A body weighs \(200\text{ N}\) on the surface of the earth. How much will it weigh halfway down to the centre of the earth?

1. \(100\text{ N}\)

2. \(150\text{ N}\)

3. \(200\text{ N}\)

4. \(250\text{ N}\)

What is the depth at which the value of acceleration due to gravity becomes \(1/n\) times the value that at the surface of the earth? (radius of earth = \(R\))

1. \(\dfrac{R}{n^2}\)

2. \(\dfrac{R(n-1)}{n}\)

3. \(\dfrac{Rn}{(n-1)}\)

4. \(\dfrac{R}{n}\)

The dependence of acceleration due to gravity \('g'\) on the distance \('r'\) from the centre of the earth, assumed to be a sphere of radius \(R\) of uniform density, is as shown in figure below:

(a)		(b)
(c)		(d)

                    
The correct figure is:
1. \(a\)

2. \(b\)

3. \(c\)

4. \(d\)

Two girls are standing on the edge of a building tossing coins over the edge. Alice is dropping her coins, each of which is \(10~\text{gm}.\) Barbara is tossing her coins horizontally at \(0.3~\text{m/s},\) and her coins are \(40~\text{gm}\) each (see figure). Ignore air resistance.

The height from the surface of the earth at which the value of \(g\) becomes one-fourth of that on the earth’s surface will be:
(\(R\) is the radius of the earth)