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A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is δT=0.01 seconds and he measures the depth of the well to be L=20 meters. Take the acceleration due to gravity g=10ms2 and the velocity of sound is  300ms1. Then the fractional error in the measurement,δL/L , is closest to

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a
1%
b
5%
c
3%
d
0.2%

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detailed solution

Correct option is A

2Lg+LV=T After differentiating, ⇒2gdL2L+0=dT dLV≈0⇒12gdLL=dT After integrating, ⇒L=12gT2⇒T=2Lg=2×2010=2sA/SO,dLL=2dTT=2×0.012=0.01⇒dLL×100=1%


Similar Questions

To estimate 'g'  from g=4π2LT2, error in measurement of L is ±2% and error in measurement of T is ±3%. The error in estimated 'g' will be:


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