Properties of Matter - Problem - Video

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http://www.youtube.com/watch?v=IdJyrbQWHPo

Current electricity problem - Youtube video - Exp-Edu

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http://www.youtube.com/watch?v=tRE-T-ZCNMo

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