Physics Test

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Physics Test - 12

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Weekly Quiz Competition

Question 1
4 / -1

A coil having 100 turns and area of 0.001 m2is free to rotate about an axis. The coil is placed perpendicular to magnetic field of 1.0 wb⁄m2. The resistance of the coil is10 $Ω$ .If the coil is rotated rapidly through an angle of 180^o, how much charge will flow through the coil?

A
0.02 coulomb

B
0.04 coulomb

C
0.08 coulomb

D
0.07 coulomb

Solution

Magnitude of the change in flux
$\Delta \phi=\left|\mathrm{nBA} \cos 180^{\circ}-\mathrm{nBA} \cos 0^{\circ}\right|=2 \mathrm{nBA}$
Amount of the charge induced in the coil $q=\frac{\Delta \phi}{R}=\frac{2 n B A}{R}$
$q=\frac{2 \times 100 \times 1 \times 0.001}{10}=0.02 C$
Question 2
4 / -1

Two electric bulbs rated P1watt V volt, and P2watt V volt are connected in parallel, and applied across V volt. The total power will be

A
(P1+ P2) watt

B
$\sqrt{P_{1} P_{2}}$ watt

C
$\frac{P_{1}+P_{2}}{P_{1}+P_{2}}$ watt

D
$\frac{P_{1}+P_{2}}{P_{1} P_{2}} w$ att

Solution

Let $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ are connected in parallel. Power across $\mathrm{R}_{1}$ is $P_{1}=\frac{V^{2}}{R_{1}}$
Power across $\mathrm{R}_{2}$ is $P_{2}=\frac{V^{2}}{R_{2}}$
Total power across the system
$P=\frac{V^{2}}{R} \Rightarrow P=V^{2}\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right) \Rightarrow P=P_{1}+P_{2}$
Where
$\frac{1}{R}=\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)$
Question 3
4 / -1

Black body emits the radiation of maximum intensity 5000 A0, when its temperature is 12270C. If its temperature is increased by 10000C, what is the maximum intensity of emitted radiation?

A
2754.8 A0

B
3000 A0

C
3500 A0

D
4000 A0

Solution

According to Wien's law
\[
\lambda_{m} T=\text { constant }(s a y b)
\]
where $\lambda_{\mathrm{m}}$ is wavelength corresponding to maximum intensity of radiation and $\mathrm{T}$ is temperature of the body in Kelvin.
$\therefore \frac{\lambda_{m}^{\prime}}{\lambda_{m}}=\frac{T}{T^{\prime}}$
Given, $T=1227+273=1500 \mathrm{K}$
$T^{\prime}=1227+1000+273=2500 K$
$\lambda_{\mathrm{m}}=5000 $
Hence, $\lambda_{\mathrm{m}} \cdot=\frac{1500}{2500} \times 5000=3000 $
Question 4
4 / -1

The thermo emf produced in a thermocouple is 3μV/oC. If the temperature of the cold junction is 20oC and the thermo emf is 0.3 mV, the temperature of the hot junction is

A
80^oC

B
100^oC

C
120^oC

D
140^oC

Solution

EMF produced = EMF produced per unit
rise in temperature $\times$ Rise in temperature
$0.3 \times 10^{-3}=3 \times 10^{-6}\left(\mathrm{T}_{\mathrm{H}}-20\right)$
$\mathrm{T}_{\mathrm{H}}-20=\frac{0.3 \times 10^{-3}}{3 \times 10^{-6}} \Rightarrow \mathrm{T}_{\mathrm{H}}=120^{\circ} \mathrm{C}$
Question 5
4 / -1

Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity $\vec{u}$ and the other from rest with uniform acceleration $f$. Let $\alpha$ be the angle between their directions of motion. The relative velocity of the

second particle w.r.t. first is least after a time of

A
$\frac{u \sin \alpha}{f}$

B
$\frac{f \cos \alpha}{u}$

C
u sin α

D
$\frac{u \cos \alpha}{f}$

Solution

After $t,$ velocity $=f x t$ $V_{B A}=\vec{f} t+\overrightarrow{(-u)}$
$\mathrm{V}_{\mathrm{BA}}=\sqrt{\mathrm{f}^{2} \mathrm{t}^{2}+\mathrm{u}^{2}-2 \mathrm{fut} \cos \alpha}$
For max. and min. $\frac{d}{d t}\left(V_{B A}^{2}\right)=2 f^{2} t-2 f u \cos \alpha=0$
$t=\frac{u \cos }{f}$
Question 6
4 / -1

Which of the following laws states that good absorbers of heat are good emitters?

