Rishabhiiitd/PhysicsQA huggingface dataset:
Topic: Centre of Mass
It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is Pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is Pc. The values of Pd and Pc are respectively :
Options: (a) (0.28,0.89)
(b) (0,0)
(c) (0,1)
(d) (0.89,0.28)
Answer: (a) (0.28,0.89)
Solution: Let the initial speed of neutron is v0 and kinetic energy is K
First collision:
By momentum conservation
mv0 = mv1 + 2mv2 ⇒ v0 = v1 + 2v2
By e = 1 v2 - v1 = v0
v2 = 2v0 / 3
v1 = -v0 / 3
Frictional loss = [1/2 mv0^2 - 1/2 m (v0 / 3)^2] / 1/2 mv0^2
Pd = 8/9 ≈ 0.89
Second Collision:
By momentum conservation
mv0 = mv1 + 12mv2
v1 + 12v2 = v0
By e = 1 v2 - v1 = v0
v2 = 2v0 / 13
v1 = -11v0 / 13
Now fraction loss of energy
Pc = [1/2 mv0^2 - 1/2 m (11v0 / 13)^2] / 1/2 mv0^2
Pc = 48/169 = 0.28
Topic: Centre of Mass
In a collinear collision, a particle with an initial speed V0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after the collision, is
Options: (a) √2 v0
(b) v0 / 2
(c) v0 √2
(d) v0 / 4
Answer: (a) √2 v0
Solution: Total kinetic energy after the collision = 1/2 mv1^2 + 1/2 mv2^2 = 3/2 (1/2 mv0^2)
v1^2 + v2^2 = (3/2) v0^2
By momentum conservation
mv0 = m(v1 + v2)
(v1 + v2)^2 = v0^2
v1^2 + v2^2 + 2v1v2 = v0^2
2v1v2 = -v0^2 / 2
(v1 - v2)^2 = v1^2 + v2^2 - 2v1v2 = (3/2) v0^2 + v0^2 / 2
(v1 - v2) = √2 v0
Topic: Electrostatics
Q1: Three charges $+Q, q,+Q$ are placed respectively, at distance $0, d / 2$ and $d$ from the origin, on the $x$-axis. If the net force experienced by $+Q$ placed at $x=0$ is zero, then value of $q$ is
Options: (a) $+\mathrm{Q} / 4$ (b) $-\mathrm{Q} / 2$ (c) $+\mathrm{Q} / 2$ (d) $-\mathrm{Q} / 4$
Answer: (d) $-\mathbf{Q} / 4$
Solution: $\mathrm{QQ} / \mathrm{d}^{2}+\mathrm{Qq} /(\mathrm{d} / 2)^{2}=0$ $Q+4 q=0$ or $\mathrm{q}=-\mathrm{Q} / 4$
Topic: Electrostatics
Q2: A parallel plate capacitor having capacitance $12 \mathrm{pF}$ is charged by a battery to a potential difference of $10 \mathrm{~V}$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is
Options: (a) $508 \mathrm{pJ}$ (b) $692 \mathrm{pJ}$ (c) $560 \mathrm{pJ}$ (d) $600 \mathrm{pJ}$
Answer: (a) 508 pJ
Solution: Initial Energy of the capacitor, $\mathrm{U}_{\mathrm{i}}=(1 / 2) \mathrm{CV}^{2}$ $=(1 / 2) \times 12 \mathrm{pF} \times 10 \times 10$ $=600 \mathrm{pJ}$ After the slab, the energy of the slab, $\mathrm{U}_{\mathrm{f}}=(1 / 2) \mathrm{Q}^{2} / \mathrm{C}^{\prime}$ $\mathrm{Q}=\mathrm{CV}=(12 \mathrm{pF})(10 \mathrm{~V})=120 \mathrm{pC}$ $\mathrm{C}^{\prime}=\mathrm{kC}=6.5 \times 120 \times 10^{-12} \mathrm{~F}$ Therefore, $\mathrm{U}_{\mathrm{f}}=\left[(1 / 2)\left(120 \times 10^{-12}\right)^{2}\right] /\left[6.5 \times 120 \times 10^{-12}\right]$ $\mathrm{U}_{\mathrm{f}}=92 \mathrm{pJ}$ $\mathrm{W}+\mathrm{U}_{\mathrm{f}}=\mathrm{U}_{\mathrm{i}}$ $\Rightarrow \mathrm{W}=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}$ $=600 \mathrm{pJ}-92 \mathrm{pJ}$ $=508 \mathrm{pJ}$
