IIT JAM Exam  >  IIT JAM Tests  >  JAM-20-PH-VTP-2 (SET-1)-FINAL - IIT JAM MCQ

JAM-20-PH-VTP-2 (SET-1)-FINAL - IIT JAM MCQ


Test Description

30 Questions MCQ Test - JAM-20-PH-VTP-2 (SET-1)-FINAL

JAM-20-PH-VTP-2 (SET-1)-FINAL for IIT JAM 2024 is part of IIT JAM preparation. The JAM-20-PH-VTP-2 (SET-1)-FINAL questions and answers have been prepared according to the IIT JAM exam syllabus.The JAM-20-PH-VTP-2 (SET-1)-FINAL MCQs are made for IIT JAM 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JAM-20-PH-VTP-2 (SET-1)-FINAL below.
Solutions of JAM-20-PH-VTP-2 (SET-1)-FINAL questions in English are available as part of our course for IIT JAM & JAM-20-PH-VTP-2 (SET-1)-FINAL solutions in Hindi for IIT JAM course. Download more important topics, notes, lectures and mock test series for IIT JAM Exam by signing up for free. Attempt JAM-20-PH-VTP-2 (SET-1)-FINAL | 60 questions in 180 minutes | Mock test for IIT JAM preparation | Free important questions MCQ to study for IIT JAM Exam | Download free PDF with solutions
JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 1

A simple pendulum is set - up in a trolley which moves to the right with an acceleration 'a' on a horizontal plane, then the thread of the pendulum in the mean position makes an angle θ with the vertical

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 1

tan θ = a/g

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 2

A combination of two thin convex lenses of equal focal lengths is kept separated along the optic axes by a distance of 20 cm between them. The combination behaves as a lens system of infinite focal length. If an object is kept at 10 cm from the first lens. Its image will be formed on the other side at a distance x from the second lens. The value of x is

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 2



for first lens

 ⇒ v = ∞

⇒ v f' = 10 cm
For second lens u = ∞

1 Crore+ students have signed up on EduRev. Have you? Download the App
JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 3

A spring mass system has undamped natural angular frequency ω0 = 100 rad S-1. The solution x(t) at critical dam ping is given by x(t) = x0 (1 + ω0t) exp(-ω0t) where x0 is a constant. The system experiences the maximum damping force at time .

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 3

Daming force 

F is maximum when dF/dt = 0

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 4

W hen two sim ple harmonic oscillations rep resented by x = A0 cos(ωt + α) and y = B0 cos (ωt + β) are superposed at right angles, the resultant is a circle of radius as shown in the figure. The condition which corresponds to this are

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 4

The lissajous figure is circle only when phase difference is π/2

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 5

The instantaneous position x(t) of a small block performing one dimensional damped oscillation is x(t) = Ae-rt sin (ωt + α). Here ω is the angular frequency,r is the damping coefficient. A is the initial amplitude and α is the initial phase . If Xt = 0 and  he values of A and α (with n = 0. 1, 2 , .........) are

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 5


0 = A sin α
sin α = 0



JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 6

A lightly damped harmonic oscillator losses energy at the rate of 4% per minute. The decrease in amplitude of the oscillator per minute will be closest to

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 6

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 7

A collimated beam of light of diameter 2nm is propagating along the x - axis. The beam is to be expanded to a collimated beam of diameter 50nm using a combination of the convex lenses. A lens of focal length of 50nm and another lens with focal length f are to be kept at a distance a between them. The value of F and d respectively are

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 7



F = 1250 nm
d = F + f = 1250 + 50 = 1300 nm

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 8

A simple pendulum suspended from the ceiling of a stationary. Cart has a time period 2 seconds when the cart accelerates in the horizontal direction with an accelerates in the horizontal direction with an acceleration of 10 m/s2, the time period of the pendulum is (g = 10 m /see2)

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 8



JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 9

A glass plate P (refractive index np, = 1.54) is coated with a dielectric material C with the refractive index nc = 1.6. In order to have enhanced reflection from this coated glass for near normal incident light of wavelength λ, the thickness of the coating material. C must be

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 9

For enhanced reflection path difference in the coating must be an integral multiple of λ.

