Physics Test

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Physics Test - 26

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Question 1
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Compute the bulk modulus of water if its volume changes from 100 litres to 99.5 litre under a pressure of 100 atmosphere.

A
2 x 10²

B
2 x 5²

C
2 x 10⁴

D
2 x 20³

Solution

By definition of bulk modulus,
\begin{aligned}
\mathrm{B}_{\mathrm{W}} &=-\mathrm{V} \frac{\Delta \mathrm{p}}{\Delta \mathrm{v}}=-100 \times \frac{\left(100 \times 1.013 \times 10^{5}\right)}{(99.5-100)} \\
&=2.026 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}
\end{aligned}
Now as isothermal elasticity of a gas is equal to its pressure,
\mathrm{B}_{\mathrm{A}}=\mathrm{E}_{\theta}=\mathrm{p}_{0}=1.013 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}
So that $\frac{\mathrm{Bw}}{\mathrm{B}_{\mathrm{A}}}=\frac{\mathrm{C}_{\mathrm{A}}}{\mathrm{C}_{\mathrm{W}}}=\frac{2.026 \times 10^{9}}{1.013 \times 10^{5}}=2 \times 10^{4} \quad\left[\right.$ as $\left.\mathrm{C}=\frac{1}{\mathrm{B}}\right]$
i.e., bulk modulus of water is very large as compared to air. This means that air is about 20,000 times more compressive than water, i.e., the average distance between air molecules is much larger than between water molecules.
Hence, the correct option is (C)
Question 2
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A stone is dropped from a height of 45 m on a horizontal level ground. There is horizontal wind blowing due to which horizontal acceleration of the stone becomes 10 m/s2.(Take g = 10 m/s2).
The time taken (t) by stone to reach the ground and the net horizontal displacement (x) of the stone from the time it is dropped and till it reaches the ground are respectively

A
t = 3 s, x = 45 m

B
t = 4 s, x = 54 m

C
t = 3 s, x = 54 m

D
t = 4 s, x = 45 m

Solution

Taking motion in vertical direction,
$\mathbf{u}=\mathbf{0}, \mathbf{g}=10 \mathbf{m} / \mathbf{s}^{2}, \mathbf{h}=45 \mathbf{m}$
$h=u t+\frac{1}{2} g t^{2}$
$\Rightarrow \mathrm{h}=0+\frac{1}{2} \mathrm{gt}^{2}$
$\Rightarrow t=\sqrt{\frac{2 h}{\mathrm{g}}}=\sqrt{\frac{2 \times 45}{10}}$
$\Rightarrow t=3 \mathrm{sec}$
(b) Taking motion in horizontal direction,
$u=0, a=10^{m} / s^{2}, \quad t=3 s c e$
$x=u t+\frac{1}{2} a t^{2}$
$\Rightarrow x=0+\frac{1}{2}(10)(3)^{2} \Rightarrow x=45 \mathrm{m}$
Hence, the correct option is (A)
Question 3
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The mean lives of a radioactive substance are 1620 years and 405 years for α - emission and β - emission respectively. Find the time during which three-fourth of a sample will decay if it is decaying both by α – emission and β – emission simultaneously.

A
449 years

B
450 years

C
365 years

D
439 years

Solution

The decay constant $\lambda$ is the reciprocal of the mean life $\tau$ Thus, Vlambda_Nalpha $y=\operatorname{lfrac}\{1\}(1620\}$ ।text $\{$ per year $\}$ and

Vlambda_nbeta $\}=\operatorname{lfrac}\{1\}(405)$ ltext $\{$ per year $\}$
$\therefore$
Total decay constant, $\lambda=\lambda_{a}+\lambda_{\beta}$
or
year $\}$ We know that $N = N _{-}(0) e ^{\wedge}\{$ Vlambda $t\}$
When $\frac{3}{4}$ th part of the sample has disintegrated, $N = N _{0} / 4$ $\therefore$
or
$e^{\wedge}($ Vlambda $t\}=4$ Taking logarithm of both sides, we get
or
$\lambda t=\log _{e} 4$
$t=\frac{1}{\lambda} \log _{e} 2^{2}=\frac{2}{\lambda} \log _{e} 2$
Hence, the correct option is (A)
Question 4
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During adiabatic change, specific heat is