Topic: Electrostatics
Q3: An electric field of $1000 \mathrm{~V} / \mathrm{m}$ is applied to an electric dipole at an angle of $45^{\circ}$. The value of the electric dipole moment is $10^{-29} \mathrm{Cm}$. What is the potential energy of the electric dipole?
Options: (a) $-10 \times 10^{-29} \mathrm{~J}$ (b) $-7 \times 10^{-27} \mathrm{~J}$ (c) $-20 \times 10^{-18} \mathrm{~J}$ (d) $-9 \times 10^{-20} \mathrm{~J}$
Answer: (b) $-\mathbf{7} \times 10^{-27} \mathrm{~J}$
Solution: $\mathrm{E}=1000 \mathrm{~V} / \mathrm{m}, \mathrm{p}=10^{-29} \mathrm{~cm}, \theta=45^{\circ}$ Potential energy stored in the dipole, $U=-p . E \cos \theta=-10^{-29} \times 1000 \times \cos 45^{\circ}$ $\mathrm{U}=-0.707 \times 10^{-26} \mathrm{~J}=-7 \times 10^{-27} \mathrm{~J}$
Topic: Electrostatics
Q4: A solid conducting sphere, having a charge $Q$, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be $V$. If the shell is now given a charge of $-4 Q$, the new potential difference between the same two surfaces is
Options: (a) $4 \mathrm{~V}$ (b) $\mathrm{V}$ (c) $2 \mathrm{~V}$ (d) $-2 \mathrm{~V}$
Answer: (b) V
Solution: $\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=\mathrm{k}\{(\mathrm{Q} / \mathrm{a})-(\mathrm{Q} / \mathrm{b})\}$ $\mathrm{V}_{\mathrm{a}}{ }^{\prime}=\mathrm{kQ} / \mathrm{a}+\mathrm{k}(-4 \mathrm{Q}) / \mathrm{b}$ $\mathrm{V}_{\mathrm{b}}{ }^{\prime}=\mathrm{kQ} / \mathrm{b}+\mathrm{k}(-4 \mathrm{Q}) / \mathrm{b}$ $\mathrm{V}_{\mathrm{a}}^{\prime}-\mathrm{V}_{\mathrm{b}}^{\prime}=\mathrm{kQ} / \mathrm{a}-\mathrm{kQ} / \mathrm{b}=\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=\mathrm{V}$
Topic: Electrostatics
Q5: Voltage rating of a parallel plate capacitor is $500 \mathrm{~V}$. Its dielectric can withstand a maximum electric field of $106 \mathrm{~V} \mathrm{~m}^{-1}$. The plate area is $10^{-4} \mathrm{~m}^{2}$. What is the dielectric constant if the capacitance is $15 \mathrm{pF}$ ? (given $\varepsilon_{0}=8.86 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ )
Options: (a) 3.8 (b) 8.5 (c) 6.2 (d) 4.5
Answer: (b) 8.5
Solution: $\mathrm{C}=\mathrm{K} \varepsilon_{0} \mathrm{~A} / \mathrm{d}$ and $\mathrm{V}=\mathrm{Ed}$ Or $\mathrm{K}=\mathrm{CV} / \varepsilon_{0} \mathrm{AE}_{\max }$ $\mathrm{K}=\left(15 \times 10^{-12} \times 500\right) /\left(8.86 \times 10^{-12} \times 10^{-4} \times 10^{6}\right)=8.5$
Topic: Electrostatics
Q6: The bob of a simple pendulum has a mass of $2 \mathrm{~g}$ and a charge of $5.0 \mathrm{C}$. It is at rest in a uniform horizontal electric field of intensity $2000 \mathrm{~V} \mathrm{~m}^{-1}$. At equilibrium, the angle that the pendulum makes with the vertical is (take $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ )
