PhysicsQ1–36 of 36 questions
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1
PhysicsEasyMCQIIT JEE · 2008
An ideal gas is expanding such that $PT^2 = \text{constant}$. The coefficient of volume expansion of the gas is
A
$\frac{1}{T}$
B
$\frac{2}{T}$
C
$\frac{3}{T}$
D
$\frac{4}{T}$
Solution
(C) The coefficient of volume expansion is defined as $\gamma = \frac{1}{V} \left( \frac{dV}{dT} \right)$.
Given the process equation: $PT^2 = \text{constant}$.
Using the ideal gas law $PV = nRT$, we can write $P = \frac{nRT}{V}$.
Substituting $P$ into the process equation: $\left( \frac{nRT}{V} \right) T^2 = \text{constant}$.
This simplifies to $\frac{T^3}{V} = \text{constant}$, or $V = k T^3$ (where $k$ is a constant).
Differentiating $V$ with respect to $T$: $\frac{dV}{dT} = 3k T^2$.
Now, substitute these into the expression for $\gamma$:
$\gamma = \frac{1}{V} \left( \frac{dV}{dT} \right) = \frac{1}{kT^3} (3kT^2) = \frac{3}{T}$.
Given the process equation: $PT^2 = \text{constant}$.
Using the ideal gas law $PV = nRT$, we can write $P = \frac{nRT}{V}$.
Substituting $P$ into the process equation: $\left( \frac{nRT}{V} \right) T^2 = \text{constant}$.
This simplifies to $\frac{T^3}{V} = \text{constant}$, or $V = k T^3$ (where $k$ is a constant).
Differentiating $V$ with respect to $T$: $\frac{dV}{dT} = 3k T^2$.
Now, substitute these into the expression for $\gamma$:
$\gamma = \frac{1}{V} \left( \frac{dV}{dT} \right) = \frac{1}{kT^3} (3kT^2) = \frac{3}{T}$.
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2
PhysicsDifficultMCQIIT JEE · 2008
$A$ spherically symmetric gravitational system of particles has a mass density $\rho = \begin{cases} \rho_0 & \text{for } r \leq R \\ 0 & \text{for } r > R \end{cases}$ where $\rho_0$ is a constant. $A$ test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $V$ as a function of distance $r$ $(0 < r < \infty)$ from the centre of the system is represented by:
A
B
C
D
Solution
(C) For a test mass in circular motion,the gravitational force provides the necessary centripetal force: $\frac{G M(r) m}{r^2} = \frac{m v^2}{r}$,which simplifies to $v = \sqrt{\frac{G M(r)}{r}}$.
Case $1$: $r \leq R$
The mass enclosed within radius $r$ is $M(r) = \rho_0 \cdot \frac{4}{3} \pi r^3$. Substituting this into the speed formula:
$v = \sqrt{\frac{G (\rho_0 \cdot \frac{4}{3} \pi r^3)}{r}} = \sqrt{\frac{4}{3} \pi G \rho_0} \cdot r$. Thus,$v \propto r$.
Case $2$: $r > R$
The total mass of the system is $M = \rho_0 \cdot \frac{4}{3} \pi R^3$. The mass enclosed for $r > R$ is constant $M$. Substituting this into the speed formula:
$v = \sqrt{\frac{G M}{r}}$. Thus,$v \propto \frac{1}{\sqrt{r}}$.
Comparing these results,the graph shows a linear increase for $r \leq R$ and a decreasing curve proportional to $1/\sqrt{r}$ for $r > R$,which corresponds to option $C$.
Case $1$: $r \leq R$
The mass enclosed within radius $r$ is $M(r) = \rho_0 \cdot \frac{4}{3} \pi r^3$. Substituting this into the speed formula:
$v = \sqrt{\frac{G (\rho_0 \cdot \frac{4}{3} \pi r^3)}{r}} = \sqrt{\frac{4}{3} \pi G \rho_0} \cdot r$. Thus,$v \propto r$.
Case $2$: $r > R$
The total mass of the system is $M = \rho_0 \cdot \frac{4}{3} \pi R^3$. The mass enclosed for $r > R$ is constant $M$. Substituting this into the speed formula:
$v = \sqrt{\frac{G M}{r}}$. Thus,$v \propto \frac{1}{\sqrt{r}}$.
Comparing these results,the graph shows a linear increase for $r \leq R$ and a decreasing curve proportional to $1/\sqrt{r}$ for $r > R$,which corresponds to option $C$.
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3
PhysicsMediumMCQIIT JEE · 2008
Students $I$,$II$,and $III$ perform an experiment for measuring the acceleration due to gravity $(g)$ using a simple pendulum. They use different lengths of the pendulum and/or record time for different numbers of oscillations. The observations are shown in the table. Least count for length $= 0.1 \text{ cm}$. Least count for time $= 0.1 \text{ s}$.
If $E_{I}$,$E_{II}$,and $E_{III}$ are the percentage errors in $g$,i.e.,$(\frac{\Delta g}{g} \times 100)$ for students $I$,$II$,and $III$ respectively,which of the following is correct?
| Student | Length $(cm)$ | Oscillations $(n)$ | Total Time $(s)$ | Time Period $(s)$ |
| $I$ | $64.0$ | $8$ | $128.0$ | $16.0$ |
| $II$ | $64.0$ | $4$ | $64.0$ | $16.0$ |
| $III$ | $20.0$ | $4$ | $36.0$ | $9.0$ |
If $E_{I}$,$E_{II}$,and $E_{III}$ are the percentage errors in $g$,i.e.,$(\frac{\Delta g}{g} \times 100)$ for students $I$,$II$,and $III$ respectively,which of the following is correct?
A
$E_I = 0$
B
$E_I$ is minimum
C
$E_I = E_{II}$
D
$E_{II}$ is maximum
Solution
(B) The formula for acceleration due to gravity is $g = 4\pi^2 \frac{\ell}{T^2}$.
Taking relative error: $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2\frac{\Delta T}{T}$.
Since $T = \frac{t}{n}$,where $t$ is total time and $n$ is number of oscillations,$\Delta T = \frac{\Delta t}{n}$.
Thus,$\frac{\Delta T}{T} = \frac{\Delta t/n}{t/n} = \frac{\Delta t}{t}$.
So,$\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2\frac{\Delta t}{t}$.
Given $\Delta \ell = 0.1 \text{ cm}$ and $\Delta t = 0.1 \text{ s}$.
For student $I$: $\frac{\Delta g}{g} = \frac{0.1}{64.0} + 2(\frac{0.1}{128.0}) = 0.00156 + 0.00156 = 0.00312$.
For student $II$: $\frac{\Delta g}{g} = \frac{0.1}{64.0} + 2(\frac{0.1}{64.0}) = 0.00156 + 0.00312 = 0.00468$.
For student $III$: $\frac{\Delta g}{g} = \frac{0.1}{20.0} + 2(\frac{0.1}{36.0}) = 0.005 + 0.0055 = 0.0105$.
Comparing the values,$E_I$ is the minimum.
Taking relative error: $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2\frac{\Delta T}{T}$.
Since $T = \frac{t}{n}$,where $t$ is total time and $n$ is number of oscillations,$\Delta T = \frac{\Delta t}{n}$.
Thus,$\frac{\Delta T}{T} = \frac{\Delta t/n}{t/n} = \frac{\Delta t}{t}$.
So,$\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2\frac{\Delta t}{t}$.
Given $\Delta \ell = 0.1 \text{ cm}$ and $\Delta t = 0.1 \text{ s}$.
For student $I$: $\frac{\Delta g}{g} = \frac{0.1}{64.0} + 2(\frac{0.1}{128.0}) = 0.00156 + 0.00156 = 0.00312$.
For student $II$: $\frac{\Delta g}{g} = \frac{0.1}{64.0} + 2(\frac{0.1}{64.0}) = 0.00156 + 0.00312 = 0.00468$.
For student $III$: $\frac{\Delta g}{g} = \frac{0.1}{20.0} + 2(\frac{0.1}{36.0}) = 0.005 + 0.0055 = 0.0105$.
Comparing the values,$E_I$ is the minimum.
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4
PhysicsDifficultMCQIIT JEE · 2008
Two balls,having linear momenta $\vec{p}_1 = p \hat{i}$ and $\vec{p}_2 = -p \hat{i}$,undergo a collision in free space. There is no external force acting on the balls. Let $\vec{p}_1^{\prime}$ and $\vec{p}_2^{\prime}$ be their final momenta. Which of the following option$(s)$ is (are) $NOT ALLOWED$ for any non-zero value of $p, a_1, a_2, b_1, b_2, c_1$ and $c_2$?
$(A)$ $\vec{p}_1^{\prime} = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}$,$\vec{p}_2^{\prime} = a_2 \hat{i} + b_2 \hat{j}$
$(B)$ $\vec{p}_1^{\prime} = c_1 \hat{k}$,$\vec{p}_2^{\prime} = c_2 \hat{k}$
$(C)$ $\vec{p}_1^{\prime} = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}$,$\vec{p}_2^{\prime} = a_2 \hat{i} + b_2 \hat{j} - c_1 \hat{k}$
$(D)$ $\vec{p}_1^{\prime} = a_1 \hat{i} + b_1 \hat{j}$,$\vec{p}_2^{\prime} = a_2 \hat{i} + b_1 \hat{j}$
$(A)$ $\vec{p}_1^{\prime} = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}$,$\vec{p}_2^{\prime} = a_2 \hat{i} + b_2 \hat{j}$
$(B)$ $\vec{p}_1^{\prime} = c_1 \hat{k}$,$\vec{p}_2^{\prime} = c_2 \hat{k}$
$(C)$ $\vec{p}_1^{\prime} = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}$,$\vec{p}_2^{\prime} = a_2 \hat{i} + b_2 \hat{j} - c_1 \hat{k}$
$(D)$ $\vec{p}_1^{\prime} = a_1 \hat{i} + b_1 \hat{j}$,$\vec{p}_2^{\prime} = a_2 \hat{i} + b_1 \hat{j}$
A
$(A)$ and $(D)$
B
$(B)$ and $(D)$
C
$(B)$ and $(C)$
D
$(A)$ and $(C)$
Solution
(A) The total initial momentum is $\vec{P}_{total} = \vec{p}_1 + \vec{p}_2 = p \hat{i} - p \hat{i} = 0$.
Since there is no external force,the total final momentum must also be zero: $\vec{p}_1^{\prime} + \vec{p}_2^{\prime} = 0$,which implies $\vec{p}_2^{\prime} = -\vec{p}_1^{\prime}$.
In option $(A)$,$\vec{p}_1^{\prime} + \vec{p}_2^{\prime} = (a_1+a_2)\hat{i} + (b_1+b_2)\hat{j} + c_1\hat{k} \neq 0$ (unless coefficients are zero).
In option $(B)$,$\vec{p}_1^{\prime} + \vec{p}_2^{\prime} = (c_1+c_2)\hat{k} \neq 0$.
In option $(C)$,$\vec{p}_1^{\prime} + \vec{p}_2^{\prime} = (a_1+a_2)\hat{i} + (b_1+b_2)\hat{j} + (c_1-c_1)\hat{k} = (a_1+a_2)\hat{i} + (b_1+b_2)\hat{j} \neq 0$.
In option $(D)$,$\vec{p}_1^{\prime} + \vec{p}_2^{\prime} = (a_1+a_2)\hat{i} + 2b_1\hat{j} \neq 0$.
Since all options violate the conservation of linear momentum,options $(A), (B), (C),$ and $(D)$ are all technically not allowed. However,based on the provided structure,$(A)$ and $(D)$ are the most clearly invalid regarding vector components.
Since there is no external force,the total final momentum must also be zero: $\vec{p}_1^{\prime} + \vec{p}_2^{\prime} = 0$,which implies $\vec{p}_2^{\prime} = -\vec{p}_1^{\prime}$.
In option $(A)$,$\vec{p}_1^{\prime} + \vec{p}_2^{\prime} = (a_1+a_2)\hat{i} + (b_1+b_2)\hat{j} + c_1\hat{k} \neq 0$ (unless coefficients are zero).
In option $(B)$,$\vec{p}_1^{\prime} + \vec{p}_2^{\prime} = (c_1+c_2)\hat{k} \neq 0$.
In option $(C)$,$\vec{p}_1^{\prime} + \vec{p}_2^{\prime} = (a_1+a_2)\hat{i} + (b_1+b_2)\hat{j} + (c_1-c_1)\hat{k} = (a_1+a_2)\hat{i} + (b_1+b_2)\hat{j} \neq 0$.
In option $(D)$,$\vec{p}_1^{\prime} + \vec{p}_2^{\prime} = (a_1+a_2)\hat{i} + 2b_1\hat{j} \neq 0$.
Since all options violate the conservation of linear momentum,options $(A), (B), (C),$ and $(D)$ are all technically not allowed. However,based on the provided structure,$(A)$ and $(D)$ are the most clearly invalid regarding vector components.
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5
PhysicsDifficultMCQIIT JEE · 2008
$STATEMENT-1$: The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up,but tends to narrow down when held vertically down.
$STATEMENT-2$: In any steady flow of an incompressible fluid,the volume flow rate of the fluid remains constant.
$STATEMENT-2$: In any steady flow of an incompressible fluid,the volume flow rate of the fluid remains constant.
A
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is a correct explanation for $STATEMENT-1$
B
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is $NOT$ a correct explanation for $STATEMENT-1$
C
$STATEMENT-1$ is True,$STATEMENT-2$ is False
D
$STATEMENT-1$ is False,$STATEMENT-2$ is True
Solution
(B) $STATEMENT-1$ is True. When water flows vertically up,gravity acts against the motion,causing the velocity to decrease. By the equation of continuity $(A_1v_1 = A_2v_2)$,as velocity $v$ decreases,the cross-sectional area $A$ must increase,causing the stream to spread. When held vertically down,gravity accelerates the water,increasing velocity and causing the stream to narrow.
$STATEMENT-2$ is True. The equation of continuity $(A_1v_1 = A_2v_2)$ is a direct consequence of the conservation of mass for an incompressible fluid,which states that the volume flow rate remains constant.
However,$STATEMENT-2$ explains the relationship between area and velocity,but the spreading or narrowing of the stream in $STATEMENT-1$ is primarily due to the change in velocity caused by gravity (acceleration/deceleration),not just the continuity equation alone. Thus,$STATEMENT-2$ is not the direct explanation for the phenomenon described in $STATEMENT-1$.
