IIT JAM Physics MCQ Test 4 - Physics MCQ

De-Broglie Wave: According to de-Broglie, every moving object is associated with a wave, known as de-Broglie wave. If m is mass of the object moving with velocity v then associated wavelength
λ is a given as
Where h is Planck's constant
The velocity of waveforms of constant phase is called phase velocity and is equal to the velocity of the wave Hence,
phase velocity = λ x frequency ........ (i)
By energy equation mc² = hv

Frequency = mc²/h....(ii)
By Eqs. (i) and (ii) 
Phase velocity = λ * v

Phase velocity = c²/V...(iii)
Now, the group velocity is given as 
Now, 
Where 
Thus, 
By differentiation we get,

So, the group velocity is given as,

We have ...(i)
The phase velocity  ...(ii)
The group velocity  ...(iii)
Differentiate equation, we get 
⇒  ...(iv)
By Eq. (ii) ω= kv_p
Eq. (iv) ⇒  ⇒   ⇒ 

The phase velocity 
where λ = wavelength
Phase velocity 
From Eqs. (i) and (ii), we have
⇒ 
⇒ 
 ...(iii)
Group velocity 
⇒ 
⇒
⇒
⇒ 
⇒ Group velocity 

Dispersion relation is


Group velocity 
phase velocity 
Given v_g =v_p

The wave equation is y = 
The particle velocity 
⇒ The peak velocity of particle =  A 2π v ...(i)
and the wave velocity =  vλ ...(ii)
By 
⇒ 
⇒ 

   or 
or 
or 
∴ 
Ans.

Distance covered by the sound to reach the person = distance covered by jet x cot 60°
∴ velocity of jet = 

The frequency of sound received by the wall is given by

Now, the wall acts as a source of sound. The apparent frequency detected by the observe is

Let source of sound moves with speed v_s and its real frequent When source approaches observer frequency is given as

v → velocity of sound 
⇒ v = 330
⇒ ...(i)
When source recedes

⇒ ...(ii)
By Eqs. (i) and (ii) we get
⇒ 
But 
So, 
⇒ 
 ⇒ 
⇒  v_s = 30 m/s

equation of an ellipse.

We know that total energy 

   

x = 3 cos 2t + 4 sin 2t
x = 3 cos 2t + 4 sin 2t





⇒ ω = 2

A = x₀

The ordinary light is a wave in which its both component E and B vibrate perpendicularly to the direction of the light. If E vibrate in a plane the light is said to plane polarised light.
The two perpendicular vibrating lightwaves form a plane polarised light if their frequencies are equal ω_x = ω_y and have a phase difference of zero. 

(i) Considering the equilibrium of m₂, we have 
...(i)
(ii) Let the mass m₂ is displaced downwards by a distance x. Now the radius of the circular path decreases by x and the angular speed of m₁ increases. Applying the conservation of angular momentum, we have 

or  ...(ii)
The tension is also increased as shown below :

∴ 
 ...(iii)
As a result, m₂ gets a restoring force, given by 
F = - [T - m₂g]
or 
or ...(iv)
According to second law of motion. F is given by
...(v)
From eq. (4) and (5). we have 

or  ...(vi)
Substituting the value of ω₀ from eq. (1). we get

or 
This represents S.H.M. The time period T is given by

Let X₀ be the equilibrium length of the spring. This is stretched by a length L So. the initial length of the spring x_i = (X₀ + l).
Potential energy stored in spring 
When the spring is released, say att = 0. the cylinder rolls without slipping being pulled by the spring. Consider the situation at time t. Letthe centre of mass be at a distance x from the oiigin O. At this instant.
K .E . (translational) of cylinder = 1/2 mv²
and K.E. (rotational) of cylinder


where v = linear velocity of cylinder 
∴ Total energy of cylinder

...(i)
Instantaneous extension of spring = (x - x₀)
Potential energy of the spring 
∴ Loss of P.E. = 
 ...(ii)
Applying the law of conservation of energy, we have gain in K.E. = loss is potential energy

Differentiating on both sides with respect to t and considering l and x₀ constant we get



As x is constant this equation can be written as

This is equation of S.H.M.
∴ 
At equilibrium position, the potential energy is completely converted to kinetic energy

or 
Further, 

Given, n = 250 Hz, v = 30 m/s

(a) Phase difference between two points at a distance λ = 2π
Phase difference between two points, unit distance apart 
∴ Phase difference for a distance of 10 cm

(b) Now 
and  
The general equation of a plane progressive wave is given by
  
Here y = 0.03 sin 2π (250 t - 25x/3)
(c) The distance between nodes in stationary wave

Equation of stationary wave is given by


here 
∴ y = 0.02 

Comparing the given equation with the general wave equation:-

we find that: -
(A) As there is negative sign between t and x terms, the wave is propagating along positive x-axis.
(B) The amplitude of the wave A = 4 cm = 0.04 m.
(C) The time period of the wave T = 0.02 s = (1/50)s.
(D) The frequency of the wave f = (1/T) = 50Hz.
(e) Angular frequency of the wave 
(f) The wavelength of the wave 
(g) The propagation const. (= wave vector =  
(h) The velocity of wave  
(i) The phase const., i.e., initial phase 
(j) The max. particle velocity 

We know for minima in newtons ring

where n=1. 2. 3 
(a) for n = 15
e = thickness of liquid film  = x²/2R
X = radius of ring
R = radius of curvature of lens one gets


Given μ = 1.5 
R = 2m
n = 15

So, 
So.  λ = 0.25 * 10^-6 m

If M is the molecular weight of the gas and T is the absolute temperature, then speed of sound in a gas.

where R is the universal gas constant.
Velocity of sound in oxygen at t°C

Velocity of sound in nitrogen at 15°C

According to the given problem


Solving we get t = 56.1°C.

We know that the speed of sound in a gas is given by

If v_a and v_h be the velocities of air and hydrogen respectively at normal pressure and temperature, then

where d_h and d_a are densities of hydrogen and air respectively.
Given that 

We know that the speed of sound is directly proportional to the square root of the absolute temperature of the gas. Let v₀ and v₈₁₉ be the speeds of sound in hydrogen at 0°C and 819°C respectively, then

 ∴ v₈₁₉ = 2 * v₀ = 2 * 1328 = 2656 m/sec

The focal length of a lens is given as

Here, n = 1.6. R₁ =0.1 m. R₂ = - 0.10 m
So, 
= (0.6) (20)
1/f =12
⇒ Power P = 1/f = +12D

So resolving power of grating 
R = Nn ≥ 10000

n ≥ 1.11 
n = 2

Resolving power required =λ/μ
= 6000/6 = 1000
R = Nn = 1000 
R = Nn = 1000 
⇒ N = 1000

D = 10 cm = 1 m 
d = 2.5 mm = 2.5 x 10^-3 m 
t = 10 x 10^-6 m , n = 1.5