Rishabhiiitd/PhysicsQA huggingface dataset:
Topic: Kinetic Theory of Gases
Q2: An ideal gas is enclosed in a cylinder at a pressure of 2 atm and temperature, $300 \mathrm{~K}$. The mean time between two successive collisions is $6 \times 10^{-8} \mathrm{~s}$. If the pressure is doubled and the temperature is increased to $500 \mathrm{~K}$, the mean time between two successive collisions will be close to
Options: (a) $4 \times 10^{-8} \mathrm{~s}$
(b) $3 \times 10^{-6} \mathrm{~s}$
(c) $0.5 \times 10^{-8} \mathrm{~S}$
(d) $2 \times 10^{-7} \mathrm{~s}$
Answer: (a) $4 \times 10^{-8} \mathrm{~s}$
Solution: Root mean square velocity
$$
\operatorname{Vrms}=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 P}{\rho}}
$$
Here,
$R$ is the universal gas constant
$M$ is the molar mass
$\mathrm{P}$ is the pressure due to gas
$\rho$ is the density
Vrms $\propto \sqrt{T}$
$\mathrm{V}_{\mathrm{rms}} \propto$ mean free path/ time between successive collisions
And mean free path,
$$
Y=\frac{k T}{\sqrt{2 \pi \sigma^{2} P}}
$$
Vrms $\propto \mathrm{Y} / \mathrm{b}$
$$
\begin{aligned}
& \text { Vrms } \propto T / \sqrt{p} \times t---(1) \\
& \text { But Vrms } \propto \sqrt{T}----(2)
\end{aligned}
$$
$\sqrt{T} \propto T / \sqrt{p} \times t$
$\mathrm{t} \propto \sqrt{T} / p \quad \frac{t_{2}}{t_{1}}=\sqrt{\frac{T_{2}}{T_{1}} \times \frac{P_{1}}{P_{2}}} \frac{t_{2}}{t_{1}}$
$=\sqrt{\frac{500}{300} \times \frac{P_{1}}{2 P_{1}}}=\sqrt{\frac{5}{6}} t_{2}=\sqrt{\frac{5}{6}} t_{1}$
$\mathrm{t}_{2} \approx 4 \times 10^{-8} \mathrm{~s}$
Topic: Kinetic Theory of Gases
Q3: The specific heats, $C_{P}$ and $C_{V}$ of gas of diatomic molecules, $A$, is given (in units of $\mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$ ) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then
Options: (a) A has a vibrational mode but $\mathrm{B}$ has none.
(b) A has one vibrational mode and $B$ has two.
(c) $A$ is rigid but $B$ has a vibrational mode.
(d) Both $A$ and $B$ have a vibrational mode each.
Answer: (a) A has a vibrational mode but $B$ has none.
Solution: Here $\mathrm{Cp}$ and $\mathrm{Cv}$ of A are 29 and 22 and $\mathrm{Cp}$ and $\mathrm{Cv}$ of $\mathrm{B}$ are 30 and 21 .
$\nu=\mathrm{Cp} / \mathrm{Cv}=1+2 / \mathrm{f}$
For $\mathrm{A}, \mathrm{Cp} / \mathrm{Cv}=1+2 / \mathrm{f} \Rightarrow \mathrm{f}=6$
Molecules $A$ has 3 translational, 2 rotational and 1 vibrational degree of freedom
For $B, C p / C v=1+2 / f \Rightarrow f=5$
i.e., $B$ has 3 translational and 2 rotational degrees of freedom.
