Physics Test

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Physics Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

There is a stream of neutrons with a kinetic energy of 0.0327 eV. If the half life of neutron is 700 second, what fraction of neutrons will decay before they travel a distance of 10 m? Given mass of neutron = 1.675 × 10^–27 kg.

A
3.96 × 10^–6

B
3.90 × 10^–6

C
3.85 × 10^–6

D
4.86 × 10^–6

Solution

From the given kinetic energy of the neutrons we first calculate their velocity. Thus
$$
\frac{1}{2} \mathrm{mv}^{2}=0.0327 \times 1.6 \times 10^{-19}
$$
$\therefore$
$$
\begin{array}{l}
v^{2}=\frac{2 \times 0.0327 \times 1.6 \times 10^{-19}}{1.675 \times 10^{-27}}=625 \times 10^{4} \\
v=2500 \mathrm{m} / \mathrm{s}
\end{array}
$$
or
With this speed, the time taken by the neutrons to travel a distance of $10 \mathrm{m}$ is
$$
\Delta t=\frac{10}{2500}=4 \times 10^{-3} s
$$
The fraction of neutrons decayed in time $\Delta t$ second is,
$$
\frac{\Delta N}{N}=\lambda \Delta t
$$
$$
\begin{array}{ll}
\text { Also, } & \lambda=\frac{0.693}{T_{1 / 2}} \\
\therefore \quad \frac{\Delta N}{N}=\frac{0.693}{T_{1 / 2}} \Delta t=\frac{0.693}{700} \times\left(4 \times 10^{-3}\right)=3.96 \times 10^{-6}
\end{array}
$$
Hence, the correct option is (A)
Question 2
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A straight segment OC of length L metre of a circuit carrying a current I amp is placed along the x-axis. Two infinitely long straight wires A and B, each extending from z = -∞, are to + ∞ are fixed at y = -a metre and each carry a current I amp. Into plane of the paper. Obtain the expression for the force acting on the segment OC.

A
$\overrightarrow{\mathrm{F}}=\frac{\mu_{0} 1^{2}}{2 \pi} \log _{\mathrm{e}} \frac{\left(\mathrm{L}^{2}+\mathrm{a}^{2}\right)}{\mathrm{a}^{2}}(-\hat{\mathrm{k}})$

B
$\overrightarrow{\mathrm{F}}=\frac{\mu_{0} I}{2 \pi} \log _{\mathrm{e}} \frac{\left(\mathrm{L}^{2}+\mathrm{a}^{2}\right)}{\left(\mathrm{a}^{2}\right)} \quad(-\widehat{\mathrm{k}})$

C
$\overrightarrow{\mathrm{F}}=\frac{\mu_{0} \mathrm{I}}{\pi} \log _{\mathrm{e}} \frac{\left(\mathrm{L}^{2}+\mathrm{a}^{2}\right)}{\left(\mathrm{a}^{2}\right)}(-\widehat{\mathrm{k}})$

D
$\overrightarrow{\mathrm{F}}=\frac{\mu_{0} \mathrm{a}^{2}}{\pi} \log _{\mathrm{e}} \frac{\left(\mathrm{L}^{2}+\mathrm{a}^{2}\right)}{\mathrm{a}^{2}}(-\widehat{\mathrm{k}})$

Solution

$\mathrm{B}=\mathrm{B}_{\mathrm{A}} \cos \theta+\mathrm{B}_{\mathrm{B}} \cos \theta=2 \mathrm{B}_{\mathrm{A}} \cos \theta$
$\mathrm{B}=2 \frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{I}}{\sqrt{\mathrm{a}^{2}+\mathrm{x}^{2}}} \times \frac{\mathrm{x}}{\sqrt{\mathrm{a}^{2}+\mathrm{x}^{2}}}$
$\mathbf{F}=\mathbf{I}(\mathbf{d x}) \mathbf{B}$
$\mathrm{F}=\frac{\mu_{0} \mathrm{I}^{2}}{4 \pi} \cdot 2 \int_{0}^{\mathrm{L}} \frac{2 \mathrm{xdx}}{\left(\mathrm{a}^{2}+\mathrm{x}^{2}\right)}$
$=\frac{\mu_{0} I^{2}}{4 \pi} 2\left[\log _{e}\left(a^{2}+x^{2}\right)\right]_{0}^{L}$
$=\frac{\mu_{0} I^{2}}{4 \pi} 2\left[\log _{e} \frac{a^{2}+L^{2}}{a^{2}}\right]$
$\overrightarrow{\mathrm{F}}=\frac{\mu_{0} \mathrm{I}^{2}}{2 \pi} \log _{\mathrm{e}} \frac{\left(\mathrm{L}^{2}+\mathrm{a}^{2}\right)}{\mathrm{a}^{2}}(-\widehat{\mathrm{k}})$
when direction is conductor B is reversed the B will be along x-axis so force on BC segment becomes zero.
Hence, the correct option is (A)
Question 3
1 / -0

