Test Index

Oscillations Part 1

Section: Physics

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Question : 16 of 100

Marks: +1, -0

The time period of a simple pendulum is given by

T=2\pi\sqrt{\frac{1}{g}}

. The measured value of the length of pendulum is

10\ \text{cm}

1\ \text{mm}

accuracy. The time for 200 oscillations of the pendulum is found to be 100 s using a clock of 1 s resolution. The percentage accuracy in the determination of

g

using this pendulum is

x

x

to the nearest integer is

[18 Mar 2021 Shift 1]

$2\%$
$3\%$
$5\%$
$4\%$

Solution:

Given, the measured value of the length of pendulum,

I=10\ \text{cm}

The time taken to complete the 200 oscillations,

T=100\ \text{s}

Time resolution,

\Delta T=1\ \text{s}

Accuracy in length,

\Delta l=1\ \text{mm}=0.1\ \text{cm}

We know that,

T=2\pi\sqrt{\;\frac{l}{g}}

\Rightarrow\;\; T^{2}=4\pi^{2}\;\frac{I}{g} \Rightarrow g=\;\frac{4\pi^{2}I}{T^{2}}

\Rightarrow\;\; \frac{\Delta g}{g}=\;\frac{\Delta l}{l}+2\;\frac{\Delta T}{T}=\;\frac{0.1}{10}+2\;\frac{1}{100}

=0.01+0.02=0.03

\Rightarrow\;\; \frac{\Delta g}{g}\times 100=0.03\times 100

=3\%

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