Dielectrics - Practice Questions & MCQ

Q. No.	Correct Answer
(i)	(a) decreases
(ii)	(a) only kinetic energy
(iii)	(c) remains same
(iv)	(d) adiabatic
(v)	(a) $\frac{1}{2 \pi} \sqrt{\frac{l \cos \theta}{g}}$
(vi)	(a) $\frac{1}{2} B \omega l^2$
(vii)	(c) 2 : 1
(viii)	(b) $2 \sqrt{3}$
(ix)	(b) 4 cm
(x)	(c) positive

Q2.

(i)

(ii) Potentiometer

(iii)

Force on a stationary charge in a uniform magnetic field:

$
F=q v B \sin \theta
$
Since the charge is stationary $(v=0)$,

$
F=0
$

(iv) 

A freely suspended bar magnet always aligns itself in the North–South direction (directive property).

(v)

In YDSE, amplitude is proportional to slit width.
Width ratio = 25:1
Therefore, Amplitude ratio = 25:1

(vi)

It is a nuclear decay in which a proton converts into a neutron with emission of a positron and a neutrino:

$
p \rightarrow n+e^{+}+\nu
$

(vii)

For a sonometer, frequency $f \propto \sqrt{T}$

If tension increases by $21 \%$ :

$
\begin{gathered}
T_2=1.21 T_1 \\
\frac{f_2}{f_1}=\sqrt{1.21}=1.1
\end{gathered}
$
So, new frequency is $\mathbf{1 . 1}$ times the initial frequency
Ratio:

$
f_1: f_2=1: 1.1
$

(viii)

A pendulum whose time period is 2 seconds (1 second for one complete to-and-fro half oscillation).

Section - B

Q3.

Eddy currents are circulating currents induced in a bulk conductor when it is placed in a changing magnetic field or when the conductor moves through a magnetic field. These currents flow in closed loops inside the conductor and oppose the change producing them (Lenz’s law).

Applications:

1. Electromagnetic braking – Used in trains and elevators to produce smooth braking.
2. Induction heating / Induction furnace – Used for melting metals.

Q4.

Sources of error:

1. End error due to resistance of copper strips and connecting wires.
2. Non-uniformity of bridge wire (variation in cross-section or material).

Minimisation:

1. Interchange the known and unknown resistances and take the mean value.
2. Use a uniform wire and take the null point near the middle (around 50 cm).
3. Ensure tight and clean connections to reduce contact resistance.

Q5.

Q6.

Let, $\mathrm{m}_{\mathrm{e}}=$ mass of electron,
$-\mathrm{e}=$ charge on electron,
$\mathrm{r}_{\mathrm{n}}=$ radius of $\mathrm{n}^{\text {th }}$ Bohr's orbit,
$+\mathrm{e}=$ charge on nucleus,
$\mathrm{v}_{\mathrm{n}}=$ linear velocity of electron in $\mathrm{n}^{\text {th }}$ orbit,
$\mathrm{Z}=$ number of electrons in an atom,
$\mathrm{n}=$ principal quantum number.
∴ The angular momentum $=\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{n}} \mathrm{r}_{\mathrm{n}}$

According to second postulate.

$
\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{n}} \mathrm{r}_{\mathrm{n}}=\frac{\mathrm{nh}}{2 \pi}
$
From Bohr's first postulate,
Centripetal force $=$ Electrostatic force

$
\begin{aligned}
& \therefore \frac{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{n}}^2}{\mathrm{r}_{\mathrm{n}}}=\frac{\mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{r}_{\mathrm{n}}^2} \\
& \therefore \mathrm{v}_{\mathrm{n}}^2=\frac{\mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{r}_{\mathrm{n}} \mathrm{~m}_{\mathrm{e}}} \ldots .
\end{aligned}
$
Squaring equation (i) we get

$
m_e^2 v_n^2 r_n^2=\frac{n^2 h^2}{4 \pi^2}
$

$
\therefore \mathrm{v}_{\mathrm{n}}^2=\frac{\mathrm{n}^2 \mathrm{~h}^2}{4 \pi^2 \mathrm{~m}_{\mathrm{e}}^2 \mathrm{r}_{\mathrm{n}}^2}
$
Equating equations (ii) and (iii), we get,

$
\begin{aligned}
& \frac{n^2 h^2}{4 \pi^2 m_e^2 r_n^2}=\frac{Z e^2}{4 \pi \varepsilon_0 r_n m_e} \\
& \therefore r_n=\frac{n^2 h^2 \varepsilon_0}{\pi m_e Z_e^2}
\end{aligned}
$
This is the required expression for radius of $\mathrm{n}^{\text {th }}$ Bohr orbit of the electron.

