Physics Test

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Physics Test - 10

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Weekly Quiz Competition

Question 1
3 / -1

Process of emission of electrons from hot metal surface is called-

A
electronic emission

B
electron capture

C
thermionic emission

D
none of the above

Solution

Thermal energy given to the charge carriers overcome the work function of the metal and thus low of charge carriers takes place from the metal surface. These charge carriers can be electrons and ions. So, process of the emission of electrons from the hot metal surface is called as thermionic emission.
Hence option C is the correct answer.
Question 2
3 / -1

A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms⁻¹. Which of the following statement(s) is (are) correct for the system of these two masses?

A
Total momentum of the system is 3 kg ms⁻¹

B
Momentum of 5 kg mass after collision is 4 kg ms⁻¹

C
Total kinetic energy of the system is 4J

D
None of these

Solution

By conservation of linear momentum v = 5v - 2 ..... (i)

By Newton's experimental law of collision u = v + 2 ....(ii)

using (i) and (ii) we have u = 1m/s and u = 3m/s

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

Light of wavelength 200nm shines on an aluminium, 4.20eV is required to eject an electron. What is the kinetic energy of the fastest ejected electrons?

A
0.5eV

B
1.00eV

C
2.00eV

D
4.00eV

Solution

Using the Einstein's equation of photoelectric effect,
$\frac{h c}{λ} = K E + ϕ$
$K E = \frac{1240}{200} - 4.2$
$K E = 2 e V$
Hence option C is the correct answer.
Question 4
3 / -1

Hologram is based on phenomenon of

A
Diffraction

B
Polarisation

C
Interference

D
Total internal reflection

Solution

Hologram is based on the phenomenon of interference.
[There are two physical phenomena as the principles of the holography: interference and diffraction of light waves. Holograms are photographs of three dimensional impressions on the surface of light waves.]
Hence option C is the correct answer.
Question 5
3 / -1

A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. Surface BC is frictionless. K_A, K_B and Kc are kinetic energies of the ball at A, B and C, respectively. Then

A
h_A>h_C; K_B>K_C

B
h_A>h_C; K_C>K_A

C
h_AC; K_B>K_C

D
All of the above

Solution

Hence option D is the correct answer.
Question 6
3 / -1

If λ is the wavelength of hydrogen atom from the transition n = 3 to n = 1 then what is the wavelength for doubly ionised lithium ion for same transition?

A
λ ÷ 3

B
3λ

C
λ ÷ 9

D
9λ

Solution

For wavelength:
$\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
Here, transition is same $\mathrm{So}, \lambda \propto \frac{1}{2 \mathrm{I}^{2}}$
$\frac{\lambda_{H}}{\lambda_{L}}=\frac{\left(Z_{L_{A}}\right)^{2}}{\left(Z_{H}\right)^{2}}=\frac{(3)^{1}}{(1)^{2}}=9$
$\lambda_{L i}=\frac{\lambda_{H}}{9}=\frac{\lambda}{9}$
Hence option C is the correct answer.
Question 7
3 / -1

Which of the following relates to photons both as wave motion and as a stream of particles?

A
Interference

B
E = mc²

C
Difraction

D
E = hv

Solution

(a) Interference exhibits the wave nature of light
(b) E = mc² exhibits the particles nature of light
(c) Difraction exhibits the wave nature of light.
(d) E = hν shows both particle nature and wave nature of light, i.e., light is having a particular wavelength and the energy is in the form of packets.
Hence option D is the correct answer.
Question 8
3 / -1

A particle is moving along a vertical circular path of radius r. When the circular path is just completed:

A
the velocity at highest position is √rg

B
the velocity at lowest position is √5rg

C
the tension in string at the lowest position is 6mg

D
All of the above

Solution

Considering the upper most point as 1 and the lowest point as $2,$ also consider $V_{t}$ and $V_{b}$ as velocity at top most and bottom most point, $T_{1}+m g=\frac{m V_{i}^{2}}{R}$
$T_{1}=0$ at highest point $V_{t}=\sqrt{g R}$ $\frac{1}{2} m V_{b}^{2}-\frac{1}{2} m V_{t}^{2}=2 m g R$
$V_{b}=\sqrt{5 R g}($ at lowest point $)$
$T_{2}-m g=\frac{m V_{b}^{2}}{R}($ at lowest point $)$
$T_{2}=m g+\frac{m V_{b}^{2}}{R}$
$T=6 m g$

Hence option D is the correct answer.
Question 9
3 / -1

The formula of kinetic mass of photon is? Where h is Planck's constant, v is frequency of the photon and c is its speed.

