Question

21. In a simple pendulum, length increases by 4% and
g increases by 2%, then time period of simple
pendulum
​

Answer 1

Explanation:

we know time period of simple pendulum is

T= 2π√(l/g)

T is proportional to √(l/g)

T1, l1 and g1 be the respective quantities for original pendulum

and T2, l2 and g2 be the respective quantities for other pendulum

Given, l2= l1+l1×4/100= l1+0.04×l1

l2= 1.04×l1

l2/l1=1.04

Now, g2= g1 + g1×2/100= g1+ g1×0.02

g2= 1.02×g1

g2/g1= 1.02

Now take ratios of time periods

T2/T1= √[l2×g1/(l1×g2)]

Putting above values

T2= √(1.04/1.02)T1

This is required answer.