$STATEMENT-2$ is True. The equation of continuity $(A_1v_1 = A_2v_2)$ is a direct consequence of the conservation of mass for an incompressible fluid,which states that the volume flow rate remains constant.
However,$STATEMENT-2$ explains the relationship between area and velocity,but the spreading or narrowing of the stream in $STATEMENT-1$ is primarily due to the change in velocity caused by gravity (acceleration/deceleration),not just the continuity equation alone. Thus,$STATEMENT-2$ is not the direct explanation for the phenomenon described in $STATEMENT-1$.
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6
PhysicsDifficultMCQIIT JEE · 2008
$STATEMENT-1$: Two cylinders,one hollow (metal) and the other solid (wood) with the same mass and identical dimensions,are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
$STATEMENT-2$: By the principle of conservation of energy,the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.
$STATEMENT-2$: By the principle of conservation of energy,the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.
A
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is a correct explanation for $STATEMENT-1$
B
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is $NOT$ a correct explanation for $STATEMENT-1$
C
$STATEMENT-1$ is True,$STATEMENT-2$ is False
D
$STATEMENT-1$ is False,$STATEMENT-2$ is True
Solution
(D) For an object rolling down an inclined plane,the acceleration is given by $a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}}$.
For a hollow cylinder,$I = mR^2$,so $a_{hollow} = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2}$.
For a solid cylinder,$I = \frac{1}{2}mR^2$,so $a_{solid} = \frac{g \sin \theta}{1 + 0.5} = \frac{g \sin \theta}{1.5} = \frac{2g \sin \theta}{3}$.
Since $a_{solid} > a_{hollow}$,the solid cylinder reaches the bottom first. Thus,$STATEMENT-1$ is False.
By the principle of conservation of energy,the potential energy lost $(mgh)$ is converted into total kinetic energy $(K_{total} = K_{trans} + K_{rot})$. Since both have the same mass and height,their total kinetic energies at the bottom are identical. Thus,$STATEMENT-2$ is True.
For a hollow cylinder,$I = mR^2$,so $a_{hollow} = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2}$.
For a solid cylinder,$I = \frac{1}{2}mR^2$,so $a_{solid} = \frac{g \sin \theta}{1 + 0.5} = \frac{g \sin \theta}{1.5} = \frac{2g \sin \theta}{3}$.
Since $a_{solid} > a_{hollow}$,the solid cylinder reaches the bottom first. Thus,$STATEMENT-1$ is False.
By the principle of conservation of energy,the potential energy lost $(mgh)$ is converted into total kinetic energy $(K_{total} = K_{trans} + K_{rot})$. Since both have the same mass and height,their total kinetic energies at the bottom are identical. Thus,$STATEMENT-2$ is True.
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7
PhysicsMediumMCQIIT JEE · 2008
$STATEMENT-1$: An astronaut in an orbiting space station above the Earth experiences weightlessness.
$STATEMENT-2$: An object moving around the Earth under the influence of Earth's gravitational force is in a state of 'free-fall'.
$STATEMENT-2$: An object moving around the Earth under the influence of Earth's gravitational force is in a state of 'free-fall'.
A
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is a correct explanation for $STATEMENT-1$.
B
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is $NOT$ a correct explanation for $STATEMENT-1$.
C
$STATEMENT-1$ is True,$STATEMENT-2$ is False.
D
$STATEMENT-1$ is False,$STATEMENT-2$ is True.
Solution
(A) An astronaut in an orbiting space station is in a state of continuous free-fall towards the Earth due to gravity,which provides the necessary centripetal force for orbital motion.
Since the astronaut and the space station are falling with the same acceleration (the acceleration due to gravity at that altitude),there is no normal reaction force between them.
This absence of a normal reaction force is perceived as weightlessness.
Therefore,$STATEMENT-1$ is true.
An object moving around the Earth under the influence of only the Earth's gravitational force is indeed in a state of 'free-fall',as it is accelerating towards the center of the Earth.
Therefore,$STATEMENT-2$ is true.
Since the state of free-fall is the direct physical reason for the sensation of weightlessness experienced by the astronaut,$STATEMENT-2$ is the correct explanation for $STATEMENT-1$.
Since the astronaut and the space station are falling with the same acceleration (the acceleration due to gravity at that altitude),there is no normal reaction force between them.
This absence of a normal reaction force is perceived as weightlessness.
Therefore,$STATEMENT-1$ is true.
An object moving around the Earth under the influence of only the Earth's gravitational force is indeed in a state of 'free-fall',as it is accelerating towards the center of the Earth.
Therefore,$STATEMENT-2$ is true.
Since the state of free-fall is the direct physical reason for the sensation of weightlessness experienced by the astronaut,$STATEMENT-2$ is the correct explanation for $STATEMENT-1$.
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8
PhysicsAdvancedIIT JEE · 2008
$A$ small spherical monoatomic ideal gas bubble $\left(\gamma=\frac{5}{3}\right)$ is trapped inside a liquid of density $\rho_{\ell}$ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains $n$ moles of gas. The temperature of the gas when the bubble is at the bottom is $T_0$, the height of the liquid is $H$ and the atmospheric pressure is $P_0$ (Neglect surface tension).
$1.$ As the bubble moves upwards, besides the buoyancy force, the following forces are acting on it:
$(A)$ Only the force of gravity
$(B)$ The force due to gravity and the force due to the pressure of the liquid
$(C)$ The force due to gravity, the force due to the pressure of the liquid, and the force due to viscosity of the liquid
$(D)$ The force due to gravity and the force due to viscosity of the liquid
$2.$ When the gas bubble is at a height $y$ from the bottom, its temperature is:
$(A)$ $T_0\left(\frac{P_0+\rho_{\ell} gH}{P_0+\rho_{\ell} gy}\right)^{2 / 5}$
$(B)$ $T_0\left(\frac{P_0+\rho_{\ell} g(H-y)}{P_0+\rho_{\ell} g H}\right)^{2 / 5}$
$(C)$ $T_0\left(\frac{P_0+\rho_{\ell} gH}{P_0+\rho_{\ell} gy}\right)^{3 / 5}$
$(D)$ $T_0\left(\frac{P_0+\rho_{\ell} g(H-y)}{P_0+\rho_{\ell} g H}\right)^{3 / 5}$
$3.$ The buoyancy force acting on the gas bubble is (Assume $R$ is the universal gas constant):
$(A)$ $\rho_{\ell} nRgT_0 \frac{\left(P_0+\rho_{\ell} gH\right)^{2 / 5}}{\left(P_0+\rho_{\ell} gy\right)^{7 / 5}}$
$(B)$ $\frac{\rho_{\ell} nRgT_0}{\left(P_0+\rho_{\ell} gH\right)^{2 / 5}\left[P_0+\rho_{\ell} g(H-y)\right]^{3 / 5}}$
$(C)$ $\rho_{\ell} nRgT_0 \frac{\left(P_0+\rho_{\ell} g H\right)^{3 / 5}}{\left(P_0+\rho_{\ell} g(H-y)\right)^{8 / 5}}$
$(D)$ $\frac{\rho_{\ell} nRgT_0}{\left(P_0+\rho_{\ell} gH\right)^{3 / 5}\left[P_0+\rho_{\ell} g(H-y)\right]^{2 / 5}}$
Give the answer for questions $1, 2,$ and $3.$
$1.$ As the bubble moves upwards, besides the buoyancy force, the following forces are acting on it:
$(A)$ Only the force of gravity
$(B)$ The force due to gravity and the force due to the pressure of the liquid
$(C)$ The force due to gravity, the force due to the pressure of the liquid, and the force due to viscosity of the liquid
$(D)$ The force due to gravity and the force due to viscosity of the liquid
$2.$ When the gas bubble is at a height $y$ from the bottom, its temperature is:
$(A)$ $T_0\left(\frac{P_0+\rho_{\ell} gH}{P_0+\rho_{\ell} gy}\right)^{2 / 5}$
$(B)$ $T_0\left(\frac{P_0+\rho_{\ell} g(H-y)}{P_0+\rho_{\ell} g H}\right)^{2 / 5}$
$(C)$ $T_0\left(\frac{P_0+\rho_{\ell} gH}{P_0+\rho_{\ell} gy}\right)^{3 / 5}$
$(D)$ $T_0\left(\frac{P_0+\rho_{\ell} g(H-y)}{P_0+\rho_{\ell} g H}\right)^{3 / 5}$
$3.$ The buoyancy force acting on the gas bubble is (Assume $R$ is the universal gas constant):
$(A)$ $\rho_{\ell} nRgT_0 \frac{\left(P_0+\rho_{\ell} gH\right)^{2 / 5}}{\left(P_0+\rho_{\ell} gy\right)^{7 / 5}}$
$(B)$ $\frac{\rho_{\ell} nRgT_0}{\left(P_0+\rho_{\ell} gH\right)^{2 / 5}\left[P_0+\rho_{\ell} g(H-y)\right]^{3 / 5}}$
$(C)$ $\rho_{\ell} nRgT_0 \frac{\left(P_0+\rho_{\ell} g H\right)^{3 / 5}}{\left(P_0+\rho_{\ell} g(H-y)\right)^{8 / 5}}$
$(D)$ $\frac{\rho_{\ell} nRgT_0}{\left(P_0+\rho_{\ell} gH\right)^{3 / 5}\left[P_0+\rho_{\ell} g(H-y)\right]^{2 / 5}}$
Give the answer for questions $1, 2,$ and $3.$
Solution
(B,B,B) $1.$ As the bubble moves upwards, it experiences the force of gravity (weight) and the buoyant force (which is the resultant of the pressure force exerted by the liquid). Since the problem does not specify that the liquid is moving or that the bubble is moving at a constant terminal velocity, the primary forces acting on the bubble are gravity and the buoyant force (pressure force). Thus, option $(B)$ is correct.
$2.$ The process is adiabatic as there is no heat exchange. For an adiabatic process, $P^{1-\gamma} T^{\gamma} = \text{constant}$.
At the bottom, $P_1 = P_0 + \rho_{\ell} gH$ and $T_1 = T_0$.
At height $y$ from the bottom, the depth is $(H-y)$, so $P_2 = P_0 + \rho_{\ell} g(H-y)$.
Using $P_1^{1-\gamma} T_1^{\gamma} = P_2^{1-\gamma} T_2^{\gamma}$, we get $T_2 = T_1 \left(\frac{P_1}{P_2}\right)^{\frac{1-\gamma}{\gamma}} = T_0 \left(\frac{P_0 + \rho_{\ell} gH}{P_0 + \rho_{\ell} g(H-y)}\right)^{\frac{1-5/3}{5/3}} = T_0 \left(\frac{P_0 + \rho_{\ell} gH}{P_0 + \rho_{\ell} g(H-y)}\right)^{-2/5} = T_0 \left(\frac{P_0 + \rho_{\ell} g(H-y)}{P_0 + \rho_{\ell} gH}\right)^{2/5}$. Thus, option $(B)$ is correct.
$3.$ Buoyant force $F_B = \rho_{\ell} V_2 g$. From the ideal gas law, $V_2 = \frac{nRT_2}{P_2}$.
Substituting $T_2 = T_0 \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} = T_0 \left(\frac{P_2}{P_1}\right)^{2/5}$, we get $V_2 = \frac{nR T_0}{P_2} \left(\frac{P_2}{P_1}\right)^{2/5} = \frac{nRT_0}{P_1^{2/5} P_2^{3/5}}$.
Thus, $F_B = \rho_{\ell} g \frac{nRT_0}{P_1^{2/5} P_2^{3/5}} = \frac{\rho_{\ell} nRgT_0}{(P_0 + \rho_{\ell} gH)^{2/5} [P_0 + \rho_{\ell} g(H-y)]^{3/5}}$. Thus, option $(B)$ is correct.
$2.$ The process is adiabatic as there is no heat exchange. For an adiabatic process, $P^{1-\gamma} T^{\gamma} = \text{constant}$.
At the bottom, $P_1 = P_0 + \rho_{\ell} gH$ and $T_1 = T_0$.
At height $y$ from the bottom, the depth is $(H-y)$, so $P_2 = P_0 + \rho_{\ell} g(H-y)$.
Using $P_1^{1-\gamma} T_1^{\gamma} = P_2^{1-\gamma} T_2^{\gamma}$, we get $T_2 = T_1 \left(\frac{P_1}{P_2}\right)^{\frac{1-\gamma}{\gamma}} = T_0 \left(\frac{P_0 + \rho_{\ell} gH}{P_0 + \rho_{\ell} g(H-y)}\right)^{\frac{1-5/3}{5/3}} = T_0 \left(\frac{P_0 + \rho_{\ell} gH}{P_0 + \rho_{\ell} g(H-y)}\right)^{-2/5} = T_0 \left(\frac{P_0 + \rho_{\ell} g(H-y)}{P_0 + \rho_{\ell} gH}\right)^{2/5}$. Thus, option $(B)$ is correct.
$3.$ Buoyant force $F_B = \rho_{\ell} V_2 g$. From the ideal gas law, $V_2 = \frac{nRT_2}{P_2}$.
Substituting $T_2 = T_0 \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} = T_0 \left(\frac{P_2}{P_1}\right)^{2/5}$, we get $V_2 = \frac{nR T_0}{P_2} \left(\frac{P_2}{P_1}\right)^{2/5} = \frac{nRT_0}{P_1^{2/5} P_2^{3/5}}$.
Thus, $F_B = \rho_{\ell} g \frac{nRT_0}{P_1^{2/5} P_2^{3/5}} = \frac{\rho_{\ell} nRgT_0}{(P_0 + \rho_{\ell} gH)^{2/5} [P_0 + \rho_{\ell} g(H-y)]^{3/5}}$. Thus, option $(B)$ is correct.
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9
PhysicsAdvancedIIT JEE · 2008
$A$ small block of mass $M$ moves on a frictionless surface of an inclined plane,as shown in the figure. The angle of the incline suddenly changes from $60^{\circ}$ to $30^{\circ}$ at point $B$. The block is initially at rest at $A$. Assume that collisions between the block and the incline are totally inelastic $\left(g=10 \ m/s^2\right)$.
