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% (use the formula of percentage error)
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Answer 1

Answer:

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

$T = 2\pi\sqrt{\dfrac{l}{g}}$

According to question

The time period of simple pendulum

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

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

Using the formula of percentage error

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

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

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

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

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

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

Answer 2

Answer:

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