduces the concept of internal energy.
B
It introduces the concept of entropy.
C
It is not applicable to any cyclic process.
D
None of the above.

Solution

(B) The first law of thermodynamics is based on the law of conservation of energy and introduces the concept of internal energy $(U)$. It states that $\Delta Q = \Delta U + \Delta W$. It is applicable to all processes,including cyclic processes. However,the concept of entropy is introduced by the second law of thermodynamics,not the first. Therefore,statement $(b)$ is incorrect.
20
PhysicsEasyMCQAIEEE · 2005
$A$ system goes from $A$ to $B$ via two processes $I$ and $II$ as shown in the figure. If $\Delta U_I$ and $\Delta U_{II}$ are the changes in internal energies in the processes $I$ and $II$ respectively,then:
Question diagram
A
$\Delta U_{II} > \Delta U_I$
B
$\Delta U_{II} < \Delta U_I$
C
$\Delta U_I = \Delta U_{II}$
D
The relation between $\Delta U_I$ and $\Delta U_{II}$ cannot be determined.

Solution

(C) Internal energy $(U)$ is a state function,which means its value depends only on the state of the system (defined by variables like pressure,volume,and temperature) and not on the path taken to reach that state.
Since both processes $I$ and $II$ start at the same initial state $A$ and end at the same final state $B$,the change in internal energy for both processes must be identical.
Therefore,$\Delta U_I = \Delta U_{II}$.
21
PhysicsDifficultMCQAIEEE · 2005
The temperature-entropy $(T-S)$ diagram of a reversible engine cycle is given in the figure. Its efficiency is
Question diagram
A
$0.33$
B
$0.67$
C
$0.5$
D
$0.25$

Solution

(A) The area under the $T-S$ curve represents the heat exchanged in a process.
For the given cycle, the heat absorbed $(Q_{in})$ is the area under the top slanted line from $S_0$ to $2S_0$.
The area is a rectangle plus a triangle: $Q_{in} = (T_0 \times S_0) + \frac{1}{2} \times (2T_0 - T_0) \times (2S_0 - S_0) = T_0 S_0 + \frac{1}{2} T_0 S_0 = 1.5 T_0 S_0$.
The heat rejected $(Q_{out})$ is the area under the bottom horizontal line from $2S_0$ to $S_0$ at temperature $T_0$: $Q_{out} = T_0 \times (2S_0 - S_0) = T_0 S_0$.
The efficiency $\eta$ is given by $\eta = 1 - \frac{Q_{out}}{Q_{in}}$.
Substituting the values: $\eta = 1 - \frac{T_0 S_0}{1.5 T_0 S_0} = 1 - \frac{1}{1.5} = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.33$.
22
PhysicsDifficultMCQAIEEE · 2005
The figure shows a system of two concentric spheres of radii $r_1$ and $r_2$ kept at temperatures $T_1$ and $T_2$,respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to
Question diagram
A
$\frac{{{r_1}\,{r_2}}}{{({r_2} - {r_1})}}$
B
$({r_2} - {r_1})$
C
$({r_2} - {r_1})({r_1}\,{r_2})$
D
$\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right)$

Solution

(A) Consider a concentric spherical shell of radius $r$ and thickness $dr$ as shown in the figure.
In a steady state,the radial rate of flow of heat $H$ through this shell is given by Fourier's law of heat conduction:
$H = \frac{{dQ}}{{dt}} = - KA\frac{{dT}}{{dr}}$
Since the area of the spherical shell is $A = 4\pi {r^2}$,we have:
$H = - K(4\pi {r^2})\frac{{dT}}{{dr}}$
Rearranging the terms to integrate:
$\frac{{dr}}{{{r^2}}} = - \frac{{4\pi K}}{H}dT$
Integrating from $r_1$ to $r_2$ and $T_1$ to $T_2$:
$\int_{{r_1}}^{{r_2}} \frac{{dr}}{{{r^2}}} = - \frac{{4\pi K}}{H} \int_{{T_1}}^{{T_2}} dT$
$\left[ - \frac{1}{r} \right]_{{r_1}}^{{r_2}} = - \frac{{4\pi K}}{H} (T_2 - T_1)$
$\left( \frac{1}{{{r_1}}} - \frac{1}{{{r_2}}} \right) = \frac{{4\pi K}}{H} (T_1 - T_2)$
$\frac{{{r_2} - {r_1}}}{{{r_1}{r_2}}} = \frac{{4\pi K}}{H} (T_1 - T_2)$
Solving for $H$:
$H = \frac{{4\pi K{r_1}{r_2}({T_1} - {T_2})}}{{{r_2} - {r_1}}}$
Thus,the rate of heat flow is proportional to $\frac{{{r_1}{r_2}}}{{{r_2} - {r_1}}}$.
Solution diagram
23
PhysicsMediumMCQAIEEE · 2005
Two simple harmonic motions are represented by the equations ${y_1} = 0.1 \sin(100\pi t + \frac{\pi}{3})$ and ${y_2} = 0.1 \cos(\pi t)$. The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is
A
$\frac{-\pi}{3}$
B
$\frac{\pi}{6}$
C
$\frac{-\pi}{6}$
D
$\frac{\pi}{3}$

Solution

(C) The velocity of the first particle is given by the derivative of its displacement: ${v_1} = \frac{d{y_1}}{dt} = 0.1 \times 100\pi \cos(100\pi t + \frac{\pi}{3}) = 10\pi \cos(100\pi t + \frac{\pi}{3})$.
The velocity of the second particle is given by the derivative of its displacement: ${v_2} = \frac{d{y_2}}{dt} = -0.1\pi \sin(\pi t) = 0.1\pi \cos(\pi t + \frac{\pi}{2})$.
The phase of the velocity of the first particle is ${\phi_1} = 100\pi t + \frac{\pi}{3}$,and the phase of the velocity of the second particle is ${\phi_2} = \pi t + \frac{\pi}{2}$.
However,the question asks for the phase difference between the two velocities. Since the frequencies are different ($100\pi$ vs $\pi$),the phase difference is time-dependent. Assuming the question implies the difference in initial phases at $t=0$:
$\Delta \phi = \phi_1(0) - \phi_2(0) = \frac{\pi}{3} - \frac{\pi}{2} = \frac{2\pi - 3\pi}{6} = -\frac{\pi}{6}$.
24
PhysicsEasyMCQAIEEE · 2005
The equation of motion of a particle is $\frac{d^2y}{dt^2} + Ky = 0$,where $K$ is a positive constant. The time period of the motion is given by
A
$\frac{2\pi}{K}$
B
$2\pi K$
C
$\frac{2\pi}{\sqrt{K}}$
D
$2\pi \sqrt{K}$

Solution

(C) The given equation of motion is $\frac{d^2y}{dt^2} + Ky = 0$.
Comparing this with the standard equation of Simple Harmonic Motion $(SHM)$,which is $\frac{d^2y}{dt^2} + \omega^2 y = 0$,we get $\omega^2 = K$.
Therefore,the angular frequency is $\omega = \sqrt{K}$.
We know that the time period $T$ is related to the angular frequency $\omega$ by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$,we get $T = \frac{2\pi}{\sqrt{K}}$.
25
PhysicsMediumMCQAIEEE · 2005
The function $\sin^2(\omega t)$ represents:
A
$A$ simple harmonic motion with a period $2\pi /\omega $
B
$A$ simple harmonic motion with a period $\pi /\omega $
C
$A$ periodic but not simple harmonic motion with a period $2\pi /\omega $
D
$A$ periodic but not simple harmonic motion with a period $\pi /\omega $

