T=2πgl
lΔl×100=(gΔg×100)+(T2ΔT×100)
⇒lΔl×100=4+(2×3)=10
E=21mω2A2=21mlgA2Therefore,EΔE×100=(gΔg×100)+(lΔl×100)=4+10=14
The acceleration due to gravity is found up to an accuracy of 4 on a planet. The energy supplied to a simple pendulum of known mass m to undertake oscillations of time period T is being estimated. If time period is measured to an accuracy of 3, the accuracy to which E is known as ________
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