Q7.

Final angular speed:

$
60 \mathrm{rpm}=1 \mathrm{rps}=2 \pi \mathrm{rad} / \mathrm{s}
$
Time $t=2 \pi s$

Angular acceleration:

$
\alpha=\frac{\omega}{t}=\frac{2 \pi}{2 \pi}=1 \mathrm{rad} / \mathrm{s}^2
$
Torque:

$
\tau=I \alpha=2 \times 1=2 N m
$
Power at maximum speed:

$
\begin{gathered}
P=\tau \omega=2 \times 2 \pi=4 \pi W \\
P \approx 12.6 W
\end{gathered}
$

Q8.

$\mathrm{q}=2 \times 10^{-6}$,

$
\begin{aligned}
& 2 \mathrm{l}=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m} \\
& E=10^5 \mathrm{~N} / \mathrm{C} \\
& \theta=180^{\circ}+\theta_0
\end{aligned}
$

Let us assume the dipole is initially aligned parallel to the field, i.e., $\theta_0=0$.

Then, $\theta=180^{\circ}$.
The work done by an external agent,

$
\begin{aligned}
& W=p E(1-\cos \theta) \\
& =q(2 \mathrm{l}) E\left(1-\cos 180^{\circ}\right) \\
& =\left(2 \times 10^{-6} \mathrm{C}\right)\left(4 \times 10^{-2} \mathrm{~m}\right)\left(10^5 \mathrm{~N} / \mathrm{C}\right)[1-(-1)] \\
& =16 \times 10^{-3} \mathrm{~J}
\end{aligned}
$

Q9.

Consider length element dl lying always perpendicular to $\vec{r}$.
Using the Biot-Savart law, the magnetic field produced at $O$ is:

$
\begin{aligned}
\overrightarrow{d B} & =\frac{\mu_0}{4 \pi} \frac{I d \vec{l} \times \vec{r}}{r^3} \\
d B & =\frac{\mu_0}{4 \pi} \frac{I d l r \sin 90^{\circ}}{r^3}=\frac{\mu_0}{4 \pi} \frac{I d l}{r^2} \ldots(1)
\end{aligned}
$

Equation (1) gives the magnitude of the field. The direction of the field is given by the right-hand rule. Thus, the direction of each of the dB is into the plane of the paper. The total field at O is

The angle subtended by element $d l$ is $d \theta$ at pt. O , therefore $d l=r d \theta$

$
\begin{aligned}
& \mathrm{B}=\int \mathrm{dB}=\frac{\mu_0}{4 \pi} I \int_0^\theta \frac{\mathrm{d} l}{\mathrm{r}^2} \\
& B=\frac{\mu_0}{4 \pi} I \int_0^\theta \frac{\mathrm{rd} \theta}{\mathrm{r}^2}=\frac{\mu_0}{4 \pi} \frac{\mathrm{I}}{\mathrm{r}} \theta \ldots
\end{aligned}
$

where the angle $\theta$ is in radians.

Q10.

Advantages:

1. Very fast response (high switching speed).
2. High sensitivity to light.
3. Small size and low power consumption.
4. Good linearity between light intensity and current.

Disadvantages:

1. Output current is very small (needs amplification).
2. Sensitive to temperature variations.
3. Requires biasing circuit.
4. Affected by noise and dark current.

Q11.

Harmonics	Overtones
Frequencies that are integer multiples of fundamental frequency	Frequencies higher than the fundamental (excluding fundamental)
First harmonic = fundamental frequency	First overtone = second harmonic
Include the fundamental	Do not include fundamental

Q12.

Use Stokes' law:

$
F=6 \pi \eta r v
$
Given:

$
\begin{gathered}
r=0.3 \mathrm{~mm}=0.3 \times 10^{-3}=3 \times 10^{-4} \mathrm{~m} \\
v=2 \mathrm{~m} / \mathrm{s} \\
\eta=0.833 \mathrm{Ns} / \mathrm{m}^2
\end{gathered}
$
Substitute:

$
\begin{gathered}
F=6 \pi(0.833)\left(3 \times 10^{-4}\right)(2) \\
F=6 \pi \times 0.833 \times 6 \times 10^{-4} \\
F=6 \pi \times 4.998 \times 10^{-4} \\
F \approx 18.85 \times 4.998 \times 10^{-4} \\
F \approx 9.4 \times 10^{-3} \mathrm{~N}
\end{gathered}
$

Q13.

