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While measuring acceleration due to gravity by a simple pendulum, a student makes a positive error of 2% in the length of the pendulum and a positive error of 1% in the value of time period. His actual percentage error in the measurement of the value of g will be

a
3%
b
0%
c
4%
d
5%

detailed solution

Correct option is B

T=2πLg or T2=4π2Lgg=4π2LT2;Δgg=ΔLL−2ΔTT Δgg×100=ΔLL×100−2ΔTT×100 Actual % error in g=ΔLL×100−2ΔTT×100=+2%−2×1%=0%

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