Physics Test

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Physics Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

In an LCR series circuit, the potential difference between the terminals of the inductance is 60 V, that between the terminals of the capacitor is 30 V and that across the resistance is 40 V. Then, the supply voltage will be ?

A
50 V

B
70 V

C
130 V

D
10 V

Solution

For series LCR circuit, effective voltage is \(V=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}\) \(v=\sqrt{40^{2}+(60-30)^{2}}=\sqrt{1600+900}\)

\(v=\sqrt{2500}-50 v\)

Hence option A is the correct answer.
Question 2
1 / -0

An electron jumps from the fourth orbit to the second orbit of hydrogen atom. If Rydberg’s constant R is equal to 10⁷ m^–1, then the frequency of the emitted radiation in Hz will be ?

A
(3/16) x 10⁵

B
(3/6) x 10⁵

C
(9/16) x 10¹⁵

D
(3/4) x 10¹⁵

Solution

\(\frac{1}{\lambda}=R\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right]\)

\(\frac{v}{c}=R\left[\frac{3}{16}\right]\)

\(\nu=3 \times 10^{8} \times 10^{7} \times \frac{3}{16}\)

\(=\frac{9}{16} \times 10^{15}\)

Hence option C is the correct answer.
Question 3
1 / -0

The door of a running refrigerator inside a closed room was left open. Which of the following observations will be seen?

A
The room will be warmed up gradually.

B
The temperature of the room will remain unaffected.

C
The room will become as cold as the refrigerator.

D
The room will be cooled slightly.

Solution

Refrigerator is a device that transfers heat from inside a box to its surroundings.

If you leave the door open, heat is merely recycled from the room into the refrigerator, then back into the room. A net room temperature increase would result from the heat of the motor that would be constantly running to move energy around in a circle.
Hence option A is correct.
Question 4
1 / -0

If the wavelength of incident light changes from 400 nm to 300 nm, then the stopping potential of photoelectrons emitted from a surface becomes (approximately).

A
1.0 V greater

B
1.0 V smaller

C
0.5 V greater

D
0.5 V smaller

Solution

Energy of light \(E=\frac{h c}{\lambda} \Rightarrow E \propto \frac{1}{\lambda}\)

\(\frac{E^{\prime}}{E}=\frac{400}{300}=1.33\)

But \(E=e V_{0}\)

thus, stopping potential for photoelectrons from a surface becomes approximately \(1.0 \mathrm{V}\) greater.

Hence option A is the correct answer.
Question 5
1 / -0

A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity s and mass m. If the temperature of the block is raised by ΔT, then the potential difference V across the capacitor will be.

A

\(\frac{m s \Delta T}{C}\)

B

\(\frac{m C \Delta T}{s}\)

C

\(\sqrt{\frac{2 m C A T}{s}}\)

D

\(\sqrt{\frac{2 m s \Delta T}{C}}\)

Solution

\(E=\frac{1}{2} C V^{2} \ldots \ldots(i)\)

The energy stored in capacitor is lost in form of heat energy.

\(H=m s \Delta T \ldots \ldots \ldots\)

From Eqs. (i) and (ii), we have

\(m s \Delta T=\frac{1}{2} C V^{2}\)

\(V=\sqrt{\frac{2 m s \Delta T}{C}}\)

Hence option D is the correct answer.
Question 6
1 / -0

The electric potential at a point on the axis of an electric dipole depends upon the distance r (>> a) of the point from centre of the dipole as :

A
∝ r

B
∝ 1/r

C
∝ 1/r²

D
∝ 1/r³

Solution

The electric potential at any point on the axis of electric dipole,

\(V=\frac{p}{4 \pi \varepsilon_{0} r^{2}}\)

or

\(V \propto \frac{1}{r^{2}}\)

Hence option C is the correct answer.
Question 7
1 / -0

Two spherical bodies of masses 'M' and '5M' and radii 'R' and '2R', respectively, are released in free space with initial separation between their centres equal to '12R'. If they attract each other due to gravitational force only, then what is the distance covered by the smaller body just before collision?

