# A mass of M kg is suspended by a weightless string . The horizontal force that is required to displace it until the string making an angle of 45o  with the initial vertical direction is Option 1) Option 2) Option 3) Option 4)

As we learnt in

Net work done by all the forces give the change in kinetic energy -

$W=\frac{1}{2}mv^{2}-\frac{1}{2}mv{_{0}}^{2}$

$W= k_{f}-k_{i}$

- wherein

$m=mass \: of\: the\: body$

$v_{0}= initial\: velocity$

$v= final\: velocity$

Height rises by the ball  $=L-\frac{L}{\sqrt{}2}$

Therefore conservation of energy $=\frac{1}{2}mv^{2}=mg.(L-\frac{L}{\sqrt{}2})$

Since kinetic energy change = Work done

Therefore work done $=F.\frac{L}{\sqrt{}2}=mg(L-\frac{L}{\sqrt{}2})$

Therefore $F=mg(\sqrt{}{2-1})$

Option 1)

This is the correct option.

Option 2)

This is an incorrect option.

Option 3)

This is an incorrect option.

Option 4)

This is an incorrect option.

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
##### Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-