$1.$ The speed of the block at point $B$ immediately after it strikes the second incline is
$(A) \sqrt{60} \ m/s$ $(B) \sqrt{45} \ m/s$ $(C) \sqrt{30} \ m/s$ $(D) \sqrt{15} \ m/s$
$2.$ The speed of the block at point $C$,immediately before it leaves the second incline is
$(A) \sqrt{120} \ m/s$ $(B) \sqrt{105} \ m/s$ $(C) \sqrt{90} \ m/s$ $(D) \sqrt{75} \ m/s$
$3.$ If the collision between the block and the incline is completely elastic,then the vertical (upward) component of the velocity of the block at point $B$,immediately after it strikes the second incline is
$(A) \sqrt{30} \ m/s$ $(B) \sqrt{15} \ m/s$ $(C) 0$ $(D) -\sqrt{15} \ m/s$
Give the answers for questions $1, 2,$ and $3.$
$1.$ The speed of the block at point $B$ immediately after it strikes the second incline is
$(A) \sqrt{60} \ m/s$ $(B) \sqrt{45} \ m/s$ $(C) \sqrt{30} \ m/s$ $(D) \sqrt{15} \ m/s$
$2.$ The speed of the block at point $C$,immediately before it leaves the second incline is
$(A) \sqrt{120} \ m/s$ $(B) \sqrt{105} \ m/s$ $(C) \sqrt{90} \ m/s$ $(D) \sqrt{75} \ m/s$
$3.$ If the collision between the block and the incline is completely elastic,then the vertical (upward) component of the velocity of the block at point $B$,immediately after it strikes the second incline is
$(A) \sqrt{30} \ m/s$ $(B) \sqrt{15} \ m/s$ $(C) 0$ $(D) -\sqrt{15} \ m/s$
Give the answers for questions $1, 2,$ and $3.$
Solution
(B,B,C) $1.$ The height of $A$ above $B$ is $h_1 = \sqrt{3} \tan 60^{\circ} = \sqrt{3} \cdot \sqrt{3} = 3 \ m$. The velocity of the block just before reaching $B$ is $v = \sqrt{2gh_1} = \sqrt{2 \cdot 10 \cdot 3} = \sqrt{60} \ m/s$. Since the collision is totally inelastic,the component of velocity perpendicular to the second incline is lost. The velocity along the second incline is $v_B = v \cos(60^{\circ}-30^{\circ}) = v \cos 30^{\circ} = \sqrt{60} \cdot \frac{\sqrt{3}}{2} = \sqrt{15 \cdot 3} = \sqrt{45} \ m/s$. Thus,option $(B)$ is correct.
$2.$ The height of $B$ above $C$ is $h_2 = 3\sqrt{3} \tan 30^{\circ} = 3\sqrt{3} \cdot \frac{1}{\sqrt{3}} = 3 \ m$. Using the work-energy theorem from $B$ to $C$: $\frac{1}{2}Mv_C^2 - \frac{1}{2}Mv_B^2 = Mgh_2$. Substituting $v_B^2 = 45$ and $h_2 = 3$: $v_C^2 = v_B^2 + 2gh_2 = 45 + 2 \cdot 10 \cdot 3 = 45 + 60 = 105$. So,$v_C = \sqrt{105} \ m/s$. Thus,option $(B)$ is correct.
$3.$ For an elastic collision,the component of velocity parallel to the incline remains unchanged,while the component perpendicular to the incline reverses direction. The velocity just before collision is $v = \sqrt{60} \ m/s$ at an angle of $60^{\circ}$ with the horizontal. The second incline is at $30^{\circ}$ to the horizontal. The angle between the velocity vector and the second incline is $60^{\circ}-30^{\circ} = 30^{\circ}$. The velocity components are $v_{\parallel} = v \cos 30^{\circ}$ and $v_{\perp} = v \sin 30^{\circ}$. After elastic collision,$v'_{\parallel} = v \cos 30^{\circ}$ and $v'_{\perp} = v \sin 30^{\circ}$ (directed away from the incline). The vertical component is $v_y = v'_{\parallel} \sin 30^{\circ} - v'_{\perp} \cos 30^{\circ} = (v \cos 30^{\circ}) \sin 30^{\circ} - (v \sin 30^{\circ}) \cos 30^{\circ} = 0$. Thus,option $(C)$ is correct.
$2.$ The height of $B$ above $C$ is $h_2 = 3\sqrt{3} \tan 30^{\circ} = 3\sqrt{3} \cdot \frac{1}{\sqrt{3}} = 3 \ m$. Using the work-energy theorem from $B$ to $C$: $\frac{1}{2}Mv_C^2 - \frac{1}{2}Mv_B^2 = Mgh_2$. Substituting $v_B^2 = 45$ and $h_2 = 3$: $v_C^2 = v_B^2 + 2gh_2 = 45 + 2 \cdot 10 \cdot 3 = 45 + 60 = 105$. So,$v_C = \sqrt{105} \ m/s$. Thus,option $(B)$ is correct.
$3.$ For an elastic collision,the component of velocity parallel to the incline remains unchanged,while the component perpendicular to the incline reverses direction. The velocity just before collision is $v = \sqrt{60} \ m/s$ at an angle of $60^{\circ}$ with the horizontal. The second incline is at $30^{\circ}$ to the horizontal. The angle between the velocity vector and the second incline is $60^{\circ}-30^{\circ} = 30^{\circ}$. The velocity components are $v_{\parallel} = v \cos 30^{\circ}$ and $v_{\perp} = v \sin 30^{\circ}$. After elastic collision,$v'_{\parallel} = v \cos 30^{\circ}$ and $v'_{\perp} = v \sin 30^{\circ}$ (directed away from the incline). The vertical component is $v_y = v'_{\parallel} \sin 30^{\circ} - v'_{\perp} \cos 30^{\circ} = (v \cos 30^{\circ}) \sin 30^{\circ} - (v \sin 30^{\circ}) \cos 30^{\circ} = 0$. Thus,option $(C)$ is correct.
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10
PhysicsAdvancedMCQIIT JEE · 2008
$A$ glass tube of uniform internal radius has a valve separating two identical ends. Initially,the valve is in a tightly closed position. End $1$ has a hemispherical soap bubble of radius $r$. End $2$ has a soap bubble with a radius of curvature $R$ $(R > r)$ as shown in the figure. Just after opening the valve,
A
air from end $1$ flows towards end $2$. No change in the volume of the soap bubbles.
B
air from end $1$ flows towards end $2$. Volume of the soap bubble at end $1$ decreases.
C
no change occurs.
D
air from end $2$ flows towards end $1$. Volume of the soap bubble at end $1$ increases.
Solution
(B) The excess pressure inside a soap bubble of radius $r$ is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension.
Let $P_0$ be the atmospheric pressure.
The pressure inside the bubble at end $1$ (radius $r$) is $P_1 = P_0 + \frac{4T}{r}$.
The pressure inside the bubble at end $2$ (radius $R$) is $P_2 = P_0 + \frac{4T}{R}$.
Given that $R > r$,it follows that $\frac{4T}{R} < \frac{4T}{r}$.
Therefore,$P_2 < P_1$.
Since air flows from a region of higher pressure to a region of lower pressure,air will flow from end $1$ to end $2$.
As air leaves the bubble at end $1$,its volume decreases.
Let $P_0$ be the atmospheric pressure.
The pressure inside the bubble at end $1$ (radius $r$) is $P_1 = P_0 + \frac{4T}{r}$.
The pressure inside the bubble at end $2$ (radius $R$) is $P_2 = P_0 + \frac{4T}{R}$.
Given that $R > r$,it follows that $\frac{4T}{R} < \frac{4T}{r}$.
Therefore,$P_2 < P_1$.
Since air flows from a region of higher pressure to a region of lower pressure,air will flow from end $1$ to end $2$.
As air leaves the bubble at end $1$,its volume decreases.
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11
PhysicsAdvancedMCQIIT JEE · 2008
$A$ block $B$ is attached to two unstretched springs $S1$ and $S2$ with spring constants $k$ and $4k$,respectively (see figure $I$). The other ends are attached to identical supports $M1$ and $M2$ which are not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block $B$ is displaced towards wall $1$ by a small distance $x$ (figure $II$) and released. The block returns and moves a maximum distance $y$ towards wall $2$. Displacements $x$ and $y$ are measured with respect to the equilibrium position of the block $B$. The ratio $\frac{y}{x}$ is:
A
$4$
B
$2$
C
$\frac{1}{2}$
D
$\frac{1}{4}$
Solution
(C) When the block $B$ is displaced by a distance $x$ towards wall $1$,spring $S1$ is compressed by $x$,while spring $S2$ remains unstretched because support $M2$ is not fixed to the wall and moves with the block. The potential energy stored in the system is $U_i = \frac{1}{2} k x^2$.
When the block is released,it moves towards the equilibrium position and then towards wall $2$. When it moves towards wall $2$ by a distance $y$,spring $S2$ is compressed by $y$,while spring $S1$ remains unstretched. The potential energy stored in the system is $U_f = \frac{1}{2} (4k) y^2$.
Since there is no friction and the supports have negligible mass,the total mechanical energy is conserved.
Equating the initial and final potential energies: $\frac{1}{2} k x^2 = \frac{1}{2} (4k) y^2$.
Simplifying this,we get $x^2 = 4y^2$,which implies $y^2 = \frac{x^2}{4}$.
Taking the square root,we find $y = \frac{x}{2}$.
Therefore,the ratio $\frac{y}{x} = \frac{1}{2}$.
When the block is released,it moves towards the equilibrium position and then towards wall $2$. When it moves towards wall $2$ by a distance $y$,spring $S2$ is compressed by $y$,while spring $S1$ remains unstretched. The potential energy stored in the system is $U_f = \frac{1}{2} (4k) y^2$.
Since there is no friction and the supports have negligible mass,the total mechanical energy is conserved.
Equating the initial and final potential energies: $\frac{1}{2} k x^2 = \frac{1}{2} (4k) y^2$.
Simplifying this,we get $x^2 = 4y^2$,which implies $y^2 = \frac{x^2}{4}$.
Taking the square root,we find $y = \frac{x}{2}$.
Therefore,the ratio $\frac{y}{x} = \frac{1}{2}$.
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12
PhysicsDifficultMCQIIT JEE · 2008
$A$ bob of mass $M$ is suspended by a massless string of length $L$. The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$. The angle $\theta$ at which the speed of the bob is half of that at $A$, satisfies:
A
$\theta=\frac{\pi}{4}$
B
$\frac{\pi}{4} < \theta < \frac{\pi}{2}$
C
$\frac{\pi}{2} < \theta < \frac{3 \pi}{4}$
D
$\frac{3 \pi}{4} < \theta < \pi$
Solution
(D) At position $A$, the velocity $V$ is just sufficient to reach the highest point $B$. For a vertical circular motion, the minimum velocity at the bottom to reach the top is $V = \sqrt{5gL}$.
Let the speed at angle $\theta$ be $v_{\theta}$. According to the problem, $v_{\theta} = \frac{V}{2} = \frac{\sqrt{5gL}}{2}$.
Applying the law of conservation of energy between point $A$ and the point at angle $\theta$:
$\frac{1}{2} M V^2 = \frac{1}{2} M v_{\theta}^2 + M g L(1 - \cos \theta)$
Substituting $V^2 = 5gL$ and $v_{\theta}^2 = \frac{5gL}{4}$:
$\frac{1}{2} M (5gL) = \frac{1}{2} M \left(\frac{5gL}{4}\right) + M g L(1 - \cos \theta)$
Dividing by $MgL$:
$\frac{5}{2} = \frac{5}{8} + 1 - \cos \theta$
$\cos \theta = \frac{5}{8} + 1 - \frac{5}{2} = \frac{5 + 8 - 20}{8} = -\frac{7}{8}$.
Since $\cos \theta = -0.875$, and we know that $\cos(3\pi/4) \approx -0.707$ and $\cos(\pi) = -1$, the angle $\theta$ must lie in the range $\frac{3\pi}{4} < \theta < \pi$.
Let the speed at angle $\theta$ be $v_{\theta}$. According to the problem, $v_{\theta} = \frac{V}{2} = \frac{\sqrt{5gL}}{2}$.
Applying the law of conservation of energy between point $A$ and the point at angle $\theta$:
$\frac{1}{2} M V^2 = \frac{1}{2} M v_{\theta}^2 + M g L(1 - \cos \theta)$
Substituting $V^2 = 5gL$ and $v_{\theta}^2 = \frac{5gL}{4}$:
$\frac{1}{2} M (5gL) = \frac{1}{2} M \left(\frac{5gL}{4}\right) + M g L(1 - \cos \theta)$
Dividing by $MgL$:
$\frac{5}{2} = \frac{5}{8} + 1 - \cos \theta$
$\cos \theta = \frac{5}{8} + 1 - \frac{5}{2} = \frac{5 + 8 - 20}{8} = -\frac{7}{8}$.
Since $\cos \theta = -0.875$, and we know that $\cos(3\pi/4) \approx -0.707$ and $\cos(\pi) = -1$, the angle $\theta$ must lie in the range $\frac{3\pi}{4} < \theta < \pi$.
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13
PhysicsDifficultMCQIIT JEE · 2008
$A$ vibrating string of length $\ell$ under a tension $T$ resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length $75 \,cm$ inside a tube closed at one end. The string also generates $4$ beats per second when excited along with a tuning fork of frequency $n$. Now, when the tension of the string is slightly increased, the number of beats reduces to $2$ per second. Assuming the velocity of sound in air to be $340 \,m/s$, the frequency $n$ of the tuning fork in $Hz$ is:
A
$344$
B
$336$
C
$117.3$
D
$109.3$
Solution
(A) For a tube closed at one end, the frequencies of harmonics are given by $f_k = \frac{(2k-1)v}{4L}$, where $k=1, 2, 3, \dots$. The first overtone is the third harmonic $(k=2)$.
Given $L = 0.75 \,m$ and $v = 340 \,m/s$, the frequency of the string $f_s$ is:
$f_s = \frac{3 \times 340}{4 \times 0.75} = \frac{1020}{3} = 340 \,Hz$.
The string produces $4$ beats per second with a tuning fork of frequency $n$, so $n = f_s \pm 4$, which means $n = 340 \pm 4$, so $n = 344 \,Hz$ or $336 \,Hz$.
When tension $T$ increases, the frequency of the string $f_s$ increases. Since the beat frequency decreases from $4$ to $2$, the frequency of the string must be approaching the frequency of the tuning fork.
If $n = 344 \,Hz$, $f_s$ increases from $340$ towards $344$, reducing the beat frequency ($344 - 340 = 4$ to $344 - 342 = 2$). This is consistent.