A
1.5R

B
2.5R

C
4.5R

D
7.5R

Solution

Relatre displacement \(=9 R\)
As masses are moving under mutuel gravitational attraction, displacement of centre of mass in zero
\(m_{1} \overrightarrow{x_{1}}+m_{2} \overrightarrow{x_{2}}=0\)
\(m_{1} \overrightarrow{x_{1}}=-m_{2} \overrightarrow{x_{2}}\)
-we sign indicates direction if displacement is opposite.
\begin{equation}\begin{array}{l}
m_{1} x_{1}=m_{2} x_{2} \\
x_{2}=L-x_{1} \\
x_{1}=\frac{m_{2}}{m_{1}+m_{2}} L \\
x_{1}=\frac{5 m}{6 m} 9 R=7.5 R
\end{array}\end{equation}

Hence option D is the correct answer.
Question 8
1 / -0

The coercivity of a small magnet where the ferromagnet gets demagnetised is 3 x 10³ Am^–1. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetised when inside the solenoid, is :

A
3A

B
6A

C
30 mA

D
60 mA

Solution

\(\mu_{0} H=\mu_{0} n i\)

\(3 \times 10^{3}=\frac{100}{0.1} \times i \Rightarrow i=3 A\)

Hence option A is the correct answer.
Question 9
1 / -0

A particle moves in xy plane according to the law x = a sin ωt and y = a(1 - cos ωt), where a and w are constant. The particle traces.

A
a straight line equally inclined to x-axis and y-axis

B
a parabola

C
a circle

D
None of these

Solution

\(x=a \sin \omega t \ldots(i)\)
\(y=a(1-\cos \omega t) \ldots(\) ii \()\)
\(x^{2}=a^{2} \sin ^{2} \omega t=a^{2}\left(1-\cos ^{2} \omega t\right)\)
\(x^{2}=a^{2}-a^{2} \cos ^{2} \omega t\)
\(a^{2} \cos ^{2} \omega t=a^{2}-x^{2} \ldots(i i i)\)
\(y=a(1-\cos \omega t)=a-a \cos \omega t\)
\(a-y=a \cos \omega t\)
\((a-y)^{2}=(a \cos \omega t)^{2}\)
\(a^{2}+y^{2}-2 a y=a^{2} \cos ^{2} \omega t \ldots(i v)\)
Put the value from equation (iii),
\(a^{2}+y^{2}-2 a y=a^{2}-x^{2}\)
\(x^{2}+y^{2}-2 a y=0\)
This is the equation of a circle.

Hence option C is the correct answer.
Question 10
1 / -0

A parachutist, after bailing out, falls 50 m without friction. When parachute opens, it retards at 2 ms^-2. He reaches the ground with a speed of 3 ms^-1. At what height did he bail out?

A
293 m

B
111 m

C
91 m

D
182 m

Solution

Let velocity at the end of \(50 m\) be \(u m / s\).
\(u^{2}=2 g h=2 \times 9.8 \times 50=980\) units
Let further displacement be s.
\(s=\frac{v^{2}-u^{2}}{2 a}=\frac{3^{2}-980}{-2 \times 2}\)
\(s=\frac{980-9}{4}=\frac{971}{4} \approx 243\)
Total height \(=h+s=50+243=293 m\)

Hence option A is the correct answer.

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Physics Test - 7

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Physics Test - 7

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SHARING IS CARING

Question 1

50 V

70 V

130 V

10 V

Solution

Question 2

(3/16) x 105

(3/6) x 105

(9/16) x 1015

(3/4) x 1015

Solution

Question 3

The room will be warmed up gradually.

The temperature of the room will remain unaffected.

The room will become as cold as the refrigerator.

The room will be cooled slightly.

Solution

Question 4

1.0 V greater

1.0 V smaller

0.5 V greater

0.5 V smaller

Solution

Question 5

\(\frac{m s \Delta T}{C}\)

\(\frac{m C \Delta T}{s}\)

\(\sqrt{\frac{2 m C A T}{s}}\)

\(\sqrt{\frac{2 m s \Delta T}{C}}\)

Solution

Question 6

∝ r

∝ 1/r

∝ 1/r2

∝ 1/r3

Solution

Question 7

1.5R

2.5R

4.5R

7.5R

Solution

Question 8

3A

6A

30 mA

60 mA

Solution

Question 9

a straight line equally inclined to x-axis and y-axis

a parabola

a circle

None of these

Solution

Question 10

293 m

111 m

91 m

182 m

Solution

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(3/16) x 10⁵

(3/6) x 10⁵

(9/16) x 10¹⁵

(3/4) x 10¹⁵

∝ 1/r²

∝ 1/r³