JEE Advanced 2016 · Paper 2 · Physics Q9

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PhysicsMultiple correct+4 / -2Question 9 of 54

In an experiment to determine the acceleration due to gravity $g$ , the formula used for the time period of a periodic motion is $T=2\pi\sqrt{\dfrac{7(R-r)}{5g}}$ . The values of $R$ and $r$ are measured to be $(60\pm1)$ mm and $(10\pm1)$ mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true?

The error in the measurement of

r

is 10%

The error in the measurement of

T

is 3.57%

The error in the measurement of

T

is 2%

The error in the determined value of

g

is 11%

Answer and worked solution

Official answer: A, B, D

(A) $\Delta r/r=1/10=10\%$ . True.

Time period. The mean is $\bar T=2.78/5=0.556\approx0.56$ s. The deviations are 0.04, 0.00, 0.01, 0.02 and 0.03, with mean 0.02 s (no smaller than the least count). So
$\frac{\Delta T}T=\frac{0.02}{0.56}=3.57\%.$
(B) is true and (C) is false.

(D) $g\propto\dfrac{R-r}{T^2}$ , with $\Delta(R-r)=2$ mm on 50 mm, which is 4%. So
$\frac{\Delta g}g=4\%+2\times3.57\%\approx11\%.$
True.

(A), (B), (D)

Topics:Errors in experiments Errors in measurement

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JEE Advanced 2016 · Paper 2, Physics question 9