A
hv/c

B
hv/c²

C
hc/v

D
c²/h

Solution

Energy of photon having frequency v is given by E = hv Energy of photon can be written as E = mkc 2 where mk is the kinetic mass of photon
∴ mkc² = hv
⟹ mk = hv/c²`
Hence option B is the correct answer.
Question 10
3 / -1

It is found that a body on an inclined plane just starts sliding down if inclination is sin⁻¹(3/5), then the angle of repose (friction) is

A
sin⁻¹3/5

B
sin⁻¹3/4

C
tan⁻¹3/5

D
All of the above

Solution

We know, angle of repose $=$ angle of friction $\mathrm{Als}$ o, $\mu=\tan \theta$ $\theta=\sin ^{-1}\left(\frac{3}{5}\right)$
Hence option A is the correct answer.
Question 11
3 / -1

A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1m/s. A small ball of mass 0.1 kg, moving with 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:

A
the ring has pure rotation about its stationary CM.

B
friction between the ring and the ground is to the left.

C
the ring comes to a complete stop.

D
there is no friction between the ring and the ground.

Solution

Lets 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. Thus on the ring at the point of contact there is a horizontal and a vertical impulse. These will have components along the tangent of the ring, which will provide angular impulses.
Using angular impulse = change in angular momentum we get:
=2cos30° 1 / 2 − 1 sin 30° 1 / 2 = 2 (1 / 4) (ω₂−ω₁)
note that we have assumed that direction of angular velocities is same before and after and since LHS of the above equation is positive ω₂>ω₁
thus 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
Question 12
3 / -1

In an inelastic collision an electron excites a hydrogen atom from its ground state to a M-shell state. A second electron collides instantaneously with the excited hydrogen atom in the M-shell state and ionizes it. At least how much energy the second electron transfers to the atom in the M-shell state?

A
+3.4eV

B
+1.51eV

C
−3.4eV

D
−1.51eV

Solution

Given that the electron is in M state.
This corresponds tothe principal quantum number n = 3
From Bohr's model, energy of a state with quantum number n is given by
$E = - \frac{13.6}{n^{2}} e V$
Thus, the energy of the electron in M shell is $E = - \frac{13.6}{3^{2}} \approx - 1.51 e V$
In order to ionise the atom, the minimum energy required is thus +1.51eV
Hence option B is the correct answer.
Question 13
3 / -1

Work function of a metal is 5.2 × 10^-18 J then its threshold wavelength will be

A
736.7A˚

B
760.7A˚

C
301Å

D
380.7Å

Solution

Given - work function of metal W = 5.2 ×10^-18 The work function of a metal is given by,
Question 14
3 / -1

In photoelectric effect, the photo current.

A
Increases with increase of frequency of incident photon

B
Decreases with increase of frequency of incident photon

C
Does not depend on the frequency of photon but depends only on intensity of incident light

D
Depends both on intensity and frequency of incident beam

Solution

When intensity of incident photon increases number of electrons emitted from the surface increases due to which current increases. Frequency of incident photon only limits the maximum kinetic energy of the photoelectron emitted and so the current is independent of frequency of photon.
Hence option C is the correct answer.
Question 15
3 / -1

The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is : (hc = 1240 eV.nm)

A
3.09 eV

B
1.41 eV

C
1.51 eV

D
1.68 eV

Solution

\begin{equation}
\begin{array}{l}
\frac{h c}{\lambda}=\phi+(K E)_{\max } \\
=\frac{1240}{400}=\phi+1.68 \\
\Rightarrow \phi=1.41 e V
\end{array}
\end{equation}
Hence option B is the correct answer.

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Physics Test - 10

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Physics Test - 10

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Question 1

electronic emission

electron capture

thermionic emission

none of the above

Solution

Question 2

Total momentum of the system is 3 kg ms−1

Momentum of 5 kg mass after collision is 4 kg ms−1

Total kinetic energy of the system is 4J

None of these

Solution

Question 3

0.5eV

1.00eV

2.00eV

4.00eV

Solution

Question 4

Diffraction

Polarisation

Interference

Total internal reflection

Solution

Question 5

hA>hC; KB>KC

hA>hC; KC>KA

hAC; KB>KC

All of the above

Solution

Question 6

λ ÷ 3

3λ

λ ÷ 9

9λ

Solution

Question 7

Interference

E = mc2

Difraction

E = hv

Solution

Question 8

the velocity at highest position is √rg

the velocity at lowest position is √5rg

the tension in string at the lowest position is 6mg

All of the above

Solution

Question 9

hv/c

hv/c2

hc/v

c2/h

Solution

Question 10

sin−13/5

sin−13/4

tan−13/5

All of the above

Solution

Question 11

the ring has pure rotation about its stationary CM.

friction between the ring and the ground is to the left.

the ring comes to a complete stop.

there is no friction between the ring and the ground.

Solution

Question 12

+3.4eV

+1.51eV

−3.4eV

−1.51eV

Solution

Total momentum of the system is 3 kg ms⁻¹

Momentum of 5 kg mass after collision is 4 kg ms⁻¹

h_A>h_C; K_B>K_C

h_A>h_C; K_C>K_A

h_AC; K_B>K_C

E = mc²

hv/c²

c²/h

sin⁻¹3/5

sin⁻¹3/4

tan⁻¹3/5