Options: (a) $\tan ^{-1}(0.2)$ (b) $\tan ^{-1}(0.5)$ (c) $\tan ^{-1}(2.0)$ (d) $\tan ^{-1}(5.0)$
Answer: (b) $\tan ^{-1}(0.5)$
Solution: The forces acting on the bob are its weight and the force due to field. At equilibrium, $\mathrm{T} \cos \theta=\mathrm{mg}$ $\mathrm{T} \sin \theta=\mathrm{qE}$ Dividing (2) by (1) $\tan \theta=\mathrm{qE} / \mathrm{mg}$ $\theta=\tan ^{-1}\left(\left(5 \times 10^{-6} \times 2 \times 10^{3}\right) /\left(2 \times 10^{-3} \times 10\right)\right)=\tan ^{-1}(0.5)$
Topic: Electrostatics
Q7: A parallel plate capacitor has $1 \mu \mathrm{F}$ capacitance. One of its two plates is given $+2 \mu \mathrm{C}$ charge and the other plate, $+4 \mu \mathrm{C}$ charge. The potential difference developed across the capacitor is
Options: (a) $3 \mathrm{~V}$ (b) $2 \mathrm{~V}$ (c) $5 \mathrm{~V}$ (d) $1 \mathrm{~V}$
Answer: (d) $1 \mathrm{~V}$
Solution: Potential difference $V_{1}-V_{2}=\left(E_{1}-E_{2}\right) d$ $\mathrm{V}_{1}-\mathrm{V}_{2}=\left[\left(\sigma_{1} / 2 \varepsilon_{0}\right)-\left(\sigma_{2} / 2 \varepsilon_{0}\right)\right] \mathrm{d}$ $\mathrm{V}_{1}-\mathrm{V}_{2}=\left(\mathrm{q}_{1} \mathrm{~d} / 2 \mathrm{~A} \varepsilon_{0}\right)-\left(\mathrm{q}_{2} \mathrm{~d} / 2 \mathrm{~A} \varepsilon_{0}\right)=(4-2) /(2 \times 1)=1 \mathrm{~V}$
Topic: Electrostatics
Q8: A capacitor with a capacitance $5 \mu \mathrm{F}$ is charged to $5 \mu \mathrm{C}$. If the plates are pulled apart to reduce the capacitance to $2 \mu \mathrm{F}$, how much work is done?
Options: (a) $6.25 \times 10^{-6} \mathrm{~J}$ (b) $3.75 \times 10^{-6} \mathrm{~J}$ (c) $2.16 \times 10^{-6} \mathrm{~J}$ (d) $2.55 \times 10^{-6} \mathrm{~J}$
Answer: (b) $3.75 \times 10^{-6} \mathrm{~J}$
Solution: Work done $=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=(1 / 2) \mathrm{q}^{2} / \mathrm{C}_{\mathrm{f}}-(1 / 2) \mathrm{q}^{2} / \mathrm{C}_{\mathrm{i}}$ Work done $=\mathrm{q}^{2} / 2\left[1 / \mathrm{C}_{\mathrm{f}}-1 / \mathrm{C}_{\mathrm{i}}\right]$ Work done $=\left[\left(5 \times 10^{-6}\right)^{2} / 2\right]\left[\left(1 /\left(2 \times 10^{-6}\right)\right)-\left(1 /\left(5 \times 10^{-6}\right)\right)\right]$ Work done $=3.75 \times 10^{-6} \mathrm{~J}$
Topic: Electrostatics
Q9: A parallel plate capacitor of capacitance $90 \mathrm{pF}$ is connected to a battery of emf $20 \mathrm{~V}$. If a dielectric material of dielectric constant $K=5 / 3$ is inserted between the plates, the magnitude of the induced charge will be
Options: (a) $1.2 \mathrm{nC}$ (b) $0.3 \mathrm{nC}$ (c) $2.4 \mathrm{nC}$ (d) $0.9 \mathrm{nC}$
Answer: (a) $1.2 \mathrm{nC}$
Solution: Induced charge on dielectric, $\mathrm{Q}_{\text {ind }}=\mathrm{Q}(1-1 / \mathrm{K})$ Final charge on capacitor, $\mathrm{Q}=\mathrm{K} \mathrm{C}_{0} \mathrm{~V}$ $\mathrm{Q}=(5 / 3) \times 90 \times 10^{-12} \times 20=3 \times 10^{-9} \mathrm{C}=3 \mathrm{nC}$ $\mathrm{Q}_{\text {ind }}=3(1-3 / 5)=3 \times 2 / 5=1.2 \mathrm{nC}$
