Questions in JEE 2008 on Young's Experiments

1. In a Young’s double slit experiment, the separation between the two slits is d and the wave length of the light is λ. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).

(A) If d = λ, the screen will contain only one maximum.

(B) If λ < d < 2λ, at least one more maximum (besides the central maximum) will be observed on the screen.

(C) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase.

(D) If the intensity of light falling on slit 2 is reduced so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase


2. Suppose the wave length λ and the double slit separation d in a Young’s double slit experiment are such that the 6th dark fringe is obtained at point P shown in the figure. The path difference (S₂P – S₁P) will be

(a) 5 λ

(b) 5 λ/2

(c) 6 λ

(d) 3 λ

(e) 11 λ/2



Visit for the answers

http://physicsplus.blogspot.com/2008/09/iit-jee-2008-question-mcq-on-youngs.html

Spring - MQ

An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. The maximum extension in the spring is

a. 4Mg/k
b. 2Mg/k
c. Mg/k
d. Mg/2k

(JEE 2002)

Answer (b)

Solution

Change in gravitational energy = Energy stored in the spring

Mgx = ½ kx²

=> x = 2Mg/k

Kepler’s Laws - MQ

According to Kepler’s second law, the radius vector to a plant form the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of -----------------.

(JEE 1985)

Answer: Angular momentum

Collisions - Model questions - JEE

Collisions – Psat JEE Problems

1. A ball hits the floor and rebounds after an inelastic collision. In this case

a. the momentum of the ball just after the collision is the same as that just before the collision.
b. the mechanical energy of the ball remains the same in the collision.
c. the total momentum of the ball and the earth is conserved.
d. the total energy of the ball and the earth is conserved.

(More than one alternative may be correct)
JEE 1986

Answer ( c ) and (d)

Note: total kinetic energy is not conserved but total energy is conserved.

2. A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m, are found to move with a speed v, each in mutually perpendicular directions. The total energy released in the process of explosion is _________ .


(JEE 1987)

answer (3/2)mv²

Solution;

As the two fragments went in mutual perpendicular directions with velocity v the resultant momentum

2mv cos 45° = 2mv(1/√2) = √2 mv

if the velocity of fragment with 2m mass is V

then 2 mV =/√2mv

V = v/√2

Hence energy released = ½ mv² + ½ mv² + ½ *2 m (v/√2) ²

= (3/2)mv²

Moment of inertia - Model Problems

1. A circular disc of radius r/3 is removed from the outer-edge of a bigger circular disc of mass 9M and radius r. the moment of inertia of the remaining portion of the disc about the centre O of the disc perpendicular to the plane of the disc is

a. 4Mr²
b. (37/9)Mr²
c. (40/9)Mr²
d. 9Mr²

JEE 2005

Answer (a)

Solution

Mass of the disc removed

Total mass – M
Total area = π r²
Area of piece removed π (r/3) ²

Hence mass of the disc removed = M[π (r/3) ²]/( π r²) = M

Moment of inertia of removed about its own centre is = ½ (M)*(r/3) ²

Parallel axis theorem

Moment of inertia of a body about an axis parallel to its centre of mass is equal to
Moment of the body about the axis passing through its centre of mass plus Mr² (where M is the mass of body and r is the perpendicular distance between two axes.

According to the parallel axis theorem

Moment of inertia of the removed piece about an axis passing through the centre of the big disc = ½ (M)*(r/3) ² + M(2r/3) ²
=1/2(M) (r²/9) + M(4r²/9) = (½)(M) (r²/9 + 8r²/9) = (½)Mr²

Moment of inertia of bigger disc before cutting the piece

= ½ (9M)(r²)

Hence moment of inertia after removal of the piece
=( ½) (9M)(r²) - (½)Mr² = 4Mr²

Work and Energy Past JEE Questions

1. A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to

a. t^1/2
b. t^3/4
c. t^3/2
d. t²

(1984)

2. A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is:

a. MgL
b. MgL/3
c. MgL/9
d. MgL/18

(1985)

3. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:

a. its velocity is constant.
b. its acceleration is constant
c. its kinetic energy is constant
d. it moves in a circular path

(1987)

4. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with t as ac = k²rt² where k is a constant. The power delivered to the particle by the force acting on it is

a. 2 πmk²r²t
b. mk²r²t
c. (mk⁴r²t⁵)/3
d. zero

(1994)

5. Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of

a. 0.42 m from mass of 0.3 kg
b. 0.7 m form mass of 0.7 kg
c. 0.98 m from mass of 0.3 kg
d. 0.98 m from mass of 0.7 kg

(1995)

6. A force F = -K(yi + xj) (where K is a positive constant) acts on a particle moving in the xy plane. Starting from the origin, the particle is taken along the positive x axis to the point (a,0) and then parallel to the y axis to the point (a,a). The total work done by the force F on the particle is

a. -2Ka²
b. 2Ka²
c. -Ka²
d. Ka²

(1998)

solutions some time later

JEE past questions - Centre of Mass

1. Two particles initially at rest move toward eahc other under a mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2 V, then the speed at the centre of mass of the system is

a. zero
b. 1.5 V
c. V
d. 3 V

(1982)

2. State whether the following statement is true or false.

Two particles of mass 1 kg and 3 kg move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is 2m/s, their centre of mass has a velocity of 0.5 m/s. When the relative velocity of approach becomes 3 m/s, the velocity of the centre of mass is 0.75 m/s.

(1989)