Physics Test

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Physics Test - 5

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Question 1
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A pipe open at both ends and a pipe closed at one end have the same length. The ratio of frequencies of their P^thovertone is

A
A.

B
B.

C
C.

D
D.

Solution

Frequency of p^th overtone of the open organ pipe
where L = length of an organ pipe and that of closed organ pipe is
So, the ratio of p^th overtone of open to closed.
Question 2
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Maximum kinetic energy gained by the charged particle in the cyclotron is independent of

A
A. the radius of the dees

B
B. charge

C
C. mass

D
D. frequency of revolution

Solution

Maximum kinetic energy gained by a charged particle in a cyclotron is given by
where q = charge of the cyclotron,
B = intensity of the magnetic field,
R = radius of orbit
And the m = mass of the particle,
Hence, E_K is independent of frequency of revolution.
Question 3
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The magnifying power of a telescope is nine. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal length of the objective and eyepiece are respectively

A
A. 10 cm, 10 cm

B
B. 18 cm, 2 cm

C
C. 15 cm, 5 cm

D
D. 11 cm, 9cm

Solution

For the final image at infinity, magnifying power of a telescope is given by
where, m = magnification,
f_ο = focal length of objective
and f_e = focal length of eyepiece
⇒ f_ο = 9f_e …(i)
Also, distance between objective and eyepiece
⇒ f_ο = f_e = (given)
⇒ 9f_e + f_e = 20 ⇒ f_e = 2 cm
⇒ f_ο = 9f_e = 18 cm
Question 4
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In a hydrogen atom, an electron of charge e revolves in a orbit of radius r with speed v. Then, magnetic moment associated with electrons is

A
A. evr/2

B
B. 2 evr

C
C. evr

D
D. evr/3

Solution

Magnetic moment of the revolving electron,
μ_e = i x A
where i = current and A = area.
where T = time period of revolution and r = radius orbit
where, v = velocity of revolving electron
⇒ μ_e = evr/2
Hence, the magnetic moment of a revolving electron is evr/2.
Question 5
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The stopping potential of the photoelectrons from a photocell is

A
A. directly proportional to the intensity of incident light

B
B. directly proportional to the frequency of incident light

C
C. inversely proportional to the frequency of incident light

D
D. Inversely proportional to the intensity of incident light

Solution

The stopping potential of photoelectrons is the potential needed to stop the electrons from reaching the collector depends only on the frequency of incident radiation, ft is directly proportional to it.
This is because, more the frequency of incident light, more will be the energy of the photons incident photons. (E = hv)
Thus, more will be the energy (kinetic) acquired from the electrons.
The more the kinetic energy of emitted electrons, higher is the potential needed to stop them (stopping potential).
Question 6
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Consider a particle of mass m suspended by a string at the equator. Let R and M denote the radius and mass of the earth. If ω is the angular velocity of rotation of the earth about its own axis, then the tension on the string will be (cos 0º = 1)

A
A. GMm/R²

B
B. GMm/2R²

C
C.

D
D.

Solution

When a body suspended by the string situated at position P as shown in the figure, where latitude is λ, then the body is also rotated with angular frequency (ω) of earth. Hence tension on the string is given by
T = mg - mω² cos λ
When the body is suspended at the equator, then λ = 0 and r = R
∴ From Eq. (i), we have
Question 7
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The fundamental frequency of the sonometer wire Increases by 9 Hz, if its tension is increased by 69%, keeping the length constant. The frequency of the wire is

A
A. 42 Hz

B
B. 24 Hz

C
C. 30 Hz

D
D. 36 Hz

Solution

We know that, frequency of vibration of a hed string is given by
where, T = tension in string
m = mass/length
and l = length of string.
New frequency,
⇒ 13v = 10v + 90
v = 30Hz
Question 8
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A potentiometer wire has length L For a given cell of emf E, the balancing length is L/3 from the positive end of the wire. If the length of potentiometer wire is increased by 50% , then for the same cell, the balance point is obtained at length

A
A. L/2 from the positive end

B
B. L/5 from the positive end

C
C. L/3 from the positive end

D
D. L/4 from the positive end

Solution

For the cell of emf E balancing length
Increased length of wire
Since the cell is the same, therefore the new balancing length = 1/3 (new increased length)
= L/2
Therefore, for the same cell the balance point will be as length L/2 from positive end.
Question 9
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A mass is whirled in a circular path with constant angular velocity and its linear velocity is v. If the string is now halved keeping the angular momentum the same, the linear velocity is

