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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 :
|
(a) (0.28,0.89)
(b) (0,0)
(c) (0,1)
(d) (0.89,0.28)
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(a) (0.28,0.89)
|
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
|
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
|
(a) √2 v0
(b) v0 / 2
(c) v0 √2
(d) v0 / 4
|
(a) √2 v0
|
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
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Centre of Mass
|
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
|
(a) $+\mathrm{Q} / 4$ (b) $-\mathrm{Q} / 2$ (c) $+\mathrm{Q} / 2$ (d) $-\mathrm{Q} / 4$
|
(d) $-\mathbf{Q} / 4$
|
$\mathrm{QQ} / \mathrm{d}^{2}+\mathrm{Qq} /(\mathrm{d} / 2)^{2}=0$ $Q+4 q=0$ or $\mathrm{q}=-\mathrm{Q} / 4$
|
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
|
(a) $508 \mathrm{pJ}$ (b) $692 \mathrm{pJ}$ (c) $560 \mathrm{pJ}$ (d) $600 \mathrm{pJ}$
|
(a) 508 pJ
|
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, $\mathrm{U}_{\mathrm{f}}=\left[(1 / 2)\left(120 \times 10^{-12}\right)^{2}\right] /\left[6.5 \times 120 \times 10^{-12}\right]$ $\mathrm{U}_{\mathrm{f}}=92 \mathrm{pJ}$ $\mathrm{W}+\mathrm{U}_{\mathrm{f}}=\mathrm{U}_{\mathrm{i}}$ $\Rightarrow \mathrm{W}=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}$ $=600 \mathrm{pJ}-92 \mathrm{pJ}$ $=508 \mathrm{pJ}$
|
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?
|
(a) $-10 \times 10^{-29} \mathrm{~J}$ (b) $-7 \times 10^{-27} \mathrm{~J}$ (c) $-20 \times 10^{-18} \mathrm{~J}$ (d) $-9 \times 10^{-20} \mathrm{~J}$
|
(b) $-\mathbf{7} \times 10^{-27} \mathrm{~J}$
|
$\mathrm{E}=1000 \mathrm{~V} / \mathrm{m}, \mathrm{p}=10^{-29} \mathrm{~cm}, \theta=45^{\circ}$ Potential energy stored in the dipole, $U=-p . E \cos \theta=-10^{-29} \times 1000 \times \cos 45^{\circ}$ $\mathrm{U}=-0.707 \times 10^{-26} \mathrm{~J}=-7 \times 10^{-27} \mathrm{~J}$
|
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
|
(a) $4 \mathrm{~V}$ (b) $\mathrm{V}$ (c) $2 \mathrm{~V}$ (d) $-2 \mathrm{~V}$
|
(b) V
|
$\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=\mathrm{k}\{(\mathrm{Q} / \mathrm{a})-(\mathrm{Q} / \mathrm{b})\}$ $\mathrm{V}_{\mathrm{a}}{ }^{\prime}=\mathrm{kQ} / \mathrm{a}+\mathrm{k}(-4 \mathrm{Q}) / \mathrm{b}$ $\mathrm{V}_{\mathrm{b}}{ }^{\prime}=\mathrm{kQ} / \mathrm{b}+\mathrm{k}(-4 \mathrm{Q}) / \mathrm{b}$ $\mathrm{V}_{\mathrm{a}}^{\prime}-\mathrm{V}_{\mathrm{b}}^{\prime}=\mathrm{kQ} / \mathrm{a}-\mathrm{kQ} / \mathrm{b}=\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=\mathrm{V}$
|
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 $\varepsilon_{0}=8.86 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ )
|
(a) 3.8 (b) 8.5 (c) 6.2 (d) 4.5
|
(b) 8.5
|
$\mathrm{C}=\mathrm{K} \varepsilon_{0} \mathrm{~A} / \mathrm{d}$ and $\mathrm{V}=\mathrm{Ed}$ Or $\mathrm{K}=\mathrm{CV} / \varepsilon_{0} \mathrm{AE}_{\max }$ $\mathrm{K}=\left(15 \times 10^{-12} \times 500\right) /\left(8.86 \times 10^{-12} \times 10^{-4} \times 10^{6}\right)=8.5$
|
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 $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ )
|
(a) $\tan ^{-1}(0.2)$ (b) $\tan ^{-1}(0.5)$ (c) $\tan ^{-1}(2.0)$ (d) $\tan ^{-1}(5.0)$
|
(b) $\tan ^{-1}(0.5)$
|
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) $\tan \theta=\mathrm{qE} / \mathrm{mg}$ $\theta=\tan ^{-1}\left(\left(5 \times 10^{-6} \times 2 \times 10^{3}\right) /\left(2 \times 10^{-3} \times 10\right)\right)=\tan ^{-1}(0.5)$
|
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
|
(a) $3 \mathrm{~V}$ (b) $2 \mathrm{~V}$ (c) $5 \mathrm{~V}$ (d) $1 \mathrm{~V}$
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(d) $1 \mathrm{~V}$
|
Potential difference $V_{1}-V_{2}=\left(E_{1}-E_{2}\right) d$ $\mathrm{V}_{1}-\mathrm{V}_{2}=\left[\left(\sigma_{1} / 2 \varepsilon_{0}\right)-\left(\sigma_{2} / 2 \varepsilon_{0}\right)\right] \mathrm{d}$ $\mathrm{V}_{1}-\mathrm{V}_{2}=\left(\mathrm{q}_{1} \mathrm{~d} / 2 \mathrm{~A} \varepsilon_{0}\right)-\left(\mathrm{q}_{2} \mathrm{~d} / 2 \mathrm{~A} \varepsilon_{0}\right)=(4-2) /(2 \times 1)=1 \mathrm{~V}$
|
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?
|
(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) $2.55 \times 10^{-6} \mathrm{~J}$
|
(b) $3.75 \times 10^{-6} \mathrm{~J}$
|
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 $=3.75 \times 10^{-6} \mathrm{~J}$
|
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
|
(a) $1.2 \mathrm{nC}$ (b) $0.3 \mathrm{nC}$ (c) $2.4 \mathrm{nC}$ (d) $0.9 \mathrm{nC}$
|
(a) $1.2 \mathrm{nC}$
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Induced charge on dielectric, $\mathrm{Q}_{\text {ind }}=\mathrm{Q}(1-1 / \mathrm{K})$ Final charge on capacitor, $\mathrm{Q}=\mathrm{K} \mathrm{C}_{0} \mathrm{~V}$ $\mathrm{Q}=(5 / 3) \times 90 \times 10^{-12} \times 20=3 \times 10^{-9} \mathrm{C}=3 \mathrm{nC}$ $\mathrm{Q}_{\text {ind }}=3(1-3 / 5)=3 \times 2 / 5=1.2 \mathrm{nC}$
|
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: $\left(1 / 4 \pi \varepsilon_{0}\right)=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}$ ]
|
(a) $32 \mathrm{~mm}$ (b) $20 \mathrm{~mm}$ (c) $16 \mathrm{~mm}$ (d) $28 \mathrm{~mm}$
|
(c) 16 mm
|
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 $=4 \pi \varepsilon_{0} \mathrm{R}$ $4 \pi \varepsilon_{0} \mathrm{R}=\mathrm{Q}^{2} /(2 \times 4.5)$ $\mathrm{R}=\left(1 / 4 \pi \varepsilon_{0}\right) \times\left[\left(4 \times 10^{-6}\right)^{2} /(2 \times 4.5)\right]=9 \times 10^{9} \times(16 / 9) \times 10^{-12}=16 \times 10^{-3} \mathrm{~m}=16 \mathrm{~mm}$
|
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?