Energy of an electron accelerated through 1 volt

$
=1 \mathrm{eV}=1.6 \times 10^{-19} C \times 1 V=1.6 \times 10^{-19} J
$
From kinetic theory of gases, average K.E. of a gas molecule is

$
\begin{aligned}
& \frac{3}{2} k T=1.6 \times 10^{-19} J \\
& \text { } T=\frac{2 \times 1.6 \times 10^{-19}}{3 K}=\frac{2 \times 1.6 \times 10^{-19}}{3 \times 1.4 \times 10^{-23}} \\
& =7619 K
\end{aligned}
$

Q14.

Given:

$
\begin{gathered}
L=200 \mathrm{mH}=0.2 \mathrm{H} \\
V_0=220 \mathrm{~V} \\
f=50 \mathrm{~Hz}
\end{gathered}
$
Angular frequency:

$
\omega=2 \pi f=2 \pi \times 50=100 \pi \approx 314 \mathrm{rad} / \mathrm{s}
$
Inductive reactance:

$
\begin{gathered}
X_L=\omega L \\
X_L=314 \times 0.2=62.8 \Omega
\end{gathered}
$
Peak current:

$
\begin{aligned}
I_0 & =\frac{V_0}{X_L} \\
I_0 & =\frac{220}{62.8} \\
I_0 & \approx 3.5 \mathrm{~A}
\end{aligned}
$

Section - C

Q15.

(a) Mechanical equilibrium

A system is in mechanical equilibrium when there is no unbalanced force acting on it and the pressure is the same throughout, so there is no change in its state of motion.

(b) Chemical equilibrium

A system is in chemical equilibrium when the rate of forward and backward chemical reactions are equal, and the chemical composition remains constant with time.

(c) Thermal equilibrium

A system is in thermal equilibrium when there is no heat flow between its parts or between the system and surroundings, i.e., temperature is uniform throughout.

Q16.

In a series LCR circuit:

Impedance,

$
Z=\sqrt{R^2+\left(X_L-X_C\right)^2}
$
Where

$
X_L=\omega L, \quad X_C=\frac{1}{\omega C}
$
At resonance:

$
\begin{aligned}
X_L & =X_C \\
\omega L & =\frac{1}{\omega C} \\
\omega^2 & =\frac{1}{L C} \\
\omega_0 & =\frac{1}{\sqrt{L C}}
\end{aligned}
$
Since,

$
\begin{gathered}
\omega_0=2 \pi f_0 \\
f_0=\frac{1}{2 \pi \sqrt{L C}}
\end{gathered}
$

Q17.

Consider a bar magnet of magnetic dipole moment $\vec{M}$, suspended by a light twistless fibre in a uniform magnetic field $\vec{B}$ in such a way that it is free to rotate in a horizontal plane. In the rest position $\theta=0, \vec{M}$ is is parallel to $\vec{B}$.

If magnet is given a small angular displacement $\theta$ from its rest position and released, the magnet performs angular or torsional oscillations about the rest position.

Let I be the moment of inertia of the bar magnet about the axis of oscillation and $\alpha$ the angular acceleration. The deflecting torque (in magnitude) is

$
\mathrm{T}_{\mathrm{d}}=\mathrm{I} \alpha=\mathrm{I} \frac{d^2 \theta}{d t^2} -------(1)
$
However, the restoring torque tries to bring back the oscillating bar magnet in the rest position. The restoring torque (in magnitude) is,

$
\mathrm{T}_{\mathrm{r}}=-\mathrm{MB} \sin \theta --------(2)
$
The minus sign in Eq. (2) indicates that restoring torque is opposite in direction to the angular deflection.

In equilibrium, both the torques balance each other. From Eqs. (1) and (2),

$
\mathrm{I} \frac{d^2 \theta}{d t^2}=-\mathrm{MB} \sin \theta
$
For small $\theta, \sin \theta=\theta$, Thus Eq, (3) can be written as

$
\begin{aligned}
& I \frac{d^2 \theta}{d t^2}=-M B \theta \\
& \frac{d^2 \theta}{d t^2}=-\left(\frac{M B}{I}\right) \theta
\end{aligned}
$
Eq. (4) represents angular simple harmonic motion.
Writing $\omega^2=\frac{M B}{I}$, the angular frequency $\omega$ of the motion is

$
\omega=\sqrt{\frac{M B}{I}}
$
The time period of oscillations of the bar magnet is

$
\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{I}{M B}} .
$
This is the required expression.