Topic: Kinetic Theory of Gases
Q4: Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? ( $R=8.3 \mathrm{~J} / \mathrm{mol} \mathrm{K}$ )
Options: (a) $21.6 \mathrm{~J} / \mathrm{mol} \mathrm{K}$
(b) $19.7 \mathrm{~J} / \mathrm{mol} \mathrm{K}$
(c) $15.7 \mathrm{~J} / \mathrm{mol} \mathrm{K}$
(d) $17.4 \mathrm{~J} / \mathrm{mol} \mathrm{K}$
Answer: (d) $17.4 \mathrm{~J} / \mathrm{mol} \mathrm{K}$
Solution: $\mathrm{C}_{\mathrm{v} 1}$ of helium $=(3 / 2) \mathrm{R}$
$\mathrm{C}_{\mathrm{v} 2}$ of hydrogen $=(5 / 2) \mathrm{R}$
$$
\mathrm{C}_{\mathrm{v}} \text { of mixture }=\frac{2 \times \frac{3}{2} R+3 \times \frac{5}{2} R}{(2+3)}
$$
$C_{v}$ of mixture $=17.4 \mathrm{~J} / \mathrm{mol} \mathrm{K}$
Topic: Kinetic Theory of Gases
Q5: The mass of a hydrogen molecule is $3.32 \times 10^{-27} / \mathrm{kg}$. If $10^{23}$ hydrogen molecules strike, per second, a fixed wall of area $2 \mathrm{~cm}^{2}$ at an angle of $45^{\circ}$ to the normal, and rebound elastically with a speed of $10^{3} \mathrm{~m} \mathrm{~s}-1$, then the pressure on the wall is nearly
Options: (a) $2.35 \times 10^{3} \mathrm{~N} \mathrm{~m}^{-2}$
(b) $4.70 \times 10^{3} \mathrm{~N} \mathrm{~m}^{-2}$
(c) $2.35 \times 10^{2} \mathrm{~N} \mathrm{~m}^{-2}$
(d) $4.70 \times 10^{2} \mathrm{~N} \mathrm{~m}^{-2}$
Answer: (c) $2.35 \times 10^{2} \mathrm{~N} \mathrm{~m}^{-2}$
Solution: As $\mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}}$
Net force on the wall,
$\mathrm{F}=\mathrm{dp} / \mathrm{dt}=2 \mathrm{np} \mathrm{f}_{\mathrm{f}} \cos 45^{\circ}=2 \mathrm{nmv} \cos 45^{\circ}$
Here, $\mathrm{n}$ is the number of hydrogen molecules striking per second.
Pressure $=F / A=\left(2 n m v \cos 45^{\circ}\right) /$ Area
$$
=\frac{2 \times 10^{23} \times 3.32 \times 10^{-27} \times 10^{3} \times(1 / \sqrt{2})}{2 \times 10^{-4}}=2.35 \times 10^{3} \mathrm{~N} \mathrm{~m}^{-2}
$$
Topic: Kinetic Theory of Gases
Q6: In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on $\mathrm{T}$ as $\mathrm{T}^{q}$. A good estimate for $q$ is
Options: (a) 2
(b) 1
(c) $4 / 5$
(d) $4 / 7$
Answer: (b) 1
Solution: Pressure, $\mathrm{P}=1 / 3(\mathrm{mN} / \mathrm{V}) \mathrm{V}_{\text {rms }}^{2}$
$\mathrm{P}=(\mathrm{mN}) \mathrm{T} / \mathrm{V}$
If the gas mass and temperature are constant then
$\mathrm{P} \propto\left(\mathrm{V}_{\mathrm{rms}}\right)^{2} \propto \mathrm{T}$
So, force $\propto\left(\mathrm{V}_{\mathrm{rms}}\right)^{2} \propto \mathrm{T}$ I.e., Value of $q=1$
Topic: Kinetic Theory of Gases
Q7: One $\mathrm{kg}$ of a diatomic gas is at a pressure of $8 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$. The density of the gas is $4 \mathrm{~kg} / \mathrm{m}^{3}$. What is the energy of the gas due to its thermal motion?