A
Zero

B
Greater than zero but finite

C
Less than zero

D
Infinity

Solution

In adiabatic process, ΔH = 0
C =ΔH/ nΔT =0
Hence, zero is correct
Concepts :
Main Concept :
Adiabatic processWhen a thermodynamic system undergoes a change in such a way that no exchange of heat takes place between System and surroundings, the process is known as adiabatic process.
In this process P, V and T changes but $Δ Q = 0$ .
(1) Essential conditions for adiabatic process
(i) There should not be any exchange of heat between the system and its surroundings. All walls of the container and the piston must be perfectly insulating.
(ii) The system should be compressed or allowed to expand suddenly so that there is no time for the exchange of heat between the system and its surroundings. Since, these two conditions are not fully realised in practice, so no process is perfectly adiabatic.
(2) Some examples of adiabatic process
(i) Sudden compression or expansion of a gas in a container with perfectly nonconducting walls.
(ii) Sudden bursting of the tube of bicycle tyre.
(iii) Propagation of sound waves in air and other gases.
(iv) Expansion of steam in the cylinder of steam engine.
Hence, the correct option is (A)
Question 5
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A cyclist rides along the circumference of a circular horizontal plane of radius R, the friction coefficient being dependent only on distance r from the centre O of the plane as k = k₀(1 - r/R), where k₀ is a constant. What is the maximum safe velocity?

A
$\frac{1}{2} \sqrt{k_{0} \mathrm{Rg}}$

B
$-\frac{1}{2} \sqrt{k_{0} \mathrm{Rg}}$

C
$\frac{1}{3} \sqrt{k_{0} \mathrm{Rg}}$

D
$-\frac{1}{3} \sqrt{k_{0} R g}$

Solution

In this case, normal reaction on cyclist system is balanced by weight. But force of friction provides centripetal acceleration.
$\therefore \quad N = mg$
$\frac{m+2^{2}}{r}=k N$
or $\frac{m v^{2}}{r}=k m g$
given $k=k_{0}\left(1-\frac{r}{R}\right)$
or $\frac{v^{2}}{r}=k_{0}\left(1-\frac{r}{R}\right) g$
For $v_{\max }$
$v^{2}$ is also max. $\therefore \quad \frac{d\left( v ^{2}\right)}{ dr }=0 \Rightarrow r =\frac{ R }{2}$
$v^{2}=k_{0}\left(r-\frac{r^{2}}{R}\right) g$
$v_{\max }^{2}=k_{0}\left(\frac{R}{2}-\frac{(-R / 2)^{2}}{R}\right) g$
$v_{\max }=\frac{1}{2} \sqrt{k_{0} R g}$
Hence, the correct option is (A)
Question 6
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A disc of mass M and radius R is rolling with angular speed ω on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin O is

A
$\left(\frac{1}{2}\right) M R^{2} \omega$

B
MR²ω

C
$\left(\frac{3}{2}\right) M R^{2} \omega$

D
$2 \mathrm{MR}^{2} \omega$

Solution

From the theorem

$v=R w \quad$ (pure rolling)

Angular momentum about origin $\mathrm{O}, \quad L_{O}=M v R+I_{O} w$

$L_{O}=M(R w) R+\frac{1}{2} M R^{2} w$

$L_{O}=\frac{3}{2} M R^{2} w$
Hence, the correct option is (C)
Question 7
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A point object is moving with a speed v in front of an arrangement of two mirrors as shown in figure. If the velocity of image in mirror M₁ with respect to image in mirror M₂ is n V $\sin θ$ n =

A
2

B
2.5

C
3

D
3.5

Solution

Angle betweenv⁡ ₁ & v⁡ ₂ is 2 θ

Their magnitudes is v.

$\therefore v_{I_{1}} v_{I_{2}}=\left|\vec{v}_{I_{1}}-\vec{v}_{I_{2}}\right|=\sqrt{v^{2}+v^{2}-2 v v \cos 2 \theta}$

$=2 v \sin \theta$
Hence, the correct option is (A)
Question 8
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A conducting ring of mass 2 kg and radius 0.5 m is placed on a smooth horizontal plane. The ring carries a current i = 4A. A horizontal magnetic field B = 10 T is switched on at time t = 0 as shown in figure. The initial angular acceleration of the ring will be