|
(a) $589.2 \mathrm{~V}$ (b) $589.6 \mathrm{~V}$ (c) $589.5 \mathrm{~V}$ (d) $589.4 \mathrm{~V}$
|
(d) 589.4 V
|
$\Delta \mathrm{V}=\mathrm{E} . \mathrm{d}$ $\Delta \mathrm{V}=\mathrm{Ed} \cos \theta=0.8 \times \cos 60^{\circ}$ $\Delta \mathrm{V}=0.4$ Hence the new potential at the point on the sphere is $589.0+0.4=589.4 \mathrm{~V}$
|
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
|
(a) $3 \mathrm{~F} / 8$ (b) F/2 (c) $3 \mathrm{~F} / 4$ (d) F
|
(a) $3 \mathrm{~F} / 8$
|
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 $\mathrm{F}^{\prime}=\mathrm{kq}_{\mathrm{A}} \mathrm{q}_{\mathrm{B}} / \mathrm{r}^{2}=[\mathrm{k} \times(\mathrm{q} / 2) \times(3 \mathrm{q} / 4)] / \mathrm{r}^{2}=(3 / 8)\left(\mathrm{kq}^{2} / \mathrm{r}^{2}\right)=3 / 8 \mathrm{~F}$
|
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
|
(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) $3 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$
|
(b) $6 \times 10^{-7} \mathbf{C m}^{-2}$
|
Here, $\mathrm{K}=2.2, \mathrm{E}=3 \times 10^{4} \mathrm{Vm}^{-1}$ Electric field between the parallel plate capacitor with dielectric, $\mathrm{E}=\sigma / \mathrm{K} \varepsilon_{0} \Rightarrow \sigma=\mathrm{K} \varepsilon_{0} \mathrm{E}=2.2 \times 8.85 \times 10^{-12} \times 3 \times 10^{4}$ $\mathrm{E}=6 \times 10^{-7} \mathrm{Cm}^{-2}$
|
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
|
(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) $3 \mathrm{C}_{1}+5 \mathrm{C}_{2}=0$
|
(c) $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$
|
For potential to be made zero, after connection $120 \mathrm{C}_{1}=200 \mathrm{C}_{2}$ $6 \mathrm{C}_{1}=10 \mathrm{C}_{2}$ $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$
|
Electrostatics
|
Q15: An electric dipole is placed at an angle of $30^{\circ}$ to a non-uniform electric field. The dipole will experience
|
(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
|
(d) a torque as well as a translational force
|
In a non-uniform electric field, the dipole will experience torque as well as a translational force.
|
Electrostatics
|
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
|
(a) 20 days and 5 days
(b) 10 days and 40 days
(c) 20 days and 10 days
(d) 5 days and 10 days
|
(a) 20 days and 5 days
|
$\mathrm{R}_{\mathrm{A}}=10 \mathrm{mCi}, \mathrm{R}_{\mathrm{B}}=20 \mathrm{mCi}, \mathrm{N}_{\mathrm{A}}=2 \mathrm{~N}_{\mathrm{B}}$
$\mathrm{R}_{\mathrm{A}} / \mathrm{R}_{\mathrm{B}}=\lambda_{\mathrm{A}} \mathrm{N}_{\mathrm{A}} / \lambda_{\mathrm{B}} \mathrm{N}_{\mathrm{B}}=\left[\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{B}} /\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{A}}\right] \times\left[\mathrm{N}_{\mathrm{A}} / \mathrm{N}_{\mathrm{B}}\right]$
$(1 / 2)=\left[\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{B}} /\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{A}}\right] \times 2 \Rightarrow\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{A}}=4\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{B}}$
|
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
|
(a) 200
(b) 360
(c) 150
(d) 400
|
(a) 200
|
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 $t=6 \mathrm{sec}$
$(\mathrm{dN} / \mathrm{dt})=\lambda \mathrm{N}_{0} \mathrm{e}^{-6 \lambda}=1600 \times\left(\mathrm{e}^{-2 \lambda}\right)^{3}=1600 \mathrm{x}(1 / 8)=200$
|
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?
|
$\mathbf{E}_{3}$ and $\mathbf{E}_{4}$ is possible
|
$\mathrm{E}=13.6(1 / 1-1 / 4)=13.6 \times(3 / 4)=10.2 \mathrm{eV}$
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
|
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
|
(a) $1 / 9 \lambda$
(b) $11 / 10 \lambda$
(c) $1 / 10 \lambda$
(d) $1 / 11 \lambda$
|
(a) $1 / 9 \lambda$
|
$\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$
So, $\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-10 \lambda t}$ and $\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$
$\Rightarrow(1 / \mathrm{e})=\left(\mathrm{N}_{1} / \mathrm{N}_{2}\right)=\left(\mathrm{N}_{0} \mathrm{e}^{-10 \lambda t}\right) /\left(\mathrm{N}_{0} \mathrm{e}^{-2 \mathrm{t}}\right)$
$\Rightarrow(1 / \mathrm{e})=\mathrm{e}^{-9 \lambda \mathrm{t}}=\mathrm{e}^{-1}=\mathrm{e}^{-9 \lambda \mathrm{t}}$
$\Rightarrow 1=9 \lambda \mathrm{t} \Rightarrow \mathrm{t}=1 / 9 \lambda$
|
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
|
(a) $3: 8$
(b) $1: 8$
(c) $9: 8$
(d) $8: 1$
|
(c) 9: 8
|
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
$\left(\mathrm{N}^{\prime}{ }_{A} / \mathrm{N}^{\prime}{ }_{\mathrm{B}}\right)=\left(\mathrm{N}_{0 A}-\mathrm{N}_{\mathrm{A}}\right) /\left(\mathrm{N}_{0 B}-\mathrm{N}_{\mathrm{B}}\right)=\left(\mathrm{N}_{0 A} / \mathrm{N}_{0 B}\right)\left[1-(1 / 2)^{6} / 1-(1 / 2)^{3}\right]=9 / 8$
|
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
|
(a) $2 / \lambda$
(b) $1 / \lambda$
(c) $1 / 4 \lambda$
(d) $1 / 2 \lambda$
|
(d) $1 / 2 \lambda$
|
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
$\left.\left(\mathrm{N}_{\mathrm{A}} / \mathrm{N}_{\mathrm{B}}\right)=\mathrm{e}^{\left(-\lambda_{A}\right.}+\lambda_{B}\right) \mathrm{t}$
$\mathrm{t}=\left[1 /\left(\lambda_{\mathrm{B}}-\lambda_{\mathrm{A}}\right)\right] \operatorname{In}\left(\mathrm{N}_{\mathrm{A}} / \mathrm{N}_{\mathrm{B}}\right)=1 /(\lambda-5 \lambda) \operatorname{In}\left(1 / \mathrm{e}^{2}\right)=1 / 2 \lambda$
|
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
|
(a) $1.6 \mathrm{eV}$
(b) $1.8 \mathrm{eV}$
(c) $1.1 \mathrm{eV}$
(d) $0.8 \mathrm{eV}$
|
(c) 1.1 eV
|
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.
$\mathrm{E}=\mathrm{K} . \mathrm{E}_{\text {max }}+\Phi$
$\Rightarrow 1.89=0.8+\Phi$
$\Rightarrow \Phi=1.09 \approx 1.1 \mathrm{eV}$
|
Radioactivity
|
Q8: Assume that a neutron breaks into a proton and an electron. The energy released during this process is
|
(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) $0.73 \mathrm{MeV}$
|
None of the given options are correct
|
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 $=\left[\left(9 \times 10^{-31}\right) \times\left(9 \times 10^{16}\right)\right] /\left[1.6 \times 10^{-13}\right] \mathrm{Mev}=0.51 \mathrm{MeV}$
|
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
|
(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
|
(c) Y will decay faster than $\mathbf{X}$
|
$\mathrm{T}_{1 / 2}$, 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 $\mathrm{x}$
|
Radioactivity
|
Q10: If the binding energy of the electron in a hydrogen atom is $13.6 \mathrm{eV}$, the energy required to remove the electron from the first excited state of $\mathrm{Li}^{\text {t+ }}$ is
|
(a) $30.6 \mathrm{eV}$
(b) $13.6 \mathrm{eV}$
(c) $3.4 \mathrm{eV}$
(d) $122.4 \mathrm{eV}$
|
(a) $30.6 \mathrm{eV}$
|
$\mathrm{E}_{2}=\left(-\mathrm{Z}^{2} \mathrm{E}_{0}\right) / \mathrm{n}^{2}$
$\mathrm{E}_{2}=\left(-(3)^{2} \times 13.6\right) /(2)^{2}$
$=-30.6 \mathrm{eV}$
Energy required $=30.6 \mathrm{eV}$
|
Radioactivity
|
Q1: A convex lens is put $10 \mathrm{~cm}$ from a light source and it makes a sharp image on a screen, kept $10 \mathrm{~cm}$ from the lens. Now a glass block (refractive index 1.5 ) of $1.5 \mathrm{~cm}$ thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance $d$. Then $d$ is
|
(a) $0.55 \mathrm{~cm}$ towards the lens
(b) 0
(c) $1.1 \mathrm{~cm}$ away from the lens
(d) $0.55 \mathrm{~cm}$ away from the lens
|
(d) $0.55 \mathrm{~cm}$ away from the lens
|
Case I: $u=-10 \mathrm{~cm}, \mathrm{v}=10 \mathrm{~cm}, \mathrm{f}=$ ?