Topic: Electrostatics
Q10: The energy stored in the electric field produced by a metal sphere is $4.5 \mathrm{~J}$. If the sphere contains $4 \mu \mathrm{C}$ charges, its radius will be [Take: $\left(1 / 4 \pi \varepsilon_{0}\right)=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}$ ]
Options: (a) $32 \mathrm{~mm}$ (b) $20 \mathrm{~mm}$ (c) $16 \mathrm{~mm}$ (d) $28 \mathrm{~mm}$
Answer: (c) 16 mm
Solution: The energy stored in the electric field produced by a metal sphere $=4.5 \mathrm{~J}$ $\Rightarrow \mathrm{Q}^{2} / 2 \mathrm{C}=4.5$ or $\mathrm{C}=\mathrm{Q}^{2} / 2 \times 4.5$ Capacitance of spherical conductor $=4 \pi \varepsilon_{0} \mathrm{R}$ $4 \pi \varepsilon_{0} \mathrm{R}=\mathrm{Q}^{2} /(2 \times 4.5)$ $\mathrm{R}=\left(1 / 4 \pi \varepsilon_{0}\right) \times\left[\left(4 \times 10^{-6}\right)^{2} /(2 \times 4.5)\right]=9 \times 10^{9} \times(16 / 9) \times 10^{-12}=16 \times 10^{-3} \mathrm{~m}=16 \mathrm{~mm}$
Topic: Electrostatics
Q11: There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at $P$, in the region, is found to vary between the limits $589.0 \mathrm{~V}$ to $589.8 \mathrm{~V}$. What is the potential at a point on the sphere whose radius vector makes an angle of $60^{\circ}$ with the direction of the field?
Options: (a) $589.2 \mathrm{~V}$ (b) $589.6 \mathrm{~V}$ (c) $589.5 \mathrm{~V}$ (d) $589.4 \mathrm{~V}$
Answer: (d) 589.4 V
Solution: $\Delta \mathrm{V}=\mathrm{E} . \mathrm{d}$ $\Delta \mathrm{V}=\mathrm{Ed} \cos \theta=0.8 \times \cos 60^{\circ}$ $\Delta \mathrm{V}=0.4$ Hence the new potential at the point on the sphere is $589.0+0.4=589.4 \mathrm{~V}$
Topic: Electrostatics
Q12: Two identical conducting spheres $A$ and $B$, carry equal charge. They are separated by a distance much larger than their diameters, and the force between them is $\mathbf{F}$. A third identical conducting sphere, $\mathbf{C}$, is uncharged. Sphere $C$ is first touched to $A$, then to $B$, and then removed. As a result, the force between $A$ and $B$ would be equal to
Options: (a) $3 \mathrm{~F} / 8$ (b) F/2 (c) $3 \mathrm{~F} / 4$ (d) F
Answer: (a) $3 \mathrm{~F} / 8$
Solution: Initially force between spheres $\mathrm{A}$ and $\mathrm{B}, \mathrm{F}=\mathrm{kq}^{2} / \mathrm{r}$ When A and C are touched, charge on both will be $\mathrm{q} / 2$ Again $\mathrm{C}$ is touched with $\mathrm{B}$ the charge on $\mathrm{B}$ is given by $\mathrm{q}_{\mathrm{B}}=((\mathrm{q} / 2)+\mathrm{q}) / 2=3 \mathrm{q} / 4$ Required force between spheres A and B is given by $\mathrm{F}^{\prime}=\mathrm{kq}_{\mathrm{A}} \mathrm{q}_{\mathrm{B}} / \mathrm{r}^{2}=[\mathrm{k} \times(\mathrm{q} / 2) \times(3 \mathrm{q} / 4)] / \mathrm{r}^{2}=(3 / 8)\left(\mathrm{kq}^{2} / \mathrm{r}^{2}\right)=3 / 8 \mathrm{~F}$
