question:
A piston can freely move inside a horizontal cylinder
closed from both ends. Initially, the piston separates the inside space
of the cylinder into two equal parts each of volume $V_{0}$, in which an
ideal gas is contained under the same pressure $p_{0}$ and at the same
temperature. What work has to be performed in order to increase
isothermally the volume of one part of gas $\eta$ times compared to that
of the other by slowly moving the piston?

answer: $p_{0}V_{0}\ln\frac{(\eta + 1)^{2}}{4\eta}$

solution:
Let the agent move the piston to the right by $x$. In
equilibrium position,
$p_{1}S + F_{agent} = p_{2}S,$ or,
$F_{agent} = \left( p_{2} - p_{1} \right)S$
Work done by the agent in an infinitesimal change $dx$ is
$$F_{agent} \bullet dx = \left( p_{2} - p_{1} \right)Sdx = \left( p_{2} - p_{1} \right)dV$$
By applying $pV =$ constant, for the two parts,
$p_{1}\left( V_{0} + Sx \right) = p_{0}V_{0}$ and
$p_{2}\left( V_{0} - Sx \right) = p_{0}V_{0}$
So,
$p_{2} - p_{1} = \frac{p_{0}V_{0}2Sx}{V_{0}^{2} - S^{2}x^{2}} = \frac{2p_{0}V_{0}V}{V_{1}^{2} - V^{2}}$
(where $Sx = V$)
When the volume of the left end is $\eta$ times the volume of the
right end
$\left( V_{0} + V \right) = \eta\left( V_{0} - V \right),$ or,
$V = \frac{\eta - 1}{\eta + 1}V_{0}$.
$$A = \int_{0}^{V}\left( p_{2} - p_{1} \right)dv = \int_{0}^{V}\frac{2p_{0}V_{0}V}{V_{0}^{2} - V^{2}}dV = - p_{0}V_{0}\left\lbrack \ln\left( V_{0}^{2} - V^{2} \right) \right\rbrack_{0}^{V}$$
$$= - p_{0}V_{0}\left\lbrack \ln{\left( V_{0}^{2} - V^{2} \right) - \ln V_{0}^{2}} \right\rbrack$$
$$= - p_{0}V_{0}\left\lbrack \ln{\left\{ V_{0}^{2} - \left( \frac{\eta - 1}{\eta + 1^{2}} \right)V_{0}^{2} \right\} - \ln V_{0}^{2}} \right\rbrack$$
$$= - p_{0}V_{0}\left( \ln\frac{4\eta}{(\eta + 1)^{2}} \right) = p_{0}V_{0}\ln\frac{(\eta + 1)^{2}}{4\eta}$$

subject: Physics

topic: Thermodynamics

question_type: subjective

exam: JEE Advanced


question:
A pendulum clock is mounted in an elevator car which
starts going up with a constant acceleration $w$, with $w < g$. At a
height $h$ the acceleration of the car reverses, its magnitude remaining
constant. How soon after the start of the motion will the clock show the
right time again?

answer: $\sqrt{\frac{2h}{w}}\frac{\sqrt{1 + \beta} - \sqrt{1 - \beta}}{1 - \sqrt{1 - \beta}}$

solution:
Let the downward acceleration of the elevator car has
continued for time $t'$, then the sought time
$t = \sqrt{\frac{2h}{w}} + t'$, where obviously $\sqrt{\frac{2h}{w}}$
is the time of upward acceleration of the elevator. One should note
that if the point of suspension of a mathematical pendulum moves with
an acceleration $\overrightarrow{w}$, then the time period of the
pendulum becomes
$$2\pi\sqrt{\frac{l}{\left| \overrightarrow{g} - \overrightarrow{w} \right|}}$$
In this problem the time period of the pendulum while it is moving
upward with acceleration $w$ becomes $2\pi\sqrt{\frac{l}{g + w}}$ and
its time period while the elevator moves downward with the same
magnitude of acceleration becomes
$$2\pi\sqrt{\frac{l}{g - w}}$$
As the time of upward acceleration equal $\sqrt{\frac{2h}{w}}$, the
total number of oscillations during this time equals
$$\frac{\sqrt{2h/w}}{2\pi\sqrt{l/(g + w)}}$$
Thus, the indicated time
$= \frac{\sqrt{2h/w}}{2\pi\sqrt{l/(g + w)}} \bullet 2\pi\sqrt{l/g} = \sqrt{2h/w}\sqrt{(g + w)/g}$
Similarly, the indicated time for the time interval $t'$
$$= \frac{t'}{2\pi\sqrt{l/(g - w)}}2\pi\sqrt{l/g} = t^{'}\sqrt{(g - w)/g}$$
we demand that
$$\sqrt{2h/w}\ \sqrt{(g + w)/g} + t^{'}\sqrt{(g - w)/g} = \sqrt{2h/w + t'}$$
Or,
$t^{'} = \sqrt{2h/w}\frac{\sqrt{g + w} - \sqrt{g}}{\sqrt{g} - \sqrt{g - w}}$
Hence the sought time
$$t = \sqrt{\frac{2h}{w}} + t^{'} = \sqrt{\frac{2h}{w}}\frac{\sqrt{g + w} - \sqrt{g - w}}{\sqrt{g} - \sqrt{g - w}}$$
$= \sqrt{\frac{2h}{w}}\frac{\sqrt{1 + \beta} - \sqrt{1 - \beta}}{1 - \sqrt{1 - \beta}}$,
where $\beta = w/g$