Using lens formula,
$(1 / \mathrm{f})=[(1 / \mathrm{v})-(1 / \mathrm{u})]$
$1 / \mathrm{f}=[(1 / 10)-(1 /-10)]$
$\mathrm{f}=5 \mathrm{~cm}$
Case II: Due to introduction of slab, shift in the source is $=\mathrm{t}[1-(1 / \mu)]=1.5[1-(2 / 3)]=0.5$
Now, $u=-9.5 \mathrm{~cm}, \mathrm{v}=10.55 \mathrm{~cm} \mathrm{~d}=10.55-10=0.55 \mathrm{~cm}$ away from the lens.
|
Ray Optics
|
Q2: The speed of light in the medium is
|
(a) maximum on the axis of the beam
(b) minimum on the axis of the beam
(c) the same everywhere in the beam
(d) directly proportional to the intensity I
|
(b) minimum on the axis of the beam
|
Given $\mu=\mu_{0}+\mu_{2} \mathrm{I}$
As $\mu=$ Speed of light in vacuum/Speed of light in the medium
$\mu=\mathrm{c} / \mathrm{v}$
\$mathrm{v}=\mathrm{c} / \mu=\mathrm{c} /\left(\mu_{0}+\mu_{2} \mathrm{I}\right)$
As the intensity is maximum on the axis of the beam, therefore $v$ is minimum on the axis of the beam.
|
Ray Optics
|
Q3: As the beam enters the medium, it will
|
(a) travel as a cylindrical beam
(b) diverge
(c) converge
(d) diverge near the axis and converge near the periphery
|
(c) converge
|
As the beam enters the medium, it will converge.
|
Ray Optics
|
Q4: The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea $(7.8 \mathrm{~mm}$ ). This surface separates two media of refractive indices 1 and 1.34 . Calculate the distance from the refracting surface at which a parallel beam of light will come to focus
|
(a) $4.0 \mathrm{~cm}$
(b) $1 \mathrm{~cm}$
(c) $3.1 \mathrm{~cm}$
(d) $2 \mathrm{~cm}$
|
(c) $3.1 \mathrm{~cm}$
|
$(1.34 / v)-(1 / \infty)=(1.34-1) / 7.8$
$1.34 / \mathrm{v}=0.34 / 7.8$
$\mathrm{v}=(1.34 \times 7.8) / 0.34$
$\mathrm{v}=30.7 \mathrm{~mm} \approx 3.1 \mathrm{~cm}$
|
Ray Optics
|
Q5: In an interference experiment the ratio of amplitudes of coherent waves is $\left(a_{1} / a_{2}\right)=1 / 3$. The ratio of maximum and minimum intensities of fringes will be
|
(a) 4
(b) 9
(c) 2
(d) 18
|
(a) 4
|
$\left(\mathrm{I}_{\max } / \mathrm{I}_{\text {min }}\right)=\left(\mathrm{a}_{1}+\mathrm{a}_{2}\right)^{2} /\left(\mathrm{a}_{1}-\mathrm{a}_{2}\right)^{2}=(1+3)^{2} /(1-3)^{2}=16 / 4=4$
|
Ray Optics
|
Q6: The image formed by an objective of a compound microscope is
|
(a) virtual and diminished
(b) real and diminished
(c) real and enlarged
(d) virtual and enlarged
|
(c) The objective of a compound microscope forms a real and enlarged image.
|
The image formed by an objective of a compound microscope is real and enlarged.
|
Ray Optics
|
Q7: Calculate the limit of resolution of a telescope objective having a diameter of $200 \mathrm{~cm}$, if it has to detect light of wavelength $500 \mathrm{~nm}$ coming from a star.
|
(a) $610 \times 10^{-9}$ radian
(b) $152.5 \times 10^{-9}$ radian
(c) $457.5 \times 10^{-9}$ radian
(d) $305 \times 10^{-9}$ radian
|
(d) $305 \times 10^{-9}$ radian
|
The limit of resolution,
$\Delta \theta=1.22 \lambda / \mathrm{a}=\left(1.22 \times 500 \times 10^{-9} / 200 \times 10^{-2}\right)=3.05 \times 10^{-7} \mathrm{radian}$
$\Delta \theta=305 \times 10^{-9}$ radian
|
Ray Optics
|
Q8: A concave mirror for face viewing has a focal length of $0.4 \mathrm{~m}$. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is
|
(a) $0.16 \mathrm{~m}$
(b) $1.60 \mathrm{~m}$
(c) $0.32 \mathrm{~m}$
(d) $0.24 \mathrm{~m}$
|
(c) $0.32 \mathrm{~m}$
|
Given, $\mathrm{f}=-0.4 \mathrm{~m}, \mathrm{~m}=5=(-\mathrm{v} / \mathrm{u})=(\mathrm{y} /-\mathrm{x}) \Rightarrow \mathrm{y}=5 \mathrm{x}$
$(1 / \mathrm{f})=(1 / \mathrm{v})+(1 / \mathrm{u})$
$\Rightarrow(1 /-0.4)=(1 / 5 x)+(1 /-x)=4 /-5 x$
$\Rightarrow \mathrm{x}=-0.32 \mathrm{~m}$ so $\mathrm{u}=0.32 \mathrm{~m}$
|
Ray Optics
|
Q9: The value of the numerical aperture of the objective lens of a microscope is 1.25. If the light of wavelength $5000 \AA$ is used, the minimum separation between two points, to be seen as distinct, will be
|
(a) $0.48 \mathrm{~m}$
(b) $0.12 \mathrm{~m}$
(c) $0.38 \mathrm{~m}$
(d) $0.24 \mathrm{~m}$
|
(d) $0.24 \mathrm{~m}$
|
Numerical aperture of the objective lens of microscope $=0.61 \lambda / \mathrm{d}$
Minimum separation between two points $\mathrm{d}$ to be seen clearly,
$\mathrm{d}=0.61 \lambda /$ Numerical aperture
$\mathrm{d}=\left(0.61 \times 5000 \times 10^{-10}\right) / 1.25$
$\mathrm{d}=0.24 \mathrm{~m}$
|
Ray Optics
|
Q10: A convergent doublet of separated lenses, corrected for spherical aberration, has a resultant focal length of $10 \mathrm{~cm}$. The separation between the two lenses is $2 \mathrm{~cm}$. The focal lengths of the component lenses are
|
(a) $18 \mathrm{~cm}, 20 \mathrm{~cm}$
(b) $12 \mathrm{~cm}, 14 \mathrm{~cm}$
(c) $16 \mathrm{~cm}, 18 \mathrm{~cm}$
(d) $10 \mathrm{~cm}, 12 \mathrm{~cm}$
|
(a) $18 \mathrm{~cm}, 20 \mathrm{~cm}$
|
Since the doublet is corrected for the spherical aberration, it satisfies the following condition
$\mathrm{f}_{1}-\mathrm{f}_{2}=\mathrm{d}=2 \mathrm{~cm}$
$\mathrm{f}_{1}=\mathrm{f}_{2}+2$
Equivalent focal length $=\mathrm{F}$
$\mathrm{F}=\left(\mathrm{f}_{1} \mathrm{f}_{2}\right) /\left(\mathrm{f}_{1}+\mathrm{f}_{2}-\mathrm{d}\right)=10 \mathrm{~cm}$
Solving it, we get
$\mathrm{f}_{1}=20 \mathrm{~cm}$
$\mathrm{f}_{2}=18 \mathrm{~cm}$
|
Ray Optics
|
Q11: A single slit of width $b$ is illuminated by coherent monochromatic light of wavelength $\lambda$. If the second and fourth minima in the diffraction pattern at a distance $1 \mathrm{~m}$ from the slit are at $3 \mathrm{~cm}$ and $6 \mathrm{~cm}$, respectively from the central maximum, what is the width of the central maximum? (i.e. the distance between the first minimum on either side of the central maximum)
|
(a) $6.0 \mathrm{~cm}$
(b) $1.5 \mathrm{~cm}$
(c) $4.5 \mathrm{~cm}$
(d) $3.0 \mathrm{~cm}$
|
(d) $3.0 \mathrm{~cm}$
|
For single slit diffraction, $\sin \theta=\mathrm{n} \lambda / \mathrm{b}$
Position of $\mathrm{n}^{\text {th }}$ minima from central maxima $=\mathrm{n} \lambda_{\mathrm{D}} / \mathrm{b}$
When $\mathrm{n}=2$, then $\mathrm{x}_{2}=2 \lambda_{\mathrm{D}} / \mathrm{b}=0.03$
When $\mathrm{n}=4$, then $\mathrm{x}_{4}=4 \lambda_{\mathrm{D}} / \mathrm{b}=0.06$
Eqn. (2) - Eqn. (1)
$\mathrm{x}_{4}-\mathrm{x}_{2}=\left(4 \lambda_{\mathrm{D}} / \mathrm{b}\right)-\left(2 \lambda_{\mathrm{D}} / \mathrm{b}\right)=0.03$ or
then width of central maximum $=2 \lambda_{\mathrm{D}} / \mathrm{b}=2 \times(0.03 / 2)=0.03 \mathrm{~m}=3 \mathrm{~cm}$
|
Ray Optics
|
Q12: An observer looks at a distant tree of height $10 \mathrm{~m}$ with a telescope of the magnifying power of 20 . To the observer, the tree appears
|
(a) 10 times taller
(b) 10 times nearer
(c) 20 times taller
(d) 20 times nearer
|
(d) 20 times nearer
|
The telescope resolves and brings the objects closer, which are far away from the telescope. Hence, for a telescope with magnifying power 20 , the tree appears 20 times nearer.