Topic: Electrostatics
Q13: A parallel plate capacitor is made of two circular plates separated by a distance of $5 \mathrm{~mm}$ and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3 \times 10^{4} \mathrm{~V} / \mathrm{m}$, the charge density of the positive plate will be close to
Options: (a) $6 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$ (b) $6 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$ (c) $3 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$ (d) $3 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$
Answer: (b) $6 \times 10^{-7} \mathbf{C m}^{-2}$
Solution: Here, $\mathrm{K}=2.2, \mathrm{E}=3 \times 10^{4} \mathrm{Vm}^{-1}$ Electric field between the parallel plate capacitor with dielectric, $\mathrm{E}=\sigma / \mathrm{K} \varepsilon_{0} \Rightarrow \sigma=\mathrm{K} \varepsilon_{0} \mathrm{E}=2.2 \times 8.85 \times 10^{-12} \times 3 \times 10^{4}$ $\mathrm{E}=6 \times 10^{-7} \mathrm{Cm}^{-2}$
Topic: Electrostatics
Q14: Two capacitors $C_{1}$ and $C_{2}$ are charged to $120 \mathrm{~V}$ and $200 \mathrm{~V}$, respectively. It is found that by connecting them together the potential on each one can be made zero. Then
Options: (a) $9 \mathrm{C}_{1}=4 \mathrm{C}_{2}$ (b) $5 \mathrm{C}_{1}=3 \mathrm{C}_{2}$ (c) $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$ (d) $3 \mathrm{C}_{1}+5 \mathrm{C}_{2}=0$
Answer: (c) $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$
Solution: For potential to be made zero, after connection $120 \mathrm{C}_{1}=200 \mathrm{C}_{2}$ $6 \mathrm{C}_{1}=10 \mathrm{C}_{2}$ $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$
Topic: Electrostatics
Q15: An electric dipole is placed at an angle of $30^{\circ}$ to a non-uniform electric field. The dipole will experience
Options: (a) a torque only (b) a translational force only in the direction of the field (c) a translational force only in a direction normal to the direction of the field (d) a torque as well as a translational force
Answer: (d) a torque as well as a translational force
Solution: In a non-uniform electric field, the dipole will experience torque as well as a translational force.
Topic: Radioactivity
Q1: A sample of radioactive material $\mathrm{A}$, that has an activity of $10 \mathrm{mCi}\left(1 \mathbf{C i}=3.7 \times 10^{10}\right.$ decays/s) has twice the number of nuclei as another sample of a different radioactive material $B$, which has an activity of 20 $\mathrm{mCi}$. The correct choices for half-lives of $A$ and $B$ would then be respectively
Options: (a) 20 days and 5 days
(b) 10 days and 40 days
(c) 20 days and 10 days
(d) 5 days and 10 days
Answer: (a) 20 days and 5 days
Solution: $\mathrm{R}_{\mathrm{A}}=10 \mathrm{mCi}, \mathrm{R}_{\mathrm{B}}=20 \mathrm{mCi}, \mathrm{N}_{\mathrm{A}}=2 \mathrm{~N}_{\mathrm{B}}$

