In an experiment to find acceleration due to gravity g using simple pendulum, time period of 0 . 5 s is measured from time of 100 oscillation with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy. The accuracy in the determination of g is found to be x % . The value of x is

Question

In an experiment to find acceleration due to gravity g using simple pendulum, time period of 0 . 5 s is measured from time of 100 oscillation with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy. The accuracy in the determination of g is found to be x % . The value of x is,

MARKS App · Accepted Answer

Time period of oscillation is given by T = 2 π l g , here, l is length and g is acceleration due to gravity. From above relation, we have g = 1 4 π 2 T 2 l . Fractional error in g is stated as Δ g g = 2 Δ T T + Δ l l ⇒ Δ g g × 100 = 2 × 1 100 × 0 . 5 × 100 + 1 mm 10 cm × 100 ⇒ Δ g g × 100 = 5 100 × 100 = 5 % Hence, value of x = 5 .

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