Q18.

Magnetization $M$ is the magnetic moment per unit volume of a material.

$
M=\frac{\text { Magnetic moment }}{\text { Volume }}
$
SI unit

$
\mathrm{A} / \mathrm{m}
$
Dimensions
Magnetic moment $=A \cdot m^2$

$
\begin{gathered}
{[M]=\frac{A \cdot m^2}{m^3}=A m^{-1}} \\
{\left[M^0 L^{-1} T^0 I^1\right]}
\end{gathered}
$

Relation between permeability and susceptibility

$
B=\mu_0(H+M)
$
But,

$
M=\chi_m H
$
So,

$
\begin{gathered}
B=\mu_0\left(1+\chi_m\right) H \\
\mu=\mu_0\left(1+\chi_m\right)
\end{gathered}
$
Relative permeability:

$
\mu_r=1+\chi_m
$

Q19.

Consider a positive charge ' $q$ ' kept fixed at the origin. Let $P$ be a point at distance $r$ from the charge $q$.

The electric potential at the point A is

$
V=\int_{\infty}^r(-\vec{E}) \cdot d \vec{r}=-\int_{\infty}^r \vec{E} \cdot \overrightarrow{d r}
$

Electric field due to positive point charge is

$
\begin{aligned}
& \overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}} \\
& \mathrm{~V}=-\frac{1}{4 \pi \varepsilon_0} \int_{\infty}^{\mathrm{r}} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}} \cdot \mathrm{~d} \overrightarrow{\mathrm{r}}
\end{aligned}
$

The infinitesimal displacement vector, $\mathrm{d} \overrightarrow{\mathrm{r}}=\mathrm{dr} \hat{\mathrm{r}}$ and using $\hat{\mathrm{r}}\hat{\mathrm{r}}=1$, we have

$
\mathrm{V}=-\frac{1}{4 \pi \varepsilon_0} \int_{\infty}^{\mathrm{r}} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}} \cdot \mathrm{dr} \hat{r}=-\frac{1}{4 \pi \varepsilon_0} \int_{\infty}^{\mathrm{r}} \frac{\mathrm{q}}{\mathrm{r}^2} \mathrm{dr}
$

After the integration,

$
V=-\frac{1}{4 \pi \varepsilon_0} q\left\{-\frac{1}{r}\right\}_{\infty}^r=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r}
$
Hence the electric potential due to a point charge q at a distance r is

$
\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}}
$

Q20.

An electron of mass $m$ is accelerated through a potential difference of $V$ volt. The kinetic energy acquired by the electron is given by

$
\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}
$
Therefore, the speed v of the electron is

$
\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{~m}}}
$
Hence, the de Broglie wavelength of the electron is

$
\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{emV}}}
$

Substituting the known values in the above equation, we get

$
\begin{aligned}
& \lambda=\frac{6.626 \times 10^{-34}}{\sqrt{2 \mathrm{~V} \times 1.6 \times 10^{-19} \times 9.11 \times 10^{-31}}} \\
& \lambda=\frac{12.27 \times 10^{-10}}{\sqrt{\mathrm{~V}}} \text { meter (or) } \lambda=\frac{12.27}{\sqrt{\mathrm{~V}}} Å
\end{aligned}
$
For example, if an electron is accelerated through a potential difference of 100 V , then its de Broglie wavelength is 1.227 Å.

Since the kinetic energy of the electron, $\mathrm{K}=\mathrm{eV}$, then the de Broglie wavelength associated with electron can be also written as

$
\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}
$

Q21.

A device or a circuit which rectifies both halves of each cycle of an alternating voltage is called a full-wave rectifier.
Electric circuit: The alternating voltage to be rectified is applied across the primary coil ( $P_1 P_2$ ) of a transformer with a center-tapped secondary coil $\left(S_1 S_2\right)$. The terminals $S_1$ and $S_2$ of the secondary are connected to the two p-regions of two junction diodes $\mathrm{D}_1$ and $\mathrm{D}_2$, respectively. The center-tap Tis connected to the ground. The load resistance RL is connected across the common n-regions and the ground.

