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(B) The observed values of time period are $T_1=0.52 	ext{ s}, T_2=0.56 	ext{ s}, T_3=0.57 	ext{ s}, T_4=0.54 	ext{ s}, T_5=0.59 	ext{ s}$.The me

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$A, B, C$

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(B) The observed values of time period are $T_1=0.52 	ext{ s}, T_2=0.56 	ext{ s}, T_3=0.57 	ext{ s}, T_4=0.54 	ext{ s}, T_5=0.59 	ext{ s}$.The mean value of time period is $T = \frac{0.52+0.56+0.57+0.54+0.59}{5} = \frac{2.78}{5} = 0.556 	ext{ s} \approx 0.56 	ext{ s}$.The mean absolute error in time period is $\Delta T_m = \frac{|0.56-0.52| + |0.56-0.56| + |0.56-0.57| + |0.56-0.54| + |0.56-0.59|}{5} = \frac{0.04+0+0.01+0.02+0.03}{5} = \frac{0.10}{5} = 0.02 	ext{ s}$.Percentage error in $T = \frac{\Delta T_m}{T} 	imes 100 = \frac{0.02}{0.56} 	imes 100 \approx 3.57 \%$. Thus,$(B)$ is correct.Percentage error in $r = \frac{\Delta r}{r} 	imes 100 = \frac{1}{10} 	imes 100 = 10 \%$. Thus,$(A)$ is correct.From $T^2 = \frac{4 \pi^2 \cdot 7(R-r)}{5g}$,we get $g = \frac{28 \pi^2 (R-r)}{5 T^2}$.The relative error in $g$ is $\frac{\Delta g}{g} = \frac{\Delta R + \Delta r}{R-r} + 2 \frac{\Delta T_m}{T}$.Percentage error in $g = \left( \frac{1+1}{60-10} ight) 	imes 100 + 2 	imes 3.57 \% = \frac{2}{50} 	imes 100 + 7.14 \% = 4 \% + 7.14 \% = 11.14 \% \approx 11 \%$. Thus,$(D)$ is correct.Therefore,statements $(A)$,$(B)$,and $(D)$ are true.

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