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While measuring acceleration due to gravity by simpe pendulum a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of the time period. His percentage error in the measurement of the value of g will be -A. `5%`B. `7%`C. `4%`D. `2%` |
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Answer» Correct Answer - B `T=2pil^(1//2) g^(-1//2)` `g=4pi^2T^(-2) l % (Deltag)/g=(2DeltaT)/T+(Deltal)/l` `%Deltag=2% DeltaT+%l` `%Deltag`=6 +1 = 7% |
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