$P_1 P_2, S_1 S_2$ - Primary and secondary transformer,

T - Centre-tap on secondary,
$\mathrm{D}_1, \mathrm{D}_2$ - Junction diodes,
$\mathrm{R}_{\mathrm{L}}$ - Load resistance,
$\mathrm{I}_{\mathrm{L}}$ - Load current,
$\mathrm{V}_{\mathrm{i}}$ - AC input voltage,
$\mathrm{V}_{\mathrm{o}}$ - DC output voltage

Working: During the one-half cycle of the input, terminal $S_1$ of the secondary is positive while $S_2$ is negative with respect to the ground (the centre-tap T ). During this half cycle, diode $\mathrm{D}_1$ is forward biased and conducts, while diode $\mathrm{D}_2$ is reverse biased and does not conduct. The direction of current $I_L$ through $R_L$ is in the sense shown.

During the next half-cycle of the input voltage, $S_2$ becomes positive while $S$, is negative with respect to $T$. Diode $D_2$ now conducts, sending a current $I_L$ through $R_L$ in the same sense as before. $D_1$ now does not conduct. Thus, the current through $\mathrm{R}_{\mathrm{L}}$ flows in the same direction, i.e., it is unidirectional, for both halves or the full wave of the input. This is called full-wave rectification. The output voltage has a fixed polarity but varies periodically with the time between zero and a maximum value. The above figure shows the input and output voltage waveforms. The pulsating de output voltage of a full-wave rectifier has twice the frequency of the input.

Q22.

$\mathrm{L}_1=80 \mathrm{~cm}, \mathrm{n}_1=112 \mathrm{~Hz}, \mathrm{n}_2=160 \mathrm{~Hz}$
According to the law of length, $\mathrm{n}_1 \mathrm{~L}_1=\mathrm{n}_2 \mathrm{~L}_2$.
∴ The vibrating length to produce the fundamental frequency of 160 Hz,

$
\mathrm{L}_2=\frac{\mathrm{n}_1 \mathrm{~L}_1}{\mathrm{n}_2}=\frac{112(80)}{160}=56 \mathrm{~cm}
$

Q23.

Number of moles of gas $(\mathrm{n})=0.5$ moles
Temperature $(\mathrm{T})=300 \mathrm{~K}$
Initial Volume $\left(V_1\right)=2.0 \mathrm{~L}=2.0 \times 10^{-3} \mathrm{~m}^3$

Final Volume $\left(V_2\right)=6.0 \mathrm{~L}=6.0 \times 10^{-3} \mathrm{~m}^3$

Universal Gas Constant $(\mathrm{R})=8.314 \mathrm{~J} / \mathrm{mol} \mathrm{K}$

(a)
$\mathrm{W}=n R T \ln \left(\frac{V_2}{V_1}\right)$
$\mathrm{W}=0.5 \times 8.314 \times 300 \times \ln \left(\frac{6.0}{2.0}\right)$
$\mathrm{W}=0.5 \times 8.314 \times 300 \times \ln 3$
$\mathrm{W}=0.5 \times 8.314 \times 300 \times 1.0986$
$\mathrm{W}=1370 \mathrm{~J}$

(b)

For an isothermal process:

$
\Delta U=0 \quad \text { (temperature constant) }
$
From the first law:

$
\begin{gathered}
Q=W \\
Q \approx 1369 J
\end{gathered}
$

Q24.

Given:
Galvanometer resistance, $G=40 \Omega$
Full scale current, $I_g=4 m A=0.004 A$
(a) Conversion into an ammeter of range 0.4 A

To convert a galvanometer into an ammeter, a shunt resistance $(S)$ is connected in parallel.
Total current required:

$
I=0.4 A
$
Current through shunt:

$
I_s=I-I_g=0.4-0.004=0.396 \mathrm{~A}
$

Since the potential difference across the galvanometer and the shunt is the same:

$
\begin{gathered}
I_g G=I_s S \\
0.004 \times 40=0.396 \times S \\
0.16=0.396 S \\
S=\frac{0.16}{0.396} \approx 0.404 \Omega
\end{gathered}
$

(b) Conversion into a Voltmeter of range 5 V

To convert a galvanometer into a voltmeter, a high resistance (R) is connected in series.

Voltage range:

$
V=5 V
$
Total resistance required:

$
R_{\text {total }}=\frac{V}{I_g}=\frac{5}{0.004}=1250 \Omega
$
Series resistance:

$
\begin{gathered}
R=R_{\text {total }}-G \\
R=1250-40=1210 \Omega
\end{gathered}
$

Q25.