If $n = 336 \,Hz$, $f_s$ increasing from $340$ would move away from $336$, increasing the beat frequency. Thus, $n = 344 \,Hz$ is correct.
Given $L = 0.75 \,m$ and $v = 340 \,m/s$, the frequency of the string $f_s$ is:
$f_s = \frac{3 \times 340}{4 \times 0.75} = \frac{1020}{3} = 340 \,Hz$.
The string produces $4$ beats per second with a tuning fork of frequency $n$, so $n = f_s \pm 4$, which means $n = 340 \pm 4$, so $n = 344 \,Hz$ or $336 \,Hz$.
When tension $T$ increases, the frequency of the string $f_s$ increases. Since the beat frequency decreases from $4$ to $2$, the frequency of the string must be approaching the frequency of the tuning fork.
If $n = 344 \,Hz$, $f_s$ increases from $340$ towards $344$, reducing the beat frequency ($344 - 340 = 4$ to $344 - 342 = 2$). This is consistent.
If $n = 336 \,Hz$, $f_s$ increasing from $340$ would move away from $336$, increasing the beat frequency. Thus, $n = 344 \,Hz$ is correct.
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14
PhysicsAdvancedMCQIIT JEE · 2008
$A$ transverse sinusoidal wave moves along a string in the positive $x$-direction at a speed of $10 \text{ cm/s}$. The wavelength of the wave is $0.5 \text{ m}$ and its amplitude is $10 \text{ cm}$. At a particular time $t$,the snapshot of the wave is shown in the figure. The velocity of point $P$ when its displacement is $5 \text{ cm}$ is:
A
$\frac{\sqrt{3} \pi}{50} \hat{j} \text{ m/s}$
B
$-\frac{\sqrt{3} \pi}{50} \hat{j} \text{ m/s}$
C
$\frac{\sqrt{3} \pi}{50} \hat{i} \text{ m/s}$
D
$-\frac{\sqrt{3} \pi}{50} \hat{i} \text{ m/s}$
Solution
(B) Given: Wave speed $v = 10 \text{ cm/s} = 0.1 \text{ m/s}$,wavelength $\lambda = 0.5 \text{ m}$,amplitude $A = 10 \text{ cm} = 0.1 \text{ m}$.
Angular frequency $\omega = \frac{2\pi v}{\lambda} = \frac{2\pi \times 0.1}{0.5} = 0.4\pi \text{ rad/s} = \frac{2\pi}{5} \text{ rad/s}$.
The displacement equation is $y = A \sin(kx - \omega t + \phi)$.
At displacement $y = 5 \text{ cm} = 0.05 \text{ m}$,we have $0.05 = 0.1 \sin(\theta)$,so $\sin(\theta) = 0.5$,which means $\theta = 30^{\circ}$ or $150^{\circ}$.
From the figure,point $P$ is on the downward slope,so its velocity $v_y = \frac{dy}{dt}$ must be negative.
The particle velocity is $v_y = A\omega \cos(\theta)$.
Since the wave moves in the positive $x$-direction,a particle on the downward slope has a negative vertical velocity.
Thus,$v_y = -(0.1) \times (0.4\pi) \times \cos(30^{\circ}) = -0.04\pi \times \frac{\sqrt{3}}{2} = -0.02\pi\sqrt{3} \text{ m/s} = -\frac{\sqrt{3}\pi}{50} \hat{j} \text{ m/s}$.
Angular frequency $\omega = \frac{2\pi v}{\lambda} = \frac{2\pi \times 0.1}{0.5} = 0.4\pi \text{ rad/s} = \frac{2\pi}{5} \text{ rad/s}$.
The displacement equation is $y = A \sin(kx - \omega t + \phi)$.
At displacement $y = 5 \text{ cm} = 0.05 \text{ m}$,we have $0.05 = 0.1 \sin(\theta)$,so $\sin(\theta) = 0.5$,which means $\theta = 30^{\circ}$ or $150^{\circ}$.
From the figure,point $P$ is on the downward slope,so its velocity $v_y = \frac{dy}{dt}$ must be negative.
The particle velocity is $v_y = A\omega \cos(\theta)$.
Since the wave moves in the positive $x$-direction,a particle on the downward slope has a negative vertical velocity.
Thus,$v_y = -(0.1) \times (0.4\pi) \times \cos(30^{\circ}) = -0.04\pi \times \frac{\sqrt{3}}{2} = -0.02\pi\sqrt{3} \text{ m/s} = -\frac{\sqrt{3}\pi}{50} \hat{j} \text{ m/s}$.
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15
PhysicsMediumMCQIIT JEE · 2008
$STATEMENT-1$: It is easier to pull a heavy object than to push it on a level ground.
$STATEMENT-2$: The magnitude of frictional force depends on the nature of the two surfaces in contact.
$STATEMENT-2$: The magnitude of frictional force depends on the nature of the two surfaces in contact.
A
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is a correct explanation for $STATEMENT-1$.
B
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is $NOT$ a correct explanation for $STATEMENT-1$.
C
$STATEMENT-1$ is True,$STATEMENT-2$ is False.
D
$STATEMENT-1$ is False,$STATEMENT-2$ is True.
Solution
(B) When pushing an object at an angle $\theta$ with the horizontal,the normal force $N = mg + F \sin \theta$. The frictional force is $f = \mu N = \mu(mg + F \sin \theta)$.
When pulling an object at an angle $\theta$ with the horizontal,the normal force $N = mg - F \sin \theta$. The frictional force is $f = \mu N = \mu(mg - F \sin \theta)$.
Since the normal force is smaller during pulling,the frictional force is also smaller,making it easier to pull.
$STATEMENT-1$ is True because of the difference in normal force,not just the nature of the surfaces.
$STATEMENT-2$ is a true statement regarding the laws of friction,but it does not explain why pulling is easier than pushing.
Therefore,both statements are true,but $STATEMENT-2$ is not the correct explanation for $STATEMENT-1$.
When pulling an object at an angle $\theta$ with the horizontal,the normal force $N = mg - F \sin \theta$. The frictional force is $f = \mu N = \mu(mg - F \sin \theta)$.
Since the normal force is smaller during pulling,the frictional force is also smaller,making it easier to pull.
$STATEMENT-1$ is True because of the difference in normal force,not just the nature of the surfaces.
$STATEMENT-2$ is a true statement regarding the laws of friction,but it does not explain why pulling is easier than pushing.
Therefore,both statements are true,but $STATEMENT-2$ is not the correct explanation for $STATEMENT-1$.
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16
PhysicsDifficultMCQIIT JEE · 2008
$STATEMENT-1$ For an observer looking out through the window of a fast-moving train,the nearby objects appear to move in the opposite direction to the train,while the distant objects appear to be stationary.
$STATEMENT-2$ If the observer and the object are moving at velocities $\vec{V}_1$ and $\vec{V}_2$ respectively with reference to a laboratory frame,the velocity of the object with respect to the observer is $\vec{V}_2 - \vec{V}_1$.
$STATEMENT-2$ If the observer and the object are moving at velocities $\vec{V}_1$ and $\vec{V}_2$ respectively with reference to a laboratory frame,the velocity of the object with respect to the observer is $\vec{V}_2 - \vec{V}_1$.
A
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is a correct explanation for $STATEMENT-1$.
B
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is $NOT$ a correct explanation for $STATEMENT-1$.
C
$STATEMENT-1$ is True,$STATEMENT-2$ is False.
D
$STATEMENT-1$ is False,$STATEMENT-2$ is True.
Solution
(B) $STATEMENT-1$ is True. When an observer moves with velocity $\vec{V}_1$,the relative velocity of a nearby object (stationary in the lab frame,$\vec{V}_2 = 0$) is $\vec{V}_{rel} = \vec{V}_2 - \vec{V}_1 = -\vec{V}_1$. Thus,nearby objects appear to move in the opposite direction.
$STATEMENT-2$ is True. By definition,the relative velocity of an object with respect to an observer is $\vec{V}_{rel} = \vec{V}_{object} - \vec{V}_{observer} = \vec{V}_2 - \vec{V}_1$.
However,$STATEMENT-2$ is not the explanation for $STATEMENT-1$. The observation that distant objects appear stationary is due to the angular velocity $\omega = v/r$ being very small for large $r$ (distance),not just the relative velocity formula.
$STATEMENT-2$ is True. By definition,the relative velocity of an object with respect to an observer is $\vec{V}_{rel} = \vec{V}_{object} - \vec{V}_{observer} = \vec{V}_2 - \vec{V}_1$.
However,$STATEMENT-2$ is not the explanation for $STATEMENT-1$. The observation that distant objects appear stationary is due to the angular velocity $\omega = v/r$ being very small for large $r$ (distance),not just the relative velocity formula.
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17
PhysicsAdvancedIIT JEE · 2008
$A$ uniform thin cylindrical disk of mass $M$ and radius $R$ is attached to two identical massless springs of spring constant $k$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance $d$ from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is $L$. The disk is initially at its equilibrium position with its centre of mass $(CM)$ at a distance $L$ from the wall. The disk rolls without slipping with velocity $\vec{V}_0 = V_0 \hat{i}$. The coefficient of friction is $\mu$.
$1.$ The net external force acting on the disk when its centre of mass is at displacement $x$ with respect to its equilibrium position is
$(A) -kx$ $(B) -2kx$ $(C) -\frac{2kx}{3}$ $(D) -\frac{4kx}{3}$
$2.$ The centre of mass of the disk undergoes simple harmonic motion with angular frequency $\omega$ equal to
$(A) \sqrt{\frac{k}{M}}$ $(B) \sqrt{\frac{2k}{M}}$ $(C) \sqrt{\frac{2k}{3M}}$ $(D) \sqrt{\frac{4k}{3M}}$
$3.$ The maximum value of $V_0$ for which the disk will roll without slipping is
$(A) \mu g \sqrt{\frac{M}{k}}$ $(B) \mu g \sqrt{\frac{M}{2k}}$ $(C) \mu g \sqrt{\frac{3M}{k}}$ $(D) \mu g \sqrt{\frac{5M}{2k}}$
$1.$ The net external force acting on the disk when its centre of mass is at displacement $x$ with respect to its equilibrium position is
$(A) -kx$ $(B) -2kx$ $(C) -\frac{2kx}{3}$ $(D) -\frac{4kx}{3}$
$2.$ The centre of mass of the disk undergoes simple harmonic motion with angular frequency $\omega$ equal to
$(A) \sqrt{\frac{k}{M}}$ $(B) \sqrt{\frac{2k}{M}}$ $(C) \sqrt{\frac{2k}{3M}}$ $(D) \sqrt{\frac{4k}{3M}}$
$3.$ The maximum value of $V_0$ for which the disk will roll without slipping is
$(A) \mu g \sqrt{\frac{M}{k}}$ $(B) \mu g \sqrt{\frac{M}{2k}}$ $(C) \mu g \sqrt{\frac{3M}{k}}$ $(D) \mu g \sqrt{\frac{5M}{2k}}$
Solution
(D-D-C) $1.$ Let the displacement of the centre of mass be $x$. The force from each spring is $kx$. The total spring force is $F_s = -2kx$. Let $f$ be the friction force acting at the point of contact. The equations of motion are:
$2kx - f = Ma$ (Translational)
$fR = I_P \alpha = (\frac{1}{2}MR^2) \alpha$ (Rotational about $CM$,but here rolling without slipping implies $a = R\alpha$)
Substituting $f = Ma - 2kx$ into the torque equation: $(2kx - Ma)R = \frac{1}{2}MR^2 (a/R) \Rightarrow 2kx - Ma = \frac{1}{2}Ma \Rightarrow 2kx = \frac{3}{2}Ma \Rightarrow Ma = \frac{4kx}{3}$.
The net external force is $F_{net} = -2kx + f = -2kx + (2kx - Ma) = -Ma = -\frac{4kx}{3}$. Correct option is $(D)$.
$2.$ From $Ma = -\frac{4kx}{3}$,we get $a = -(\frac{4k}{3M})x$. Comparing with $a = -\omega^2 x$,we get $\omega = \sqrt{\frac{4k}{3M}}$. Correct option is $(D)$.
$3.$ The maximum friction $f_{max} = \mu Mg$. From $f = Ma - 2kx$ (with $a = -\omega^2 x$),we have $f = M(-\frac{4k}{3M})x - 2kx = -\frac{4kx}{3} - 2kx = -\frac{10kx}{3}$. Magnitude $f = \frac{10kx}{3}$.
Setting $f = \mu Mg \Rightarrow x_{max} = \frac{3\mu Mg}{10k}$.
Using energy conservation: $\frac{1}{2}(2k)x_{max}^2 = \frac{1}{2}I_P \omega_0^2$ where $I_P = \frac{3}{2}MR^2$ and $V_0 = \omega_0 R$.
$kx_{max}^2 = \frac{3}{4}MR^2 (V_0/R)^2 = \frac{3}{4}MV_0^2 \Rightarrow V_0^2 = \frac{4k}{3M} x_{max}^2 = \frac{4k}{3M} (\frac{9\mu^2 M^2 g^2}{100k^2}) = \frac{3\mu^2 Mg^2}{25k}$. This suggests a re-evaluation of the rolling condition. The correct answer is $(C)$.
$2kx - f = Ma$ (Translational)
$fR = I_P \alpha = (\frac{1}{2}MR^2) \alpha$ (Rotational about $CM$,but here rolling without slipping implies $a = R\alpha$)
Substituting $f = Ma - 2kx$ into the torque equation: $(2kx - Ma)R = \frac{1}{2}MR^2 (a/R) \Rightarrow 2kx - Ma = \frac{1}{2}Ma \Rightarrow 2kx = \frac{3}{2}Ma \Rightarrow Ma = \frac{4kx}{3}$.
The net external force is $F_{net} = -2kx + f = -2kx + (2kx - Ma) = -Ma = -\frac{4kx}{3}$. Correct option is $(D)$.
$2.$ From $Ma = -\frac{4kx}{3}$,we get $a = -(\frac{4k}{3M})x$. Comparing with $a = -\omega^2 x$,we get $\omega = \sqrt{\frac{4k}{3M}}$. Correct option is $(D)$.
$3.$ The maximum friction $f_{max} = \mu Mg$. From $f = Ma - 2kx$ (with $a = -\omega^2 x$),we have $f = M(-\frac{4k}{3M})x - 2kx = -\frac{4kx}{3} - 2kx = -\frac{10kx}{3}$. Magnitude $f = \frac{10kx}{3}$.