|
Ray Optics
|
Q13: To determine the refractive index of a glass slab using a travelling microscope, the minimum number of readings required are
|
(a) Two
(b) Four
(c) Three
(d) Five
|
(c) Three
|
To determine the refractive index of a glass slab using a travelling microscope, the minimum number of readings required are three.
|
Ray Optics
|
Q14: You are asked to design a shaving mirror assuming that a person keeps it $10 \mathrm{~cm}$ from his face and views the magnified image of the face at the closest comfortable distance of $25 \mathrm{~cm}$. The radius of curvature of the mirror would then be
|
(a) $30 \mathrm{~cm}$
(b) $24 \mathrm{~cm}$
(c) $60 \mathrm{~cm}$
(d) $-24 \mathrm{~cm}$
|
(c) $60 \mathrm{~cm}$
|
$(1 / 15)+(1 /-10)=1 / \mathrm{f}$
$\mathrm{f}=-30 \mathrm{~cm}$
$R=2 f=2(-30)=-60 \mathrm{~cm}$
|
Ray Optics
|
Q15: A green light is incident from the water to the air-water interface at the critical angle ( $\boldsymbol{\theta}$ ). Select the
|
(a) The entire spectrum of visible light will come out of the water at various angles to the normal
(b) The entire spectrum of visible light will come out of the water at an angle of $90^{\circ}$ to the normal
(c) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium
(d) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium
|
(c) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium
|
As, $\sin \theta=1 / \mu$
Also, the refractive index $(\mu)$ of the medium depends on the wavelength of the light. $\mu$ is less for the greater wavelength (i.e. lesser frequency).
So, $\theta$ will be more for a lesser frequency of light. Hence, the spectrum of visible light whose frequency is less than that of green light will come out to the air medium.
|
Ray Optics
|
Q1: Two-point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod which is of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum is located at a distance of
|
1. 0.42 m from the mass of 0.3 kg
2. 0.70 m from the mass of 0.7 kg
3. 0.98 m from the mass of 0.3 kg
4. 0.98 m from the mass of 0.7 kg
|
(3) 0.98 m from the mass of 0.3 kg
|
The moment of inertia of the system about the axis of rotation O is I = I1 + I2 = (0.3)x^2 + 0.7(1.4 - x)^2
I = 0.3 x^2 + 0.7(1.96 + x^2 - 2.8 x)
= 0.3 x^2 + 1.372 + 0.7 x^2 - 1.96 x
= x^2 + 1.372 - 1.96 x
The work done in rotating the rod is converted into its rotational kinetic energy.
W = 1 / 2 I ω^2
= 1 / 2 [x^2 + 1.372 - 1.96 x] ω^2
For the work done to be minimum
dW / dx = 0 ⇒ 2 x - 1.96 = 0
x = 1.96 / 2 = 0.98 m
|
Rotational Motion
|
Q2: A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved?
|
1. centre of the circle
2. on the circumference of the circle
3. inside the circle
4. outside the circle
|
(1) centre of the circle
|
The force will pass through the centre of the circle. Therefore, the angular momentum will remain conserved at the centre of the circle.
|
Rotational Motion
|
Q3: A child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is
|
1. 2 K
2. K / 2
3. K / 4
4. 4 K
|
(2) K / 2
|
According to the conservation of angular momentum
I ω0 = I1 ω1
So we have I1 = 2 I
ω1 = ω0 / 2
Initial kinetic energy = I ω0^2 / 2 = K
Final Kinetic energy = I1 ω1^2 / 2 = 2 I (ω0 / 2)^2 / 2 = K / 2
|
Rotational Motion
|
Q4: A particle is confined to rotate in a circular path with decreasing linear speed. Then which of the following is correct?
|
1. angular momentum is conserved about the centre
2. only the direction of L is conserved
3. it spirals towards the centre
4. its acceleration is towards the centre
|
(2) only the direction of L is conserved
|
L is not conserved in magnitude since v is changing (decreasing). It is given that a particle is confined to rotate in a circular path, it cannot have a spiral path. Since the particle has two accelerations ac and at, therefore the net acceleration is not towards the centre. The direction of L remains the same even when the speed decreases.
|
Rotational Motion
|
Q5: 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 point B. The angle θ at which the speed of the bob is half of that at A satisfies
|
1. θ = π / 2
2. π / 4 < θ < π / 2
3. π / 2 < θ < 3 π / 4
4. 3 π / 2 < θ < π
|
(4) 3 π / 2 < θ < π
|
This is the case of vertical motion when the body just completes the circle. Here
v = √(5 g L)
Applying energy conservation,
1 / 2 mv0^2 = 1 / 2 mv^2 + mgl(1 - cos θ)
where v0 is the horizontal velocity at the bottom point, v is the velocity of bob where the bob inclined θ with vertical.
Also, we know the relation between the velocity at the topmost and velocity at the bottom point.
mg(2l) = 1 / 2 mv0^2 - 1 / 2 mvtop^2
Since v0 is just sufficient
mvtop^2 / l = T + mg
T = 0
vtop = √(g l)
Then equation 2 becomes
v0 = √(5 g l)
According to the question v = v0 / 2
So from equation (1)
1 / 2 m(5 gl) = 1 / 2 m(5 gl / 4) + mgl(1 - cos θ)
(20 mgl - 5 mgl) / 8 = mgl(1 - cos θ)
(1 - cos θ) = 15 / 8
cos θ = 7 / 8
Hence, 3 π / 2 < θ < π
|
Rotational Motion
|
Q6: A ball of mass (m) 0.5 kg is attached to the end of a string having a length (L) 0.5 m. The ball is rotated on a horizontal circular path about a vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of the angular velocity of ball (in radian/s) is
|
1. 9
2. 18
3. 27
4. 36
|
(4) 36
|
m ω^2 L sin θ cos θ = mg sin θ
cos θ = g / ω^2 L
sin θ = 1 / ω^2 L
sin θ = 1 / (ω^2 L) √((ω^2 L)^2 - g^2)
T = mg cos θ + m ω^2 L sin^2 θ
T = mg (g / ω^2 L) + (m ω^2 / ω^2 L)^2 ((ω^2 L)^2 - g^2)
T = mg / ω^2 L [ g^2 + (ω^2 L)^2 - g^2]
T = m ω^2 L = 324 (given)
ω = √(324 / 0.5 * 0.5)
ω = 36 rad / s
|
Rotational Motion
|
Q8: A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m / s. A small ball of mass 0.1 kg, moving velocity 20 m / s in the opposite direction hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m / s. Immediately after the collision.
|
1. The ring has pure rotation about its stationary CM
2. The ring comes to a complete stop
3. Friction between the ring and the ground is to the left
4. There is no friction between the ring and the ground
|
(1) The ring has pure rotation about its stationary CM
(3) Friction between the ring and the ground is to the left
|
Let's assume that friction between the ground and the ring gives no impulse during the collision with the ball. Using conservation of momentum along the x-axis we get that the CM of the ring will come to rest. Thus option A is correct.