Options: (a) $3 \times 10^{4} \mathrm{~J}$
(b) $5 \times 10^{4} \mathrm{~J}$
(c) $6 \times 10^{4} \mathrm{~J}$
(d) $7 \times 10^{4} \mathrm{~J}$
Answer: (b) $5 \times 10^{4} \mathrm{~J}$
Solution: The thermal energy or internal energy is $52 \mathrm{U} T$ for diatomic gases. (degree of freedom for diatomic gas $=5$ )
But PV $=$ RT
$V=$ mass $/$ density $=1 \mathrm{~kg} /\left(4 \mathrm{~kg} / \mathrm{m}^{3}\right)=(1 / 4) \mathrm{m}^{3}$
$P=8 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$
$U=(5 / 2) \times 8 \times 10^{4} \times 1 / 4=5 \times 10^{4} \mathrm{~J}$
Topic: Kinetic Theory of Gases
Q8: A gaseous mixture consists of $16 \mathrm{~g}$ of helium and $16 \mathrm{~g}$ of oxygen. The ratio $\mathrm{C}_{\mathrm{P}} / \mathrm{C}_{V}$ of the mixture is
Options: (a) 1.4
(b) 1.54
(c) 1.59
(d) 1.62
Answer: (d) 1.62
Solution: For $16 \mathrm{~g}$ of helium, $\mathrm{n}_{1}=16 / 4=4$
For $16 \mathrm{~g}$ of oxygen, $\mathrm{n}_{2}=16 / 32=1 / 2$
For a mixture of gases
$\mathrm{C}_{\mathrm{v}}=\left(\mathrm{n}_{1} \mathrm{Cv}_{1}+\mathrm{n}_{2} \mathrm{Cv}_{2}\right) /\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)$ where $\mathrm{C}_{\mathrm{v}}=(\mathrm{f} / 2) \mathrm{R}$
$C_{p}=\left(n_{1} C_{1}+n_{2} C p_{2}\right) /\left(n_{1}+n_{2}\right)$ where $C_{p}=([f / 2]+1) R$
For helium, $f=3, n_{1}=4$
For oxygen, $f=5, n_{2}=1 / 2$
$$
\frac{C p}{C v}=\frac{\left(4 \times \frac{5}{2} R\right)+\left(\frac{1}{2} \times \frac{7}{2} R\right)}{\left(4 \times \frac{3}{2} R\right)+\left(\frac{1}{2} \times \frac{5}{2} R\right)}=\frac{47}{29}=1.62
$$
Topic: Kinetic Theory of Gases
Q9: Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will
Options: (a) increase
(b) decrease
(c) remain the same
d) decrease for some, while the increase for others.
Answer: (c) remain the same
Solution: It is the relative velocities between molecules that are important. Root mean square velocities are different from lateral translation.
Topic: Kinetic Theory of Gases
Q10: At what temperature is the root mean square velocity of a hydrogen molecule equal to that of an oxygen molecule at $47^{\circ} \mathrm{C}$ ?
Options: (a) $80 \mathrm{~K}$
(b) $-73 K$
(c) $3 \mathrm{~K}$
(d) $20 \mathrm{~K}$.
