IIT JAM Physics MCQ Test 3 - Physics MCQ

Amplitude of motion A = 2x₀
Time to cover from extreme position to mean positon  
(i.e. from compressed position to norma position) = T/4
y = A sin ωt


∴ So, total time taken to hit the wall.

we know that 




further  
ω= 3

The total energy E of a particle vibrating in S.H.M. is given by
 ...(1)
The kinetic energy K is given by

where y =  displacement of the particle.



Hence the kinetic energy is half of the total energy when the displacement of the particle is  Given that a = 4cm.

Let x be the displacement of spring C. The displacement of A = x cos45⁰ = xl√2
Similarly, the displacement of B 

Now, forces acting on m in the displacement position of m are kx in the direction C,

Net force in direction C

Let the rod be depressed by a small amount x. Both the springs are compressed by x. When the rod is released, the restoring torque is given by


where θ is expressed in raidan. Thus 

If is the moment of inertia of the rod about O, then

Since  the motion is simple harmonic whose angular frequency is given by

Now  Therefore we have 

First two springs are connected parallel hence displacement in both the spring will be equal say x then by hook's law

F = -kx 
By Eqs (i) and (ii)

⇒  (since K₁ = K₂)
In third case equivalent spring constant becomes k/2
We know frequency ω = 

The displacement y = 10 sin 2π/5 (1800 t -x)
Comparing this y equation 

we get  ...(1)
and  ..(2)
By Eq. (i), 
⇒ F = 3600/10
= 360
⇒  1/T = 360
⇒  T = 1/360 s

We know the S.H.M. can be written as y = 1 sin ωt = a sin θ 
For first we have,

For second,
a/2= a sinθ₂ ⇒ θ₂ = 30⁰
Phase difference = 

We have x =  a sin (ωt + φ) ...(i)
x = a sin ωt and φ = 90⁰
Thus, when φ = 90⁰ the Eq. (i) becomes
x = a cos ωt
y = b sin ωt


Squaring and adding 

Displacement x of curve -B is positive at all instance and the system return to is equilibrium position without large change in oscillations. (and returns is slower)

The equation of force, 
F =F₀ sin ωt
where F₀ → a constant 
ω → angular velocity (a constant)
Now by Newton's second law,




Integrating 
when 


So, 




To draw the graph, 
(i) when t = 0, s = 0
(ii) initially as t increases from 0, the rate of increase of s is slower.
So, graph is:

and total mass oscillating = M + m

For damped oscillation amplitude of oscillation is proportional to e^-yt • Where γ —► damping parameter. And this is exponentially decreases with time.

Given 

Here, m = 50kg, v = 2s^-1
a = 5 cm = 0.05m
Max. acceleration, a_max = ω₂a


∴ Max. force on the man

Mini. force on the man

Consider a tunnel dug along the diameter of the earth. A particle of mass m is placed at a distance y from the center of the earth. There will be a gravitational attraction of the earth experienced by this particle due to the mass of matter contained in a sphere of radius y. Force acting on particle at distance y from center.

Given that U(x) = k[1-e^-x2]
The graph of U(x) is shown in fig.(1)

Now F(x) = -
Thus force is zero only at following three points:

At any point away from origin (excluding ) the particle is not even in equilibrium.
Further, at finite nonzero values of x, the force is directed towards origin.


If displaced a little about origin, the particle will execute s.H.M. Again, Total mechanical energy = K.E. + P.E.

or 
The graph of K.E. is also shown in fig. 1 it is obvious from the graph that K.E. is not the minimum at the origin.
 

Here as k =  k₁ +k₂ = 0.5 + 0.5 = 1N/m and m = 0.01kg,


As the block is displaced by 0.02m, amplitude will be 0.02m. So the equation of motion along the x-axis will be
 ..(i)
Where φ is phase constant.
But along the y-axis as paper is moving down, the pointer relative to paper is moving up with constant velocity 0.1 m/s, so that 
y = vt = 0.1t
The equation of the path will be obtained by eliminating t between Eqns. (1) and (2) i.e.
x= 0.02 sin(100y+
Now x will be maximum when
sin (100y +  = max = 1, 
which implies 
So that  

The radius of zone plate 
where b = distance of point from zone plate
⇒ r ∝ √n
focus f  
=> focal length   => 
Hence, first statements is true.
The highest intensity occurs at a point for which zone plate is produced. => (iv) statement is wrong for a particular wavelength a number of foci are formed, and (iii) statement is wrong.
Thus, statements (i) and (ii) are right.

When an unpolarised light beam passes through a double refracting medium it splits into two beams viz extraordinary ray which has different velocity in different directions and other is ordinary ray whose velocity in all direction is same. Both these are plane polarised and the plane of polarisations of these are perpendicular to each other. Thus, refractive index of the ordinary ray remains constant while refractive index of the extraordinary ray does not remain constant.

Here if R = normal reaction then R-mg
= -mg
or R = mg-ma = m(g-a)
Also Max. acceleration  a= Aω²
∴ R = mg - mAω²
For object, not to leave the platform R = 0
∴ mg = - mAω²

As the string has mass and it is suspended vertically, tension in it will be different at different points. For a point at a distance x from the free end, tension will be due to the weight of the string below it .So if M is the mass of string of length L, the mass of length x of the string will be (M/L)x.

So, 
Here x 

At point x the wave travels a distance dx in time dt.
 [from eq. 1]

i.e [from eq. 2]
Hence L = 2.45 m so 
 

When the source is coming to stationary observer

When the source is moving away from stationary observer

For seconds pendulum

= T’ = 2√2 s = 2.83 s

 Comparing this with the standard equation of standing wave given as

where velocity of wave v = ω / k
We have, 
⇒  = 150 cm/s

If the astronaut (source) at speed v is a approaching the moon, the frequency ‘heard’ by the moon will be
...(1)
Now this frequency will be reflected back by the moon (now source ) towards the astronaut which is observer and moving towards the moon . So the astronaut will hear a frequency
...(2)
Substituting the value of f, from Eq (1) in (2) 

or  i.e
So substituting the given data

The sound from the back-directed speaker has Doppler frequency

where c and v are the speeds of sound and the car respectively, and v is the frequency of the sound emitted. As the wall is stationary with respect to the obseiver, v_b is also the frequency as heard by the latter. The sound from the front-directed speaker has Doppler frequency

Hence the beat frequency is

∴ R = 2 x 1.75 x 0.3 = 1.05 m

The power of the lens is given by

= (1.5 — 1) (2 / R) = 1/R
The power of each water lens (plano-concave)

= 1/3 R
The power of combination