subject: Physics

topic: Waves and Oscillations

question_type: subjective

exam: JEE Advanced


question:
There is an infinite wire grid with square cells Fig.
The resistance of each wire between neighbouring joint connections is
equal to $R_{0}$. Find the resistance $R$ of the whole grid between
points $A$ and $B$.
Instruction: Make use of principles of symmetry ad superposition.
[IMAGE]

answer: $R = \frac{R_{0}}{2}$

solution:
Suppose that the voltage $V$ is applied between the points
$A$ and $B$ then
$$V = IR = I_{0}R_{0}$$
where $R$ is resistance of whole the grid, $I$, the current through
the grid and $I_{0}$, the current through the segment $AB$. Now from
symmetry, $I/4$ is the part of the current, flowing through all the
four wire segments, meeting at the point $A$ and similarly the amount
of current flowing through the wires, meeting at $B$ is also $I/4$.
Thus, a current $I/2$ flows through the conductor $AB$, i.e.
$$I_{0} = \frac{I}{2}$$
Hence, $R = \frac{R_{0}}{2}$

subject: Physics

topic: Current Electricity

question_type: subjective

exam: JEE Advanced


question:
A metal ball of radius $a$ is located at a distance $l$
from an infinite ideally conducting plane. The space around the ball is
filled with a homogeneous poorly conducting medium with resistivity ρ.
In the case of $a \ll l$ find:
\(a\) the current density at the conducting plane as a function of
distance $r$ from the ball if the
potential difference between the ball and the plane is equal to V;
\(b\) the electric resistance of the medium between the ball and the
plane.

answer:
(a) $j = \frac{1}{\rho}E = \frac{2alV}{\rho r^{3}}$
\(b\) $R = \frac{V}{I} = \frac{\rho}{2\pi a}$

solution:
(a) The potential in the unshaded region beyond the
conductor as the potential of the given charge and its image and has the
form
$$\varphi = A\left( \frac{1}{r_{1}} - \frac{1}{r_{2}} \right)$$
where $r_{1},r_{2}$ are the distances of the point from the charge and
its image. The potential has been
taken to be zero on the conducting plane and on the ball
$$\varphi \approx A\left( \frac{1}{a} - \frac{1}{2l} \right) = V$$
[IMAGE] So $A \approx Va$. In this calculation the conditions $a \ll l$ is
used to ignore the variation of $\varphi$ over
the ball.
The electric field at $P$ can be calculated similarly. The charge on
the ball is
$$Q = 4\pi\varepsilon_{0}Va$$
and $E_{P} = \frac{Va}{r^{2}}2\cos\theta = \frac{2alV}{r^{3}}$
Then $j = \frac{1}{\rho}E = \frac{2alV}{\rho r^{3}}$ normal to the
plane.
\(b\) The total current flowing into the conducting plane is
$I = \int_{0}^{\infty}{2\pi x}dxj = \int_{0}^{\infty}{2\pi x}dx\frac{2alV}{\rho\left( \pi^{2} + l^{2} \right)^{3/2}}$
(On putting $y = x^{2} + l^{2}$)
$I = \frac{2\pi alV}{\rho}\int_{l^{2}}^{\infty}\frac{dy}{y^{3/2}} = \frac{4\pi aV}{\rho}$
Hence $R = \frac{V}{I} = \frac{\rho}{2\pi a}$

subject: Physics

topic: Electrostatics

question_type: subjective

exam: JEE Advanced


question:
A cylindrical capacitor connected to a dc voltage
source V touches the surface of water with its end Fig. The separation d
between the capacitor electrodes is substantially less than their mean
radius. Find a height $h$ to which the water level in the will rise. The
capillary effects are to be neglected.
[IMAGE]