Answer: (d) $20 \mathrm{~K}$
Solution: $$
v_{r m s}=\sqrt{R T / M}\left(\mathrm{~V}_{\mathrm{rms}}\right) \mathrm{O}_{2}=\left(\mathrm{V}_{\mathrm{rms}}\right) \mathrm{H}_{2}
$$
$$\sqrt{\frac{273+47}{32}}=\sqrt{\frac{T}{2}}$$
$\mathrm{T}=20 \mathrm{~K}$
Topic: Kinetic Theory of Gases
Q11:The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to [Boltzmann constant, $\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$, Avogadro Number, $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{26} / \mathrm{kg}$, Radius of Earth: $6.4 \times 10^{6} \mathrm{~m}$, Gravitational acceleration on Earth $=10 \mathrm{~m} \mathrm{~s}^{-2}$ ]
Options: (a) $10^{4} \mathrm{~K}$
(b) $800 \mathrm{~K}$
(c) $650 \mathrm{~K}$
(d) $3 \times 10^{5} \mathrm{~K}$
Answer: (a) $10^{4} \mathrm{~K}$
Solution: The escape speed of the molecule, $v_{e}=\sqrt{2 g R}$
Root mean square velocity, $v_{r m s}=\sqrt{\frac{\left.3\left(k_{B} N\right) T\right)}{m}}$
So, for $v_{e}=v_{r m s} \Rightarrow 2 g R=3\left(k_{B} \mathrm{~N}\right) \mathrm{T} / \mathrm{m}$
$\mathrm{T}=2 \mathrm{gRm} / 3 \mathrm{k}_{\mathrm{B}} \mathrm{N}=\left(2 \times 10 \times 6.4 \times 10^{6} \times 2 \times 10^{-3}\right) /\left(3 \times 1.38 \times 10^{-23} \times 6.02 \times 10^{23}\right)=10^{4} \mathrm{~K}$
Topic: Kinetic Theory of Gases
Q13: The value closest to the thermal velocity of a Helium atom at room temperature ( $300 \mathrm{~K}$ ) in $\mathrm{m} \mathrm{s}$ ${ }^{1}$ is $\left[k_{B}=1.4 \times 10^{-23} \mathrm{~J} / \mathrm{K} ; \mathrm{m}_{\mathrm{He}}=7 \times 10^{-27} \mathrm{~kg}\right]$
Options: (a) $1.3 \times 10^{3}$
(b) $1.3 \times 10^{5}$
(c) $1.3 \times 10^{2}$
(d) $1.3 \times 10^{4}$
Answer: (a) $1.3 \times 10^{3}$
Solution: $$
\begin{aligned}
& \text { (3/2) } \mathrm{k}_{\mathrm{B}} \mathrm{T}=1 / 2 \mathrm{mv} \\
& v=\sqrt{\frac{3 k_{B} T}{m}}=\sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{7 \times 10^{-27}}} \\
& =1.3 \times 10^{3} \mathrm{~ms}^{-1}
\end{aligned}
$$
Topic: Kinetic Theory of Gases
Q14: An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure ( $C_{P}$ ) and at constant volume ( $C_{V}$ ) is
Options: (a) 6
(b) $7 / 2$
(c) $5 / 2$
(d) $7 / 5$
Answer: (d) $7 / 5$
Solution: An ideal gas has molecules with 5 degrees of freedom, then
$\mathrm{Cv}=(5 / 2) \mathrm{R}$ and $\mathrm{Cp}=(7 / 2) \mathrm{R}$
$\mathrm{Cp} / \mathrm{Cv}=(7 / 2) \mathrm{R} /(5 / 2) \mathrm{R}$
$\mathrm{Cp} / \mathrm{Cv}=7 / 5$
Topic: Kinetic Theory of Gases
Q15: $\mathbf{N}$ moles of a diatomic gas in a cylinder is at a temperature $\mathrm{T}$. Heat is supplied to the cylinder such that the temperature remains constant but $n$ moles of the diatomic gas get converted into monoatomic gas. What is the change in the total kinetic energy of the gas?
