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The acceleration due to gravity is found upto 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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alok kumar singh | Contributor-Level 10

2 months ago

$L e t, E \propto m^{a} g^{b} T^{c}$

So, $[E] = {[m]}^{a} {[g]}^{b} {[T]}^{c}$

$\Rightarrow M L^{2} T^{- 2} = M^{a} {(L T^{- 2})}^{b} T^{c}$

Thus, E = k mg² T²

$\frac{Δ E}{E} = 2 . \frac{Δ g}{g} + 2 . \frac{Δ T}{T}$

Percentage error in E = 2 × 4 + 2 × 3 = 14%

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The acceleration due to gravity is found upto 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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