Question

2.
A mass falls from a height 'h' and its time of fall
't is recorded in terms of time period T of a
simple pendulum. On the surface of earth it is
found that t = 2T. The entire set up is taken on
the surface of another planet whose mass is half
of that of earth and radius the same. Same
experiment is repeated and corresponding times
noted as ť and T'. [NEET-2019 (Odisha)]
Then we can say
(1) ť = 2T'
(2) t' = VET
(3) t'> 2T
(4) t' <2T​

Answer 1

Answer:

1.t'=2T'

Explanation:

We are given that

Height =h

On the surface of earth ,t=2 T

Let mass of earth=M

Mass of another planet=$\frac{1}{2}M$

Radius of earth=Radius of another planet=R

We have to find t' when the experiment is repeated.

We know that at the surface of earth,t=$\sqrt{\frac{2h}{g}}$

and T=$2\pi \sqrt{\frac{l}{g}}$

t=2T(Given )

For the surface of another planet , g'=$\frac{g}{2}$

because mass of another planet is half of that of earth.

Time taking  in falling h distance

$t'=\sqrt{\frac{2h}{g'}}=\sqrt{\frac{4h}{g}}=\sqrt 2t$

$T'=2\pi\sqrt{\frac{l}{g'}}=2\pi\sqrt{\frac{2l}{g}}=\sqrt2 T$

$\frac{t'}{T'}=\frac{\sqrt 2 t}{\sqrt 2 T}=\frac{2T}{T}=2$

Hence, t'=2T'

Answer:1.t'=2T'

Answer 2

Explanation:

See the picture

ask if you have doubt