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While measuring acceleration due to gravity bysimple pendulum a student makes a positive error of1% in the length of the pendulum,and a negative errorof 3% in the value,of the time period. His percentageerror in the measurement of the value of g will be-a |
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Answer» time period is given by T = 2π√l/gfor finding error in case of g is given byΔg/g = Δl/l + 2ΔT/T ⇒ Δg/g × 100 = Δl/l × 100 + 2 × ΔT/T × 100⇒ % error in g = % error in l + 2 × % error in T⇒ % error in g = 1 % + 2 × 3 % = 1% + 6% = 7%hence the error in measurement of value of g is 7% |
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