$\mathrm{R}_{\mathrm{A}} / \mathrm{R}_{\mathrm{B}}=\lambda_{\mathrm{A}} \mathrm{N}_{\mathrm{A}} / \lambda_{\mathrm{B}} \mathrm{N}_{\mathrm{B}}=\left[\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{B}} /\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{A}}\right] \times\left[\mathrm{N}_{\mathrm{A}} / \mathrm{N}_{\mathrm{B}}\right]$

$(1 / 2)=\left[\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{B}} /\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{A}}\right] \times 2 \Rightarrow\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{A}}=4\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{B}}$
Topic: Radioactivity
Q2: Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At $t$ $=0$ it was 1600 counts per second and at $t=8$ seconds it was 100 counts per second. The count rate observed, as counts per second, at $t=6$ seconds is close to
Options: (a) 200
(b) 360
(c) 150
(d) 400
Answer: (a) 200
Solution: According to the law of radioactivity, the count rate at $\mathrm{t}=8$ seconds is

$\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$

$\mathrm{dN} / \mathrm{dt}=\lambda \mathrm{N}=\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$

At $\mathrm{t}=0,1600=\lambda \mathrm{N}_{0} \mathrm{e}^{0}=\lambda \mathrm{N}_{0}-----(1)$

At $\mathrm{t}=8 \mathrm{~s}, 100=\lambda \mathrm{N}_{0} \mathrm{e}^{-8 \lambda}$

$\Rightarrow 100=1600 \mathrm{e}^{-8 \lambda}$

$\mathrm{e}^{8 \lambda}=16$

Therefore half life is $\mathrm{t} 1 / 2=2 \mathrm{sec}$

At $t=6 \mathrm{sec}$

$(\mathrm{dN} / \mathrm{dt})=\lambda \mathrm{N}_{0} \mathrm{e}^{-6 \lambda}=1600 \times\left(\mathrm{e}^{-2 \lambda}\right)^{3}=1600 \mathrm{x}(1 / 8)=200$
Topic: Radioactivity
Q3: Radiation coming from transitions $n=2$ to $n=1$ of hydrogen atoms fall on $H e+$ ions in $n=1$ and $n=2$ states. The possible transition of helium ions as they absorb energy from the radiation?
Answer: $\mathbf{E}_{3}$ and $\mathbf{E}_{4}$ is possible
Solution: $\mathrm{E}=13.6(1 / 1-1 / 4)=13.6 \times(3 / 4)=10.2 \mathrm{eV}$

Let us check the transitions possible on $\mathrm{He}$

$\mathrm{n}=1$ or $\mathrm{n}=2$

$\mathrm{E}_{1}=4 \times 13.6(1-1 / 4)=40.8 \mathrm{eV}\left[\mathrm{E}_{1}>\mathrm{E}\right.$, hence not possible $]$

$\mathrm{n}=1$ or $\mathrm{n}=3$

$\mathrm{E}_{2}=4 \times 13.6(1-(1 / 9))=48.3 \mathrm{eV}\left[\mathrm{E}_{2}>\mathrm{E}\right.$, hence not possible $]$

$\mathrm{n}=2$ or $\mathrm{n}=3$

$\mathrm{E}_{3}=4 \times 13.6((1 / 4)-(1 / 9))=7.56 \mathrm{eV}\left[\mathrm{E}_{3}<\mathrm{E}\right.$, hence it is possible $]$

$\mathrm{n}=2$ or $\mathrm{n}=4$

$\mathrm{E}_{4}=4 \times 13.6((1 / 4)-(1 / 6))=10.2 \mathrm{eV}\left[\mathrm{E}_{4}=\mathrm{E}\right.$, hence it is possible $]$

Hence $\mathrm{E}_{3}$ and $\mathrm{E}_{4}$ can be possible
Topic: Radioactivity
Q4: Two radioactive materials $A$ and $B$ have decay constants $10 \lambda$ and $\lambda$, respectively. If initially, they have the same number of nuclei, then the ratio of the number of nuclei of $A$ to that of $B$ will be $1 / e$ after a time
Options: (a) $1 / 9 \lambda$
(b) $11 / 10 \lambda$
(c) $1 / 10 \lambda$
(d) $1 / 11 \lambda$
Answer: (a) $1 / 9 \lambda$
Solution: $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$

So, $\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-10 \lambda t}$ and $\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$

$\Rightarrow(1 / \mathrm{e})=\left(\mathrm{N}_{1} / \mathrm{N}_{2}\right)=\left(\mathrm{N}_{0} \mathrm{e}^{-10 \lambda t}\right) /\left(\mathrm{N}_{0} \mathrm{e}^{-2 \mathrm{t}}\right)$

$\Rightarrow(1 / \mathrm{e})=\mathrm{e}^{-9 \lambda \mathrm{t}}=\mathrm{e}^{-1}=\mathrm{e}^{-9 \lambda \mathrm{t}}$