answer: $\frac{\varepsilon_{0}(\varepsilon - 1)V^{2}}{2\rho gd^{2}}$

solution:
If $C_{0}$ is the initial capacitance of the condenser
before water rises in it then
$U_{i} = \frac{1}{2}C_{0}V^{2}$, where
$C_{0} = \frac{\varepsilon_{0}2l\pi R}{d}$
($R$ is the mean radius and $l$ is the length of the capacitor
plates.)
Suppose the liquid rises to a height $h$ in it. Then the capacitor of
the condenser is
$$C = \frac{\varepsilon\varepsilon_{0}h2\pi R}{d} + \frac{\varepsilon(l - h)2\pi R}{d} = \ \frac{\varepsilon_{0}2\pi R}{d}\left( l + (\varepsilon - 1)h \right)$$
and energy of the capacitor and the liquid (including both
gravitational and electrostatic contribution) is
$$\frac{1}{2}\frac{\varepsilon_{0}2\pi R}{d}\left( l + (\varepsilon - 1)h \right)V^{2} + \rho g(2\pi R\ hd)\frac{h}{2}$$
If the capacitor were not connected to a battery this energy would
have to be minimized. But the capacitor is connected to the battery
and, in effect, the potential energy of the whole system has to be
minimized. Suppose we increase $h$ by $\delta h$. Then the energy of
the capacitor and the liquid increases by
$$\delta h\left( \frac{\varepsilon_{0}2\pi R}{2d}(\varepsilon - 1)V^{2} + \rho g(2\pi Rd)h \right)$$
and that of the cell diminishes by the quantity $A_{cell}$ which is
the product of charge flown and $V$
$$\delta h\frac{\varepsilon_{0}(2\pi R)}{d}(\varepsilon - 1)V^{2}$$
In equilibrium, the two must balance; so
$$\rho gdh = \frac{\varepsilon_{0}(\varepsilon - 1)V^{2}}{2d}$$
Hence $h = \frac{\varepsilon_{0}(\varepsilon - 1)V^{2}}{2\rho gd^{2}}$

subject: Physics

topic: Fluid Mechanics

question_type: subjective

exam: JEE Advanced


question:
One end of a steel rectangular girder is embedded into a
wall Fig. Due to gravity it sags slightly. Find the radius of curvature
of the neutral layer (see the dotted line in the figure) in the vicinity
of the point O if the length of the protruding section of the girder is
equal to $l = 6.0\ m$ and the thickness of the girder equals
$h = 10\ cm$.
[IMAGE]

answer: $(0.121\ km)^{- 1}$

solution:
A beam clamped at one end and supporting an applied load
at the free end is called a cantilever. The theory of cantilevers is
discussed in advanced text book on mechanics. The key result is that
elastic forces in the beam generate a couple, whose moment, called the
moment of resistances, balances the external bending moment due to
weight of the beam, load etc. The moment of resistance, also called
internal bending moment (I.B.M) is given by
$$I.B.M. = EI/R$$
Here $R$ is the radius of curvature of the beam at the representative
point $(x,y)$. I is called the geometrical moment of inertia
$$I = \int_{}^{}z^{2}dS$$
of the cross section relative to the axis passing through the netural
layer which remains unstretched. Fig. The section of the beam beyond
$P$ exerts the bending moment $N(x)$ and we have,
$$\frac{EI}{R} = N(x)$$
[IMAGE] If there is no load other than that due to the weight of the beam,
then
$$N(x) = \frac{1}{2}\rho g(l - x)^{2}bh$$
where $\rho =$ density of steel.
Hence, at $x = 0$
$$\left( \frac{I}{R} \right)_{0} = \frac{\rho gl^{2}bh}{2EI}$$
Here $b =$ width of the beam perpendicular to paper.
Also, $I = \int_{- h/2}^{h/2}z^{2}bdz = \frac{bh^{3}}{12}$.
[IMAGE]

subject: Physics

topic: Elasticity

question_type: subjective

exam: JEE Advanced


question:
A heat-conducting piston can freely move inside a closed
thermally insulated cylinder with an ideal gas. In equilibrium the
piston divides the cylinder into two equal parts, the gas temperature
being equal to $T_{0}$. The piston is slowly displaced. Find the gas
temperature as a function of the ratio $\eta$ of the volume of the
greater and smaller sections. The adiabatic exponent of the gas is equal
to $\gamma$.