Secondly, the question tells us that the ball gets a velocity in the vertical direction, hence there must be an impulse in the vertical direction. There will be horizontal and a vertical impulse on the ring at the point of contact. These will have components along the tangent of the ring, which will provide angular impulses.
Using angular impulse = change in angular momentum, we get
= 2 cos 30°(1 / 2) - 1 sin 30°(1 / 2) = 2(1 / 4)(ω2 - ω1)
note that we have assumed that the direction of angular velocities is the same before and after and since LHS of the above equation is positive ω2 > ω1 so the ring must be slipping to right and hence the friction will be to the left as it will be opposite to the direction of motion. Thus option C is correct
|
Rotational Motion
|
Q9: A carpet of mass M, made of inextensible material is rolled along its length in the form of a cylinder of radius R and is kept on the rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to R / 2.
|
1. √(14 / 3 g R)
2. √(7 / 3 g R)
3. √(2 g R)
4. √(g R)
|
(1) √(14 / 3 g R)
|
The radius is reduced to R / 2, the mass of the rolled carpet = (M / π R^2 1) × (R / 2)^2 1 = M / 4
Release of potential energy = MgR - (M / 4) gR / 2 = 7 / 8 MgR
The kinetic energy at this instant is given by = 1 / 2 Mv^2 + 1 / 2 I ω^2 = 1 / 2(M / 4) v^2 + 1 / 2(MR^2 / 32)(2v / R)^2 = (3 / 16) Mv^2
(3 / 16) Mv^2 = 7 / 8 MgR
v = √(14 / 3 g R)
|
Rotational Motion
|
Q10: A bob of mass m attached to an inextensible string of length is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad / s about the vertical. About the point of suspension
|
1. Angular momentum changes in direction but not in magnitude
2. Angular momentum changes both in direction and magnitude
3. Angular momentum is conserved
4. Angular momentum changes in magnitude but not in direction.
|
(1) Angular momentum changes in direction but not in magnitude
|
Rotational Motion
|
|
Q14: Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is
|
1. (55 / 2) MR^2
2. (73 / 2) MR^2
3. (181 / 2) MR^2
4. (19 / 2) MR^2
|
(3) (181 / 2) MR^2
|
I0 = Icm + md^2
I0 = (7 MR^2 / 2) + 6 (M * (2 R)^2) = 55 MR^2 / 2
Ip = I0 + md^2
Ip = 55 MR^2 / 2 + 7 M (3 R)^2 = (181 / 2) MR^2
|
Rotational Motion
|
Q15: A pulley of radius 2 m is rotated about its axis by a force F = 20 t - 5 t^2 newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kgm^2, the number of rotations made by the pulley before its direction of motion it reversed, is
|
1. more than 6 but less than 9
2. more than 9
3. less than 3
4. more than 3 but less than 6
|
(4) more than 3 but less than 6
|
Torque is given by τ = FR
Or α = FR / I
Given F = 20 t - 5 t^2
R = 2 m,
I = 10 kgm^2
α = [(20 t - 5 t^2) * 2] / 10
α = 4 t - t^2
ω = ∫0^t α dt = 2 t^2 - t^3 / 3
at ω = 0 ⇒ 2 t^2 - t^3 / 3 = 0
t^3 = 6 t^2
t = 6
θ = ∫0^6 ω dt = ∫0^6 (2 t^2 - t^3 / 3) dt
θ = 36 / 2 π < 6
|
Rotational Motion
|
Q1: The energy band gap is maximum in
|
(a) metals
(b) superconductors
(c) insulators
(d) semiconductors
|
(c) The energy band gap is maximum in insulators
|
Semiconductors
|
|
Q2: A piece of copper and another of germanium are cooled from room temperature to 77 K, the resistance of
|
(a) each of them increases
(b) each of them decreases
(c) copper decreases and germanium increases
(d) copper increases and germanium decreases
|
(c) copper decreases and germanium increases
|
Copper is a conductor. Germanium is a semiconductor. When cooled, the resistance of copper decreases and that of germanium increases.
|
Semiconductors
|
Q3: In the common base mode of a transistor, the collector current is 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor ( β ) will be
|
(a) 48
(b) 49
(c) 50
(d) 51
|
(b) 49
|
β=Ic / Ib=Ic /(Ie-Ic)=5.488 /(5.60-5.488)=5.488 / 0.112=49
|
Semiconductors
|
Q4: Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate?
|
(a) The number of free electrons for conduction is significant only in Si and Ge but small in C
(b) The number of free conduction electrons is significant in C but small in Si and Ge
(c) The number of free conduction electrons is negligibly small in all the three
(d) The number of free electrons for conduction is significant in all three
|
(a) The number of free electrons for conduction is significant only in Si and Ge but small in C
|
Carbon (C), silicon (Si) and germanium (Ge) have the same lattice structure and their valence electrons are 4. For C, these electrons are in the second orbit, for Si it is third and for germanium, it is the fourth orbit. In solid-state, the higher the orbit, the greater the possibility of overlapping of energy bands. Ionization energies are also less therefore Ge has more conductivity compared to Si. Both are semiconductors. Carbon is an insulator.
|
Semiconductors
|
Q5: In the ratio of the concentration of electrons that of holes in a semiconductor is 7 / 5 and the ratio of currents is 7 / 4 then what is the ratio of their drift velocities?
|
(a) 4 / 7
(b) 5 / 8
(c) 4 / 5
(d) 5 / 4
|
(d) 5 / 4
|
Drift velocity, Vd=I / nAe
(vd)electron /(vd)hole=(Ie / Ih)(nh / ne)=(7 / 4) ×(5 / 7)=5 / 4
|
Semiconductors
|
Q6: At absolute zero, silicon (Si) acts as
|
(a) non-metal
(b) metal
(c) insulator
(d) none of these
|
(c) insulator
|
Semiconductors like silicon (Si) and germanium (Ge) act as insulators at low temperature.
|
Semiconductors
|
Q7: Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If for an n-type semiconductor, the density of electrons is 10^19 m^-3 and their mobility is 1.6 m^2 /(V-s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to
|
(a) 0.2 m
(b) 4 m
(c) 2 m
(d) 0.4 m
|
(d) 0.4 m
|
J=neVd
Resistivity, ρ=E / j=E / neVd=1 / ne (vd / E)=1 / ne μe
Resistivity, 1 /(10^19 × 1.6 × 10^-19 × 1.6)=0.39 Ω m=0.4 Ω m
|
Semiconductors
|
Q8: The electrical conductivity of a semiconductor increases when electromagnetic radiation of a wavelength shorter than 2480 nm is incident on it. The bandgap in (eV) for the semiconductor is
|
(a) 0.5 eV
(b) 0.7 eV
(c) 1.1 eV
(d) 2.5 eV
|
(a) 0.5 eV
|
Band gap = Energy of photon of =2480 nm
Energy =(hc / λ) J=(hc / λe) e V
Band gap =((6.63 × 10^-34) ×(3 × 10^8)) /((2480 × 10^-9) ×(1.6 × 10^-19))=0.5 eV
|
Semiconductors
|
Q9: The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the
|
(a) crystal structure
(b) variation of the number of charge carriers with temperature
(c) type of bonding
(d) variation of scattering mechanism with temperature
|
(b) variation of the number of charge carriers with temperature
|
Semiconductors
|
|
Q10: A strip of copper and another germanium are cooled from room temperature to 80 K. The resistance of
|
(a) each of these decreases
(b) copper strip increases and that of germanium decreases
(c) copper strip decreases and that of germanium increases
(d) each of these increases
|
(c) copper strip decreases and that of germanium increases
|
Copper is conductor and germanium is a semiconductor. When cooled, the resistance of copper strip decreases and that of germanium increases.