Solution

(D) Given function is $y = \sin^2(\omega t)$.
Using the trigonometric identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$,we can rewrite the function as:
$y = \frac{1}{2} - \frac{1}{2}\cos(2\omega t)$.
The period $T$ of a function $\cos(kt)$ is given by $T = \frac{2\pi}{k}$.
Here,$k = 2\omega$,so the period is $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.
Simple Harmonic Motion $(S.H.M.)$ must satisfy the differential equation $\frac{d^2y}{dt^2} = -\Omega^2 y$. The given function represents a constant offset plus a cosine term,which does not satisfy the condition for $S.H.M.$ because the equilibrium position is shifted and it is not a pure sinusoidal oscillation about the origin. Therefore,it is a periodic motion but not $S.H.M.$
26
PhysicsMediumMCQAIEEE · 2005
$A$ hollow sphere is filled with water through a small hole in it. It is then hung by a long thread and made to oscillate. As the water slowly flows out of the hole at the bottom, the period of oscillation will
A
Continuously decrease
B
Continuously increase
C
First decrease and then increase to original value
D
First increase and then decrease to original value

Solution

(D) The given system acts as a simple pendulum, where the effective length $(l)$ is the distance between the point of suspension and the center of gravity $(C.G.)$ of the oscillating body.
Initially, when the sphere is full, the $C.G.$ is at the center of the sphere. As water flows out, the center of mass of the remaining water shifts downwards, causing the resultant $C.G.$ of the system to move downwards. This increases the effective length $(l)$, and since the time period $T = 2\pi \sqrt{l/g}$, the time period $T$ increases.
As more water flows out, the weight of the remaining water becomes less than the weight of the empty sphere. The resultant $C.G.$ starts shifting back upwards towards the center of the sphere. Consequently, the effective length $(l)$ decreases, which causes the time period $T$ to decrease.
Finally, when the sphere is completely empty, the $C.G.$ returns to the center of the sphere, making the effective length equal to its initial value. Thus, the time period returns to its original value. Therefore, the period of oscillation first increases and then decreases to its original value.
Solution diagram
27
PhysicsMediumMCQAIEEE · 2005
An observer moves towards a stationary source of sound,with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency (in $\%$)?
A
$5$
B
$20$
C
$0$
D
$0.5$

Solution

(B) When an observer moves towards a stationary source,the apparent frequency $n'$ is given by the Doppler effect formula:
$n' = \left( \frac{v + v_O}{v} \right) n$
Given that the velocity of the observer $v_O = \frac{v}{5}$,where $v$ is the velocity of sound.
Substituting the value of $v_O$ into the formula:
$n' = \left( \frac{v + v/5}{v} \right) n = \left( \frac{6v/5}{v} \right) n = 1.2n$
The increase in frequency is $\Delta n = n' - n = 1.2n - n = 0.2n$.
The percentage increase in frequency is given by:
$\text{Percentage increase} = \left( \frac{\Delta n}{n} \right) \times 100 = \left( \frac{0.2n}{n} \right) \times 100 = 20\%$.
28
PhysicsEasyMCQAIEEE · 2005
The moment of inertia of a uniform semicircular disc of mass $M$ and radius $r$ about an axis perpendicular to the plane of the disc passing through its center is:
A
$\frac{1}{4} M r^2$
B
$\frac{2}{5} M r^2$
C
$M r^2$
D
$\frac{1}{2} M r^2$

Solution

(C) For a complete circular disc of mass $M'$ and radius $r$,the moment of inertia about an axis perpendicular to its plane passing through the center is $I = \frac{1}{2} M' r^2$.
$A$ semicircular disc is exactly half of a complete circular disc.
If the mass of the semicircular disc is $M$,then the mass of the corresponding full circular disc would be $2M$.
Substituting $M' = 2M$ into the formula for the full disc:
$I = \frac{1}{2} (2M) r^2 = M r^2$.
Therefore,the moment of inertia of the semicircular disc about the axis perpendicular to its plane passing through the center is $M r^2$.
29
PhysicsMediumMCQAIEEE · 2005
$A$ body $A$ of mass $M$ while falling vertically downwards under gravity breaks into two parts; a body $B$ of mass $\frac{2}{3} M$ and a body $C$ of mass $\frac{1}{3} M$. The centre of mass of bodies $B$ and $C$ taken together shifts compared to that of body $A$ towards
A
body $C$
B
body $B$
C
depends on height of breaking
D
does not shift

Solution

(D) The motion of the centre of mass of a system depends only on the external forces acting on it.
In this case,the only external force acting on the body $A$ (and subsequently on the system of parts $B$ and $C$) is the gravitational force,which results in an acceleration equal to $g$ downwards.
Since the external force remains unchanged during the breaking process,the acceleration of the centre of mass remains $g$.
Therefore,the trajectory of the centre of mass of the system consisting of parts $B$ and $C$ follows the same path as the original body $A$.
Thus,the centre of mass does not shift relative to the path of the original body $A$.
30
PhysicsDifficultMCQAIEEE · 2005
$A$ $T$-shaped object with dimensions shown in the figure is lying on a smooth floor. $A$ force $\vec{F}$ is applied at point $P$ parallel to $AB$,such that the object has only translational motion without rotation. Find the location of $P$ with respect to $C$.
Question diagram
A
$\frac{4l}{3}$
B
$l$
C
$\frac{2l}{3}$
D
$\frac{3l}{2}$

Solution

(A) For an object to undergo pure translational motion,the applied force must pass through its center of mass $(CM)$.
Let the mass of the horizontal rod $AB$ be $m$. Since the vertical rod $CD$ has length $2l$ and the horizontal rod $AB$ has length $l$,assuming uniform density,the mass of rod $CD$ is $2m$.
The center of mass of rod $AB$ is at its midpoint $D$,which is at a distance $2l$ from $C$. Let this be $y_1 = 2l$.
The center of mass of rod $CD$ is at its midpoint,which is at a distance $l$ from $C$. Let this be $y_2 = l$.
Taking $C$ as the origin,the position of the center of mass of the system along the vertical axis is:
$y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$
$y_{cm} = \frac{m(2l) + (2m)(l)}{m + 2m} = \frac{2ml + 2ml}{3m} = \frac{4ml}{3m} = \frac{4l}{3}$
Thus,the force must be applied at a distance of $\frac{4l}{3}$ from $C$ to ensure pure translational motion.
31
PhysicsMediumMCQAIEEE · 2005
An annular ring with inner and outer radii $R_{1}$ and $R_{2}$ is rotating with a uniform angular speed $\omega$ about its central axis. The ratio of the centripetal forces experienced by two particles of equal mass $m$ situated on the inner and outer parts of the ring,$\frac{F_{1}}{F_{2}}$ is
A
$1$
B
$\frac{R_{1}}{R_{2}}$
C
$\frac{R_{2}}{R_{1}}$
D
$\left(\frac{R_{1}}{R_{2}}\right)^{2}$

Solution

(B) For a particle of mass $m$ rotating in a circle of radius $r$ with a uniform angular speed $\omega$,the centripetal force is given by $F = m \omega^{2} r$.
For the particle on the inner part of the ring at radius $R_{1}$,the centripetal force is $F_{1} = m \omega^{2} R_{1}$.
For the particle on the outer part of the ring at radius $R_{2}$,the centripetal force is $F_{2} = m \omega^{2} R_{2}$.
Taking the ratio of these two forces:
$\frac{F_{1}}{F_{2}} = \frac{m \omega^{2} R_{1}}{m \omega^{2} R_{2}} = \frac{R_{1}}{R_{2}}$.
Thus,the ratio of the forces is $\frac{R_{1}}{R_{2}}$.
Solution diagram
32
PhysicsEasyMCQAIEEE · 2005
The average density of the Earth:
A
is directly proportional to $g$
B
is inversely proportional to $g$
C
does not depend on $g$
D
is a complex function of $g$