A

$40 \pi r a d / 5^{2}$

B

$20 \pi rad / s ^{2}$

C

$5 \pi \operatorname{rad} / 5^{2}$

D

$15 \pi \operatorname{rad} / 5^{2}$

Solution

Concepts :
Main Concept :
Magnetic dipole and dipole moment

Magnetic Dipole Moment
From the expression for the torque on a current loop, the characteristics of the current loop are summarized in its magnetic moment

If there are N loops , thenμ=NIA

If there are $N$ loops, then $\mu = NIA$
The magnetic moment can be considered to be a vector quantity with direction perpendicular to the current loop in the right-handrule direction.
Torque on a current loop in magnetic field
i. Torque: Consider a rectangular current carrying coil PQRs having N tums and area A, placed in a uniform field $\vec{B}$, in such a way that the normal to the coil makes an angle $\theta$ with the direction of $\vec{B}$ the coil experiences a torque given by $\tau=N B i A \sin \theta .$ Vectorially $\vec{\tau}=\vec{M} \times \vec{B}$
a) $\tau$ is zero when $\theta=0$. i.e. when the plane of the coil is perpendicular to the field.
b) $\tau$ is maximum when $\theta=90^{\circ}$, i.e. the plane of the coil is parallel to the field $\tau_{\max }=N B i A$

Force on a current carrying loop in a uniform magnetic field
When an arbitrary current carrying loop placed in a magnetic field 1 to the plane of loop), each element of loop experiences a magnetic force due to which loop stretches and open into circular loop and tension developed in it's each part.

Therefore net force on a current carrying loop in a uniform magnetic field is zero.

Hence, the correct option is (A)
Question 9
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As per the diagram a point charge +q is placed at the origin O. Work done in taking another point charge -Q from the point A [coordinates (0, a)] to another point B [coordinates (a, 0)] along the straight path AB is

A
Zero

B
1

C
2

D
3

Solution

Potential energy of two charge system is $U=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0} r_{12}}$
When charge $-Q$ at $A,$ potential energy $U_{A}=\frac{(-Q) q}{4 \pi \epsilon_{0}(O A)}=\frac{(-Q) q}{4 \pi \epsilon_{0} a}$
When charge $-Q$ at $\mathrm{B},$ potential energy $U_{B}=\frac{(-Q) q}{4 \pi \epsilon_{0}(O B)}=\frac{(-Q) q}{4 \pi \epsilon_{0} a}$
Work done $, W_{A B}=-Q\left(V_{B}-V_{A}\right)=U_{B}-U_{A}=\frac{(-Q) q}{4 \pi \epsilon_{0} a}-\frac{(-Q) q}{4 \pi \epsilon_{0} a}=0$
Hence, the correct option is (A)
Question 10
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When a train is at a distance of 2km, its engine sounds a whistle. A man near the railway track hears the whistle directly and by placing his ear against the track of the train. If the two sounds are heard at an interval of 5.2 s, find the speed of the sound in iron (material of the rail track). given that velocity of sound in air is 330 m s ^-1.

A
2,325.6 m s^-1

B
1,932.4 m s^-1

C
2,132.5 m s^-1

D
3,213.2 m s^-1

Solution

Here, S = 2 km = 2,000 m ; v_air = 330 m s^-1
Therefore, time taken by sound to travel through air,
$t_{\text {air }}=\frac{S}{v_{\text {air }}}=\frac{2,000}{330}=6.06 \mathrm{s}$
As the two sounds (through air and iron rails) are heard at an interval of 5.2 s, time taken by sound to travel through iron rails,
t_iron = t_air - 5.2
= 6.06 - 5.2 = 0.86 s
Therefore, velocity of sound in iron,
$\begin{aligned} V_{\text {iron }} &=\frac{S}{t_{\text {iron }}}=\frac{2,000}{0.86} \\ &=2,325.6 \mathrm{m} \mathrm{s}^{-1} \end{aligned}$
= 2,325.6 m s^-1
Concepts : Main Concept : Examples on Speed of Sound
The speed of sound is the distance travelled per unit time by a soundwave propagating through an elastic medium. The SI unit of the speed of sound is the metre per second (m/s). In dry air at $20^{ο} C$ , the speed of sound is 343.2 metres per second (1,126 ft/s).
The speed of sound is the distance travelled per unit time by a soundwave propagating through an elastic medium. The SI unit of the speed of sound is the metre per second (m/s). In dry air at $20^{ο} C$ , the speed of sound is 343.2 metres per second (1,126 ft/s).
Hence, the correct option is (A)