Question

In a simple pendulum, length increases by 4% and
g increases by 2%, then time period of simple
pendulum
(1) Increases by 4% (2) Increases by 3%
(3) Decreases by 3% (4) Increases by 1%​

Answer 1

Explanation:

The time period of simple pendulum is Increases by 1%.

Explanation:

Given that,

The length increases by 4% and g increases by 2%.

Using the formula of time period of simple pendulum

T = 2\pi\sqrt{\dfrac{l}{g}}T=2π

g

l

According to question

The time period of simple pendulum

T'=2\pi\sqrt{\dfrac{1.04l}{1.02g}}T

′

=2π

1.02g

1.04l

T'=\sqrt{\dfrac{1.04}{1.02}}TT

′

=

1.02

1.04

T

Using the formula of percentage error

\dfrac{\DeltaT}{T}=\dfrac{T'-T}{T}

T

\DeltaT

=

T

T

′

−T

\dfrac{\Delta T}{T}=\dfrac{\sqrt{\dfrac{1.04}{1.02}}T-T}{T}

T

ΔT

=

T

1.02

1.04

T−T

\dfrac{\Delta T}{T}=\sqrt{\dfrac{1.04}{1.02}}-1

T

ΔT

=

1.02

1.04

−1

\dfrac{\Delta T}{T}=0.009

T

ΔT

=0.009

\dfrac{\Delta T}{T}\times100=0.9 = 1\%

T

ΔT

×100=0.9=1%