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I. Multiple Choice Questions · Q3

Q.If the length and time period of an oscillating pendulum have errors of 1% and 3% respectively then the error in measurement of acceleration due to gravity is [Related to AIPMT 2008]

(a) 4%
(b) 5%
(c) 6%
(d) 7%
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Step 1. Simple pendulum: T=2πl/g⇒g=4π2lT2T=2\pi\sqrt{l/g}\Rightarrow g=\dfrac{4\pi^2l}{T^2}.

Step 2. Applying the propagation rule for a quotient with a power: Δgg=Δll+2ΔTT\dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+2\dfrac{\Delta T}{T} (4π24\pi^2 is an exact constant, contributing no error).

Step 3. Given Δll×100=1%\dfrac{\Delta l}{l}\times100=1\% and ΔTT×100=3%\dfrac{\Delta T}{T}\times100=3\%: Δgg×100=1%+2×3%=1%+6%=7%\dfrac{\Delta g}{g}\times100=1\%+2\times3\%=1\%+6\%=7\%.

✓Final answer

(d) 7%

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