Mathematics in Physics

JEE Main level, numerical answer, medium

86 of 121

Question 86 of 121

The acceleration due to gravity is found up to an accuracy of 44% on a planet. The energy supplied to a simple pendulum of known mass mm to undertake oscillations of time period TT is being estimated. If time period is measured to an accuracy of 33%, the accuracy to which EE is known as ________%

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Answer: 14

T=2πlgT=2\pi \sqrt{\dfrac{l}{g}}
Δll×100=(Δgg×100)+(2ΔTT×100)\dfrac{\Delta l}{l}\times 100=\left(\dfrac{\Delta g}{g}\times 100\right)+\left(\dfrac{2\Delta T}{T}\times 100\right)
⇒Δll×100=4+(2×3)=10\Rightarrow \dfrac{\Delta l}{l}\times 100=4+\left(2\times 3\right)=10%
E=12mω2A2=12mglA2Therefore,ΔEE×100=(Δgg×100)+(Δll×100)=4+10=14E=\dfrac{1}{2}m{\omega }^{2}{A}^{2}=\dfrac{1}{2}m\dfrac{g}{l}{A}^{2} \text{Therefore},\dfrac{\Delta E}{E}\times 100=\left(\dfrac{\Delta g}{g}\times 100\right)+\left(\dfrac{\Delta l}{l}\times 100\right)=4+10=14%

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