$\Rightarrow 1=9 \lambda \mathrm{t} \Rightarrow \mathrm{t}=1 / 9 \lambda$
Topic: Radioactivity
Q5: Half-lives of two radioactive nuclei $A$ and $B$ are 10 minutes and 20 minutes, respectively. If, initially a sample has an equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei $A$ and B will be
Options: (a) $3: 8$
(b) $1: 8$
(c) $9: 8$
(d) $8: 1$
Answer: (c) 9: 8
Solution: By the law of radioactivity $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \lambda}$

For nuclei A,

$\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0 \mathrm{~A}} \mathrm{e}^{-2 \mathrm{t}}$

$\operatorname{Or}\left(\mathrm{N}_{\mathrm{A}} / \mathrm{N}_{0 \mathrm{~A}}\right)=(1 / 2)^{\mathrm{n}}=(1 / 2)^{t / 10}=(1 / 2)^{6}$

$\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0 \mathrm{~A}} / 2^{6}$

For nuclei B,

$\left(\mathrm{N}_{\mathrm{B}} / \mathrm{N}_{0 \mathrm{~B}}\right)=(1 / 2)^{\mathrm{n}}=(1 / 2)^{t 20}=(1 / 2)^{3}$

$\Rightarrow \mathrm{N}_{\mathrm{B}}=\left(\mathrm{N}_{0 \mathrm{~B}}\right) / 2^{3}$

Ratio of nuclei decayed will be

$\left(\mathrm{N}^{\prime}{ }_{A} / \mathrm{N}^{\prime}{ }_{\mathrm{B}}\right)=\left(\mathrm{N}_{0 A}-\mathrm{N}_{\mathrm{A}}\right) /\left(\mathrm{N}_{0 B}-\mathrm{N}_{\mathrm{B}}\right)=\left(\mathrm{N}_{0 A} / \mathrm{N}_{0 B}\right)\left[1-(1 / 2)^{6} / 1-(1 / 2)^{3}\right]=9 / 8$
Topic: Radioactivity
Q6: Two radioactive substances $A$ and $B$ have decay constant $5 \lambda$ and $\lambda$ respectively. At $t=0$, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become $(1 / e)^{2}$ will be
Options: (a) $2 / \lambda$
(b) $1 / \lambda$
(c) $1 / 4 \lambda$
(d) $1 / 2 \lambda$
Answer: (d) $1 / 2 \lambda$
Solution: The number of undecayed nuclei at any time $t$,

$\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$

As $\mathrm{N}_{0 \mathrm{~A}}=\mathrm{N}_{0 \mathrm{~B}}$ (given)

So, for nuclei A and B

$\left.\left(\mathrm{N}_{\mathrm{A}} / \mathrm{N}_{\mathrm{B}}\right)=\mathrm{e}^{\left(-\lambda_{A}\right.}+\lambda_{B}\right) \mathrm{t}$

$\mathrm{t}=\left[1 /\left(\lambda_{\mathrm{B}}-\lambda_{\mathrm{A}}\right)\right] \operatorname{In}\left(\mathrm{N}_{\mathrm{A}} / \mathrm{N}_{\mathrm{B}}\right)=1 /(\lambda-5 \lambda) \operatorname{In}\left(1 / \mathrm{e}^{2}\right)=1 / 2 \lambda$
Topic: Radioactivity
Q7: The radiation corresponding to $3 \rightarrow 2$ transitions of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of $3 \times 10^{-4} \mathrm{~T}$. If the radius of the largest circular path followed by these electrons is $10.0 \mathrm{~mm}$, the work function of the metal is close to
Options: (a) $1.6 \mathrm{eV}$
(b) $1.8 \mathrm{eV}$
(c) $1.1 \mathrm{eV}$
(d) $0.8 \mathrm{eV}$
Answer: (c) 1.1 eV
Solution: Radius of a charged particle moving in a constant magnetic field is given by

$\mathrm{R}=(\mathrm{mv} / \mathrm{qB})$ or $\mathrm{R}^{2}=\mathrm{m}^{2} \mathrm{v}^{2} / \mathrm{q}^{2} \mathrm{~B}^{2}$