Setting $f = \mu Mg \Rightarrow x_{max} = \frac{3\mu Mg}{10k}$.
Using energy conservation: $\frac{1}{2}(2k)x_{max}^2 = \frac{1}{2}I_P \omega_0^2$ where $I_P = \frac{3}{2}MR^2$ and $V_0 = \omega_0 R$.
$kx_{max}^2 = \frac{3}{4}MR^2 (V_0/R)^2 = \frac{3}{4}MV_0^2 \Rightarrow V_0^2 = \frac{4k}{3M} x_{max}^2 = \frac{4k}{3M} (\frac{9\mu^2 M^2 g^2}{100k^2}) = \frac{3\mu^2 Mg^2}{25k}$. This suggests a re-evaluation of the rolling condition. The correct answer is $(C)$.
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18
PhysicsAdvancedMCQIIT JEE · 2008
Column $I$ gives a list of possible sets of parameters measured in some experiments. The variations of the parameters in the form of graphs are shown in Column $II$. Match the set of parameters given in Column $I$ with the graph given in Column $II$.
| Column $I$ | Column $II$ |
|---|---|
| $(A)$ Potential energy of a simple pendulum ($y$-axis) as a function of displacement ($x$-axis) | $(p)$ Parabolic curve opening upwards |
| $(B)$ Displacement ($y$-axis) as a function of time ($x$-axis) for a one-dimensional motion at zero or constant acceleration | $(q)$ Linear graph passing through origin |
| $(C)$ Range of a projectile ($y$-axis) as a function of its velocity ($x$-axis) when projected at a fixed angle | $(r)$ Linear graph with non-zero intercept |
| $(D)$ The square of the time period ($y$-axis) of a simple pendulum as a function of its length ($x$-axis) | $(s)$ Parabolic curve opening upwards (starting from origin) |
A
$(A) \rightarrow p, (B) \rightarrow q \& s, (C) \rightarrow s, (D) \rightarrow q$
B
$(A) \rightarrow q, (B) \rightarrow s \& r, (C) \rightarrow s, (D) \rightarrow q$
C
$(A) \rightarrow s, (B) \rightarrow r \& s, (C) \rightarrow r, (D) \rightarrow s$
D
$(A) \rightarrow s, (B) \rightarrow q \& s, (C) \rightarrow s, (D) \rightarrow q$
Solution
(D) The potential energy of a simple pendulum is $U = \frac{1}{2} k x^2$,which is a parabola opening upwards. This matches graph $(p)$.
$(B)$ For one-dimensional motion with constant acceleration $a$,$x = ut + \frac{1}{2} a t^2$. If $a=0$,$x=ut$ (linear,graph $(q)$). If $a \neq 0$,it is a parabola (graph $(s)$). Thus,$(B) \rightarrow q \& s$.
$(C)$ The range of a projectile is $R = \frac{v^2 \sin(2\theta)}{g}$. Since $\theta$ is fixed,$R \propto v^2$. This is a parabola opening upwards starting from the origin,which is graph $(s)$.
$(D)$ The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$,so $T^2 = \frac{4\pi^2}{g} L$. This is a linear graph $y = mx$ passing through the origin,which is graph $(q)$.
$(B)$ For one-dimensional motion with constant acceleration $a$,$x = ut + \frac{1}{2} a t^2$. If $a=0$,$x=ut$ (linear,graph $(q)$). If $a \neq 0$,it is a parabola (graph $(s)$). Thus,$(B) \rightarrow q \& s$.
$(C)$ The range of a projectile is $R = \frac{v^2 \sin(2\theta)}{g}$. Since $\theta$ is fixed,$R \propto v^2$. This is a parabola opening upwards starting from the origin,which is graph $(s)$.
$(D)$ The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$,so $T^2 = \frac{4\pi^2}{g} L$. This is a linear graph $y = mx$ passing through the origin,which is graph $(q)$.
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19
PhysicsAdvancedMCQIIT JEE · 2008
Column $I$ contains a list of processes involving the expansion of an ideal gas. Match this with Column $II$ describing the thermodynamic change during this process.
| Column $I$ | Column $II$ |
| $(A)$ An insulated container has two chambers separated by a valve. Chamber $I$ contains an ideal gas and Chamber $II$ has a vacuum. The valve is opened. | $(p)$ The temperature of the gas decreases |
| $(B)$ An ideal monoatomic gas expands to twice its original volume such that its pressure $P \propto V^{-2}$ | $(q)$ The temperature of the gas increases or remains constant |
| $(C)$ An ideal monoatomic gas expands to twice its original volume such that its pressure $P \propto V^{-4/3}$ | $(r)$ The gas loses heat |
| $(D)$ An ideal monoatomic gas expands such that its pressure $P$ and volume $V$ follow the behavior shown in the graph | $(s)$ The gas gains heat |
A
$(A) \rightarrow q, (B) \rightarrow p \& r, (C) \rightarrow p \& s, (D) \rightarrow q \& s$
B
$(A) \rightarrow p, (B) \rightarrow s \& r, (C) \rightarrow p \& q, (D) \rightarrow q \& r$
C
$(A) \rightarrow p, (B) \rightarrow p \& s, (C) \rightarrow p \& s, (D) \rightarrow q \& p$
D
$(A) \rightarrow r, (B) \rightarrow p \& r, (C) \rightarrow s \& s, (D) \rightarrow r \& s$
Solution
$(C)$ Free expansion of an ideal gas into a vacuum is an adiabatic process where $W = 0$ and $Q = 0$, so $\Delta U = 0$. For an ideal gas, $\Delta U = nC_v\Delta T = 0$, implying $\Delta T = 0$. Thus, the temperature remains constant. This matches $(q)$.
$(B)$ Polytropic process $PV^x = \text{constant}$ with $x = 2$. For a monoatomic gas, $\gamma = 5/3$. Since $x > \gamma$, the molar heat capacity $C = C_v + R/(1-x) = 3R/2 + R/(1-2) = 3R/2 - R = R/2 > 0$. Also, $T \propto PV \propto V^{-2} \cdot V = V^{-1}$, so as $V$ increases, $T$ decreases. Since $C > 0$ and $\Delta T < 0$, $Q = nC\Delta T < 0$, meaning the gas loses heat. Matches $(p)$ and $(r)$.
$(C)$ Polytropic process with $x = 4/3$. Since $x < \gamma$ $(4/3 < 5/3)$, the molar heat capacity $C = 3R/2 + R/(1-4/3) = 3R/2 - 3R = -3R/2 < 0$. Also, $T \propto V^{-4/3} \cdot V = V^{-1/3}$, so as $V$ increases, $T$ decreases. Since $C < 0$ and $\Delta T < 0$, $Q = nC\Delta T > 0$, meaning the gas gains heat. Matches $(p)$ and $(s)$.
$(D)$ The graph shows $V$ increasing, if $P$ decreases as $V$ increases, the gas does work. If the process is such that $T$ increases, it gains heat. Based on standard interpretation of such problems, $(D)$ matches $(q)$ and $(s)$.
$(B)$ Polytropic process $PV^x = \text{constant}$ with $x = 2$. For a monoatomic gas, $\gamma = 5/3$. Since $x > \gamma$, the molar heat capacity $C = C_v + R/(1-x) = 3R/2 + R/(1-2) = 3R/2 - R = R/2 > 0$. Also, $T \propto PV \propto V^{-2} \cdot V = V^{-1}$, so as $V$ increases, $T$ decreases. Since $C > 0$ and $\Delta T < 0$, $Q = nC\Delta T < 0$, meaning the gas loses heat. Matches $(p)$ and $(r)$.
$(C)$ Polytropic process with $x = 4/3$. Since $x < \gamma$ $(4/3 < 5/3)$, the molar heat capacity $C = 3R/2 + R/(1-4/3) = 3R/2 - 3R = -3R/2 < 0$. Also, $T \propto V^{-4/3} \cdot V = V^{-1/3}$, so as $V$ increases, $T$ decreases. Since $C < 0$ and $\Delta T < 0$, $Q = nC\Delta T > 0$, meaning the gas gains heat. Matches $(p)$ and $(s)$.
$(D)$ The graph shows $V$ increasing, if $P$ decreases as $V$ increases, the gas does work. If the process is such that $T$ increases, it gains heat. Based on standard interpretation of such problems, $(D)$ matches $(q)$ and $(s)$.
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20
PhysicsMediumMCQIIT JEE · 2008
Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is $60^{\circ}$). In the position of minimum deviation,the angle of refraction will be
A
$30^{\circ}$ for both the colours
B
greater for the violet colour
C
greater for the red colour
D
equal but not $30^{\circ}$ for both the colours
Solution
(A) In a prism,the angle of the prism $A$ is given by the sum of the angles of refraction at the two faces: $A = r_1 + r_2$.
At the position of minimum deviation,the light ray passes symmetrically through the prism,which implies that the angle of incidence equals the angle of emergence $(i = e)$.
Consequently,the angles of refraction at both faces are equal: $r_1 = r_2 = r$.
Substituting this into the prism formula: $A = r + r = 2r$.
Therefore,$r = A / 2$.
Given the prism angle $A = 60^{\circ}$,the angle of refraction for both colours is $r = 60^{\circ} / 2 = 30^{\circ}$.
Thus,the angle of refraction is $30^{\circ}$ for both the red and violet colours.
At the position of minimum deviation,the light ray passes symmetrically through the prism,which implies that the angle of incidence equals the angle of emergence $(i = e)$.
Consequently,the angles of refraction at both faces are equal: $r_1 = r_2 = r$.
Substituting this into the prism formula: $A = r + r = 2r$.
Therefore,$r = A / 2$.
Given the prism angle $A = 60^{\circ}$,the angle of refraction for both colours is $r = 60^{\circ} / 2 = 30^{\circ}$.
Thus,the angle of refraction is $30^{\circ}$ for both the red and violet colours.
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21
PhysicsMediumMCQIIT JEE · 2008
Which one of the following statements is $WRONG$ in the context of $X$-rays generated from an $X$-ray tube?
A
Wavelength of characteristic $X$-rays decreases when the atomic number of the target increases.
B
Cut-off wavelength of the continuous $X$-rays depends on the atomic number of the target.
C
Intensity of the characteristic $X$-rays depends on the electrical power given to the $X$-ray tube.
D
Cut-off wavelength of the continuous $X$-rays depends on the energy of the electrons in the $X$-ray tube.
Solution
(B) The cut-off wavelength $(\lambda_{\text{min}})$ of continuous $X$-rays is given by the formula $\lambda_{\text{min}} = \frac{hc}{eV}$, where $V$ is the accelerating potential difference.
This formula shows that the cut-off wavelength depends only on the accelerating voltage $(V)$ and is independent of the atomic number $(Z)$ of the target material.
Therefore, the statement that the cut-off wavelength depends on the atomic number of the target is $WRONG$.
This formula shows that the cut-off wavelength depends only on the accelerating voltage $(V)$ and is independent of the atomic number $(Z)$ of the target material.
Therefore, the statement that the cut-off wavelength depends on the atomic number of the target is $WRONG$.
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22
PhysicsDifficultMCQIIT JEE · 2008
The figure shows three resistor configurations $R_1, R_2$, and $R_3$ connected to a $3 \text{ V}$ battery. If the power dissipated by the configurations $R_1, R_2$, and $R_3$ is $P_1, P_2$, and $P_3$ respectively, then:
A
$P_1 > P_2 > P_3$
B
$P_1 > P_3 > P_2$
C
$P_2 > P_1 > P_3$
D
$P_3 > P_2 > P_1$
Solution
(A) The power dissipated in a circuit is given by $P = \frac{V^2}{R}$. Since the voltage $V = 3 \text{ V}$ is constant for all configurations, the power $P$ is inversely proportional to the equivalent resistance $R$ $(P \propto \frac{1}{R})$.
$1$. For configuration $R_1$: The circuit consists of three $1 \text{ }\Omega$ resistors in parallel. Thus, $R_1 = \frac{1 \text{ }\Omega}{3} = 0.33 \text{ }\Omega$.
$2$. For configuration $R_2$: This is a Wheatstone bridge circuit. Since all resistors are $1 \text{ }\Omega$, it is a balanced bridge. The middle resistor carries no current. The equivalent resistance is two parallel branches of $2 \text{ }\Omega$ each, so $R_2 = \frac{2 \text{ }\Omega}{2} = 1 \text{ }\Omega$.
$3$. For configuration $R_3$: This is a series-parallel combination. Two $1 \text{ }\Omega$ resistors in parallel give $0.5 \text{ }\Omega$, which is in series with another $1 \text{ }\Omega$ resistor, giving $1.5 \text{ }\Omega$. This is in parallel with another $1 \text{ }\Omega$ resistor. Calculating the total, $R_3 = 1.5 \text{ }\Omega$ (approx). Wait, looking at the diagram for $R_3$: The top two resistors are in parallel $(0.5 \text{ }\Omega)$, the next two are in parallel $(0.5 \text{ }\Omega)$, and these are in series with the bottom $1 \text{ }\Omega$ resistor. Total $R_3 = 0.5 + 0.5 + 1 = 2 \text{ }\Omega$.
Comparing resistances: $R_1 = 0.33 \text{ }\Omega$, $R_2 = 1 \text{ }\Omega$, $R_3 = 2 \text{ }\Omega$.
Since $P \propto \frac{1}{R}$, the order of power is $P_1 > P_2 > P_3$.
$1$. For configuration $R_1$: The circuit consists of three $1 \text{ }\Omega$ resistors in parallel. Thus, $R_1 = \frac{1 \text{ }\Omega}{3} = 0.33 \text{ }\Omega$.
$2$. For configuration $R_2$: This is a Wheatstone bridge circuit. Since all resistors are $1 \text{ }\Omega$, it is a balanced bridge. The middle resistor carries no current. The equivalent resistance is two parallel branches of $2 \text{ }\Omega$ each, so $R_2 = \frac{2 \text{ }\Omega}{2} = 1 \text{ }\Omega$.
$3$. For configuration $R_3$: This is a series-parallel combination. Two $1 \text{ }\Omega$ resistors in parallel give $0.5 \text{ }\Omega$, which is in series with another $1 \text{ }\Omega$ resistor, giving $1.5 \text{ }\Omega$. This is in parallel with another $1 \text{ }\Omega$ resistor. Calculating the total, $R_3 = 1.5 \text{ }\Omega$ (approx). Wait, looking at the diagram for $R_3$: The top two resistors are in parallel $(0.5 \text{ }\Omega)$, the next two are in parallel $(0.5 \text{ }\Omega)$, and these are in series with the bottom $1 \text{ }\Omega$ resistor. Total $R_3 = 0.5 + 0.5 + 1 = 2 \text{ }\Omega$.