Options: (a) 0
(b) $(5 / 2) n R T$
(c) $\left(\frac{1}{2}\right) \mathrm{nRT}$
(d) $(3 / 2) n R T$
Answer: (c) $(1 / 2)$ nRT
Solution: Initial kinetic energy of the system $K_{i}=(5 / 2)$ RTN
Final kinetic energy of the system $K_{f}=(5 / 2) R T(N-n)+(3 / 2) R T(2 n)$
$K_{f}-K_{i}=\Delta K=n R T(3-5 / 2)=(1 / 2) n R T$
Topic: Magnetism
A paramagnetic material has $10^{28}$ atoms $/ \mathrm{m}^{3}$. It's magnetic susceptibility at temperature $350 \mathrm{~K}$ is $2.8 \times 10^{-4}$. Its susceptibility at $300 \mathrm{~K}$ is
Options: (a) $3.267 \times 10^{-4}$, (b) $3.672 \times 10^{-4}$, (c) $2.672 \times 10^{-4}$, (d) $3.726 \times 10^{-4}$
Answer: (a) $3.267 \times 10^{-4}$
Solution: For a paramagnetic material, magnetic susceptibility $\chi \propto 1 / T$ $\chi_{2}=\chi_{1}\left(T_{1} / T_{2}\right)=2.8 \times 10^{-4} \times(350 / 300)=3.267 \times 10^{-4}$
Topic: Magnetism
A magnetic compass needle oscillates 30 times per minute at a place where the dip is $45^{\circ}$, and 40 times per minute where the dip is $30^{\circ}$. If $B_{1}$ and $B_{2}$ are respectively the total magnetic field due to the earth at the two places, then the ratio $B_{1} / B_{2}$ is best given by
Options: (a) 0.7, (b) 3.6, (c) 1.8, (d) 2.2
Answer: (a) 0.7
Solution: Frequency of oscillation $=\sqrt{\text { magnetic field }}$ $\frac{30}{40}=\sqrt{\frac{B_{1} \cos 45^{\circ}}{B_{2} \cos 30^{\circ}}}$ (Horizontal component of magnetic field $=\mathrm{B} \cos \delta$ ) $\left(B_{1} / B_{2}\right)=0.7$
Topic: Magnetism
A magnetic needle of magnetic moment $6.7 \times 10^{-2} \mathrm{~A} \mathrm{~m}^{2}$ and moment of inertia $7.5 \times 10^{-6} \mathrm{~kg} \mathrm{~m}^{2}$ is performing simple harmonic oscillations in a magnetic field of $0.01 \mathrm{~T}$. Time taken for 10 complete oscillations is
Options: (a) $6.65 \mathrm{~s}$, (b) $8.89 \mathrm{~s}$, (c) $6.98 \mathrm{~s}$, (d) $8.76 \mathrm{~s}$
Answer: (a) $6.65 \mathrm{~s}$
Solution: Time period of magnetic needle oscillating simple harmonically is given by $T=2 \pi \sqrt{\frac{I}{M B}} T=2 \pi \sqrt{\frac{7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times 0.01}}$ $\mathrm{T}=(2 \pi / 10) \times 1.05 \mathrm{~s}$ For 10 oscillations, total time taken $\mathrm{T}^{\prime}=10 \mathrm{~T}=2 \pi \times 1.05=6.65 \mathrm{~s}$
Topic: Magnetism
A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of $75^{\circ}$. One of the fields has a magnitude of $15 \mathrm{mT}$. The dipole attains stable equilibrium at an angle of $30^{\circ}$ with this field. The magnitude of the other field (in $\mathrm{mT}$ ) is close to
Options: (a) 1, (b) 11, (c) 36, (d) 1060
Answer: (b) 11
Solution: The magnetic dipole attains stable equilibrium under the influence of these two fields making an angle $\theta_{1}=30^{\circ}$ with $B_{1}$ and $\theta_{2}=75^{\circ}-30^{\circ}=45^{\circ}$ with $B_{2}$. For stable equilibrium, net torque acting on dipole must be zero, I.e., $\tau_{1}+\tau_{2}=0$ $\tau_{1}=\tau_{2}$ $m B_{1} \sin \theta_{1}=m B_{2} \sin \theta_{2}$ $B_{2}=B_{1}\left(\sin \theta_{1} / \sin \theta_{2}\right)=15 \mathrm{mT} \times\left(\sin 30^{\circ} / \sin 45^{\circ}\right)$ $=15 \mathrm{mT} \times(1 / 2) \times \sqrt{2}=10.6 \mathrm{mT}=$
Topic: Magnetism
A short bar magnet is placed in the magnetic meridian of the earth with a north pole pointing north. Neutral points are found at a distance of $30 \mathrm{~cm}$ from the magnet on the East - West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in $\mathrm{Am}^{2}$ is close to (Given $\left(\mu_{0} / 4 \pi\right)=10^{-7}$ in SI units and $B_{H}=$ Horizontal component of earth's magnetic field $=3.6 \times 10^{-}$ ${ }^{5}$ Tesla.)