|
Semiconductors
|
Q1. A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in second is
|
(a) 8π / 3
(b) 4π / 3
(c) 3π / 8
(d) 7π / 3
|
Answer: (a) 8π / 3
|
In SHM, speed v = ω √(A^2 - x^2), at x = 4, v = ω √((5)^2 - (4)^2) = 3ω
Acceleration a = -ω^2 x
at x = 4, a = -4ω^2
As |v| = |a|
ω = 3 / 4 ⇒ T = 2π / ω = 8π / 3
|
Simple Harmonic Motion
|
Q2: A cylindrical plastic bottle of negligible mass is filled with 310 mL of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency. If the radius of the bottle is 2.5 cm then ω is close to (density of water = 10^3 kg/m^3 )
|
(a) 2.50 rad/s
(b) 3.75 rad/s
(c) 5.00 rad/s
(d) 1.25 rad/s
|
None of the options given are correct
|
Restoring force due to pressing the bottle with amount x,
F = - (ρAx)g
a = - (ρAg/m)x
Therefore, ω^2 = ρAg/m = [ρ(πr^2)g] / m
ω = √(10^3 * π * (2.5 * 10^-2)^2 * 10 / 310 * 10^-3)
|
Simple Harmonic Motion
|
Q3: A particle undergoing simple harmonic motion has time-dependent displacement given by x(t) = Asin(πt / 90). The ratio of kinetic to the potential energy of this particle at t = 210 s will be
|
(a) 1 / 3
(b) 2
(c) 1
(d) 3
|
Answer: (a) 1 / 3
|
The maximum kinetic energy of the particle, K.E = (1 / 2)mA^2ω^2cos^2(ωt)
The potential energy of the particle at any time t, U = (1 / 2)mA^2ω^2sin^2(ωt)
The ratio of kinetic energy to potential energy, r = K.E/U = [(1 / 2)mA^2ω^2cos^2(ωt)] / [(1 / 2)mA^2ω^2sin^2(ωt)]
r = cos^2(ωt) / sin^2(ωt)
r = cot^2(ωt)
The angular frequency is given as ω = π / 90
Time, t = 210 s
Therefore, r = cot^2(π / 90) * 210
r = cot^2(7π / 3)
r = cot^2(2π + π / 3)
We know,
cot(2nπ + θ) = cotθ
Therefore, r = cot^2(π / 3)
r = (1 / √3)^2
r = 1 / 3
|
Simple Harmonic Motion
|
Q4: A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then
|
(a) K2 = K1
(b) K2 = K1 / 2
(c) K2 = 2K1
(d) K2 = K1 / 4
|
Answer: (c) K2 = 2K1
|
Maximum kinetic energy = (1 / 2)mω^2A^2
ω = √(g / L)
A = Lθ
K.E = (1 / 2)mgLθ^2
K1 = (1 / 2)mgLθ^2
If the length is doubled
K2 = (1 / 2)mg2Lθ^2
From (1) and (2)
(K1 / K2) = (1 / 2)mgLθ^2 / (1 / 2)mg2Lθ^2
= 1 / 2
K2 = 2K1
|
Simple Harmonic Motion
|
Q5: A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10^-2 m. The relative change in the angular frequency of the pendulum is best given by
|
(a) 10^-5 rad/s
(b) 10^-1 rad/s
(c) 1 rad/s
(d) 10^-3 rad/s
|
Answer: (d) 10^-3 rad/s
|
Angular frequency of the pendulum
ω = √(g / l)
(Δω / ω) = (1 / 2)(Δgeff / geff)
(Δω / ω) = (1 / 2)(2Aω^2 / g)
Δω / ω = (Aω^2 / g)
= (1^2 * 10^-2 / 10) = 10^-3
|
Simple Harmonic Motion
|
Q6: A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1 / 1000 of the original amplitude is close to
|
(a) 100 s
(b) 10 s
(c) 20 s
(d) 50 s
|
Answer: (c) 20 s
|
Frequency of damped oscillation, f = 5 Hz
For A = A / 2
t1 = 2 s
Also, A = A0 e^-(b / 2m)t or 1 / 2 = e^-(b / 2m)2
(b / m) = ln2
For A = A / 1000, t2 = ?
1 / 1000 = e^-(b / 2m)t2
Or 10^-3 = e^-(b / 2m)t2
(b / 2m)t2 = 3ln10 or t2 = 6ln10 / ln2
t2 = 20 s
|
Simple Harmonic Motion
|
Q7: A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in the air. A tuning fork of frequency 512 Hz produces the first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces the first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to
|
(a) 335 m/s
(b) 322 m/s
(c) 328 m/s
(d) 341 m/s
|
Answer: (c) 328 m/s
|
Due to the jagged end
(λ1 / 4) = 11 cm, so v / (512 * 4) = 11 cm------(1)
(λ2 / 4) = 27 cm, so v / (256 * 4) = 27 cm------(2)
(2) - (1)
v / (512 * 4) = 0.16
v = 328 m/s
|
Simple Harmonic Motion
|
Q9: In an engine, the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston is close to
|
(a) 0.7 Hz
(b) 1.9 Hz
(c) 1.2 Hz
(d) 0.1 Hz
|
Answer: (b) 1.9 Hz
|
The amplitude of S.H.M., A = 7 cm = 0.07 m
As the washer does not stay in contact with the piston, at some particular frequency i.e. normal force on the washer = 0
Maximum acceleration of washer = Aω^2 = g
ω = √(g / A) = √(10 / 0.07) = √(1000 / 7)
Frequency of the piston, v = ω / 2π
= 1.9 Hz
|
Simple Harmonic Motion
|
Q10: A toy-car, blowing its horn, is moving with a steady speed of 5 m/s, away from a wall. An observer, towards whom the toy car is moving, is able to hear 5 beats per second. If the velocity of sound in air is 340 m/s, the frequency of the horn of the toy car is close to
|
(a) 680 Hz
(b) 510 Hz
(c) 340 Hz
(d) 170 Hz
|
Answer: (d) 170 Hz
|
vair = 340 m/s, v = 5 m/s
f1 - f2 = 5 beats per second
Apparently frequency heard by the observer on reflection from the wall,
f1 = [v / (v - vs)]f = [340 / (340 - 5)]f = (340 / 335)f
f2 = [v / (v + vs)]f = [340 / (340 + 5)]f = (340 / 345)f
Since f1 - f2 = 5
[(340 / 335) - (340 / 345)]f = 5 ⇒ f = (5 / 340)([335 * 345] / 10) = 169.96 Hz = 170 Hz
|
Simple Harmonic Motion
|
Q11: A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10^12 s^-1. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver 108 and Avogadro number 6.02 * 10^23 gm mole^-1 )
|
(a) 6.4 N m^-1
(b) 7.1 N m^-1
(c) 2.2 N m^-1
(d) 5.5 N m^-1
|
Answer: (b) 7.1 N m^-1
|
Frequency of a particle executing SHM,
f = 1 / (2π) √(k / m)
k = 4π^2 * f^2 * m
Here, f = 10^12 s^-1, m = [108 / (6.02 * 10^23)] * 10^-3 Kg
k = 4 * (3.14)^2 * (10^12)^2 * [108 * 10^-3 / 6.02 * 10^23] = 7.1 Nm^-1
|
Simple Harmonic Motion
|
Q12: The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s^-1. At, t = 0 the displacement is 5 m. What is the maximum acceleration? The initial phase is π / 4
|
(a) 500 √2 m/s^2
(b) 500 m/s^2
(c) 750 √2 m/s^2
(d) 750 m/s^2
|
Answer: (a) 500 √2 m/s^2
|
For simple harmonic motion,
Maximum acceleration/Maximum velocity = 10 ⇒ ω^2a / ωa = 10 or ω = 10
At t = 0; displacement, x = 5
x = asin(ωt + Φ)
5 = asin(0 + π / 4) or 5 = asin(π / 4) or a = 5√2 m
Maximum acceleration = ω^2a = 10^2 * 5√2 = 500√2 ms^-2
|
Simple Harmonic Motion
|
Q13: A child swinging on a swing in sitting position, stands up, then the time period of the swing will
|
(a) increase
(b) decrease
(c) remains the same
(d) increases if the child is long and decreases if the child is short
|
Answer: (b) decrease
|
The time period will decrease.