A
A. 2v

B
B. v/2

C
C. v

D
D. v√2

Solution

Angular velocity = ω
linear velocity = v
Length of string = Radius = r
If string is halved, r’ = r/2
Angular momentum. L = r x p
where , r = radius vector
and p = linear momentum
= r x mv
= mvt
As L remain constant,
mvr = constant
⇒ r' → r/2 , v' = 2v such that
mv' r' = m 2v x (r/2)
= m v r
Hence, linear velocity will be 2 v.
Question 10
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In Young’s double slit experiment, a fifth dark fringe is formed opposite to one of the slit. If D is the distance between the slits and the screen and d is the separation between toe slits, then toe wavelength of fight used is

A
A. d² /5D

B
B. d² /9D

C
C. d² /6D

D
D. d² /15D

Solution

Position of nth dark fringes in Young's double slit experiment is given
⇒
Since, fifth dark fringe is formed opposite to one of the slit.
∴ 2x₅ = d (distance between slits]
Put n = 5 in Eq. (i). we get
Question 11
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A 220 V input is supplied to a transformer. The output circuit draws a current of 2.0 A at 440 V. If the ratio of output to input power is 0.8, then the current drawn by primary winding is

A
A. 3.6 A

B
B. 5.0 A

C
C. 2.5 A

D
D. 2.8 A

Solution

Given, input voltage supplied to the transformer, V₁ = 220 V
Current at output circuit of transformer, i₂ = 2A
V₂ = 440V
Hence, the current drawn by the primary windings is 5A.
Question 12
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If the speed of an electron of hydrogen atom in the ground state is 2.2 x 10⁶ m/s, then its speed in the third excited state will be

A
5.5 x 10⁶ m/s

B
5.5 x 10⁵ m/s

C
88 x 10⁵ m/s

D
6.8 x 10⁶ m/s

Solution

speed of electron of the hydrogen atom in the ground state,
v₁ = 2.2 x 10⁶ m/s
Since
∴
[for third excited state. n₂ = 4]
⇒
5.5 x 10⁵ m/s
Question 13
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The vectors (A + B) and (A - B) are at right angles to each other. This Is possible under the condition

A
A. |A| = |B|

B
B. A·B = 0

C
C. A·B = 1

D
D. A x B = 0

Solution

Vectors (A + B) and (A - B) are at the right angle to each other, therefore
(A + B) · (A - B) = 0
A·A + B·A - A·B - B·B = 0
|A| + BAcosθ - AB cosθ - |B|² = 0
|A|² = |B|²
Hence. |A| = |B|
Question 14
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A person measures a time period of a simple pendulum inside a stationary lift and finds it to be T. If the lit starts accelerating upwards with an acceleration (g/3) the time period of the pendulum will be

A
T/√3

B

C
√3 T

D
T/3

Solution

Time period of simple pendulum inside a stationary lift
..(i)
where, l = length of string and
g = gravitational acceleration
Acceleration of lift in upward direction,
a = g/3
∴Time period of the pendulum
Question 15
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In damped SHM, the Sl unit of damping constant is

A
A. N/s

B
B. kg/s

C
C. kg/m

D
D. N/m

Solution

In damped SHM, damping force is proportional to the velocity of the oscillator
i.e., F_d ∝ v ⇒ F_d = bv
where, b is a damping constant.
∴ Damping constant. b = F_d /v
SI unit of damping constant
Hence, the SI unit of damping constant is kg/s.
Question 16
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When a certain metal surface is illuminated with a light of wavelength λ, the stopping potential is V, When the same surface is illuminated by light of wavelength 2λ, the stopping potential is (V/3). The threshold wavelength for the surface is

A
A. 8λ/3

B
B. 4λ/3

C
C. 4λ

D
D. 6λ

Solution

Given for a metal, the wavelength of light used = λ,
Stopping potential = V
If λ₀ be the threshold wavelength, then the maximum kinetic energy of emitted electrons.
Again, the wavelength of used light
λ' = 2λ
Stopping potential V' = V/3
Then
⇒
From Eqs (i) and (ii), we have
⇒ λ0 = 4λ
So, the threshold wavelength is 4 times of wavelength of light.
Question 17
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A hole is drilled half way to the centre of the earth. A body weighs 300 N on the surface of the earth. How much wil, it weigh at the bottom of the hole?

A
A. 200N

B
B. 250N

C
C. 120 N

D
D. 150N

Solution

Given, the distance of the bottom of the hole from the surface of the earth (d) = half the radius of earth = R_ο /2
If g be the value of gravitational acceleration on the surface of the earth, then weight of body mg = 300N
If g' be the gravitational acceleration at the bottom of the hole, then
∴ Weight of the body on the bottom erf hole,
mg' = mg/2 = 300/2 = 150N