## Thursday, May 09, 2013

### NEET 2013 Questions (MCQ) on Rotational Motion

God used beautiful mathematics in creating the world.

The following two questions on rotational motion, which were included in the NEET 2013 question paper, are worth noting:

(1) A small object of uniform density rolls up a curved surface with an initial velocity “v”. It reaches up to a maximum height of 3v2/4g with respect to the initial position. The object is

(1) solid sphere

(2) hollow sphere

(3) disc

(4) ring

Initially the object has rotational and translational kinetic energy but zero gravitational potential energy. At the maximum height of 3v2/4g it has zero kinetic energy since the entire kinetic energy is converted into gravitational potential energy. Thus we have

½ I ω2 + ½ mv2 + 0 = mg(3v2/4g) where ‘I’ is the moment of inertia of the object about the axis of rotation, ‘ω’ is the angular velocity, ‘m’ is the mass and ‘v’ is the linear velocity of the object.

Simplifying, ½ I ω2 = ¼  mv2

Since ω =v/R the above equation becomes

Iv2/R2 = ½ mv2

Therefore I = mR2/2

This means that the object is a disc [Option (3)].

(2) A rod PQ of M  and length L is hinged at end P. The rod is held horizontally by a massless string tied to point Q as shown in the figure. When the string is cut, the initial acceleration of the rod is (1) g/L

(2) 2g/L

(3) 2g/3L

(4) 3g/2L

When the string is cut, the rod rotates about the end P and the torque responsible for the rotation is MgL/2.

[The weight Mg of the rod acts through the centre of gravity located at the middle of the rod. The distance of the line of action of the weight from the axis of rotation (the lever arm for the torque) is L/2]

Since torque is equal to Iα where I is the moment of inertia and α is the angular acceleration we have

Iα = MgL/2 ……………..(i)

The moment of inertia of the rod about the axis of rotation through its end is ML2/3 as given by the parallel axes theorem.

[Moment of inertia of a uniform rod about a central axis perpendicular to its length is ML2/12. Moment of inertia about a parallel axis through the middle is ML2/12 + M(L/2)2 = ML2/3].

Substituting for I in equation (i),  we have

(ML2/3)α = MgL/2

Therefore α = 3g/2L

## Sunday, May 05, 2013

### JEE Main 2013 Questions on Geometric Optics

Science distinguishes a Man of Honor from one of those Athletic Brutes whom undeservedly we call Heroes.
– John Dryden (English poet, critic, dramatist)

The following questions on geometrical optics appeared in JEE Main 2013 question paper:
(1) Diameter of a plano - convex lens is 6 cm and thickness at the centre is 3mm. If speed of light in material of lens is 2 × 108 m/s, the focal length of the lens is :
(1) 15 cm
(3) 30 cm
(4) 10 cm

In the adjoining figure the plano-convex lens is represented by ABCDA. The thickness of the lens is BD (which is equal to 0.3 cm) and the radius of the curved surface of the lens is R.
We have
R2 = (R – 0.3)2 + 32
This gives R ≈ 15 cm.
The refractive index n of the material of the lens is the ratio of the speed of light in free space to the speed in the material of the lens:
n = (3×108) /(2×108) = 1.5
The focal length f of the lens is given by the lens maker’s equation.
1/f = (n – 1)(1/R  – 0)
[Since the second surface is plane, the reciprocal of its radius of curvature is zero]
Therefore 1/f = (1.5 – 1)(1/15) from which f = 30 cm
(2) The graph between angle of deviation (δ) and angle of incidence (i) for a triangular prism is represented by :

This is a very simple question. Most of you might have drawn the i - δ  curve by experimentally determining the values of δ for various values of i during your lab session. On increasing the angle of incidence i from a small value, the angle of deviation δ  decreases, attains a minimum value and then increases. The non-linear increase and decrease are in accordance with the relation,
δ = i + e A where e is the angle of emergence and A is the angle of the prism.
The correct option is (3).