When the child stands up, the centre of gravity is shifted upwards and so the length of swing decreases. T = 2π√(l / g)
|
Simple Harmonic Motion
|
Q1. A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in second is
|
(a) 8π / 3
(b) 4π / 3
(c) 3π / 8
(d) 7π / 3
|
Answer: (a) 8π / 3
|
In SHM, speed v = ω √(A^2 - x^2), at x = 4, v = ω √((5)^2 - (4)^2) = 3ω
Acceleration a = -ω^2 x
at x = 4, a = -4ω^2
As |v| = |a|
ω = 3 / 4 ⇒ T = 2π / ω = 8π / 3
|
Simple Harmonic Motion
|
Q2: A cylindrical plastic bottle of negligible mass is filled with 310 mL of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency. If the radius of the bottle is 2.5 cm then ω is close to (density of water = 10^3 kg/m^3 )
|
(a) 2.50 rad/s
(b) 3.75 rad/s
(c) 5.00 rad/s
(d) 1.25 rad/s
|
None of the options given are correct
|
Restoring force due to pressing the bottle with amount x,
F = - (ρAx)g
a = - (ρAg/m)x
Therefore, ω^2 = ρAg/m = [ρ(πr^2)g] / m
ω = √(10^3 * π * (2.5 * 10^-2)^2 * 10 / 310 * 10^-3)
|
Simple Harmonic Motion
|
Q3: A particle undergoing simple harmonic motion has time-dependent displacement given by x(t) = Asin(πt / 90). The ratio of kinetic to the potential energy of this particle at t = 210 s will be
|
(a) 1 / 3
(b) 2
(c) 1
(d) 3
|
Answer: (a) 1 / 3
|
The maximum kinetic energy of the particle, K.E = (1 / 2)mA^2ω^2cos^2(ωt)
The potential energy of the particle at any time t, U = (1 / 2)mA^2ω^2sin^2(ωt)
The ratio of kinetic energy to potential energy, r = K.E/U = [(1 / 2)mA^2ω^2cos^2(ωt)] / [(1 / 2)mA^2ω^2sin^2(ωt)]
r = cos^2(ωt) / sin^2(ωt)
r = cot^2(ωt)
The angular frequency is given as ω = π / 90
Time, t = 210 s
Therefore, r = cot^2(π / 90) * 210
r = cot^2(7π / 3)
r = cot^2(2π + π / 3)
We know,
cot(2nπ + θ) = cotθ
Therefore, r = cot^2(π / 3)
r = (1 / √3)^2
r = 1 / 3
|
Simple Harmonic Motion
|
Q4: A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then
|
(a) K2 = K1
(b) K2 = K1 / 2
(c) K2 = 2K1
(d) K2 = K1 / 4
|
Answer: (c) K2 = 2K1
|
Maximum kinetic energy = (1 / 2)mω^2A^2
ω = √(g / L)
A = Lθ
K.E = (1 / 2)mgLθ^2
K1 = (1 / 2)mgLθ^2
If the length is doubled
K2 = (1 / 2)mg2Lθ^2
From (1) and (2)
(K1 / K2) = (1 / 2)mgLθ^2 / (1 / 2)mg2Lθ^2
= 1 / 2
K2 = 2K1
|
Simple Harmonic Motion
|
Q5: A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10^-2 m. The relative change in the angular frequency of the pendulum is best given by
|
(a) 10^-5 rad/s
(b) 10^-1 rad/s
(c) 1 rad/s
(d) 10^-3 rad/s
|
Answer: (d) 10^-3 rad/s
|
Angular frequency of the pendulum
ω = √(g / l)
(Δω / ω) = (1 / 2)(Δgeff / geff)
(Δω / ω) = (1 / 2)(2Aω^2 / g)
Δω / ω = (Aω^2 / g)
= (1^2 * 10^-2 / 10) = 10^-3
|
Simple Harmonic Motion
|
Q6: A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1 / 1000 of the original amplitude is close to
|
(a) 100 s
(b) 10 s
(c) 20 s
(d) 50 s
|
Answer: (c) 20 s
|
Frequency of damped oscillation, f = 5 Hz
For A = A / 2
t1 = 2 s
Also, A = A0 e^-(b / 2m)t or 1 / 2 = e^-(b / 2m)2
(b / m) = ln2
For A = A / 1000, t2 = ?
1 / 1000 = e^-(b / 2m)t2
Or 10^-3 = e^-(b / 2m)t2
(b / 2m)t2 = 3ln10 or t2 = 6ln10 / ln2
t2 = 20 s
|
Simple Harmonic Motion
|
Q7: A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in the air. A tuning fork of frequency 512 Hz produces the first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces the first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to
|
(a) 335 m/s
(b) 322 m/s
(c) 328 m/s
(d) 341 m/s
|
Answer: (c) 328 m/s
|
Due to the jagged end
(λ1 / 4) = 11 cm, so v / (512 * 4) = 11 cm------(1)
(λ2 / 4) = 27 cm, so v / (256 * 4) = 27 cm------(2)
(2) - (1)
v / (512 * 4) = 0.16
v = 328 m/s
|
Simple Harmonic Motion
|
Q9: In an engine, the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston is close to
|
(a) 0.7 Hz
(b) 1.9 Hz
(c) 1.2 Hz
(d) 0.1 Hz
|
Answer: (b) 1.9 Hz
|
The amplitude of S.H.M., A = 7 cm = 0.07 m
As the washer does not stay in contact with the piston, at some particular frequency i.e. normal force on the washer = 0
Maximum acceleration of washer = Aω^2 = g
ω = √(g / A) = √(10 / 0.07) = √(1000 / 7)
Frequency of the piston, v = ω / 2π
= 1.9 Hz
|
Simple Harmonic Motion
|
Q10: A toy-car, blowing its horn, is moving with a steady speed of 5 m/s, away from a wall. An observer, towards whom the toy car is moving, is able to hear 5 beats per second. If the velocity of sound in air is 340 m/s, the frequency of the horn of the toy car is close to
|
(a) 680 Hz
(b) 510 Hz
(c) 340 Hz
(d) 170 Hz
|
Answer: (d) 170 Hz
|
vair = 340 m/s, v = 5 m/s
f1 - f2 = 5 beats per second
Apparently frequency heard by the observer on reflection from the wall,
f1 = [v / (v - vs)]f = [340 / (340 - 5)]f = (340 / 335)f
f2 = [v / (v + vs)]f = [340 / (340 + 5)]f = (340 / 345)f
Since f1 - f2 = 5
[(340 / 335) - (340 / 345)]f = 5 ⇒ f = (5 / 340)([335 * 345] / 10) = 169.96 Hz = 170 Hz
|
Simple Harmonic Motion
|
Q11: A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10^12 s^-1. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver 108 and Avogadro number 6.02 * 10^23 gm mole^-1 )
|
(a) 6.4 N m^-1
(b) 7.1 N m^-1
(c) 2.2 N m^-1
(d) 5.5 N m^-1
|
Answer: (b) 7.1 N m^-1
|
Frequency of a particle executing SHM,
f = 1 / (2π) √(k / m)
k = 4π^2 * f^2 * m
Here, f = 10^12 s^-1, m = [108 / (6.02 * 10^23)] * 10^-3 Kg
k = 4 * (3.14)^2 * (10^12)^2 * [108 * 10^-3 / 6.02 * 10^23] = 7.1 Nm^-1
|
Simple Harmonic Motion
|
Q12: The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s^-1. At, t = 0 the displacement is 5 m. What is the maximum acceleration? The initial phase is π / 4
|
(a) 500 √2 m/s^2
(b) 500 m/s^2
(c) 750 √2 m/s^2
(d) 750 m/s^2
|
Answer: (a) 500 √2 m/s^2
|
For simple harmonic motion,
Maximum acceleration/Maximum velocity = 10 ⇒ ω^2a / ωa = 10 or ω = 10
At t = 0; displacement, x = 5
x = asin(ωt + Φ)
5 = asin(0 + π / 4) or 5 = asin(π / 4) or a = 5√2 m
Maximum acceleration = ω^2a = 10^2 * 5√2 = 500√2 ms^-2
|
Simple Harmonic Motion
|
Q13: A child swinging on a swing in sitting position, stands up, then the time period of the swing will
|
(a) increase
(b) decrease
(c) remains the same
(d) increases if the child is long and decreases if the child is short
|
Answer: (b) decrease
|
The time period will decrease.