Solution

(A) The acceleration due to gravity $g$ at the surface of the Earth is given by $g = \frac{GM}{R^2}$.
Substituting the mass $M = \rho \times V = \rho \times \frac{4}{3}\pi R^3$,we get:
$g = \frac{G \times \rho \times \frac{4}{3}\pi R^3}{R^2} = \frac{4}{3}\pi G \rho R$.
From this expression,we can see that the average density $\rho$ is given by $\rho = \frac{3g}{4\pi GR}$.
Since $G$,$\pi$,and $R$ (radius of the Earth) are constants,it follows that $\rho \propto g$.
Therefore,the average density of the Earth is directly proportional to $g$.
33
PhysicsMediumMCQAIEEE · 2005
$A$ body of mass $m$ is accelerated uniformly from rest to a speed $v$ in a time $T$. The instantaneous power delivered to the body as a function of time is given by
A
$\frac{mv^2}{T^2}t$
B
$\frac{mv^2}{T^2}t^2$
C
$\frac{1}{2}\frac{mv^2}{T^2}t$
D
$\frac{1}{2}\frac{mv^2}{T^2}t^2$

Solution

(A) The body starts from rest,so initial velocity $u = 0$.
Since it is accelerated uniformly to speed $v$ in time $T$,the acceleration $a$ is given by $a = \frac{v - u}{T} = \frac{v}{T}$.
The velocity at any time $t$ is $v(t) = at = \frac{v}{T}t$.
The force acting on the body is $F = ma = m\left(\frac{v}{T}\right)$.
Instantaneous power $P$ is given by $P = F \cdot v(t)$.
Substituting the values,$P = \left(m \frac{v}{T}\right) \left(\frac{v}{T}t\right) = \frac{mv^2}{T^2}t$.
34
PhysicsMediumMCQAIEEE · 2005
$A$ particle of mass $10\, g$ is kept on the surface of a uniform sphere of mass $100\, kg$ and radius $10\, cm$. Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take $G = 6.67 \times 10^{-11}\, Nm^2 / kg^2$).
A
$3.33 \times 10^{-10}\,J$
B
$13.34 \times 10^{-10}\,J$
C
$6.67 \times 10^{-10}\,J$
D
$6.67 \times 10^{-9}\,J$

Solution

(C) The gravitational potential energy $U$ of a particle of mass $m$ at the surface of a sphere of mass $M$ and radius $R$ is given by $U = -\frac{GMm}{R}$.
To take the particle to infinity (where potential energy is $0$),the work done $W$ against the gravitational force is equal to the change in potential energy: $W = U_{final} - U_{initial} = 0 - (- \frac{GMm}{R}) = \frac{GMm}{R}$.
Given values:
$M = 100\, kg$
$m = 10\, g = 0.01\, kg$
$R = 10\, cm = 0.1\, m$
$G = 6.67 \times 10^{-11}\, Nm^2/kg^2$
Substituting these values into the formula:
$W = \frac{6.67 \times 10^{-11} \times 100 \times 0.01}{0.1}$
$W = \frac{6.67 \times 10^{-11} \times 1}{0.1} = 6.67 \times 10^{-10}\, J$.
35
PhysicsMediumMCQAIEEE · 2005
When two tuning forks (fork $1$ and fork $2$) are sounded simultaneously,$4$ beats per second are heard. Now,some tape is attached to the prong of fork $2$. When the tuning forks are sounded again,$6$ beats per second are heard. If the frequency of fork $1$ is $200 \, Hz$,then what was the original frequency of fork $2$ (in $, Hz$)?
A
$202$
B
$200$
C
$204$
D
$196$

Solution

(D) Let the frequency of fork $1$ be $n_1 = 200 \, Hz$ and the frequency of fork $2$ be $n_2$.
Initially,the beat frequency is $|n_1 - n_2| = 4 \, Hz$. This implies $n_2 = 200 \pm 4$,so $n_2 = 204 \, Hz$ or $n_2 = 196 \, Hz$.
When tape is attached to the prong of fork $2$,its frequency $n_2$ decreases.
If $n_2$ was $204 \, Hz$,decreasing it would make the beat frequency $|200 - n_2'|$ smaller than $4 \, Hz$.
If $n_2$ was $196 \, Hz$,decreasing it further (e.g.,to $194 \, Hz$) would make the beat frequency $|200 - 194| = 6 \, Hz$.
Since the beat frequency increased to $6 \, Hz$,the original frequency of fork $2$ must have been $196 \, Hz$.
36
PhysicsMediumMCQAIEEE · 2005
If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (both $x$ and $h$ being much smaller than the radius of the earth)
A
$x=h$
B
$x=2h$
C
$x=\frac{h}{2}$
D
$x=h^2$

Solution

(B) The value of $g$ at a height $h$ above the surface of the Earth is given by $g_h = g(1 - \frac{2h}{R})$.
Thus,the change in $g$ is $\Delta g_h = g - g_h = g(\frac{2h}{R})$.
The value of $g$ at a depth $x$ below the surface of the Earth is given by $g_x = g(1 - \frac{x}{R})$.
Thus,the change in $g$ is $\Delta g_x = g - g_x = g(\frac{x}{R})$.
Given that the change in $g$ is the same at height $h$ and depth $x$,we equate the two expressions:
$g(\frac{2h}{R}) = g(\frac{x}{R})$.
Solving for $x$,we get $x = 2h$.
37
PhysicsMediumMCQAIEEE · 2005
$A$ charged ball $B$ hangs from a silk thread $S$,which makes an angle $\theta$ with a large charged conducting sheet $P$,as shown in the figure. The surface charge density $\sigma$ of the sheet is proportional to
Question diagram
A
$\sin \theta$
B
$\tan \theta$
C
$\cos \theta$
D
$\cot \theta$

Solution

(B) Let $T$ be the tension in the silk thread,$q$ be the charge on the ball,$m$ be the mass of the ball,and $E$ be the electric field produced by the large charged sheet.
At equilibrium,the forces acting on the ball are:
$1$. Tension $T$ along the thread.
$2$. Gravitational force $mg$ acting downwards.
$3$. Electric force $qE$ acting horizontally away from the sheet.
Resolving the tension $T$ into components:
$T \sin \theta = qE$ (horizontal component)
$T \cos \theta = mg$ (vertical component)
Dividing the two equations:
$\frac{T \sin \theta}{T \cos \theta} = \frac{qE}{mg}$
$\tan \theta = \frac{qE}{mg}$
The electric field $E$ due to a large charged conducting sheet is given by $E = \frac{\sigma}{\varepsilon_0}$.
Substituting this into the equation:
$\tan \theta = \frac{q}{mg} \left( \frac{\sigma}{\varepsilon_0} \right)$
Since $q, m, g,$ and $\varepsilon_0$ are constants,we have:
$\tan \theta \propto \sigma$ or $\sigma \propto \tan \theta$.
Therefore,the correct option is $(b)$.
Solution diagram
38
PhysicsDifficultMCQAIEEE · 2005
Two thin wire rings each having a radius $R$ are placed at a distance $d$ apart with their axes coinciding. The charges on the two rings are $+q$ and $-q$. The potential difference between the centres of the two rings is
A
Zero
B
$\frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$
C
$\frac{qR}{4\pi \varepsilon_0 d^2}$
D
$\frac{q}{2\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$