Options: (a) 9.7, (b) 4.9, (c) 19.4, (d) 14.6
Answer: (a) 9.7
Solution: $\left(\mu_{0} / 4 \pi\right)\left(M / r^{3}\right)=3.6 \times 10^{-5}$ $M=\left[\left(3.6 \times 10^{-5}\right) /\left(10^{-7}\right)\right](0.3)^{3}$ $M=9.7 \mathrm{Am}^{2}$
Topic: Magnetism
Needles $\mathrm{N}_{1}, \mathrm{~N}_{2}$ and $\mathrm{N}_{3}$ are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will
Options: (a) attract all three of them, (b) attract $\mathrm{N}_{1}$ and $\mathrm{N}_{2}$ strongly but repel $\mathrm{N}_{3}$, (c) attract $\mathrm{N}_{1}$ strongly, $\mathrm{N}_{2}$ weakly and repel $\mathrm{N}_{3}$ weakly, (d) attract $\mathrm{N}_{1}$ strongly, but repel $\mathrm{N}_{2}$ and $\mathrm{N}_{3}$ weakly
Answer: (c) Magnet will attract $\mathrm{N}_{1}$ strongly, $\mathrm{N}_{2}$ weakly and repel $\mathrm{N}_{3}$ weakly
Topic: Magnetism
A magnetic needle is kept in a non-uniform magnetic field. It experiences
Options: (a) a force and a torque, (b) a force but not a torque, (c) a torque but not a force, (d) neither a force nor a torque
Answer: (a) A force and a torque act on a magnetic needle kept in a non-uniform magnetic field
Topic: Magnetism
The magnetic lines of force inside a bar magnet
Options: (a) are from north-pole to south-pole of the magnet, (b) do not exist, (c) depend upon the area of cross-section of the bar magnet, (d) are from south-pole to north-pole of the magnet
Answer: (b) Materials of low retentivity and low coercivity are suitable for making electromagnets
Topic: Magnetism
Curie temperature is the temperature above which
Options: (a) a ferromagnetic material becomes paramagnetic, (b) a paramagnetic material becomes diamagnetic, (c) a ferromagnetic material becomes diamagnetic, (d) a paramagnetic material becomes ferromagnetic
Answer: (a) a ferromagnetic material becomes paramagnetic
Topic: Magnetism
The materials suitable for making electromagnets should have
Options: (a) high retentivity and high coercivity, (b) low retentivity and low coercivity, (c) high retentivity and low coercivity, (d) low retentivity and high coercivity
Answer: (b) Materials of low retentivity and low coercivity are suitable for making electromagnets.
Topic: Magnetism
Relative permittivity and permeability of a material are $\epsilon_{r}$ and $\mu_{r}$, respectively. Which of the following values of these quantities are allowed for a diamagnetic material?
Options: (a) $\epsilon_{r}=1.5, \mu_{r}=1.5$, (b) $\epsilon_{r}=0.5, \mu_{r}=1.5$, (c) $\epsilon_{r}=1.5, \mu_{r}=0.5$, (d) $\epsilon_{r}=0.5, \mu_{r}=0.5$
Answer: (c) $\epsilon_{r}=1.5, \mu_{r}=0.5$
Solution: The values of relative permeability of diamagnetic materials are slightly less than 1 and $\varepsilon_{r}$ is quite high. Therefore $\varepsilon_{\mathrm{r}}=1.5$ and $\mu_{\mathrm{r}}=0.5$ could be the allowed values
Topic: Magnetism
In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a
Options: (a) circle, (b) helix, (c) straight line, (d) ellipse
Answer: (c) straight line
Solution: Magnetic field exerts a force $=$ Bevsin0 $=0$ Electric field exerts force along a straight line The path of the charged particle will be a straight line
Topic: Magnetism
A charged particle moves through a magnetic field perpendicular to its direction. Then
Options: (a) kinetic energy changes but the momentum is constant, (b) the momentum changes but the kinetic energy is constant, (c) both momentum and kinetic energy of the particle are not constant, (d) both momentum and kinetic energy of the particle are constant.