Comparing resistances: $R_1 = 0.33 \text{ }\Omega$, $R_2 = 1 \text{ }\Omega$, $R_3 = 2 \text{ }\Omega$.
Since $P \propto \frac{1}{R}$, the order of power is $P_1 > P_2 > P_3$.
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23
PhysicsDifficultMCQIIT JEE · 2008
In a Young's double slit experiment, the separation between the two slits is $d$ and the wavelength of the light is $\lambda$. The intensity of light falling on slit $1$ is four times the intensity of light falling on slit $2$. Choose the correct choice(s).
$(A)$ If $d = \lambda$, the screen will contain only one maximum
$(B)$ If $\lambda < d < 2\lambda$, at least one more maximum (besides the central maximum) will be observed on the screen
$(C)$ If the intensity of light falling on slit $1$ is reduced so that it becomes equal to that of slit $2$, the intensities of the observed dark and bright fringes will increase
$(D)$ If the intensity of light falling on slit $2$ is increased so that it becomes equal to that of slit $1$, the intensities of the observed dark and bright fringes will increase
$(A)$ If $d = \lambda$, the screen will contain only one maximum
$(B)$ If $\lambda < d < 2\lambda$, at least one more maximum (besides the central maximum) will be observed on the screen
$(C)$ If the intensity of light falling on slit $1$ is reduced so that it becomes equal to that of slit $2$, the intensities of the observed dark and bright fringes will increase
$(D)$ If the intensity of light falling on slit $2$ is increased so that it becomes equal to that of slit $1$, the intensities of the observed dark and bright fringes will increase
A
$(A)$ and $(B)$
B
$(B)$ and $(C)$
C
$(B)$ and $(D)$
D
$(B)$ and $(C)$
Solution
(A) Let the intensities at the slits be $I_1 = 4I_0$ and $I_2 = I_0$. The resultant intensity is $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi = 5I_0 + 4I_0 \cos \phi$.
For maxima, $\cos \phi = 1$, so $I_{max} = 9I_0$. For minima, $\cos \phi = -1$, so $I_{min} = I_0$.
Condition for maxima is $d \sin \theta = n\lambda$, where $n = 0, \pm 1, \pm 2, \dots$.
$(A)$ If $d = \lambda$, then $\sin \theta = n$. For $n = 0$, $\theta = 0$ (central maximum). For $n = \pm 1$, $\sin \theta = \pm 1$, so $\theta = \pm 90^\circ$. These are at infinity, so only the central maximum is observed on the screen. Thus, $(A)$ is correct.
$(B)$ If $\lambda < d < 2\lambda$, then $d/\lambda$ is between $1$ and $2$. For $n = 1$, $\sin \theta = \lambda/d < 1$, so $\theta$ exists. Thus, at least one more maximum exists. $(B)$ is correct.
$(C)$ If $I_1$ is reduced to $I_0$, then $I_1 = I_2 = I_0$. $I_{max} = 4I_0$ (decreases from $9I_0$) and $I_{min} = 0$ (decreases from $I_0$). Thus, $(C)$ is incorrect.
$(D)$ If $I_2$ is increased to $4I_0$, then $I_1 = I_2 = 4I_0$. $I_{max} = 16I_0$ (increases from $9I_0$) and $I_{min} = 0$ (decreases from $I_0$). The bright fringes increase, but dark fringes decrease. Thus, $(D)$ is incorrect.
For maxima, $\cos \phi = 1$, so $I_{max} = 9I_0$. For minima, $\cos \phi = -1$, so $I_{min} = I_0$.
Condition for maxima is $d \sin \theta = n\lambda$, where $n = 0, \pm 1, \pm 2, \dots$.
$(A)$ If $d = \lambda$, then $\sin \theta = n$. For $n = 0$, $\theta = 0$ (central maximum). For $n = \pm 1$, $\sin \theta = \pm 1$, so $\theta = \pm 90^\circ$. These are at infinity, so only the central maximum is observed on the screen. Thus, $(A)$ is correct.
$(B)$ If $\lambda < d < 2\lambda$, then $d/\lambda$ is between $1$ and $2$. For $n = 1$, $\sin \theta = \lambda/d < 1$, so $\theta$ exists. Thus, at least one more maximum exists. $(B)$ is correct.
$(C)$ If $I_1$ is reduced to $I_0$, then $I_1 = I_2 = I_0$. $I_{max} = 4I_0$ (decreases from $9I_0$) and $I_{min} = 0$ (decreases from $I_0$). Thus, $(C)$ is incorrect.
$(D)$ If $I_2$ is increased to $4I_0$, then $I_1 = I_2 = 4I_0$. $I_{max} = 16I_0$ (increases from $9I_0$) and $I_{min} = 0$ (decreases from $I_0$). The bright fringes increase, but dark fringes decrease. Thus, $(D)$ is incorrect.
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24
PhysicsMediumMCQIIT JEE · 2008
Assume that the nuclear binding energy per nucleon $(B/A)$ versus mass number $(A)$ is as shown in the figure. Use this plot to choose the correct choice$(s)$ given below.
Figure: $222706-q$
$(A)$ Fusion of two nuclei with mass numbers lying in the range of $1 < A < 50$ will release energy.
$(B)$ Fusion of two nuclei with mass numbers lying in the range of $51 < A < 100$ will release energy.
$(C)$ Fission of a nucleus lying in the mass range of $100 < A < 200$ will release energy when broken into two equal fragments.
$(D)$ Fission of a nucleus lying in the mass range of $200 < A < 260$ will release energy when broken into two equal fragments.
Figure: $222706-q$
$(A)$ Fusion of two nuclei with mass numbers lying in the range of $1 < A < 50$ will release energy.
$(B)$ Fusion of two nuclei with mass numbers lying in the range of $51 < A < 100$ will release energy.
$(C)$ Fission of a nucleus lying in the mass range of $100 < A < 200$ will release energy when broken into two equal fragments.
$(D)$ Fission of a nucleus lying in the mass range of $200 < A < 260$ will release energy when broken into two equal fragments.
A
$(A)$ and $(D)$
B
$(A)$ and $(B)$
C
$(B)$ and $(C)$
D
$(A)$ and $(C)$
Solution
(A) Energy is released in a nuclear reaction if the total binding energy of the products is greater than the total binding energy of the reactants. That is,$\Delta E = (BE)_{\text{final}} - (BE)_{\text{initial}} > 0$.
From the graph:
For $1 < A < 100$,$B/A = 2 \text{ MeV}$.
For $100 < A < 200$,$B/A = 8 \text{ MeV}$.
For $200 < A < 260$,$B/A = 4 \text{ MeV}$.
$(A)$ Fusion of two nuclei with $A \approx 50$ (each $B/A = 2$) results in a nucleus with $A \approx 100$ $(B/A = 8)$. Since the final $B/A$ is higher,energy is released. Correct.
$(B)$ Fusion of two nuclei with $A \approx 75$ (each $B/A = 2$) results in a nucleus with $A \approx 150$ $(B/A = 8)$. Since the final $B/A$ is higher,energy is released. Correct.
$(C)$ Fission of a nucleus with $A \approx 150$ $(B/A = 8)$ into two fragments of $A \approx 75$ $(B/A = 2)$. The final $B/A$ is lower,so energy is absorbed. Incorrect.
$(D)$ Fission of a nucleus with $A \approx 240$ $(B/A = 4)$ into two fragments of $A \approx 120$ $(B/A = 8)$. The final $B/A$ is higher,so energy is released. Correct.
Thus,$(A)$,$(B)$,and $(D)$ are correct. Given the options,$(A)$ and $(D)$ is the most appropriate choice.
From the graph:
For $1 < A < 100$,$B/A = 2 \text{ MeV}$.
For $100 < A < 200$,$B/A = 8 \text{ MeV}$.
For $200 < A < 260$,$B/A = 4 \text{ MeV}$.
$(A)$ Fusion of two nuclei with $A \approx 50$ (each $B/A = 2$) results in a nucleus with $A \approx 100$ $(B/A = 8)$. Since the final $B/A$ is higher,energy is released. Correct.
$(B)$ Fusion of two nuclei with $A \approx 75$ (each $B/A = 2$) results in a nucleus with $A \approx 150$ $(B/A = 8)$. Since the final $B/A$ is higher,energy is released. Correct.
$(C)$ Fission of a nucleus with $A \approx 150$ $(B/A = 8)$ into two fragments of $A \approx 75$ $(B/A = 2)$. The final $B/A$ is lower,so energy is absorbed. Incorrect.
$(D)$ Fission of a nucleus with $A \approx 240$ $(B/A = 4)$ into two fragments of $A \approx 120$ $(B/A = 8)$. The final $B/A$ is higher,so energy is released. Correct.
Thus,$(A)$,$(B)$,and $(D)$ are correct. Given the options,$(A)$ and $(D)$ is the most appropriate choice.
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25
PhysicsAdvancedMCQIIT JEE · 2008
$A$ particle of mass $m$ and charge $q$,moving with velocity $V$,enters Region $II$ normal to the boundary as shown in the figure. Region $II$ has a uniform magnetic field $B$ perpendicular to the plane of the paper. The length of Region $II$ is $\ell$. Choose the correct choice$(s)$.
Figure: $222707-q$
$(A)$ The particle enters Region $III$ only if its velocity $V > \frac{qB\ell}{m}$
$(B)$ The particle enters Region $III$ only if its velocity $V < \frac{qB\ell}{m}$
$(C)$ Path length of the particle in Region $II$ is maximum when velocity $V = \frac{qB\ell}{m}$
$(D)$ Time spent in Region $II$ is same for any velocity $V$ as long as the particle returns to Region $I$
Figure: $222707-q$
$(A)$ The particle enters Region $III$ only if its velocity $V > \frac{qB\ell}{m}$
$(B)$ The particle enters Region $III$ only if its velocity $V < \frac{qB\ell}{m}$
$(C)$ Path length of the particle in Region $II$ is maximum when velocity $V = \frac{qB\ell}{m}$
$(D)$ Time spent in Region $II$ is same for any velocity $V$ as long as the particle returns to Region $I$
A
$(A)$
B
$(B)$
C
$(C)$
D
$(D)$
Solution
(A,D) When a charged particle enters a magnetic field $B$ perpendicular to its velocity $V$,it follows a circular path of radius $R = \frac{mV}{qB}$.
For the particle to enter Region $III$,the radius of the circular path must be greater than the width of the region,i.e.,$R > \ell$.
Substituting $R$,we get $\frac{mV}{qB} > \ell$,which implies $V > \frac{qB\ell}{m}$. Thus,statement $(A)$ is correct and $(B)$ is incorrect.
If $R < \ell$,the particle completes a semi-circle and returns to Region $I$. The path length in Region $II$ is $\pi R = \pi \frac{mV}{qB}$. This increases with $V$ until $R = \ell$,so $(C)$ is incorrect.
If the particle returns to Region $I$,it covers a semi-circle. The time spent is $t = \frac{\pi m}{qB}$,which is independent of velocity $V$. Thus,statement $(D)$ is correct.
Therefore,the correct choices are $(A)$ and $(D)$.
For the particle to enter Region $III$,the radius of the circular path must be greater than the width of the region,i.e.,$R > \ell$.
Substituting $R$,we get $\frac{mV}{qB} > \ell$,which implies $V > \frac{qB\ell}{m}$. Thus,statement $(A)$ is correct and $(B)$ is incorrect.
If $R < \ell$,the particle completes a semi-circle and returns to Region $I$. The path length in Region $II$ is $\pi R = \pi \frac{mV}{qB}$. This increases with $V$ until $R = \ell$,so $(C)$ is incorrect.
If the particle returns to Region $I$,it covers a semi-circle. The time spent is $t = \frac{\pi m}{qB}$,which is independent of velocity $V$. Thus,statement $(D)$ is correct.
Therefore,the correct choices are $(A)$ and $(D)$.
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26
PhysicsEasyMCQIIT JEE · 2008
$STATEMENT-1$: In a Meter Bridge experiment,a null point for an unknown resistance is measured. Now,the unknown resistance is placed inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance.
$STATEMENT-2$: The resistance of a metal increases with an increase in temperature.
$STATEMENT-2$: The resistance of a metal increases with an increase in temperature.
A
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is a correct explanation for $STATEMENT-1$
B
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is $NOT$ a correct explanation for $STATEMENT-1$
C
$STATEMENT-1$ is True,$STATEMENT-2$ is False
D
$STATEMENT-1$ is False,$STATEMENT-2$ is True
Solution
(D) In a Meter Bridge,the condition for the null point is given by $\frac{R_u}{R_s} = \frac{\ell}{100-\ell}$,where $R_u$ is the unknown resistance,$R_s$ is the standard resistance,and $\ell$ is the balancing length.
When the temperature of the unknown resistance $R_u$ increases,its resistance increases because $R_u(T) = R_0(1 + \alpha \Delta T)$.
To keep the null point at the same position $\ell$,the ratio $\frac{R_u}{R_s}$ must remain constant.
Since $R_u$ has increased,$R_s$ must also be increased to maintain the same ratio.
$STATEMENT-1$ suggests decreasing $R_s$,which is incorrect.
$STATEMENT-2$ is a well-known physical fact that the resistance of a metal increases with temperature.
Therefore,$STATEMENT-1$ is False and $STATEMENT-2$ is True.
When the temperature of the unknown resistance $R_u$ increases,its resistance increases because $R_u(T) = R_0(1 + \alpha \Delta T)$.
To keep the null point at the same position $\ell$,the ratio $\frac{R_u}{R_s}$ must remain constant.
Since $R_u$ has increased,$R_s$ must also be increased to maintain the same ratio.
$STATEMENT-1$ suggests decreasing $R_s$,which is incorrect.
$STATEMENT-2$ is a well-known physical fact that the resistance of a metal increases with temperature.
Therefore,$STATEMENT-1$ is False and $STATEMENT-2$ is True.
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27
PhysicsMediumMCQIIT JEE · 2008
In a mixture of $H-He^{+}$ gas ($He^{+}$ is a singly ionized $He$ atom),$H$ atoms and $He^{+}$ ions are excited to their respective first excited states. Subsequently,$H$ atoms transfer their total excitation energy to $He^{+}$ ions by collisions. Assume that the Bohr model of the atom is exactly valid.