θ = 0 for near normal incident ray 

⇒ even multiples of 

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 10

Two thin convex lenses ℓ1, and ℓ2 with focal lengths 1 cm and 2 cm respectively are separated by a distance of 4 cm along their axis as shown below.

An object is placed at a distance of 1.5 cm before the first lens. The ratio of the final image size to the object size is

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 10

For first lens u = -1.5 , f = 1 cm


For second lens u = -1 cm , f = 2 cm From lens maker formula

Magnification m2 
So final magnification m = m1 x m2
= -2 x 2 = -4
So, Ratio of final image size to the object size = 4

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 11

The dispersion relation for deep water is given by ω2 = gk + ak3, where g and a are constants.

The phase velocity in terms of λ is

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 11

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 12

Two beams of light having Intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point A and π at point B ,then the difference between the resultant intensities A and B is

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 12

Intensities at point A is


JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 13

Consider a plane transmission diffraction grating, let d be the distance between ruled lines, m the order number and θ the observation angle. The angular dispersion for incident light of wavelength λ is

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 13


JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 14

Which one of the following is a elliptically perpendicularly polarized standing wave ?

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 14


It is the combination of
X1 = A2 cos ωt sin kz

These are the equation of ellipse.

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 15

In a double slit experiment two parallel slits are illuminated first by light of wavelength 400 nm and then by light of unknown wavelength. The fourth order dark fringe resulting from the known wavelength of light falls in the same place on the screen as the second order bright fringe from the unknown wave length. The value of unknown wavelength of the light is

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 15


JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 16

The equation of motion is  = 0.124 sin100t, where ail quantities are in S.l. units, then natural frequency of undamped oscillation is

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 16


We get m = 2 * 10-4 kg

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 17

The velocity of extraordinary ray in calcite crystal in a plane perpendicular to the optic axis. Given μ0 =1.658, μe = 1.486 and c = 3x1010 cm/s.

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 17

The velocity of the O-wave in the crystal is

The velocity of the E-wave in the crystal normal to be optic axis is

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 18

The critical angle of light in a certain substance is 45°. What is the polarizing angle?

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 18

The incident light is moving in the denser medium. If c be the critical angle, then we have sin c = 1/μ 

where μ is the refractive index of the denser medium. Here c = 45°.

If p be the polarising angle, then by Brewster’s law, we have tan p = μ.
or p = tan-1 (μ)

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 19

How many orders will be visible if the wavelength of incident radiation is 5000 Å and the number of lines on the grating is 2620 to an inch?

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 19

We know that, (b + d) sinθ = mλ, the maximum value of θ is 90°, that is sin θ = 1. Therefore, the number of orders visible with the grating is given as (b+d) = nλ

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 20

How many lines per centimetre are there in a plane transmission grating which gives 1st order of light of wavelength 6000 Å at angle of diffraction 30°?

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 20

The plane grating equation is, 
(b+d) sin θ = mλ

Here θ = 30°
So sinθ = 0.5
n = 1, λ = 6000Å = 6000 x 10-8cm
∴ Grating element,

Since the getting element is equal to the reciprocal of the number of lines per cm. Therefore, number of lines per cm

Hence, the maximum number of orders visible in the spectrum are 19.

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 21

In Fraunhofer diffraction due to a narrow slit, a screen is placed 2 m away from the lens to obtain the pattern. If the slit width is 0.2 mm and first minima lie 5 mm on either side of the central maximum, find the wavelength of light.