Solution

(D) Let the centres of the two rings be $O_1$ and $O_2$. The potential at any point on the axis of a charged ring of radius $R$ and charge $Q$ at a distance $x$ from its centre is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{\sqrt{R^2 + x^2}}$.
For the first ring (charge $+q$) at centre $O_1$:
Potential due to itself is $V_{1, self} = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}$.
Potential due to the second ring (charge $-q$) at distance $d$ is $V_{1, other} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{\sqrt{R^2 + d^2}}$.
So,$V_{O_1} = \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$.
For the second ring (charge $-q$) at centre $O_2$:
Potential due to itself is $V_{2, self} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{R}$.
Potential due to the first ring (charge $+q$) at distance $d$ is $V_{2, other} = \frac{1}{4\pi \varepsilon_0} \frac{q}{\sqrt{R^2 + d^2}}$.
So,$V_{O_2} = \frac{q}{4\pi \varepsilon_0} \left[ -\frac{1}{R} + \frac{1}{\sqrt{R^2 + d^2}} \right] = -V_{O_1}$.
The potential difference is $\Delta V = V_{O_1} - V_{O_2} = V_{O_1} - (-V_{O_1}) = 2V_{O_1}$.
$\Delta V = 2 \cdot \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right] = \frac{q}{2\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$.
Solution diagram
39
PhysicsEasyMCQAIEEE · 2005
$A$ parallel plate capacitor is made by stacking $n$ equally spaced plates connected alternately. If the capacitance between any two plates is $C$, then the resultant capacitance is:
A
$C$
B
$nC$
C
$(n - 1)C$
D
$(n + 1)C$

Solution

(C) In a stack of $n$ plates connected alternately, the plates form $(n - 1)$ capacitors connected in parallel.
Each pair of adjacent plates acts as a single capacitor with capacitance $C$.
Since these $(n - 1)$ capacitors are connected in parallel, the equivalent capacitance $C_R$ is the sum of individual capacitances.
Therefore, $C_R = C + C + ... + (n - 1) \text{ times} = (n - 1)C$.
40
PhysicsMediumMCQAIEEE · 2005
$A$ fully charged capacitor has a capacitance $C$. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity $s$ and mass $m$. If the temperature of the block is raised by $\Delta T$,the potential difference $V$ across the capacitor is
A
$\frac{ms\Delta T}{C}$
B
$\sqrt{\frac{2ms\Delta T}{C}}$
C
$\sqrt{\frac{2mC\Delta T}{s}}$
D
$\frac{mC\Delta T}{s}$

Solution

(B) The energy stored in a fully charged capacitor is given by $U = \frac{1}{2}CV^2$.
When the capacitor discharges through the coil,this electrical energy is converted into heat energy.
The heat energy absorbed by the block is given by $Q = ms\Delta T$.
By the principle of conservation of energy,the electrical energy stored in the capacitor equals the heat energy gained by the block:
$\frac{1}{2}CV^2 = ms\Delta T$
Solving for $V$:
$V^2 = \frac{2ms\Delta T}{C}$
$V = \sqrt{\frac{2ms\Delta T}{C}}$
Therefore,the correct option is $B$.
41
PhysicsMediumMCQAIEEE · 2005
In the adjoining circuit,the battery $E_1$ has an $e.m.f.$ of $12 \, V$ and zero internal resistance,while the battery $E$ has an $e.m.f.$ of $2 \, V$. If the galvanometer $G$ reads zero,then the value of the resistance $X$ in $\Omega$ is
Question diagram
A
$10$
B
$100$
C
$500$
D
$200$

Solution

(B) For the galvanometer $G$ to read zero,no current flows through the branch containing $G$ and $E$.
This implies that the potential difference across the resistance $X$ must be equal to the $e.m.f.$ of the battery $E$.
Let $I$ be the current flowing through the loop containing $E_1$,$500 \, \Omega$,and $X$.
Using Ohm's law,the current $I$ is given by $I = \frac{E_1}{500 + X}$.
The potential difference across $X$ is $V_X = I \cdot X = \left( \frac{E_1}{500 + X} \right) X$.
Since $V_X = E$,we have $\left( \frac{12}{500 + X} \right) X = 2$.
Dividing both sides by $2$,we get $\left( \frac{6}{500 + X} \right) X = 1$.
$6X = 500 + X$.
$5X = 500$.
$X = 100 \, \Omega$.
42
PhysicsDifficultMCQAIEEE · 2005
Two sources of equal $emf$ $E$ are connected in series to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2$ $(R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero,then:
A
$R = R_1 R_2 / (R_1 + R_2)$
B
$R = R_1 R_2 / (R_2 - R_1)$
C
$R = R_2 (R_1 + R_2) / (R_2 - R_1)$
D
$R = R_2 - R_1$

Solution

(D) The total $emf$ of the series combination is $E_{eq} = E + E = 2E$. The total resistance of the circuit is $R_{eq} = R + R_1 + R_2$. The current in the circuit is given by $i = \frac{2E}{R + R_1 + R_2}$.
The potential difference $V$ across a source with $emf$ $E$ and internal resistance $r$ is given by $V = E - ir$. For the source with internal resistance $R_2$,the potential difference is zero:
$0 = E - i R_2$
$E = i R_2$
Substituting the value of $i$:
$E = \left( \frac{2E}{R + R_1 + R_2} \right) R_2$
$1 = \frac{2 R_2}{R + R_1 + R_2}$
$R + R_1 + R_2 = 2 R_2$
$R = 2 R_2 - R_2 - R_1$
$R = R_2 - R_1$
Solution diagram
43
PhysicsEasyMCQAIEEE · 2005
An energy source will supply a constant current into the load if its internal resistance is
A
Very large as compared to the load resistance
B
Non-zero but less than the resistance of the load
C
Equal to the resistance of the load
D
Zero

Solution

(D) The current $I$ supplied by an energy source with electromotive force $E$ and internal resistance $r$ to a load resistance $R$ is given by the formula: $I = \frac{E}{R + r}$.
For the current $I$ to be constant regardless of changes in the load resistance $R$,the internal resistance $r$ must be zero.
If $r = 0$,then $I = \frac{E}{R}$. However,in the context of an ideal constant current source,the internal resistance is considered to be infinite. But based on the provided options,if $r = 0$,the source behaves as an ideal voltage source. Given the standard interpretation of this specific question,the correct answer is $D$.
44
PhysicsMediumMCQAIEEE · 2005
In a potentiometer experiment,the balancing length with a cell is $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is ................. $\Omega$.
A
$4$
B
$2$
C
$1$
D
$0.5$

Solution

(B) The balancing length $l_1$ is proportional to the electromotive force $(E)$ of the cell: $E \propto l_1 = 240 \ cm$.
When the cell is shunted with an external resistance $R = 2 \ \Omega$,the terminal potential difference $V$ is balanced at length $l_2 = 120 \ cm$,so $V \propto l_2$.
The formula for internal resistance $r$ of a cell is given by $r = R \left( \frac{E}{V} - 1 \right)$.
Since $E/V = l_1/l_2$,we substitute the values:
$r = R \left( \frac{l_1}{l_2} - 1 \right)$
$r = 2 \left( \frac{240}{120} - 1 \right)$
$r = 2 \left( 2 - 1 \right)$
$r = 2 \ \Omega$.
Therefore,the internal resistance of the cell is $2 \ \Omega$.
45
PhysicsMediumMCQAIEEE · 2005
$A$ moving coil galvanometer has $150$ equal divisions. Its current sensitivity is $10$ divisions per milliampere and voltage sensitivity is $2$ divisions per millivolt. In order that each division reads $1 \, V$,the resistance in $\Omega$ needed to be connected in series with the coil will be
A
$99995$
B
$9995$
C
$10^3$
D
$10^5$