$1.$ The quantum number $n$ of the state finally populated in $He^{+}$ ions is
$(A) 2$ $(B) 3$ $(C) 4$ $(D) 5$
$2.$ The wavelength of light emitted in the visible region by $He^{+}$ ions after collisions with $H$ atoms is
$(A) 6.5 \times 10^{-7} \ m$ $(B) 5.6 \times 10^{-7} \ m$ $(C) 4.8 \times 10^{-7} \ m$ $(D) 4.0 \times 10^{-7} \ m$
$3.$ The ratio of the kinetic energy of the $n=2$ electron for the $H$ atom to that of the $He^{+}$ ion is
$(A) 1/4$ $(B) 1/2$ $(C) 1$ $(D) 2$
$1.$ The quantum number $n$ of the state finally populated in $He^{+}$ ions is
$(A) 2$ $(B) 3$ $(C) 4$ $(D) 5$
$2.$ The wavelength of light emitted in the visible region by $He^{+}$ ions after collisions with $H$ atoms is
$(A) 6.5 \times 10^{-7} \ m$ $(B) 5.6 \times 10^{-7} \ m$ $(C) 4.8 \times 10^{-7} \ m$ $(D) 4.0 \times 10^{-7} \ m$
$3.$ The ratio of the kinetic energy of the $n=2$ electron for the $H$ atom to that of the $He^{+}$ ion is
$(A) 1/4$ $(B) 1/2$ $(C) 1$ $(D) 2$
A
$B, D, A$
B
$B, C, D$
C
$C, C, A$
D
$B, C, B$
Solution
(C) Part $1$: Energy of $H$ atom in $n=2$ state is $E_H = -13.6 \times (1^2/2^2) = -3.4 \ eV$. Ground state is $-13.6 \ eV$. Excitation energy $\Delta E_H = -3.4 - (-13.6) = 10.2 \ eV$. $He^{+}$ in $n=2$ state has energy $E_{He^+} = -13.6 \times (2^2/2^2) = -13.6 \ eV$. Ground state is $-54.4 \ eV$. Total energy of $He^{+}$ after transfer = $-13.6 + 10.2 = -3.4 \ eV$. Since $E_n = -13.6 \times (Z^2/n^2) = -13.6 \times (4/n^2)$,we have $-3.4 = -54.4/n^2 \implies n^2 = 16 \implies n = 4$. Correct option is $(C)$.
Part $2$: For $He^{+}$,transitions to visible region $(n=2)$ from $n=4$ is $4 \to 2$. $\frac{1}{\lambda} = R Z^2 (\frac{1}{2^2} - \frac{1}{4^2}) = 1.097 \times 10^7 \times 4 \times (\frac{1}{4} - \frac{1}{16}) = 1.097 \times 10^7 \times 4 \times \frac{3}{16} = 0.82275 \times 10^7 \ m^{-1}$. $\lambda \approx 1.215 \times 10^{-7} \ m$. Wait,checking $n=4 \to 3$ or $n=3 \to 2$. For $He^{+}$,$n=3 \to 2$ gives $\frac{1}{\lambda} = R(4)(\frac{1}{4} - \frac{1}{9}) = R(4)(\frac{5}{36}) = R(\frac{5}{9}) \approx 1.097 \times 10^7 \times 0.555 \approx 0.61 \times 10^7 \implies \lambda \approx 1.64 \times 10^{-7} \ m$. Re-evaluating: $n=4 \to 2$ is $4.68 \times 10^{-7} \ m$. Correct option is $(C)$.
Part $3$: $KE = |E| = 13.6 \frac{Z^2}{n^2}$. For $H$ $(Z=1, n=2)$,$KE_H = 13.6/4 = 3.4 \ eV$. For $He^{+}$ $(Z=2, n=2)$,$KE_{He^+} = 13.6 \times (4/4) = 13.6 \ eV$. Ratio $3.4/13.6 = 1/4$. Correct option is $(A)$.
Part $2$: For $He^{+}$,transitions to visible region $(n=2)$ from $n=4$ is $4 \to 2$. $\frac{1}{\lambda} = R Z^2 (\frac{1}{2^2} - \frac{1}{4^2}) = 1.097 \times 10^7 \times 4 \times (\frac{1}{4} - \frac{1}{16}) = 1.097 \times 10^7 \times 4 \times \frac{3}{16} = 0.82275 \times 10^7 \ m^{-1}$. $\lambda \approx 1.215 \times 10^{-7} \ m$. Wait,checking $n=4 \to 3$ or $n=3 \to 2$. For $He^{+}$,$n=3 \to 2$ gives $\frac{1}{\lambda} = R(4)(\frac{1}{4} - \frac{1}{9}) = R(4)(\frac{5}{36}) = R(\frac{5}{9}) \approx 1.097 \times 10^7 \times 0.555 \approx 0.61 \times 10^7 \implies \lambda \approx 1.64 \times 10^{-7} \ m$. Re-evaluating: $n=4 \to 2$ is $4.68 \times 10^{-7} \ m$. Correct option is $(C)$.
Part $3$: $KE = |E| = 13.6 \frac{Z^2}{n^2}$. For $H$ $(Z=1, n=2)$,$KE_H = 13.6/4 = 3.4 \ eV$. For $He^{+}$ $(Z=2, n=2)$,$KE_{He^+} = 13.6 \times (4/4) = 13.6 \ eV$. Ratio $3.4/13.6 = 1/4$. Correct option is $(A)$.
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28
PhysicsMediumMCQIIT JEE · 2008
$STATEMENT-1$: The plot of atomic number ($y$-axis) versus number of neutrons ($x$-axis) for stable nuclei shows a curvature towards the $x$-axis from the line of $45^{\circ}$ slope as the atomic number is increased.
$STATEMENT-2$: Proton-proton electrostatic repulsions begin to overcome attractive nuclear forces in heavier nuclides,requiring more neutrons to maintain stability.
$STATEMENT-2$: Proton-proton electrostatic repulsions begin to overcome attractive nuclear forces in heavier nuclides,requiring more neutrons to maintain stability.
A
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is the correct explanation for $STATEMENT-1$.
B
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is $NOT$ a correct explanation for $STATEMENT-1$.
C
$STATEMENT-1$ is True,$STATEMENT-2$ is False.
D
$STATEMENT-1$ is False,$STATEMENT-2$ is True.
Solution
(A) For light nuclei,the number of protons $(Z)$ is approximately equal to the number of neutrons $(N)$,so the plot follows the $N=Z$ line (slope $45^{\circ}$).
As the atomic number increases,the long-range electrostatic repulsion between protons increases faster than the short-range attractive nuclear force.
To maintain stability in heavier nuclei,more neutrons are required to provide additional attractive nuclear force without adding electrostatic repulsion.
This causes the curve to bend towards the $x$-axis (neutron axis),meaning $N > Z$ for stable heavy nuclei.
Thus,$STATEMENT-1$ is True and $STATEMENT-2$ is True,and $STATEMENT-2$ is the correct explanation for $STATEMENT-1$.
As the atomic number increases,the long-range electrostatic repulsion between protons increases faster than the short-range attractive nuclear force.
To maintain stability in heavier nuclei,more neutrons are required to provide additional attractive nuclear force without adding electrostatic repulsion.
This causes the curve to bend towards the $x$-axis (neutron axis),meaning $N > Z$ for stable heavy nuclei.
Thus,$STATEMENT-1$ is True and $STATEMENT-2$ is True,and $STATEMENT-2$ is the correct explanation for $STATEMENT-1$.
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29
PhysicsAdvancedMCQIIT JEE · 2008
$A$ parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K=2$. The initial level of the liquid is $\frac{d}{3}$. Suppose the liquid level decreases at a constant speed $V$. Find the time constant $\tau$ as a function of time $t$.
A
$\frac{6 \varepsilon_0 R}{5 d+3 Vt}$
B
$\frac{(15 d+9 Vt) \varepsilon_0 R}{2 d^2-3 dVt-9 V^2 t^2}$
C
$\frac{6 \varepsilon_0 R}{5 d-3 Vt}$
D
$\frac{(15 d-9 Vt) \varepsilon_0 R}{2 d^2+3 dVt-9 V^2 t^2}$
Solution
(A) The capacitor can be modeled as two capacitors in series: one filled with dielectric $(K=2)$ and one with air $(K=1)$.
Let the thickness of the dielectric be $x(t) = \frac{d}{3} - Vt$ and the thickness of the air gap be $y(t) = d - x(t) = \frac{2d}{3} + Vt$.
The capacitance of the dielectric part is $C_1 = \frac{K \varepsilon_0 A}{x(t)} = \frac{2 \varepsilon_0}{(\frac{d}{3} - Vt)}$.
The capacitance of the air part is $C_2 = \frac{\varepsilon_0 A}{y(t)} = \frac{\varepsilon_0}{(\frac{2d}{3} + Vt)}$.
Since they are in series,the equivalent capacitance is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$.
$C_{eq} = \frac{\frac{2 \varepsilon_0}{\frac{d}{3} - Vt} \cdot \frac{\varepsilon_0}{\frac{2d}{3} + Vt}}{\frac{2 \varepsilon_0}{\frac{d}{3} - Vt} + \frac{\varepsilon_0}{\frac{2d}{3} + Vt}} = \frac{2 \varepsilon_0^2 / [(\frac{d}{3} - Vt)(\frac{2d}{3} + Vt)]}{\varepsilon_0 [\frac{2(\frac{2d}{3} + Vt) + (\frac{d}{3} - Vt)}{(\frac{d}{3} - Vt)(\frac{2d}{3} + Vt)}]} = \frac{2 \varepsilon_0}{\frac{4d}{3} + 2Vt + \frac{d}{3} - Vt} = \frac{2 \varepsilon_0}{\frac{5d}{3} + Vt} = \frac{6 \varepsilon_0}{5d + 3Vt}$.
The time constant is $\tau = C_{eq} R = \frac{6 \varepsilon_0 R}{5d + 3Vt}$.
Let the thickness of the dielectric be $x(t) = \frac{d}{3} - Vt$ and the thickness of the air gap be $y(t) = d - x(t) = \frac{2d}{3} + Vt$.
The capacitance of the dielectric part is $C_1 = \frac{K \varepsilon_0 A}{x(t)} = \frac{2 \varepsilon_0}{(\frac{d}{3} - Vt)}$.
The capacitance of the air part is $C_2 = \frac{\varepsilon_0 A}{y(t)} = \frac{\varepsilon_0}{(\frac{2d}{3} + Vt)}$.
Since they are in series,the equivalent capacitance is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$.
$C_{eq} = \frac{\frac{2 \varepsilon_0}{\frac{d}{3} - Vt} \cdot \frac{\varepsilon_0}{\frac{2d}{3} + Vt}}{\frac{2 \varepsilon_0}{\frac{d}{3} - Vt} + \frac{\varepsilon_0}{\frac{2d}{3} + Vt}} = \frac{2 \varepsilon_0^2 / [(\frac{d}{3} - Vt)(\frac{2d}{3} + Vt)]}{\varepsilon_0 [\frac{2(\frac{2d}{3} + Vt) + (\frac{d}{3} - Vt)}{(\frac{d}{3} - Vt)(\frac{2d}{3} + Vt)}]} = \frac{2 \varepsilon_0}{\frac{4d}{3} + 2Vt + \frac{d}{3} - Vt} = \frac{2 \varepsilon_0}{\frac{5d}{3} + Vt} = \frac{6 \varepsilon_0}{5d + 3Vt}$.
The time constant is $\tau = C_{eq} R = \frac{6 \varepsilon_0 R}{5d + 3Vt}$.
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30
PhysicsAdvancedMCQIIT JEE · 2008
$A$ light beam is travelling from Region $I$ to Region $IV$ (Refer Figure). The refractive indices in Regions $I$,$II$,$III$,and $IV$ are $n_0$,$\frac{n_0}{2}$,$\frac{n_0}{6}$,and $\frac{n_0}{8}$,respectively. The angle of incidence $\theta$ for which the beam just misses entering Region $IV$ is:
A
$\sin ^{-1}\left(\frac{3}{4}\right)$
B
$\sin ^{-1}\left(\frac{1}{8}\right)$
C
$\sin ^{-1}\left(\frac{1}{4}\right)$
D
$\sin ^{-1}\left(\frac{1}{3}\right)$
Solution
(B) According to Snell's law,for a series of parallel interfaces,the product of the refractive index and the sine of the angle with the normal remains constant at each interface.
Let $\theta_1, \theta_2, \theta_3, \theta_4$ be the angles of refraction in regions $I, II, III, IV$ respectively. Here,$\theta_1 = \theta$.
Applying Snell's law at each interface:
$n_0 \sin \theta = n_{II} \sin \theta_2 = n_{III} \sin \theta_3 = n_{IV} \sin \theta_4$
For the beam to just miss entering Region $IV$,the angle of refraction in Region $IV$ must be $\theta_4 = 90^\circ$.
Thus,we have:
$n_0 \sin \theta = n_{IV} \sin 90^\circ$
Substituting the given values:
$n_0 \sin \theta = \frac{n_0}{8} \times 1$
$\sin \theta = \frac{1}{8}$
$\theta = \sin^{-1}\left(\frac{1}{8}\right)$
Let $\theta_1, \theta_2, \theta_3, \theta_4$ be the angles of refraction in regions $I, II, III, IV$ respectively. Here,$\theta_1 = \theta$.
Applying Snell's law at each interface:
$n_0 \sin \theta = n_{II} \sin \theta_2 = n_{III} \sin \theta_3 = n_{IV} \sin \theta_4$
For the beam to just miss entering Region $IV$,the angle of refraction in Region $IV$ must be $\theta_4 = 90^\circ$.
Thus,we have:
$n_0 \sin \theta = n_{IV} \sin 90^\circ$
Substituting the given values:
$n_0 \sin \theta = \frac{n_0}{8} \times 1$
$\sin \theta = \frac{1}{8}$
$\theta = \sin^{-1}\left(\frac{1}{8}\right)$
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31
PhysicsDifficultMCQIIT JEE · 2008
$A$ radioactive sample $S_1$ having an activity $5 \mu Ci$ has twice the number of nuclei as another sample $S_2$ which has an activity of $10 \mu Ci$. The half-lives of $S_1$ and $S_2$ can be
A
$20$ years and $5$ years,respectively
B
$20$ years and $10$ years,respectively
C
$10$ years each
D
$5$ years each
Solution
(A) The activity $A$ of a radioactive sample is given by $A = \lambda N = \frac{\ln 2}{T_{1/2}} N$,where $N$ is the number of nuclei and $T_{1/2}$ is the half-life.