Answer: (b) the momentum changes but the kinetic energy is constant
Solution: The magnetic force will act perpendicular to velocity at every instant, the path will be circular. Due to change in direction momentum will change but the total energy will remain the same.
Topic: Magnetism
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating it is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to
Options: (a) induction of electrical charge on the plate, (b) shielding of magnetic lines of force as aluminium is a paramagnetic material, (c) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping, (d) development of air current when the plate is placed
Answer: (c) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.
Topic: Nuclear Physics
An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of 8: 27 . The ratio of the radii of the nuclei (assumed to be spherical) is
Options: (a) $3: 2$
(b) $2: 3$
(c) 4: 9
(d) 8: 27
Answer: (a) 3: 2
Solution: $\operatorname{As}\left(v_{1} / v_{2}\right)=8 / 27 ;\left(r_{1} / r_{2}\right)=$ ?
Using law of conservation of linear momentum, $0=m_{1} \mathrm{~V}_{1}-\mathrm{m}_{2} \mathrm{~V}_{2}$
(As both are moving in opposite directions)
Or $m_{1} / m_{2}=v_{2} / v_{1}=27 / 8$
$$
\frac{\rho\left(\frac{4}{3}\right) \pi r_{1}^{3}}{\rho\left(\frac{4}{3}\right) \pi r_{2}^{3}}=\frac{27}{8}
$$
Therefore, $\left(r_{1} / r_{2}\right)=3 / 2$
Topic: Nuclear Physics
According to Bohr's theory, the time-averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the $\mathrm{n}^{\text {th }}$ orbit is proportional to ( $\mathrm{n}=$ principal quantum number)
Options: (a) $\mathrm{n}^{-2}$
(b) $\mathrm{n}^{-3}$
(c) $\mathrm{n}^{-4}$
(d) $n^{-5}$
Answer: (d) $\mathrm{n}^{-5}$
Solution: Magnetic field at the centre, $B_{n}=\mu_{0} / / 2 r_{n}$
For a hydrogen atom, the radius of $\mathrm{n}^{\text {th }}$ orbit is given by
$r_{n}=\left(n^{2} / m\right)(h / 2 \pi)^{2}\left(4 \pi \varepsilon_{0} / e^{2}\right)$
$r_{n} \propto n^{2}$
$\mathrm{I}=\mathrm{e} / \mathrm{T}=\mathrm{e} /\left(2 \pi r_{n} / v_{n}\right)=e v_{n} / 2 \pi r_{n}$
Also, $\mathrm{v}_{\mathrm{n}} \propto \mathrm{n}^{-1} \therefore 1 \propto \mathrm{n}^{-3}$
Hence, $B_{n} \propto n^{-5}$
Topic: Nuclear Physics
Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is (given binding energy per nucleon for deuteron $=1.1 \mathrm{MeV}$ and for helium $=7.0 \mathrm{MeV}$ )
Options: (a) $25.8 \mathrm{MeV}$
(b) $32.4 \mathrm{MeV}$
(c) $30.2 \mathrm{MeV}$
(d) $23.6 \mathrm{MeV}$
Answer: (d) 23.6 MeV
Solution: ${ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{2} \rightarrow{ }_{2} \mathrm{He}^{4}$
Energy released $=4\left(\right.$ Binding Energy of $\left.\left(1_{1}^{2} \mathrm{H}\right)\right)-4\left(\right.$ Binding Energy of $\left.\left(2{ }^{4} \mathrm{He}\right)\right)=4 \times 7-4 \times 1.1=23.6 \mathrm{MeV}$
Topic: Nuclear Physics
As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion
Options: (a) kinetic energy decreases, potential energy increases but the total energy remains the same
(b) kinetic energy and total energy decrease but potential energy increases
(c) its kinetic energy increases but potential energy and total energy decrease
(d) kinetic energy, potential energy and total energy decrease.