Given

• Inner conductor radius $=a$
• Outer conductor radius $=b$
• Equal currents $I$ flow in opposite directions

Ampere's Law:

$
\oint \vec{B} \cdot d \vec{l}=\mu_0 I_{\mathrm{enc}}
$
For circular symmetry:

$
\begin{gathered}
B(2 \pi r)=\mu_0 I_{\mathrm{enc}} \\
B=\frac{\mu_0 I_{\mathrm{enc}}}{2 \pi r}
\end{gathered}
$

(i) For $a<r<b$

The Amperian loop lies between the inner and outer conductors.

Enclosed current:

$
I_{\mathrm{enc}}=I
$
Therefore,

$
B=\frac{\mu_0 I}{2 \pi r}
$
Direction
Using righthe t-hand thumb rule:

• Thumb along the current in the inner conductor
• The magnetic field is circular (tangential) around the axis.

(ii) For $r>b$

The loop encloses both conductors.

Inner current $=I$
Outer current $=-I$ (opposite)

$
I_{\mathrm{enc}}=I-I=0
$
Hence,

$
B=0
$
Direction
Since $B=0$, no magnetic field exists outside the coaxial cable.

Q26.

For a hydrogen atom in the Bohr model:

$
E_n=-\frac{13.6}{n^2} \mathrm{eV}
$

For $n=3$, total energy in third orbit:

$
\begin{gathered}
E_3=-\frac{13.6}{9} \\
E_3=-1.51 \mathrm{eV} \text { (approx) }
\end{gathered}
$

Kinetic energy in the third orbit

$
\begin{gathered}
K_3=-E_3 \\
K_3=+1.51 \mathrm{eV}
\end{gathered}
$
Potential energy in the third orbit

$
\begin{gathered}
U_3=2 E_3 \\
U_3=2(-1.51) \\
U_3=-3.02 \mathrm{eV}
\end{gathered}
$

Section - D

Q27.

Assume that the radius of the drop is increasing from $r$ to $r+\Delta r$ where $\Delta r$ is negligibly small, for the pressure inside could be thought of as constant.

Then, the initial surface area,

$
A_1=4 \pi r^2
$
Final surface area,

$
\begin{aligned}
& A_2=4 \pi(r+\Delta r)^2 \\
& \Rightarrow A_2=4 \pi r^2+8 \pi r \Delta r+4 \pi \Delta r^2
\end{aligned}
$
We could neglect $4 \pi \Delta r^2$ as $\Delta r^2$ will be negligibly small.

$
A_2=4 \pi r^2+8 \pi r \Delta r
$

Change in area,

$
\Delta A=A_2-A_1=8 \pi r \Delta r
$
We could express the work done in increasing the surface area as,

$
d W=T \times d A=T(8 \pi r \Delta r)
$
But we know that work done is normally expressed as the product of force and displacement, that is,

$
d W=F \times S=F \times \Delta r
$
Also, $P=\frac{F}{A}$
Therefore, $\mathrm{F}=$ excess pressure × area

$
F=\left(P_i-P_o\right) \times 4 \pi r^2
$
Substituting this in equation (2), we get,

$
d W=\left(P_i-P_o\right) \times 4 \pi r^2 \Delta r
$
Equating equations (2) and (3), we get,

$
\begin{aligned}
& \left(P_i-P_o\right) \times 4 \pi r^2 \times \Delta r=T \times 8 \pi r \Delta r \\
& \therefore\left(P_i-P_o\right)=\frac{2 T}{r}
\end{aligned}
$

Q28.

Consider a vertical section of a car moving on a horizontal circular track having a radius ' r ' with ' C ' as the centre of the track.

Forces acting on the car (considered to be a particle):
a. Weight (mg), vertically downwards,
b. Normal reaction ( N ), vertically upwards, that balances the weight
c. Force of static friction $\left(f_S\right)$ between the road and the tyres.

Since normal reaction balances the weight

$
\therefore \mathrm{N}=\mathrm{mg}
$
While working in the frame of reference attached to the vehicle, the frictional force balances the centrifugal force.