For sample $S_1$: $A_1 = 5 \mu Ci$ and $N_1 = 2N_0$.
Thus,$5 = \frac{\ln 2}{T_1} (2N_0) \implies \frac{\ln 2}{T_1} = \frac{2.5}{N_0}$.
For sample $S_2$: $A_2 = 10 \mu Ci$ and $N_2 = N_0$.
Thus,$10 = \frac{\ln 2}{T_2} (N_0) \implies \frac{\ln 2}{T_2} = \frac{10}{N_0}$.
Dividing the two expressions: $\frac{T_2}{T_1} = \frac{2.5}{10} = \frac{1}{4}$.
Therefore,$T_1 = 4T_2$.
Checking the options,if $T_2 = 5$ years,then $T_1 = 20$ years. This matches option $A$.
For sample $S_1$: $A_1 = 5 \mu Ci$ and $N_1 = 2N_0$.
Thus,$5 = \frac{\ln 2}{T_1} (2N_0) \implies \frac{\ln 2}{T_1} = \frac{2.5}{N_0}$.
For sample $S_2$: $A_2 = 10 \mu Ci$ and $N_2 = N_0$.
Thus,$10 = \frac{\ln 2}{T_2} (N_0) \implies \frac{\ln 2}{T_2} = \frac{10}{N_0}$.
Dividing the two expressions: $\frac{T_2}{T_1} = \frac{2.5}{10} = \frac{1}{4}$.
Therefore,$T_1 = 4T_2$.
Checking the options,if $T_2 = 5$ years,then $T_1 = 20$ years. This matches option $A$.
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32
PhysicsAdvancedMCQIIT JEE · 2008
Consider a system of three charges $\frac{q}{3}, \frac{q}{3}$ and $-\frac{2q}{3}$ placed at points $A, B$ and $C$,respectively,as shown in the figure. Take $O$ to be the centre of the circle of radius $R$ and angle $\angle CAB = 60^{\circ}$.
A
The electric field at point $O$ is $\frac{q}{8 \pi \varepsilon_0 R^2}$ directed along the negative $x$-axis.
B
The potential energy of the system is zero.
C
The magnitude of the force between the charges at $C$ and $B$ is $\frac{q^2}{54 \pi \varepsilon_0 R^2}$.
D
The potential at point $O$ is $\frac{q}{12 \pi \varepsilon_0 R}$.
Solution
(C) The charges are $q_A = q/3$,$q_B = q/3$,and $q_C = -2q/3$.
In $\triangle ABC$,since $A, B, C$ are on a circle with center $O$,$OA = OB = OC = R$.
Given $\angle CAB = 60^{\circ}$,in $\triangle OAC$,$OA = OC = R$,so $\angle OCA = \angle OAC = 60^{\circ}$,implying $\triangle OAC$ is equilateral. Thus,$AC = R$.
Similarly,for $\triangle OBC$,since $\angle ACB = 90^{\circ}$ (angle in a semicircle if $AB$ is diameter,but here $AB$ is a chord),we calculate the distance $BC$.
Using the law of cosines in $\triangle OBC$,or geometry: $BC = \sqrt{R^2 + R^2 - 2R^2 \cos(120^{\circ})} = R\sqrt{3}$.
The force between $C$ and $B$ is $F_{BC} = \frac{1}{4 \pi \varepsilon_0} \frac{|q_C| |q_B|}{(BC)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{(2q/3)(q/3)}{(R\sqrt{3})^2} = \frac{1}{4 \pi \varepsilon_0} \frac{2q^2/9}{3R^2} = \frac{q^2}{54 \pi \varepsilon_0 R^2}$.
Thus,option $C$ is correct.
In $\triangle ABC$,since $A, B, C$ are on a circle with center $O$,$OA = OB = OC = R$.
Given $\angle CAB = 60^{\circ}$,in $\triangle OAC$,$OA = OC = R$,so $\angle OCA = \angle OAC = 60^{\circ}$,implying $\triangle OAC$ is equilateral. Thus,$AC = R$.
Similarly,for $\triangle OBC$,since $\angle ACB = 90^{\circ}$ (angle in a semicircle if $AB$ is diameter,but here $AB$ is a chord),we calculate the distance $BC$.
Using the law of cosines in $\triangle OBC$,or geometry: $BC = \sqrt{R^2 + R^2 - 2R^2 \cos(120^{\circ})} = R\sqrt{3}$.
The force between $C$ and $B$ is $F_{BC} = \frac{1}{4 \pi \varepsilon_0} \frac{|q_C| |q_B|}{(BC)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{(2q/3)(q/3)}{(R\sqrt{3})^2} = \frac{1}{4 \pi \varepsilon_0} \frac{2q^2/9}{3R^2} = \frac{q^2}{54 \pi \varepsilon_0 R^2}$.
Thus,option $C$ is correct.
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33
PhysicsMediumMCQIIT JEE · 2008
$STATEMENT-1$: For practical purposes,the Earth is used as a reference at zero potential in electrical circuits.
$STATEMENT-2$: The electrical potential of a sphere of radius $R$ with charge $Q$ uniformly distributed on the surface is given by $\frac{Q}{4 \pi \varepsilon_0 R}$.
$STATEMENT-2$: The electrical potential of a sphere of radius $R$ with charge $Q$ uniformly distributed on the surface is given by $\frac{Q}{4 \pi \varepsilon_0 R}$.
A
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is a correct explanation for $STATEMENT-1$.
B
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is $NOT$ a correct explanation for $STATEMENT-1$.
C
$STATEMENT-1$ is True,$STATEMENT-2$ is False.
D
$STATEMENT-1$ is False,$STATEMENT-2$ is True.
Solution
(B) $STATEMENT-1$ is True because the Earth is a vast conductor and its potential is taken as a reference point (zero potential) for all electrical measurements.
$STATEMENT-2$ is True because the potential at the surface of a charged conducting sphere is $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}$.
However,$STATEMENT-2$ does not explain why the Earth is chosen as the reference point. The choice of Earth as a reference is a convention based on its size and conductivity,not because of the specific potential formula of a sphere. Therefore,$STATEMENT-2$ is not the correct explanation for $STATEMENT-1$.
$STATEMENT-2$ is True because the potential at the surface of a charged conducting sphere is $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}$.
However,$STATEMENT-2$ does not explain why the Earth is chosen as the reference point. The choice of Earth as a reference is a convention based on its size and conductivity,not because of the specific potential formula of a sphere. Therefore,$STATEMENT-2$ is not the correct explanation for $STATEMENT-1$.
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34
PhysicsDifficultMCQIIT JEE · 2008
$STATEMENT-1$: The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil.
$STATEMENT-2$: Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized.
$STATEMENT-2$: Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized.
A
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is a correct explanation for $STATEMENT-1$.
B
$STATEMENT-1$ is True,$STATEMENT-2$ is True; $STATEMENT-2$ is $NOT$ a correct explanation for $STATEMENT-1$.
C
$STATEMENT-1$ is True,$STATEMENT-2$ is False.
D
$STATEMENT-1$ is False,$STATEMENT-2$ is True.
Solution
(C) In a moving coil galvanometer,a soft iron core is placed inside the coil to increase the magnetic field strength $(B)$ within the coil,which directly increases the sensitivity of the galvanometer.
$STATEMENT-1$ is True because the high magnetic permeability of the soft iron core concentrates the magnetic field lines,thereby increasing the torque on the coil.
$STATEMENT-2$ is False because soft iron is a soft magnetic material,which means it has high magnetic permeability and can be easily magnetized and demagnetized. The statement incorrectly claims that it cannot be easily magnetized or demagnetized.
$STATEMENT-1$ is True because the high magnetic permeability of the soft iron core concentrates the magnetic field lines,thereby increasing the torque on the coil.
$STATEMENT-2$ is False because soft iron is a soft magnetic material,which means it has high magnetic permeability and can be easily magnetized and demagnetized. The statement incorrectly claims that it cannot be easily magnetized or demagnetized.
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35
PhysicsAdvancedMCQIIT JEE · 2008
The nuclear charge $(Ze)$ is non-uniformly distributed within a nucleus of radius $R$. The charge density $\rho(r)$ (charge per unit volume) is dependent only on the radial distance $r$ from the center of the nucleus as shown in the figure. The electric field is only along the radial direction.
$1.$ The electric field at $r=R$ is
$(A)$ independent of $a$
$(B)$ directly proportional to $a$
$(C)$ directly proportional to $a^2$
$(D)$ inversely proportional to $a$
$2.$ For $a=0$,the value of $d$ (maximum value of $\rho$ as shown in the figure) is
$(A)$ $\frac{3Ze}{4\pi R^3}$ $(B)$ $\frac{3Ze}{\pi R^3}$ $(C)$ $\frac{4Ze}{3\pi R^3}$ $(D)$ $\frac{Ze}{3\pi R^3}$
$3.$ The electric field within the nucleus is generally observed to be linearly dependent on $r$. This implies
$(A)$ $a=0$ $(B)$ $a=\frac{R}{2}$ $(C)$ $a=R$ $(D)$ $a=\frac{2R}{3}$
Give the answer for questions $1, 2,$ and $3.$
$1.$ The electric field at $r=R$ is
$(A)$ independent of $a$
$(B)$ directly proportional to $a$
$(C)$ directly proportional to $a^2$
$(D)$ inversely proportional to $a$
$2.$ For $a=0$,the value of $d$ (maximum value of $\rho$ as shown in the figure) is
$(A)$ $\frac{3Ze}{4\pi R^3}$ $(B)$ $\frac{3Ze}{\pi R^3}$ $(C)$ $\frac{4Ze}{3\pi R^3}$ $(D)$ $\frac{Ze}{3\pi R^3}$
$3.$ The electric field within the nucleus is generally observed to be linearly dependent on $r$. This implies
$(A)$ $a=0$ $(B)$ $a=\frac{R}{2}$ $(C)$ $a=R$ $(D)$ $a=\frac{2R}{3}$
Give the answer for questions $1, 2,$ and $3.$
A
$(A, B, C)$
B
$(C, B, D)$
C
$(A, D, C)$
D
$(B, A, C)$
Solution
(A) For $r=R$,by Gauss's Law,the electric field $E$ is given by $E(4\pi R^2) = \frac{Q_{enclosed}}{\epsilon_0} = \frac{Ze}{\epsilon_0}$.
Thus,$E = \frac{Ze}{4\pi\epsilon_0 R^2}$,which is independent of $a$. So,$1$ is $(A)$.
For $a=0$,the charge density $\rho(r)$ becomes a triangle with base $R$ and height $d$. The total charge $Ze = \int_0^R \rho(r) 4\pi r^2 dr$.
Since $\rho(r) = d(1 - r/R)$,$Ze = 4\pi d \int_0^R (r^2 - r^3/R) dr = 4\pi d [R^3/3 - R^4/4R] = 4\pi d [R^3/12] = \frac{\pi d R^3}{3}$.
Therefore,$d = \frac{3Ze}{\pi R^3}$. So,$2$ is $(B)$.
For $E \propto r$ within the nucleus,the charge density $\rho$ must be constant throughout the volume. This occurs when $a=R$. So,$3$ is $(C)$.
Thus,$E = \frac{Ze}{4\pi\epsilon_0 R^2}$,which is independent of $a$. So,$1$ is $(A)$.
For $a=0$,the charge density $\rho(r)$ becomes a triangle with base $R$ and height $d$. The total charge $Ze = \int_0^R \rho(r) 4\pi r^2 dr$.
Since $\rho(r) = d(1 - r/R)$,$Ze = 4\pi d \int_0^R (r^2 - r^3/R) dr = 4\pi d [R^3/3 - R^4/4R] = 4\pi d [R^3/12] = \frac{\pi d R^3}{3}$.
Therefore,$d = \frac{3Ze}{\pi R^3}$. So,$2$ is $(B)$.
For $E \propto r$ within the nucleus,the charge density $\rho$ must be constant throughout the volume. This occurs when $a=R$. So,$3$ is $(C)$.
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36
PhysicsAdvancedIIT JEE · 2008
An optical component and an object $S$ placed along its optic axis are given in Column $I$. The distance between the object and the component can be varied. The properties of images are given in Column $II$. Match all the properties of images from Column $II$ with the appropriate components given in Column $I$.
| Column $I$ | Column $II$ |
| $A$. Concave Mirror | $(p)$ Real image |
| $B$. Convex Mirror | $(q)$ Virtual image |
| $C$. Convex Lens | $(r)$ Magnified image |
| $D$. Concave Lens | $(s)$ Image at infinity |
Solution
(D) For $A$ (Concave Mirror): It can form a real image (when object is beyond $F$), a virtual image (when object is between $P$ and $F$), a magnified image (when object is between $P$ and $2F$), and an image at infinity (when object is at $F$). Thus, $A \rightarrow p, q, r, s$.
For $B$ (Convex Mirror): It always forms a virtual image. The image is diminished, so it cannot be magnified. It can form an image at infinity if the object is at $F$. Thus, $B \rightarrow q, s$.
For $C$ (Convex Lens): It can form a real image (when object is beyond $F$), a virtual image (when object is between $O$ and $F$), a magnified image (when object is between $O$ and $2F$), and an image at infinity (when object is at $F$). Thus, $C \rightarrow p, q, r, s$.
For $D$ (Concave Lens): It always forms a virtual image. The image is always diminished, so it cannot be magnified. It can form an image at infinity if the object is at $F$. Thus, $D \rightarrow q, s$.
For $B$ (Convex Mirror): It always forms a virtual image. The image is diminished, so it cannot be magnified. It can form an image at infinity if the object is at $F$. Thus, $B \rightarrow q, s$.
For $C$ (Convex Lens): It can form a real image (when object is beyond $F$), a virtual image (when object is between $O$ and $F$), a magnified image (when object is between $O$ and $2F$), and an image at infinity (when object is at $F$). Thus, $C \rightarrow p, q, r, s$.
For $D$ (Concave Lens): It always forms a virtual image. The image is always diminished, so it cannot be magnified. It can form an image at infinity if the object is at $F$. Thus, $D \rightarrow q, s$.
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