$\mathrm{R}^{2}=\left[2 \mathrm{~m}\left((1 / 2) \mathrm{mv} v^{2}\right)\right] / \mathrm{q}^{2} \mathrm{~B}^{2}$

$\mathrm{R}^{2}=2 \mathrm{~m}(\mathrm{~K} . \mathrm{E}) / \mathrm{q}^{2} \mathrm{~B}^{2}$

$\Rightarrow K . E=\left(q^{2} B^{2} R^{2}\right) / 2 m \Rightarrow K . E_{\max }=\left(q^{2} B^{2} R^{2}{ }_{\max }\right) / 2 m=0.80 \mathrm{eV}$

Energy of photon corresponding transition from orbit $3 \rightarrow 2$ in hydrogen atom.

$\left.\mathrm{E}=13.6\left(1 / 2^{2}\right)-\left(1 / 3^{2}\right)\right)=1.89 \mathrm{eV}$

Using Einstein photoelectric equation.

$\mathrm{E}=\mathrm{K} . \mathrm{E}_{\text {max }}+\Phi$

$\Rightarrow 1.89=0.8+\Phi$

$\Rightarrow \Phi=1.09 \approx 1.1 \mathrm{eV}$
Topic: Radioactivity
Q8: Assume that a neutron breaks into a proton and an electron. The energy released during this process is
Options: (Mass of neutron $=1.6725 \times 10^{-27} \mathrm{~kg}$
Mass of proton $=1.6725 \times 10^{-27} \mathrm{~kg}$
Mass of electron $=9 \times 10^{-31} \mathrm{~kg}$ )

(a) $7.10 \mathrm{MeV}$
(b) $6.30 \mathrm{MeV}$
(c) $5.4 \mathrm{MeV}$
(d) $0.73 \mathrm{MeV}$
Answer: None of the given options are correct
Solution: Mass defect, $\Delta \mathrm{m}=\mathrm{m}_{\mathrm{p}}+\mathrm{m}_{\mathrm{e}}-\mathrm{m}_{\mathrm{n}}$

$\Delta \mathrm{m}=\left(1.6725 \times 10^{-27}\right)+\left(9 \times 10^{-31}\right)-\left(1.6725 \times 10^{-27}\right) \mathrm{Kg}$

$\Delta \mathrm{m}=9 \times 10^{-31} \mathrm{~kg}$

Energy released $=\Delta \mathrm{mc}^{2}$

Energy released $=\left(9 \times 10^{-31}\right) \times\left(3 \times 10^{8}\right)^{2} \mathrm{~J}$

Energy released $=\left[\left(9 \times 10^{-31}\right) \times\left(9 \times 10^{16}\right)\right] /\left[1.6 \times 10^{-13}\right] \mathrm{Mev}=0.51 \mathrm{MeV}$
Topic: Radioactivity
Q9: The half-life period of a radioactive element $X$ is the same as the mean lifetime of another radioactive element Y. Initially, they have the same number of atoms. Then
Options: (a) $\mathrm{X}$ and $\mathrm{Y}$ decay at the same rate always
(b) $\mathrm{X}$ will decay faster than $\mathrm{Y}$
(c) $\mathrm{Y}$ will decay faster than $\mathrm{X}$
(d) $\mathrm{X}$ and $\mathrm{Y}$ have the same decay rate initially
Answer: (c) Y will decay faster than $\mathbf{X}$
Solution: $\mathrm{T}_{1 / 2}$, half-life of $\mathrm{X}=\mathrm{T}_{\text {mean }}$, mean life of $\mathrm{y}$

Or $0.693 / \lambda_{\mathrm{x}}=1 / \lambda_{\mathrm{y}}$

$\lambda_{\mathrm{x}}=0.693 \lambda_{\mathrm{y}}$

$\lambda_{\mathrm{x}}<\lambda_{\mathrm{y}}$

Rate of decay $=\lambda \mathrm{N}$

Initially, number of atoms $(\mathrm{N})$ of both are equal but since $\lambda_{\mathrm{x}}<\lambda_{\mathrm{y}}$, therefore $\mathrm{y}$ will decay at a faster rate than $\mathrm{x}$