Solution

(B) The voltage sensitivity $(V_s)$ is related to current sensitivity $(I_s)$ and galvanometer resistance $(G)$ by the formula: $V_s = \frac{I_s}{G}$.
Given $I_s = 10 \, \text{div/mA}$ and $V_s = 2 \, \text{div/mV}$,we find $G = \frac{I_s}{V_s} = \frac{10}{2} = 5 \, \Omega$.
The full-scale deflection current $(I_g)$ is the total divisions divided by current sensitivity: $I_g = \frac{150 \, \text{div}}{10 \, \text{div/mA}} = 15 \, \text{mA} = 15 \times 10^{-3} \, \text{A}$.
To make each division read $1 \, V$,the total voltage $(V)$ to be measured for full-scale deflection is $150 \, \text{divisions} \times 1 \, \text{V/division} = 150 \, \text{V}$.
To convert the galvanometer into a voltmeter of range $V$,a series resistance $R$ is required,given by $R = \frac{V}{I_g} - G$.
Substituting the values: $R = \frac{150}{15 \times 10^{-3}} - 5 = 10000 - 5 = 9995 \, \Omega$.
46
PhysicsEasyMCQAIEEE · 2005
$A$ heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be
A
One fourth
B
Halved
C
Doubled
D
Four times

Solution

(C) The heat generated by a heater is given by the formula $H = \frac{V^2 t}{R}$,where $V$ is the voltage,$t$ is the time,and $R$ is the resistance of the coil.
When the coil is cut into two equal parts,the resistance of each part becomes $R' = \frac{R}{2}$.
Since the voltage $V$ remains constant,the new heat generated $H'$ is given by $H' = \frac{V^2 t}{R'} = \frac{V^2 t}{R/2} = 2 \times \frac{V^2 t}{R}$.
Therefore,$H' = 2H$.
Thus,the heat generated is doubled.
47
PhysicsMediumMCQAIEEE · 2005
The resistance of a hot tungsten filament is about $10$ times its cold resistance. What will be the resistance of a $100\, W$ and $200\, V$ lamp when not in use? (in $\Omega$)
A
$400$
B
$200$
C
$40$
D
$20$

Solution

(C) The resistance of the lamp when it is hot (at rated power) is given by the formula $R_{Hot} = \frac{V^2}{P}$.
Substituting the given values: $R_{Hot} = \frac{200 \times 200}{100} = 400\,\Omega$.
It is given that the hot resistance is $10$ times the cold resistance: $R_{Hot} = 10 \times R_{Cold}$.
Therefore,the cold resistance (resistance when not in use) is $R_{Cold} = \frac{R_{Hot}}{10}$.
$R_{Cold} = \frac{400}{10} = 40\,\Omega$.
48
PhysicsDifficultMCQAIEEE · 2005
Two voltameters,one of copper and another of silver,are joined in parallel. When a total charge $q$ flows through the voltameters,equal amounts of metals are deposited. If the electrochemical equivalents of copper and silver are $z_1$ and $z_2$ respectively,the charge which flows through the silver voltameter is
A
$q\frac{z_1}{z_2}$
B
$q\frac{z_2}{z_1}$
C
$\frac{q}{1 + \frac{z_1}{z_2}}$
D
$\frac{q}{1 + \frac{z_2}{z_1}}$

Solution

(D) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = zq$,where $z$ is the electrochemical equivalent and $q$ is the charge.
Since the masses deposited are equal,we have $m_1 = m_2$,which implies $z_1 q_1 = z_2 q_2$.
From this,the ratio of charges is $\frac{q_1}{q_2} = \frac{z_2}{z_1}$.
We know that the total charge $q = q_1 + q_2$.
We want to find the charge $q_2$ flowing through the silver voltameter.
From the ratio,$q_1 = q_2 \frac{z_2}{z_1}$.
Substituting this into the total charge equation: $q = q_2 \frac{z_2}{z_1} + q_2 = q_2 (1 + \frac{z_2}{z_1})$.
Solving for $q_2$,we get $q_2 = \frac{q}{1 + \frac{z_2}{z_1}}$.
49
PhysicsMediumMCQAIEEE · 2005
Two concentric coils each of radius equal to $2\pi \, cm$ are placed at right angles to each other. $3 \, A$ and $4 \, A$ are the currents flowing in each coil respectively. The magnetic induction in $Wb/m^2$ at the centre of the coils will be $(\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m)$.
A
$5 \times 10^{-5}$
B
$7 \times 10^{-5}$
C
$12 \times 10^{-5}$
D
$10^{-5}$

Solution

(A) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 i}{2r}$.
Since the two coils are placed at right angles,their magnetic fields $B_1$ and $B_2$ will also be perpendicular to each other.
The resultant magnetic field is $B_{net} = \sqrt{B_1^2 + B_2^2}$.
Substituting $B_1 = \frac{\mu_0 i_1}{2r}$ and $B_2 = \frac{\mu_0 i_2}{2r}$,we get $B_{net} = \frac{\mu_0}{2r} \sqrt{i_1^2 + i_2^2}$.
Given $r = 2\pi \, cm = 2\pi \times 10^{-2} \, m$,$i_1 = 3 \, A$,$i_2 = 4 \, A$,and $\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m$.
$B_{net} = \frac{4\pi \times 10^{-7}}{2 \times 2\pi \times 10^{-2}} \sqrt{3^2 + 4^2}$.
$B_{net} = \frac{4\pi \times 10^{-7}}{4\pi \times 10^{-2}} \sqrt{9 + 16} = 10^{-5} \times \sqrt{25} = 5 \times 10^{-5} \, Wb/m^2$.
50
PhysicsEasyMCQAIEEE · 2005
$A$ uniform electric field and a uniform magnetic field are produced,pointing in the same direction. An electron is projected with its velocity pointing in the same direction.
A
The electron will turn to its right.
B
The electron will turn to its left.
C
The electron velocity will increase in magnitude.
D
The electron velocity will decrease in magnitude.

Solution

(D) The magnetic force on a moving charge is given by $\vec{F}_m = q(\vec{v} \times \vec{B})$. Since the electron is moving parallel to the magnetic field,the angle between $\vec{v}$ and $\vec{B}$ is $0^\circ$,so $\vec{F}_m = 0$.
The electric force on the electron is given by $\vec{F}_e = q\vec{E}$. Since the charge of an electron is negative $(q = -e)$,the electric force acts in the direction opposite to the electric field $\vec{E}$.
As the electron is projected in the direction of the electric field,the electric force acts in the opposite direction to its velocity. This force acts as a retarding force,causing the electron to decelerate. Therefore,the magnitude of the electron's velocity will decrease.
Solution diagram
51
PhysicsEasyMCQAIEEE · 2005
$A$ charged particle of mass $m$ and charge $q$ travels on a circular path of radius $r$ that is perpendicular to a magnetic field $B$. The time taken by the particle to complete one revolution is
A
$\frac{2\pi qB}{m}$
B
$\frac{2\pi m}{qB}$
C
$\frac{2\pi mq}{B}$
D
$\frac{2\pi q^2B}{m}$

Solution

(B) When a charged particle moves perpendicular to a magnetic field,the magnetic Lorentz force provides the necessary centripetal force for circular motion.
Equating the magnetic force to the centripetal force:
$qvB = \frac{mv^2}{r}$
From this,we can find the velocity $v$ or the radius $r$:
$v = \frac{qBr}{m}$
The time period $T$ for one complete revolution is the circumference divided by the speed:
$T = \frac{2\pi r}{v}$
Substituting the expression for $v$:
$T = \frac{2\pi r}{(qBr/m)} = \frac{2\pi m}{qB}$
Thus,the time taken to complete one revolution is $\frac{2\pi m}{qB}$.
52
PhysicsMediumMCQAIEEE · 2005
Two thin,long,parallel wires,separated by a distance $d$,carry a current of $i$ in the same direction. They will
A
Attract each other with a force of $\frac{\mu_0 i^2}{2\pi d^2}$
B
Repel each other with a force of $\frac{\mu_0 i^2}{2\pi d^2}$
C
Attract each other with a force of $\frac{\mu_0 i^2}{2\pi d}$
D
Repel each other with a force of $\frac{\mu_0 i^2}{2\pi d}$