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 21

In the Fraunhofer diffraction pattern due to a single slit of width b, the directions of minima are given by
b sin θ = ±m λ wher n = 12.3.....
or  
If θ is measured in radian, than sin θ = 0
θ = mλ/b

The angular separation θ between the first minimum on either side of the central maximum is

θ = λ/b

Here n = 1

In the given problem, b = 0.2 mm = 0.02 cm

Linear separation between the first minimum and central maximum is as 

CP = 5 mm = 0.5 cm 

Comparing equation (1) and (2), we get 

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 22

Newton’s rings are observed by keeping a spherical surface of 100 cm radius on a plane glass plate. If the diameter of the 15th bright ring is 0.590 cm and the diameter of the 5th ring is 0.336 cm, what is the wavelength of light used?

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 22

If Dn+p and Dn be the diameters of (n+p)th and nth bright ring, then 

Here D15 = 0.590 cm, D5 = 0.336 cm,
p = 10 and R = 100 cm

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 23

A particle of mass m moves in one dimension under the influence of a potential energy where a and b are positive constants and ℓ is a characteristic length. The frequency of small oscillation about a point of stable equilibrium is

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 23


For stable equilibrium V(x) = 0


JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 24

In an interference pattern with two coherent sources, the amplitude of intensity variation is found to be 5% of the average intensity. Calculate the relative intensities of the interfering sources.

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 24

Let us suppose that the amplitude ratio of the two interfering sources be a :1, where a > 1. If lmax and lmin be the maximum and minimum intensities, then in the interference pattern, the intensity ratio is

Here, if average intensity is 100 units, then the maximum is 105 minimum is 95. Therefore, the ratio is

or  
or  a +1 = 1.051(a - 1)
or  

Therefore, the intensity ratio of the two interfering sources is 

or I1 : l2 = 1600 : 1 

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 25

A string under a tension of 129.6 N produces 10 beats per sec when is vibrated along with a tuning fork. When the tension in the string is increased to 160 N. it sounds in unison with the same tuning fork. Calculate the fundamental frequency of the tuning fork.

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 25

As here the tension in the wire in charged, so its fundamental frequency

Now with increase in tension, k√T will increase and beats and decreasing to zero when T = 160 N (as unison means the frequencies are equal).

Substituting the value of K from second equation in first.

or f - 0.9 f = 10
i.e., f = 100 Hz

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 26

A beam of plane-polarized light falls on a polarizer which rotates about the axis of the ray with angular velocity ω = 22 rad/sec. Find the energy (mJ) of light passing through the polarizer per one revolution if the flux of energy of the incident ray is equal to φ = 14 mω.

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 26

When the polarizer rotates with angular velocity ω its instantaneous principal direction makes angle cot from a reference direction which we choose to be along the direction of vibration of the plane polarized incident light. The transmitted flux at this instant is φ0cos2ωt and the total energy passing through the polarizer per revolution is

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 27

The radius of curvature of curved surface of a plano-convex thin lens of refractive index n = 3 of focal length 0.4 m is

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 27

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 28

A pulse of electromagnetic radiation propagates through a normally dispersive medium. As it propagates, the pulse gets broadened. This is because

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 28

W e know for normally dispersive medium refractive index « frequency (f) and velocity in a medium c = c/n

So the lower frequency components in the pulse travel faster than the higher frequency ones.

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 29

The dispersion relation for electromagnetic waves in a certain medium is given by ω2 = ak where a is a constant, ω the frequency and k the magnitude of the wave vector. The velocity of energy propagation in this medium is

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 29

ω2 - ak
Velocity of energy propagation is 
ω2 = ak
2ω dω = adk

JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 30

A fast charged particle passes perpendicularly through a thin glass sheet of refractive index 1.5. Inside the glass, the particle emits light. The minimum speed of the particle is

Detailed Solution for JAM-20-PH-VTP-2 (SET-1)-FINAL - Question 30

∵ refractive index n = c/v

View more questions
Information about JAM-20-PH-VTP-2 (SET-1)-FINAL Page
In this test you can find the Exam questions for JAM-20-PH-VTP-2 (SET-1)-FINAL solved & explained in the simplest way possible. Besides giving Questions and answers for JAM-20-PH-VTP-2 (SET-1)-FINAL , EduRev gives you an ample number of Online tests for practice
Download as PDF