Ultraviolet beam of wavelength 280 nm is incident on lithium surface of work function 2.5 eV. The maximum velocity of electron emitted from metal surface is.......(Mass of electron is 9.1 x 10^-31 kg)

A
8.2 x 10⁵ m/s

B
10⁶ m/s

C
7 x 10⁵ m/s

D
None of these

Solution

$\frac{\mathrm{hc}}{\lambda}=\phi+\frac{1}{2} \mathrm{mv}_{\mathrm{max}}^{2}$
$\therefore \quad \mathrm{v}_{\mathrm{max}}=\sqrt{\frac{2\left(\frac{\mathrm{bc}}{\lambda}-\phi\right)}{\mathrm{m}}}$
$=\sqrt{\frac{2\left(\frac{1250 \mathrm{v}}{200}-2.5 \mathrm{eV}\right)}{9.1 \times 10^{-31}}}$
$=\sqrt{\left(\frac{2 \times 1.9 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}\right)}$
$=8.2 \times 10^{5} \mathrm{m} / \mathrm{s}$
Hence, the correct option is (A)
Question 4
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Two microwave coherent point sources emitting waves of wavelengths λ are placed at 5λ distance apart. The interference is being observed on a flat non-reflecting surface along a line passing through one source, in a direction perpendicular to the line joining the two sources (refer figure) Considering λ as 4mm, calculate the positions of maxima. Take k initial phase difference between the two sources to be zero.

A
48, 21, 32/3,9/2,0 mm

B
50, 22, 32/2,92,1 mm

C
46, 23, 22/3,92,2 mm

D
47, 20, 33/3,92,1mm

Solution

Path difference at $P=S_{2} P-S_{1} P$
$=\sqrt{(5 \lambda)^{2}+x^{2}}-x$
For $P$ to be maximum
Path difference $=n \lambda$
From (1)$\&(2)$ and
$\sqrt{(5 \lambda)^{2}+\mathrm{x}^{2}}-\mathrm{x}=\mathrm{n} \lambda$
$\sqrt{(5 \lambda)^{2}+\mathbf{x}^{2}}=(\mathrm{n} \lambda+\mathrm{x})$
$25 \lambda^{2}+x^{2}=n^{2} \lambda^{2}+x^{2}+2 n \lambda x$
$\left(25-n^{2}\right) \lambda^{2}=2 n \lambda x$
$x=\frac{\left(25-n^{2}\right) \lambda}{2 n}$
$x=\frac{\left(25-n^{2}\right) \times 4}{2 n} m m$
For $n=1, \quad x_{1}=48 m m$
$n=2, \quad x_{2}=21 m m$
$n=3, \quad x_{3}=\frac{32}{3} m m$
$n=4, \quad x=\frac{9}{2} m m$
$n=5, \quad x=0$
Hence, the correct option is (A)
Question 5
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An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by 100 Hz than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is

A
200

B
300

C
240

D
480

Solution

Length of the organ pipe is same in both the cases. Fundamental frequency of open pipe is $f_{1}=\frac{v}{2 l}$ Frequency of third harmonic of closed pipe will be
$$
f_{2}=3\left(\frac{v}{4 l}\right)
$$
Given that, $f_{2}=f_{1}+100$
or
$$
f_{2}-f_{1}=100
$$
or $\frac{3}{4}\left(\frac{v}{l}\right)-\left(\frac{1}{2}\right)\left(\frac{v}{l}\right)=100 \Rightarrow \frac{v}{4 l}=100$
$\therefore \quad \frac{v}{2 l}$ or $f_{1}=200 \mathrm{Hz}$
Therefore, fundamental frequency of the open pipe is $200 \mathrm{Hz}$.
Hence, the correct option is (A)
Question 6
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Calculate the magnetic field induction at the centre of a hydrogen atom caused by an electron moving along the first Bohr orbit.

A
125

B
100

C
150

D
200

Solution

The magnetic field at a centre of H-atom due to moving electron is
$$
\begin{aligned}
\mathrm{B} &=\frac{\mu_{0}}{4 \pi}\left|\frac{-e^{\rightarrow} \overrightarrow{\mathbf{v}}_{n} \times \overrightarrow{\mathrm{r}}_{n}}{\mathrm{r}_{n}^{3}}\right|=\frac{\mu_{0} e v_{n} \sin 90^{\circ}}{4 \pi r_{n}^{2}} \\
\therefore \quad & \mathrm{B}=\frac{\mu_{0} e v_{n}}{4 \pi r_{n}^{2}} & \ldots(\mathrm{i})
\end{aligned}
$$

