In a simple pendulum experiment for determination of acceleration due to gravity $(g)$ , time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be $30 s$ . The length of pendulum is measured by using a meter scale of least count $1 m m$ and the value obtained is $55.0 c m$ . The percentage error in the determination of $g$ is close to

0.7 %

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6.8 %

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3.5 %

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0.2 %

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Solution

The correct option is B $6.8 %$
Given:
Time period, $T = \frac{30 s e c}{20}$ , $Δ T = \frac{1}{20} sec$
$L = 55 c m, Δ L = 1 m m = 0.1 c m$

As we know,
$T = 2 π \sqrt{\frac{L}{g}} . . . (i)$

From equation (i) we get,
$g = \frac{4 π^{2} L}{T^{2}}$

Percentage error in 𝑔 is

$\frac{Δ g}{g} \times 100 = (\frac{Δ L}{L} + \frac{2 Δ T}{T}) 100$

$\frac{Δ g}{g} \times 100 = (\frac{0.1}{55} + \frac{2 (\frac{1}{20})}{\frac{30}{20}}) \times 100$

$\frac{Δ g}{g} % ≃ 6.8$ %

Final Answer: (d)

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Q. Time period of 10 oscillations of a simple pendulum is found to be 20 sec by a stopwatch of resolution 0.25 sec. Length of the string is measured to be 30 cm by a scale of least count 0.1 cm. Percentage error in measurement of acceleration due to gravity g by simple pendulum is?

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The correct option is B 6.8%Given: Time period, T=30 sec20, ΔT=120 sec L=55 cm, ΔL=1 mm=0.1 cm As we know, T=2π√Lg...(i) From equation (i) we get, g=4π2LT2 Percentage error in 𝑔 is Δgg×100=(ΔLL+2ΔTT)100 Δgg×100=(0.155+2(120)3020)×100 Δgg%≃6.8% Final Answer: (d)