A
Stefan's law

B
Kirchhoff's law

C
Planck's law

D
Wien's law

Solution

Kirchhoff's law states that good absorbers of heat are good emitters.
Question 7
4 / -1

A nucleus decays by $\beta$ + emission, followed by $\gamma$ -emission. If the atomic and mass numbers of the parent nucleus are $Z$ and $A$, respectively, then the corresponding numbers for the daughter nucleus are

A
Z - 1 and A - 1

B
Z + 1 and A

C
Z - 1 and A

D
Z + 1 and A - 1

Solution

For the positron emission, the mass number decreases by 1 and atomic number remains same. Hence, the corresponding numbers for the daughter nucleus are Z - 1 and A.
Question 8
4 / -1

An engine pumps up 100 kg of water through a height of 10 m in 5s. Given that the efficiency of engine is 60%. If g = 10 ms^-2, the power of the engine is :

A
3.3 kW

B
0.33 kW

C
0.033 kW

D
33 kW

Solution

Power is the rate at which a force does work.
Efficiency of engine, $n=60 \%$ Thus, power $=\frac{\text { work/time }}{\eta}=\frac{100}{60} \times \frac{m g h}{t}$
Given, $m=100 \mathrm{kg}, h=10 \mathrm{m}, t=5 \mathrm{s} \quad$ and
$g=10 \mathrm{ms}^{-2}$
\[
\begin{aligned}
\text { Hence, power } &=\frac{100}{60} \times \frac{100 \times 10 \times 10}{5} \\
&=3.3 \times 10^{3} \mathrm{W} \\
&=3.3 \mathrm{kW}
\end{aligned}
\]
Question 9
4 / -1

Two bodies of masses m and 4m are moving with equal momentum. The ratio of their kinetic energy is

A
1 : 4

B
4 : 1

C
1 : 1

D
1 : 2

Solution

$\frac{K E_{1}}{K E_{2}}=\frac{\frac{P_{1}^{2}}{2 m_{1}}}{\frac{P_{2}^{2}}{2 m_{2}}}=\frac{P_{1}^{2} \times 2 m_{2}}{P_{2}^{2} \times 2 m_{1}}=4: 1$
Question 10
4 / -1

A force 'F' is given by F = at + bt2, where 't' is the time. What are the dimensions of 'a' and 'b', respectively?

A
MLT-1and MLT0

B
MLT-3and ML2T4

C
MLT-4and MLT1

D
MLT-3and MLT-4

Solution

since dimension of force is MLT $^{-2}$, so dimension of each factor on the left hand side should be equal to
that of the force.
Dimension of a:
$\left[M^{1} L^{1} T^{-2}\right]=[a]\left[M^{0} L^{0} T^{1}\right] \Rightarrow[a]=\left[M^{1} L^{1} T^{-3}\right]$
Dimension of b:
\[
\left[M^{1} L^{1} T^{-2}\right]=[b]\left[M^{0} L^{0} T^{2}\right] \Rightarrow[b]=\left[M^{1} L^{1} T^{-4}\right]
\]

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Physics Test - 12

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Physics Test - 12

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Question 1

0.02 coulomb

0.04 coulomb

0.08 coulomb

0.07 coulomb

Solution

Question 2

(P1+ P2) watt

\(\sqrt{P_{1} P_{2}}\) watt

\(\frac{P_{1}+P_{2}}{P_{1}+P_{2}}\) watt

\(\frac{P_{1}+P_{2}}{P_{1} P_{2}} w\) att

Solution

Question 3

2754.8 A0

3000 A0

3500 A0

4000 A0

Solution

Question 4

80oC

100oC

120oC

140oC

Solution

Question 5

\(\frac{u \sin \alpha}{f}\)

\(\frac{f \cos \alpha}{u}\)

u sin α

\(\frac{u \cos \alpha}{f}\)

Solution

Question 6

Stefan's law

Kirchhoff's law

Planck's law

Wien's law

Solution

Question 7

Z - 1 and A - 1

Z + 1 and A

Z - 1 and A

Z + 1 and A - 1

Solution

Question 8

3.3 kW

0.33 kW

0.033 kW

33 kW

Solution

Question 9

1 : 4

4 : 1

1 : 1

1 : 2

Solution

Question 10

MLT-1and MLT0

MLT-3and ML2T4

MLT-4and MLT1

MLT-3and MLT-4

Solution

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80^oC

100^oC

120^oC

140^oC