JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Units & Measurement

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 _____ %.

Answer: 14

Solution

The energy supplied to set a pendulum swinging through a fixed angular amplitude

θ_{0}

E = m g ℓ (1 - cos θ_{0})

, so for a given bob and amplitude

E \propto g ℓ

.
The length is fixed by the measured period, since

T = 2 π \frac{ℓ}{g}

gives

ℓ = \frac{g T ^{2}}{4 π ^{2}}

, and therefore

E \propto g \cdot g T^{2} = g^{2} T^{2}

The fractional errors add:

\frac{Δ E}{E} = 2 \frac{Δ g}{g} + 2 \frac{Δ T}{T} = 2 (4%) + 2 (3%) = 14%

The energy is known to

14%

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Units & Measurement

Solution

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