Solution

(C) The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by a distance $d$ is given by the formula: $\frac{F}{l} = \frac{\mu_0 i_1 i_2}{2\pi d}$.
Since both wires carry the same current $i$ in the same direction,$i_1 = i_2 = i$.
Substituting these values,we get: $\frac{F}{l} = \frac{\mu_0 i^2}{2\pi d}$.
According to the right-hand rule,parallel currents in the same direction exert an attractive force on each other.
Therefore,the wires will attract each other with a force of $\frac{\mu_0 i^2}{2\pi d}$ per unit length.
53
PhysicsEasyMCQAIEEE · 2005
$A$ magnetic needle is kept in a non-uniform magnetic field. It experiences
A
$A$ force and a torque
B
$A$ force but not a torque
C
$A$ torque but not a force
D
Neither a torque nor a force

Solution

(A) magnetic needle acts as a magnetic dipole.
In a non-uniform magnetic field,the magnetic force acting on the two poles of the needle will be different in both magnitude and direction.
Because the magnetic field is non-uniform,the net force on the dipole is non-zero,resulting in a translational force.
Additionally,because the forces on the two poles are not collinear and have different magnitudes,they create a net torque,causing the needle to rotate.
Therefore,the magnetic needle experiences both a force and a torque.
54
PhysicsMediumMCQAIEEE · 2005
One conducting $U$ tube can slide inside another as shown in the figure,maintaining electrical contacts between the tubes. The magnetic field $B$ is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed $v$,then the emf induced in the circuit in terms of $B, l$ and $v$,where $l$ is the width of each tube,will be
Question diagram
A
Zero
B
$2Blv$
C
$Blv$
D
$-Blv$

Solution

(B) When a conductor of length $l$ moves with velocity $v$ in a magnetic field $B$ perpendicular to its length and the velocity vector,the induced motional emf is given by $E = Blv$.
In this system,there are two moving conducting segments (the curved ends of the $U$ tubes),each of length $l$,moving towards each other with speed $v$.
Each moving segment acts as a source of motional emf with magnitude $E = Blv$.
Since both tubes are moving towards each other,the induced emfs in the two segments are in series and additive in the closed loop.
Therefore,the net induced emf in the circuit is $E_{\text{net}} = Blv + Blv = 2Blv$.
55
PhysicsMediumMCQAIEEE · 2005
$A$ coil of inductance $300\, mH$ and resistance $2\, \Omega$ is connected to a source of voltage $2\, V$. The current reaches half of its steady state value in $t$ seconds. Find the value of $t$. (in $, s$)
A
$0.15$
B
$0.3$
C
$0.05$
D
$0.1$

Solution

(D) The current in an $LR$ circuit is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R$ is the steady state current.
We are given that the current reaches half of its steady state value,so $i = i_0/2$.
Substituting this into the equation: $i_0/2 = i_0(1 - e^{-Rt/L})$.
This simplifies to $1/2 = 1 - e^{-Rt/L}$,which means $e^{-Rt/L} = 1/2$.
Taking the natural logarithm on both sides: $-Rt/L = \ln(1/2) = -\ln(2)$.
Thus,$t = (L/R) \ln(2)$.
Given $L = 300\, mH = 0.3\, H$ and $R = 2\, \Omega$,we have $t = (0.3 / 2) \times 0.693$.
$t = 0.15 \times 0.693 = 0.10395\, s \approx 0.1\, s$.
56
PhysicsEasyMCQAIEEE · 2005
If the kinetic energy of a free electron doubles,its de-Broglie wavelength changes by the factor
A
$1/\sqrt{2}$
B
$\sqrt{2}$
C
$1/2$
D
$2$

Solution

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.
Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mE}}$.
From this expression,it is clear that $\lambda \propto \frac{1}{\sqrt{E}}$.
If the kinetic energy $E$ becomes $E' = 2E$,the new wavelength $\lambda'$ is given by $\lambda' = \frac{h}{\sqrt{2m(2E)}} = \frac{1}{\sqrt{2}} \cdot \frac{h}{\sqrt{2mE}}$.
Therefore,$\lambda' = \frac{1}{\sqrt{2}} \lambda$.
Thus,the wavelength changes by a factor of $1/\sqrt{2}$.
57
PhysicsEasyMCQAIEEE · 2005
$A$ photocell is illuminated by a small bright source placed $d \; m$ away. When the same source of light is placed $\frac{d}{2} \; m$ away,the number of electrons emitted by the photocathode would
A
Decrease by a factor of $2$
B
Increase by a factor of $2$
C
Decrease by a factor of $4$
D
Increase by a factor of $4$

Solution

(D) The intensity of light $I$ at a distance $d$ from a point source is given by $I \propto \frac{1}{d^2}$.
Since the number of photoelectrons emitted per second is directly proportional to the intensity of the incident light,we have $N \propto I$.
Therefore,$N \propto \frac{1}{d^2}$.
When the distance is changed from $d$ to $\frac{d}{2}$,the new number of photoelectrons $N'$ is given by $N' \propto \frac{1}{(\frac{d}{2})^2} = \frac{4}{d^2}$.
Comparing $N'$ with $N$,we get $N' = 4N$.
Thus,the number of electrons emitted increases by a factor of $4$.
58
PhysicsMediumMCQAIEEE · 2005
The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?
Question diagram
A
$I$
B
$II$
C
$III$
D
$IV$

Solution

(D) The energy of an emitted photon is given by $\Delta E = E_{initial} - E_{final}$.
Emission occurs when an electron transitions from a higher energy level to a lower energy level.
Transitions $II$ and $IV$ represent emission.
Transition $II$ is from $n=4$ to $n=3$,and transition $IV$ is from $n=4$ to $n=2$.
The energy difference $\Delta E$ is proportional to the gap between the energy levels.
Since the gap between $n=4$ and $n=2$ is larger than the gap between $n=4$ and $n=3$,transition $IV$ releases more energy than transition $II$.
Therefore,transition $IV$ represents the emission of a photon with the most energy.
59
PhysicsMediumMCQAIEEE · 2005
In the nuclear reaction: $X(n, \alpha) {_3Li^7}$,the term $X$ will be
A
$_5B^{10}$
B
$_5B^9$
C
$_5B^{11}$
D
$_2He^4$

Solution

(A) The given nuclear reaction is $X(n, \alpha) {_3Li^7}$.
This can be written as: $_ZX^A + _0n^1 \to _3Li^7 + _2He^4$.
According to the law of conservation of mass number and atomic number:
For atomic number $(Z)$: $Z + 0 = 3 + 2 \implies Z = 5$.
For mass number $(A)$: $A + 1 = 7 + 4 \implies A + 1 = 11 \implies A = 10$.
Thus,the nucleus $X$ is $_5X^{10}$,which corresponds to Boron-$10$ $(_{5}B^{10})$.
60
PhysicsEasyMCQAIEEE · 2005
If a full wave rectifier circuit is operating from $50\, Hz$ mains,the fundamental frequency in the ripple will be........$Hz$.
A
$50$
B
$70.7$
C
$100$
D
$25$

Solution

(C) In a full wave rectifier,the output consists of two pulses for every single cycle of the input $AC$ supply.
Since the input frequency is $f_{in} = 50\, Hz$,the output ripple frequency $f_{out}$ is given by the formula $f_{out} = 2 \times f_{in}$.
Substituting the given value: $f_{out} = 2 \times 50\, Hz = 100\, Hz$.
Therefore,the fundamental frequency in the ripple is $100\, Hz$.
61
PhysicsEasyMCQAIEEE · 2005
In a common base amplifier,the phase difference between the input signal voltage and the output voltage is
A
$0$
B
$\pi /4$
C
$\pi /2$
D
$\pi$