answer: $T = T_{0}\left( \frac{(\eta + 1)^{2}}{4\eta} \right)^{\frac{\gamma - 1}{2}}$

solution:
Since here the piston is conducting and it is moved slowly
the temperature on the two sides increases and maintained at the same
value.
Elementary work done by the agent = Work done in compression -- Work
done in expansion i.e.
$dA = p_{2}dV - p_{1}dV = \left( p_{2} - p_{1} \right)dV$
where $p_{1}$ and $p_{2}$ are pressures at any instant of the gas on
expansion and compression side respectively.
From the gas law $p_{1}\left( V_{0} + Sx \right) = \nu RT$ and
$p_{2}\left( V_{0} - Sx \right) = \nu RT$, for each section
($x$ is the displacement of the piston towards section 2)
So,
$p_{2} - p_{1} = \nu RT\frac{2Sx}{V_{0}^{2} - S^{2}x^{2}} = \nu RT \bullet \frac{2V}{v_{0}^{2} - V^{2}}$
(as $Sx = V$)
So $dA = \nu RT\frac{2V}{V_{0}^{2} - V^{2}}dV$
Also, from the first law of thermodynamics
$dA = - dU = - 2\nu\frac{R}{\gamma - 1}dT$ (as $dQ = 0$)
So, work done on the gas
$= - dA = 2\nu \bullet \frac{R}{\gamma - 1}dT$
Thus
$2\nu\frac{R}{\gamma - 1}dT = \nu RT\frac{2V \bullet dV}{V_{0}^{2} - V^{2}}$,
Or, $\frac{dT}{T} = \gamma - 1\frac{V\ dV}{V_{0}^{2} - V^{2}}$
[IMAGE] When the left end is $\eta$ times the volume of the right end.
$\left( V_{0} + V \right) = \eta\left( V_{0} - V \right)$ or
$V = \frac{\eta - 1}{\eta + 1}V_{0}$
On integrating
$\int_{T_{0}}^{T}\frac{dT}{T} = (\gamma - 1)\int_{0}^{V}\frac{VdV}{V_{0}^{2} - V^{2}}$
Or
$\ln\frac{T}{T_{0}} = (\gamma - 1)\left\lbrack - \frac{1}{2}\ln\left( V_{0}^{2} - V^{2} \right) \right\rbrack_{0}^{V}$
$$= - \frac{\gamma - 1}{2}\left\lbrack \ln{\left( V_{0}^{2} - V^{2} \right) - \ln V_{0}^{2} - V^{2} - \ln V_{0}^{2}} \right\rbrack$$
$$= \frac{\gamma - 1}{2}\left\lbrack \ln V_{0}^{2} - \ln V_{0}^{2}\left\{ 1 - \left( \frac{\eta - 1}{\eta + 1} \right)^{2} \right\} \right\rbrack = \frac{\gamma - 1}{2}\ln\frac{(\eta + 1)^{2}}{4\eta}$$
Hence
$T = T_{0}\left( \frac{(\eta + 1)^{2}}{4\eta} \right)^{\frac{\gamma - 1}{2}}$

subject: Physics

topic: Thermodynamics

question_type: subjective

exam: JEE Advanced


question:
From point A located on a highway (Fig.) one has to get
by car as soon as possible to point B located in the field at a distance
l from the highway. It is known that the car moves in the field η time
slower than on the highway. At what distance from point Done must turn
off the highway?
[IMAGE]

answer: $\frac{l}{\sqrt{\eta^{2} - 1}}$

solution:
Let the car turn off the highway at a distance $x$ from
the point $D$.
So, $CD = x,$ and if the speed of the car in the field is $v$, then
the time taken by the car to cover the distance $AC = AD - x$ on the
highway
[IMAGE] $t_{1} = \frac{AD - x}{\eta v}$ (1)
and the time taken to travel the distance $CB$ in the field
$t_{2} = \frac{\sqrt{l^{2} + x^{2}}}{v}$ (2)
So, the total time elapsed to move the car from point A to B
$$t = t_{1} + t - 2 = \frac{AD - x}{\eta v} + \frac{\sqrt{l^{2} - x^{2}}}{v}$$
For $t$ to be minimum
$$\frac{dt}{dx} = 0\ \ \ \ or\ \ \ \frac{1}{v}\left\lbrack - \frac{1}{\eta} + \frac{x}{\sqrt{l^{2} + x^{2}}} \right\rbrack = 0$$
Or $\eta^{2}x^{2} = l^{2} + x^{2}$ or
$x = \frac{l}{\sqrt{\eta^{2} - 1}}$

subject: Physics

topic: Mechanics

question_type: subjective

exam: JEE Advanced