Radius to H-atom, $r_{n}=\frac{4 \pi \varepsilon_{0} \hbar^{2} \quad n^{2}}{e^{2} m} \quad($ for $\mathrm{H}$ -atom $)$
and $\quad v_{n}=\frac{e^{2}}{4 \pi \varepsilon_{0} n \hbar}$
On putting the values in Eqn. (i), we get
$\mathrm{B}=\frac{\mu_{0} e}{4 \pi}\left(\frac{e^{2}}{4 \pi \varepsilon_{0} \hbar}\right) \frac{e^{4} m^{2}}{\left(4 \pi \varepsilon_{0} \hbar^{2} n^{2}\right)^{2}}$
$\mathrm{B}=\frac{m^{2} e^{7}}{c \hbar^{5}}=125 \mathrm{KG}$
Hence, the correct option is (A)
Question 7
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Two charges q₁ and q₂ are placed 30 cm apart, as shown in the figure. A third charge q₃ is moved along the arc of a circle of radius 40cm from C to D. The change in the potential energy of the system is q₃/4πε₀k , where k is

A
8q₁

B
6q₁

C
8q₂

D
6q₂

Solution

The potential energy when $\mathrm{q}_{3}$ is at point $\mathrm{C}$.
$\mathrm{U}_{1}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{0.40}+\frac{\mathrm{q}_{2} \mathrm{q}_{3}}{\sqrt{(0.40)^{2}+(0.30)^{2}}}\right]$
The potential energy when $\mathrm{q}_{3}$ is at point $\mathrm{D}$
$\mathrm{U}_{2}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{0.40}+\frac{\mathrm{q}_{2} \mathrm{q}_{3}}{0.10}\right]$
Thus change in potential energy is

$\Delta \mathrm{U}=\mathrm{U}_{2}-\mathrm{U}_{1}$
$\Rightarrow \quad \frac{\mathrm{q}_{3}}{4 \pi \varepsilon_{0}} \mathrm{k}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{0.40}+\frac{\mathrm{q}_{2} \mathrm{q}_{3}}{0.10}-\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{0.40}-\frac{\mathrm{q}_{2} \mathrm{G}_{3}}{0.50}\right]$
$\Rightarrow \mathrm{k}=\frac{5 \mathrm{q}_{2}-\mathrm{q}_{2}}{0.50}=\frac{4 \mathrm{q}_{2}}{0.50}=8 \mathrm{q}_{2}$
Hence, the correct option is (C)
Question 8
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A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equation is x² = 4ay. The wire frame is fixed and the bead can slide on it without friction. The bead is released from the point y = 4a on the wire frame from rest. The tangential acceleration of the bead when it reaches the position given by y = a is :

A
$\frac{g}{2}$

B
√3g/2

C
$\frac{g}{\sqrt{2}}$

D
$\frac{g}{\sqrt{5}}$

Solution

$x^{2}=4 a y$

Differentiating w.r.t. y, we get

$\frac{d y}{d x}=\frac{x}{2 a}$

$\therefore \quad \mathrm{At}(2 \mathrm{a}, \mathrm{a}), \frac{\mathrm{dy}}{\mathrm{dx}}=1 \Rightarrow$ hence $\theta=45^{\circ}$

the component of weight along tangential direction is $\mathrm{mg}$ sin $\theta$.

hence tangential acceleration is $\mathrm{g} \sin \theta=\frac{\mathrm{g}}{\sqrt{2}}$

Hence, the correct option is (C)
Question 9
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A gas under constant pressure of 4.5 x 10⁵ Pa when subjected to 800 kJ of heat, changes the volume from 0.5 m³ to 2.0 m³. The change in internal energy of the gas is

A
6.75 x 10⁵ J

B
5.25 x 10⁵ J

C
3.25 x 10⁵ J

D
1.25 x 10⁵ J

Solution

$P=4.5 \times 10^{5} \mathrm{Pa} ; \Delta \mathrm{Q}=800 \mathrm{kJ}$
$V_{1}=0.5 \mathrm{m}^{3} ; \mathrm{V}_{2}=2 \mathrm{m}^{3}$
$\Delta \mathrm{W}=\mathrm{P}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)=4.5 \times 10^{5}(2-0.5)$
$=6.75 \times 10^{5} \mathrm{J}$
Change in internal energy
$\Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}=800 \times 10^{3}-6.75 \times 10^{5}$
$=1.25 \times 10^{5} \mathrm{J}$
Hence, the correct option is (D)
Question 10
1 / -0

A particle executes simple harmonic motion with a frequency f. The frequency with which the potential energy oscillates is

A
f

B
f/2

C
2f

D
zero

Solution

If $x=A \sin \omega t$
Then $\quad \mathrm{P.E} .=\frac{1}{2} m A^{2} \omega^{2} \sin ^{2} \omega t$
$\therefore \quad \mathrm{P.E.}=\frac{1}{2} m A^{2} \omega^{2}\left(\frac{1-\cos 2 \omega t}{2}\right)$
$\therefore \quad \omega^{\prime}=2 \omega$
Or $2 \pi f^{\prime}=2 \times 2 \pi f$
$\therefore \quad f^{\prime}=2 f$
Hence, the correct option is (C)