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If the length of a simple pendulum is recorded as `(90 +- 0.2)cm` and period as `(1.9 +- 0.02)s`, the percentage fo error in the measurement of acceleration due to gravity isA. `4.2`B. `2.1`C. `1.5`D. `2.8` |
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Answer» Correct Answer - B `g = 4pi^(2) (l)/(T^(2)), (Delta g)/(g) xx 100 = (Delta l)/(l) xx 100 + (2Delta T)/(T) xx 100` |
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