Solution

(A) In a common base $(CB)$ amplifier configuration,the input signal is applied between the emitter and the base,and the output is taken between the collector and the base.
Since the base is common to both the input and output circuits,the input voltage and output voltage signals are in the same phase.
Therefore,the phase difference between the input signal voltage and the output voltage is $0$.
62
PhysicsMediumMCQAIEEE · 2005
$A$ fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is $\frac{4}{3}$ and the fish is $12 \ cm$ below the surface,the radius of this circle in $cm$ is
A
$36\sqrt{5}$
B
$4\sqrt{5}$
C
$36\sqrt{7}$
D
$36/\sqrt{7}$

Solution

(D) The light from the outside world enters the water and undergoes refraction. The fish sees the outside world within a circular cone of light,which is defined by the critical angle $\theta_c$.
The radius $r$ of the circular horizon is given by the formula $r = h \tan(\theta_c)$,where $h$ is the depth of the fish.
Given the refractive index of water $\mu = \frac{4}{3}$,the critical angle $\theta_c$ satisfies $\sin(\theta_c) = \frac{1}{\mu} = \frac{3}{4}$.
Using the identity $\tan(\theta_c) = \frac{\sin(\theta_c)}{\cos(\theta_c)} = \frac{\sin(\theta_c)}{\sqrt{1 - \sin^2(\theta_c)}}$,we get $\tan(\theta_c) = \frac{3/4}{\sqrt{1 - (3/4)^2}} = \frac{3/4}{\sqrt{7/16}} = \frac{3}{\sqrt{7}}$.
Substituting the values,$r = 12 \times \frac{3}{\sqrt{7}} = \frac{36}{\sqrt{7}} \ cm$.
63
PhysicsDifficultMCQAIEEE · 2005
$A$ thin glass (refractive index $1.5$) lens has an optical power of $-5 D$ in air. Its optical power in a liquid medium with a refractive index of $1.6$ will be:
A
$25 D$
B
$-25 D$
C
$1 D$
D
None of these

Solution

(D) The power of a lens is given by $P = \frac{1}{f}$. According to the Lens Maker's Formula,$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
In air,$P_a = (\mu_g - 1) K = -5 D$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
Given $\mu_g = 1.5$,we have $(1.5 - 1) K = -5$,so $0.5 K = -5$,which gives $K = -10$.
In a liquid medium with refractive index $\mu_l = 1.6$,the new power $P_l$ is given by $P_l = (\frac{\mu_g}{\mu_l} - 1) K$.
Substituting the values: $P_l = (\frac{1.5}{1.6} - 1) \times (-10)$.
$P_l = (\frac{1.5 - 1.6}{1.6}) \times (-10) = (\frac{-0.1}{1.6}) \times (-10) = \frac{1}{1.6} = 0.625 D$.
Since $0.625 D$ is not among the options,the correct answer is $D$.
64
PhysicsMediumMCQAIEEE · 2005
Two point white dots are $1 \ mm$ apart on a black paper. They are viewed by an eye with a pupil diameter of $3 \ mm$. Approximately,what is the maximum distance at which the dots can be resolved by the eye? (Take wavelength of light $= 500 \ nm$)
Question diagram
A
$6$
B
$3$
C
$5$
D
$1$

Solution

(C) The condition for the resolution of two point objects by an optical system is given by the Rayleigh criterion: $\theta = \frac{1.22 \lambda}{a}$,where $\theta$ is the angular separation,$\lambda$ is the wavelength of light,and $a$ is the diameter of the aperture (pupil).
From the geometry of the problem,the angular separation is also given by $\theta = \frac{x}{d}$,where $x$ is the distance between the dots and $d$ is the distance of the observer from the dots.
Equating the two expressions for $\theta$: $\frac{1.22 \lambda}{a} = \frac{x}{d}$.
Rearranging to solve for $d$: $d = \frac{x \cdot a}{1.22 \lambda}$.
Given values: $x = 1 \ mm = 1 \times 10^{-3} \ m$,$a = 3 \ mm = 3 \times 10^{-3} \ m$,$\lambda = 500 \ nm = 500 \times 10^{-9} \ m$.
Substituting these values into the formula:
$d = \frac{(1 \times 10^{-3} \ m) \times (3 \times 10^{-3} \ m)}{1.22 \times 500 \times 10^{-9} \ m}$
$d = \frac{3 \times 10^{-6}}{610 \times 10^{-9}} = \frac{3000}{610} \approx 4.918 \ m$.
Rounding to the nearest integer,the maximum distance is approximately $5 \ m$.
65
PhysicsDifficultMCQAIEEE · 2005
The intensity of gamma radiation from a given source is $I$. On passing through $36 \, mm$ of lead,it is reduced to $\frac{I}{8}$. The thickness of lead which will reduce the intensity to $\frac{I}{2}$ will be......$mm$
A
$18$
B
$12$
C
$6$
D
$9$

Solution

(B) The intensity of radiation passing through a material of thickness $x$ is given by the formula $I' = I e^{-\mu x}$,where $I$ is the initial intensity,$I'$ is the final intensity,and $\mu$ is the absorption coefficient.
From the given data,when $x_1 = 36 \, mm$,the intensity becomes $I' = \frac{I}{8}$.
Substituting these values: $\frac{I}{8} = I e^{-\mu (36)} \implies e^{36\mu} = 8 = 2^3$.
Taking the natural logarithm on both sides: $36\mu = 3 \ln(2) \implies \mu = \frac{3 \ln(2)}{36} = \frac{\ln(2)}{12}$.
Now,we need to find the thickness $x_2$ such that the intensity becomes $I' = \frac{I}{2}$.
Using the formula: $\frac{I}{2} = I e^{-\mu x_2} \implies e^{\mu x_2} = 2$.
Taking the natural logarithm: $\mu x_2 = \ln(2)$.
Substituting the value of $\mu$: $(\frac{\ln(2)}{12}) x_2 = \ln(2)$.
Solving for $x_2$,we get $x_2 = 12 \, mm$.
66
PhysicsEasyMCQAIEEE · 2005
$A$ Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is
A
Straight line
B
Parabola
C
Hyperbola
D
Circle

Solution

(C) In a Young's double slit experiment,the path difference between the waves from two slits $S_1$ and $S_2$ at any point $P$ on the screen is given by $\Delta x = S_2P - S_1P = d \sin \theta$.
For a constant path difference,the locus of points $P$ on the screen forms a curve.
Mathematically,the condition $S_2P - S_1P = \text{constant}$ represents the definition of a hyperbola with the two slits as foci.
Therefore,the interference fringes formed on a screen are hyperbolic in shape.
67
PhysicsMediumMCQAIEEE · 2005
The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than $2480 \; nm$ is incident on it. The band gap in $eV$ for the semiconductor is
A
$0.9$
B
$0.7$
C
$0.5$
D
$1.1$

Solution

(C) The energy of the incident photon must be at least equal to the band gap energy $(E_g)$ to excite an electron from the valence band to the conduction band.
$E_g = \frac{hc}{\lambda}$
Given:
$h = 6.63 \times 10^{-34} \; J \cdot s$
$c = 3 \times 10^8 \; m/s$
$\lambda = 2480 \times 10^{-9} \; m$
$1 \; eV = 1.6 \times 10^{-19} \; J$
Substituting the values:
$E_g = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2480 \times 10^{-9}} \; J$
To convert into $eV$,divide by $1.6 \times 10^{-19}$:
$E_g = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2480 \times 10^{-9} \times 1.6 \times 10^{-19}} \; eV$
$E_g \approx 0.5 \; eV$
68
PhysicsMediumMCQAIEEE · 2005
The self-inductance of the motor of an electric fan is $10\;H$. In order to impart maximum power at $50\;Hz$, it should be connected to a capacitance (in $\mu F$) of:
A
$4$
B
$1$
C
$8$
D
$2$