Answer: (c) its kinetic energy increases but potential energy and total energy decrease
Solution: For an electron in $\mathrm{n}^{\text {th }}$ excited state of hydrogen atom,
kinetic energy $=\mathrm{e}^{2} /\left(8 \pi \epsilon_{0} \mathrm{n}^{2} \mathrm{a}_{0}\right)$
potential energy $=-e^{2} /\left(4 \pi \epsilon_{0} n^{2} a_{0}\right)$ and total energy $=-e^{2} /\left(8 \pi \epsilon_{0} n^{2} a_{0}\right)$
where ao is Bohr radius.
As an electron makes a transition from an excited state to the ground state, $\mathrm{n}$ decreases. Therefore kinetic energy increases but potential energy and total energy decrease.
Topic: Nuclear Physics
Energy required for the electron excitation in $\mathrm{Li}^{++}$from the first to the third Bohr orbit is
Options: (a) $12.1 \mathrm{eV}$
(b) $36.3 \mathrm{eV}$
(c) $108.8 \mathrm{eV}$
(d) $122.4 \mathrm{eV}$
Answer: (c) $108.8 \mathrm{eV}$
Solution: Using, $E_{n}=\left(13.6 Z^{2} / n^{2}\right) \mathrm{eV}$
Here, $\mathrm{Z}=3$ (For $\mathrm{Li}^{++}$)
Therefore, $\mathrm{E}_{1}=\left(13.6(3)^{2} / 1^{2}\right)=-122.4 \mathrm{eV}$
and $E_{3}=\left(13.6(3)^{2} / 3^{2}\right)=-13.6 \mathrm{eV}$
$\Delta \mathrm{E}=\mathrm{E}_{3}-\mathrm{E}_{1}=-13.6+122.4=108.8 \mathrm{eV}$
Topic: Nuclear Physics
If the binding energy per nucleon in ${ }_{3} \mathrm{Li}^{7}$ and ${ }_{2} \mathrm{He}^{4}$ nuclei are $5.60 \mathrm{MeV}$ and $7.06 \mathrm{MeV}$ respectively, then in the reaction
Options: (a) $39.2 \mathrm{MeV}$
(b) $28.24 \mathrm{MeV}$
(c) $17.28 \mathrm{MeV}$
(d) $1.46 \mathrm{MeV}$
Answer: (c) 17.28 MeV
Solution: Binding energy of $3 \mathrm{Li}^{7}=7 \times 5.60=39.2 \mathrm{MeV}$
Binding energy of ${ }_{2} \mathrm{He}^{4}=4 \times 7.06=28.24 \mathrm{MeV}$
Energy of proton $=$ Energy of $\left[2\left(2 \mathrm{He}^{4}\right)-3 \mathrm{Li}^{7}\right]$
$=2 \times 28.24-39.2=17.28 \mathrm{MeV}$
Topic: Nuclear Physics
A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2: 1. The ratio of their nuclear sizes will be
Options: (a) $2^{1 / 3}: 1$
(b) $1: 3^{1 / 2}$
(c) $3^{1 / 2}: 1$
(d) $1: 2^{1 / 3}$
Answer: (d) $1: 2^{1 / 3}$
Solution: Momentum is conserved during disintegration $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}--$-(1)
For an atom $R=R_{0} A^{1 / 3}$
$\left(R_{1} / R_{2}\right)=\left(A_{1} / A_{2}\right)^{1 / 3}=\left(m_{1} / m_{2}\right)^{1 / 3}=\left(v_{2} / v_{1}\right)^{1 / 3}$ from (1)
$\left(R_{1} / R_{2}\right)=(1 / 2)^{1 / 3}=1 / 2^{1 / 3}$