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Question

In a resonance tube experiment, resonance occurs first at 32.0 cm and then at 100.0 cm. What is the percentage error in the measurement of end correction?

A
2.5 %
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B
5 %
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C
7.5 %
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D
10 %
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Solution

The correct option is D 10 %
We know that,
e=l33l12=2.0 cm
and
de=dl3+3dl12
% error=dl3+3dl12×e×100

As the values of l1 and l2, are given with 1 significant figure after decimal point, their least count will be 0.1 cm. So,
dl1=dl2=0.1 cm

% error=0.1+3×0.12×2.0×100=10 %

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