Question

A body of weight 10 N attached to one end of a
string of length lm is rotated in a vertical plane. At
the instant when the string makes an angle of 60°
with the downward vertical, the speed of the bob is
V9.8 ms. The tension in the string at that instant is
equal to
(a) 10 N
(c) 20 N
(b) 15 N
(d) 25 N​

Answer 1

Given :

Weight of body = 10 N

Angle with downward vertical = 60°

Speed of bob v = 9.8 meter per second.

Length of string r = 1m.

To find :

Tension in the string T.

Solution :

It is given that the weight of body = 10 N

Angle with downward vertical = 60°

It means the angle between upward vertical through the weight and the direction of string will also be equal to 60°.

So at the 60° angle.

The force on the string is Tension and

downward force is gravitational force.

When we make components of tension in vertical and transverse direction.

Weight of body is made in equilibrium with the Vertical component of tension.

From figure we can see that

Vertical component of tension T = Tcosθ

As

vertical component is equal tk weight of body

So

Tcosθ = 10 N (weight of body)

$T \cos \theta \: = 10 \\ T \cos 60 = 10 \\ T \frac{1}{2} = 10 \\ so \\ T = 20$

So the tension on wire = 20 N

(Second option)

Answer 2

Answer:

1. cos theta equals to 10 Newtons
2. Explanation:
3. Ti Cos 60 equals to 220