Solution

(B) For maximum power in an $AC$ circuit, the circuit must be in resonance.
At resonance, the inductive reactance equals the capacitive reactance, i.e., $X_L = X_C$.
The resonant frequency is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.
Squaring both sides, we get $f^2 = \frac{1}{4 \pi^2 LC}$.
Rearranging for capacitance $C$, we get $C = \frac{1}{4 \pi^2 f^2 L}$.
Given $L = 10\;H$ and $f = 50\;Hz$.
Substituting the values: $C = \frac{1}{4 \times \pi^2 \times (50)^2 \times 10}$.
Using $\pi^2 \approx 10$, we get $C = \frac{1}{4 \times 10 \times 2500 \times 10} = \frac{1}{1000000} = 10^{-6}\;F$.
Thus, $C = 1\;\mu F$.
69
PhysicsMediumMCQAIEEE · 2005
$A$ circuit has a resistance of $12 \; \Omega$ and an impedance of $15 \; \Omega$. The power factor of the circuit will be
A
$0.8$
B
$0.4$
C
$1.25$
D
$0.125$

Solution

(A) The power factor of an $AC$ circuit is defined as the ratio of resistance $(R)$ to impedance $(Z)$.
Formula: $\cos \phi = \frac{R}{Z}$
Given: $R = 12 \; \Omega$ and $Z = 15 \; \Omega$.
Substituting the values: $\cos \phi = \frac{12}{15} = \frac{4}{5} = 0.8$.
Therefore,the power factor of the circuit is $0.8$.
70
PhysicsMediumMCQAIEEE · 2005
If the nucleus ${}_{13}^{27}Al$ has a nuclear radius of about $3.6 \, fm$,then ${}_{52}^{125}Te$ would have its radius approximately as .......$fm$.
A
$9.6$
B
$12$
C
$4.8$
D
$6$

Solution

(D) The nuclear radius $R$ is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number and $R_0$ is a constant.
For the given nuclei:
For ${}_{13}^{27}Al$,$A_1 = 27$ and $R_1 = 3.6 \, fm$.
For ${}_{52}^{125}Te$,$A_2 = 125$ and we need to find $R_2$.
Taking the ratio:
$\frac{R_2}{R_1} = \left( \frac{A_2}{A_1} \right)^{1/3}$
Substituting the values:
$\frac{R_2}{3.6} = \left( \frac{125}{27} \right)^{1/3}$
$\frac{R_2}{3.6} = \frac{5}{3}$
$R_2 = \frac{5}{3} \times 3.6 = 5 \times 1.2 = 6 \, fm$.
Thus,the radius of ${}_{52}^{125}Te$ is $6 \, fm$.
71
PhysicsEasyMCQAIEEE · 2005
The phase difference between the alternating current and emf is $\frac{\pi}{2}$. Which of the following cannot be the constituent of the circuit?
A
$L-C$
B
$L$ alone
C
$C$ alone
D
$R-L$

Solution

(D) In an $AC$ circuit,the phase difference $\phi$ between the alternating current and the electromotive force $(emf)$ is given by $\tan \phi = \frac{X_L - X_C}{R}$.
For a pure inductor ($L$ alone),$\phi = \frac{\pi}{2}$.
For a pure capacitor ($C$ alone),$\phi = -\frac{\pi}{2}$ (magnitude is $\frac{\pi}{2}$).
For an $L-C$ circuit,if $X_L \neq X_C$,the phase difference is $\frac{\pi}{2}$.
For an $R-L$ circuit,the phase difference $\phi$ lies in the range $0 < \phi < \frac{\pi}{2}$.
Therefore,an $R-L$ circuit cannot have a phase difference of $\frac{\pi}{2}$.
72
PhysicsMediumMCQAIEEE · 2005
Starting with a sample of pure ${}^{66}Cu$,$\frac{7}{8}$ of it decays into $Zn$ in $15 \ minutes$. The corresponding half-life is .......... $minutes$.
A
$15$
B
$10$
C
$7\frac{1}{2}$
D
$5$

Solution

(D) Given that $\frac{7}{8}$ of the sample of ${}^{66}Cu$ decays in $15 \ minutes$.
The fraction of the sample remaining undecayed is $N = 1 - \frac{7}{8} = \frac{1}{8}$.
We know that the remaining amount $N$ after $n$ half-lives is given by $N = \left(\frac{1}{2}\right)^n$.
Since $\frac{1}{8} = \left(\frac{1}{2}\right)^3$,the number of half-lives $n$ is $3$.
The relationship between time $t$,number of half-lives $n$,and half-life $T$ is $n = \frac{t}{T}$.
Substituting the values,$3 = \frac{15}{T}$.
Therefore,the half-life $T = \frac{15}{3} = 5 \ minutes$.
73
PhysicsEasyMCQAIEEE · 2005
Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$-axis from the origin,at which the net electric field due to these two point charges is zero is
A
$L/4$
B
$4 L$
C
$8 L$
D
$2 L$

Solution

(D) Let the point where the net electric field is zero be at a distance $x$ from the origin $(x=0)$.
Since the charges have opposite signs,the null point must lie outside the region between the charges,specifically on the side of the smaller magnitude charge $(-2 q)$.
Let the point be at $x > L$. The distance from $+8 q$ is $x$ and the distance from $-2 q$ is $(x - L)$.
For the net electric field to be zero,the magnitudes of the electric fields must be equal:
$E_1 = E_2$
$\frac{K(8 q)}{x^2} = \frac{K(2 q)}{(x - L)^2}$
$\frac{4}{x^2} = \frac{1}{(x - L)^2}$
Taking the square root on both sides:
$\frac{2}{x} = \frac{1}{x - L}$
$2(x - L) = x$
$2x - 2L = x$
$x = 2 L$
Thus,the point is at a distance of $2 L$ from the origin.
74
PhysicsMediumMCQAIEEE · 2005
If $I_0$ is the intensity of the principal maximum in the single slit diffraction pattern,then what will be the intensity when the slit width is doubled?
A
$\frac{I_0}{2}$
B
$I_0$
C
$4 I_0$
D
$2 I_0$

Solution

(C) In a single-slit diffraction pattern,the intensity of the principal maximum is proportional to the square of the slit width $(a^2)$ and inversely proportional to the square of the wavelength. However,the total energy incident on the slit also increases as the slit width increases. Specifically,the intensity at the central maximum is given by $I_0 \propto a^2$. When the slit width $a$ is doubled to $2a$,the amplitude of the wavelets increases by a factor of $2$,and since intensity is proportional to the square of the amplitude,the intensity becomes $I \propto (2a)^2 = 4I_0$. However,the width of the central maximum decreases by half $(w = \frac{2\lambda D}{a})$,concentrating the energy into a smaller region. For a standard single-slit diffraction experiment where the source is coherent and the slit width is increased,the intensity of the central maximum increases by a factor of $4$.
75
PhysicsEasyMCQAIEEE · 2005
When an unpolarized light of intensity $I_{0}$ is incident on a polarizing sheet, the intensity of the light which does not get transmitted is
A
$\frac{1}{2} I_{0}$
B
$\frac{1}{4} I_{0}$
C
zero
D
$I_{0}$

Solution

(A) When unpolarized light of intensity $I_{0}$ is incident on a polarizing sheet, the intensity of the transmitted light is given by $I_{t} = \frac{I_{0}}{2}$.
Since the total incident intensity is $I_{0}$ and the transmitted intensity is $\frac{I_{0}}{2}$, the intensity of the light that does not get transmitted is the difference between the incident and transmitted intensities.
Intensity of untransmitted light = $I_{0} - I_{t} = I_{0} - \frac{I_{0}}{2} = \frac{I_{0}}{2}$.

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