$
f_s=\frac{m v^2}{r}
$
Dividing equation (2) by equation (1),

$
\therefore \frac{f_s}{\mathrm{~N}}=\frac{v^2}{r} g
$

However, $\mathrm{f}_{\mathrm{s}}$ has an upper limit $\left(\mathrm{f}_{\mathrm{s}}\right)_{\max }=\mu_{\mathrm{s}} \mathrm{N}$, where $\mu_{\mathrm{s}}$ is the coefficient of static friction between the road and the tyres of the vehicle. This imposes an upper limit to the speed $v$.
At the maximum possible speed,
$\frac{\left(f_s\right)_{\max }}{N}=\mu_s=\frac{v_{\max }^2}{r g} \quad \ldots[$ From equations (2) and (3) $]$
$\therefore v_{\max }=\sqrt{\mu_s r g}$

Given:
Moment of inertia of a solid sphere about its diameter

$
I_{\text {diameter }}=25 \mathrm{~kg} \mathrm{~m}^2
$

Moment of inertia about any parallel axis:

$
I=I_{\mathrm{cm}}+M d^2
$
For a sphere:
- $I_{\mathrm{cm}}$ about diameter $=\frac{2}{5} M R^2$
- Distance from centre to tangent axis $d=R$

So,

$
I_{\text {tangent }}=I_{\mathrm{cm}}+M R^2
$

Since,

$
\begin{aligned}
& I_{\mathrm{cm}}=\frac{2}{5} M R^2 \\
& M R^2=\frac{5}{2} I_{\mathrm{cm}}
\end{aligned}
$

$\begin{gathered}I_{\text {tangent }}=I_{\mathrm{cm}}+\frac{5}{2} I_{\mathrm{cm}} \\ I_{\text {tangent }}=\left(1+\frac{5}{2}\right) I_{\mathrm{cm}} \\ I_{\text {tangent }}=\frac{7}{2} I_{\mathrm{cm}}\end{gathered}$

$\begin{gathered}I_{\text {tangent }}=\frac{7}{2} \times 25 \\ I_{\text {tangent }}=87.5 \mathrm{~kg} \mathrm{~m}^2\end{gathered}$

Q29.

Let $Q$ represent the amount of radiant energy incident on a body, and let $Q_a, Q_r$, and $Q_t$ represent the amounts of radiant energy the body simultaneously absorbs, reflects, and transmits. Since there is a conservation of energy overall, we have,

$
\begin{aligned}
& \mathrm{Q}_{\mathrm{a}}+\mathrm{Q}_{\mathrm{r}}+\mathrm{Q}_{\mathrm{t}}=\mathrm{Q} \\
& \therefore \frac{Q_a}{Q}+\frac{Q_r}{Q}+\frac{Q_t}{Q}=1
\end{aligned}
$
By definition,
$\frac{Q_a}{Q}=a \quad \ldots($ coefficient of absorption $)$,
$\frac{Q_r}{Q}=r \quad \ldots$ (coefficient of reflection) and
$\frac{Q_t}{Q}=t \quad \ldots$ (coefficient of transmission)
Hence, $\mathrm{a}+\mathrm{r}+\mathrm{t}=1$

$\mathrm{M}_{01}$ (hydrogen) $=2 \mathrm{~g} / \mathrm{mol}$,
$\mathrm{M}_{02}($ oxygen $)=32 \mathrm{~g} / \mathrm{mol}$,
$\mathrm{T}_1($ hydrogen $)=273+127=400 \mathrm{~K}$,
$\mathrm{T}_2($ oxygen $)=273+27=300 \mathrm{~K}$
The rms speed, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_0}}$,
where $\mathrm{M}_0$ denotes the molar mass

$
\begin{aligned}
& \therefore \frac{\mathrm{v}_{\mathrm{rms} 1} \text { (hydrogen) }}{\mathrm{v}_{\mathrm{rms} 2} \text { (oxygen) }}=\sqrt{\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)\left(\frac{\mathrm{M}_{02}}{\mathrm{M}_{01}}\right)} \\
& =\sqrt{\left(\frac{400}{300}\right)\left(\frac{32}{2}\right)}=\sqrt{\left(\frac{4}{3}\right)(16)} \\
& =\frac{(2)(4)}{\sqrt{3}}=\frac{8}{\sqrt{3}}
\end{aligned}
$

Q30.

(i) When there is a change in the current or magnetic flux of the coil, an electromotive force is induced. This phenomenon is termed Self-Inductance. When the current starts flowing through the coil at any instant, it is found that the magnetic flux becomes directly proportional to the current passing through the circuit. 