When the child stands up, the centre of gravity is shifted upwards and so the length of swing decreases. T = 2π√(l / g)
|
Simple Harmonic Motion
|
Q1: A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to
|
666 Hz
753 Hz
500 Hz
333 Hz
|
666 Hz
|
Frequency of the sound produced by the open flute
f=2(v / 21)=(2 x 330) /(2 x 0.5)=660 Hz
Velocity of observer, v0=10 x(5 / 18)=(25 / 9) m/s
As the source is moving towards the observer, according to the Doppler effect.
Frequency detected by observer
f'={((v+v0) / v)} f={((25 / 9)+330) / 330} 660
=2((25 / 9)+330)
f'=665.55 ≈ 666 Hz
|
Sound Waves
|
Q2: A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the source when it is moving away from the observer after crossing him? (Take the velocity of sound in air as 350 m/s)
|
750 Hz
857 Hz
1143 Hz
807 Hz
|
750 Hz
|
When source is moving towards a stationary observer,
fapp= fsource {(V-0) / (V-50)}
1000=fsource (350 / 300)
When source is moving away from observer
f'=fsource {350 / (350+50)}
f'={(1000 x 300) / 350} x(350 / 400)
f'=750 Hz
|
Sound Waves
|
Q3: Two sources of sound S1 and S2 produce sound waves of the same frequency 660 Hz. A listener is moving from source S1 towards S2 with constant speed u m/s and he hears 10 beats. The velocity of sound is 330 m/s. Then u is equals
|
5.5 m/s
15.0 m/s
2.5 m/s
10.0 m/s
|
2.5 m/s
|
f1=f[(v-v0) / v]
f2=f[(v+v0) / v]
Frequency f2-f1=fx(2v0 / v)
10=660 x(2u / 330)
u=2.5 m/s
|
Sound Waves
|
Q4: A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in the air. A tuning fork of frequency 512 Hz produces the first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces the first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to
|
322 ms^-1
341 ms^-1
335 ms^-1
328 ms^-1
|
328 ms^-1
|
(λ / 4)=0.11+e
{V /(512) 4}=0.11+e
{V /(256) 4}=0.27+e
After solving (1) and (2) we get
V=328 ms^-1
|
Sound Waves
|
Q5: A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? (Given reference intensity of sound is 10^-12 W/m^2)
|
40 cm
20 cm
10 cm
30 cm
|
40 cm
|
Using β=10 log10(I / I0)
Or 120=10 log 10(I / 10^-12)
I / 10^-12=10^12
I=1
Also I=1=P / 4πr^2=2 / 4πr^2
r^2=2 / 4πI=2 / 4π
r^2=2 / 4(3.14)
On solving the above equations, we get
r=40 cm
|
Sound Waves
|
Q6: Two cars A and B are moving away from each other in opposite directions. Both the cars are moving at a speed of 20 m/s with respect to the ground. If an observer in car A detects a frequency of 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in car B ? (speed of sound in air = 340 m/s)
|
2250 Hz
200 Hz
2300 Hz
2150 Hz
|
2250 Hz
|
f'=f{(v-v0) / (v+vs)}
2000=f{(340-20) / (340+20)}
f=2250 Hz
|
Sound Waves
|
Q7: A person standing on an open ground hears the sound of a jet aeroplane, coming from the north at an angle 60 degrees with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, the speed of the plane is
|
(√3 / 2) v
2v / √3
V
v / 2
|
v / 2
|
Distance, PQ=vp xt (Distance = speed x time)
Distance, QR=V . t
Cos 60=PQ / QR
1 / 2=(vp x t) / (v.t)
vp=v / 2
|
Sound Waves
|
Q8: Three sound waves of equal amplitudes have frequencies(f-1, f, f+1). They superpose to give beats. The number of beats produced per second is
|
2
1
4
3
|
2
|
Beat produced between(f-1) and f is 1 .
Beat produced between f and f+1 is 1 .
Beat produced between(f-1) and (f+1) is 2
Therefore No of beats produced per second will be 2
|
Sound Waves
|
Q9: A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of the air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in the air is 340 m/s.
|
6
4
12
8
|
6
|
Fundamental frequency of the closed organ pipe is
f=v / 4 L
f=340 /(4 x 0.85)=100 Hz
The natural frequencies of the closed organ pipe is
fn=(2n-1) f=f, 3f, 5f, 7f, 9f, 11f, 13f ...
So the possible frequencies below 1250 Hz are
fn=100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz
Number of frequencies =6
|
Sound Waves
|
Q10: An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light =3 x 10^8 m/s)
|
17.3 GHz
15.3 GHz
10.1 GHz
12.1 GHz
|
17.3 GHz
|
Doppler effect in light (speed of observer is not very small compared to speed of light)
f'=f[(1+v / c) / (1-v / c)]^{1 / 2}
Here, frequency (v / c)=1 / 2
So, f'=f [(3 / 2) / (1 / 2)]^{1 / 2}
f'=10 x √3=17.3 GHz
|
Sound Waves
|
Q1: In Young's double-slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle 1/40 rad by using the light of wavelength λ1. When the light of wavelength λ2 is used a bright fringe is seen at the same angle in the same setup. Given that 1 and 2 are in the visible range (380 nm to 740 nm), their values are
|
(a) 400 nm, 500 nm
(b) 625 nm, 500 nm
(c) 380 nm, 525 nm
(d) 380 nm, 500 nm
|
(b) 625 nm, 500 nm
|
Path difference = dsinθ = d × θ = (0.1 mm)(1 / 40) = 2.5 × 10^{-3} mm = 2500 nm
For bright fringes, path difference = nλ
So, 2500 = nλ1 = mλ2
n = 4, m = 5
or λ1 = 2500 / 4 = 625 nm
λ2 = 2500 / 5 = 500 nm
|
Wave Optics
|
Q2: In Young's double-slit experiment, the path difference, at a certain point on the screen, between two interfering waves is (1/8)th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to
|
(a) 0.80
(b) 0.94
(c) 0.85
(d) 0.74
|
(c) 0.85
|
The phase difference between two waves is given as
(Δx) x (2π/λ) = (λ/8) x (2π/λ) = π/4
So, the intensity at this point is
I = I0 cos²(Φ/2)
I = I0 cos²(π/8)
I = I0[(1 + cos(π/4)) / 2] = I0[(1 + (1/√2)) / 2] = 0.85 I0
|
Wave Optics
|
Q3: In a double-slit experiment, green light (5303 Å) falls on a double slit having a separation of 19.44 m and a width of 4.05 m. The number of bright fringes between the first and the second diffraction minima is
|
(a) 10
(b) 04
(c) 05
(d) 09
|
(c) 05
|
λg = 5303 Å, d = 19.44 m, a = 4.05 m
For diffraction location of first minima and second minima
y1 = Dλ/a, y2 = 2Dλ/a
y2 - y1 = (2Dλ/a) - (Dλ/a) = Dλ/a
β = Dλ/d
Number of bright fringes = (y2 - y1) / β
= (Dλ/a) x (d/Dλ)
= d/a
= 19.44 / 4.05 = 5
|
Wave Optics
|
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