**Question1:**

A charge Q is divided in two parts q and Q - q separated by a distance R. If force between the two charges is maximum, find the relationship between q & Q.

Solution: F=K q(Q-q)/r

^{2}

for max. &min.dF/dq=0 , q=Q/2

**Question2**:A circular disc X of radius R is made from an iron pole of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness t/4. then the relation between the moment of inertia I

_{x}and I

_{y}is

(A) I

_{y}=32I

_{x}

(B) I

_{y}=16I

_{x}

(C) I

_{y}=64I

_{x}

(D) None of these

**Solution2:**

If t is the thickness and R is the radius of the disc, then mass = πR

^{2}tρ

ρ = density of the material of the disc.

Moment of inertia of disc X,

I

_{x}=(1/2)πR

^{4}tρ

…(i)

Moment of inertia of disc Y,

I

_{y}=32πR

^{4}tρ

…(ii)

From equation (i) and (ii)

I

_{y}=64I

_{x}

**Question3:**

On a planet a freely falling body takes 2 sec when it is dropped from a height of 8 m, the time period of simple pendulum of length 1 m on that planet is

**Solution3:**

On a planet, if a body dropped initial velocity (u = 0) from a height h and takes time t to reach the ground

then h=(1/2)g

_{p}t

^{2}

or g

_{p}=4m/s

^{2}

Using

T=2π(L/g)

^{1/2}

T=3.14 sec

**Question4:**

A car is moving with uniform velocity on a rough horizontal road. Therefore, according to Newton's first law of motion

(a) No force is being applied by its engine

(b) An acceleration is being produced in the car

(c) The kinetic energy of the car is increasing

(d) force is surely being applied by its engine

**Solution4:**d Since, force needed to overcome frictional force

**Question 5:**

A particle is projected upwards. The times corresponding to height h while ascending and while descending are a and b respectively.Find the velocity of projection

**Solution:**

If a and b are time of ascent and descent respectively then time of flight

T=a+b

Now T=2u/g

So u=g(a+b)/2

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