(ii) The phenomenon in which an emf is induced in one coil due to a change of current in the neighbouring coil is called mutual induction.

Given:
Area of loop, $A=1 \mathrm{~m}^2$
Initial magnetic field, $B_1=3 \mathrm{~Wb} / \mathrm{m}^2$
Final magnetic field, $B_2=1 \mathrm{~Wb} / \mathrm{m}^2$
Time, $\Delta t=0.5 \mathrm{~s}$
Loop is placed normal to the field $\rightarrow \theta=0^{\circ}$, so $\cos \theta=1$

$
\Phi=B A \cos \theta
$
Since $\cos 0^{\circ}=1$ :
Initial flux:

$
\Phi_1=B_1 A=3 \times 1=3 \mathrm{~Wb}
$
Final flux:

$
\Phi_2=B_2 A=1 \times 1=1 \mathrm{~Wb}
$
Change in flux:

$
\Delta \Phi=\Phi_2-\Phi_1=1-3=-2 \mathrm{~Wb}
$
Magnitude:

$
|\Delta \Phi|=2 \mathrm{~Wb}
$

$\begin{gathered}\mathcal{E}=\left|\frac{\Delta \Phi}{\Delta t}\right| \\ \mathcal{E}=\frac{2}{0.5}=4 \mathrm{~V}\end{gathered}$

Q31.

Let $S_1$ and $S_2$ be the two coherent monochromatic sources which are separated by a short distance d. They emit light waves of wavelength $\lambda$.
Let $\mathrm{D}=$ horizontal distance between screen and source.
Draw $S_1 M$ and $S_2 N \perp A B$

OP = perpendicular bisector of the slit.
Since $S_1 P=S_2 P$, the path difference between waves reaching $P$ from $S_1$ and $S_2$ is zero; there is a bright point at $P$.
d. Consider a point $Q$ on the screen which is at a distance $x$ from the central point $P$ on the screen. Light waves from $\mathrm{S}_1$ and $\mathrm{S}_2$ reach Q simultaneously by covering paths $S_1 Q$ and $S_2 Q$, where they superimpose.

In $\triangle \mathrm{S}_1 \mathrm{MQ}$,

$
\begin{aligned}
& \left(S_1 Q\right)^2=\left(S_1 M\right)^2+(M Q)^2 \\
& \left(S_1 Q\right)^2=D^2+\left[x-\frac{d}{2}\right]^2
\end{aligned}
$
In $\triangle \mathrm{S}_2 \mathrm{NQ}$,

$
\begin{aligned}
& \left(S_2 Q\right)^2=\left(S_2 N\right)^2+(N Q)^2 \\
& \left(S_2 Q\right)^2=D^2+\left[x+\frac{d}{2}\right]^2
\end{aligned}
$
Subtract equation (1) from (2),

$
\begin{aligned}
\left(S_2 Q\right)^2-\left(S_1 Q\right)^2 & =\left(D^2+\left(x+\frac{d}{2}\right)^2\right)-\left(D^2+\left(x-\frac{d}{2}\right)^2\right) \\
& =D^2+\left(x+\frac{d}{2}\right)^2-D^2-\left(x-\frac{d}{2}\right)^2 \\
& =\left(x+\frac{d}{2}\right)^2-\left(x-\frac{d}{2}\right)^2 \\
& =\left(x^2+\frac{d^2}{4}+x d\right)-\left(x^2+\frac{d^2}{4}-x d\right) \\
& =x^2+\frac{d^2}{4}+x d-x^2-\frac{d^2}{4}+x d \\
\left(S_2 Q\right)^2-\left(S_1 Q\right)^2 & =2 x d \\
\left(S_2 Q+S_1 Q\right)\left(S_2 Q-S_1 Q\right) & =2 x d \\
S_2 Q-S_1 Q & =\frac{2 x d}{S_2 Q+S_1 Q}
\end{aligned}
$

If $\mathrm{x} \ll \mathrm{D}$ and $\mathrm{d} \ll \mathrm{D}$ then,

$
\begin{aligned}
& S_1 Q \approx S_2 Q \approx D \\
& S_2 Q+S_1 Q=2 D
\end{aligned}
$
Equation (3) becomes,

$
\begin{aligned}
& S_2 Q-S_1 Q=\frac{2 x d}{2 D} \\
& S_2 Q-S_1 Q=\frac{x d}{D} \\
& \Delta x=\frac{x d}{D}
\end{aligned}
$