गुरुत्वीय त्वरण g के निर्धारण के एक प्रयोग मे | vedclass

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Question

The observed values of time period, $T _{ i }=0.52 s, 0.56 s, 0.57 s, 0.54 s$ and $0.59$ s

The mean value of time period, $T =\frac{\sum T _{ i }}{

Vedclass · Accepted Answer

$A,B,D$

Vedclass · Answer

The observed values of time period, $T _{ i }=0.52 s, 0.56 s, 0.57 s, 0.54 s$ and $0.59$ s

The mean value of time period, $T =\frac{\sum T _{ i }}{5}=\frac{2.78}{5}=0.56 s$

Magnitude of absolute error in each observation,

$\left|\Delta T _1ight|=|0.56-0.52|=|0.04| s$

Similarly, $\left|\Delta T _2ight|=|0.0| s \left|\Delta T _3ight|=|0.01| s \left|\Delta T _4ight|=|0.02| s \left|\Delta T _5ight|=|0.03| s$

Mean absolute error in time period,

$\Delta T _{ m }=\frac{0.04+0.00+0.01+0.02+0.03}{5}=0.02 s$

$	herefore$ Error in $T , \frac{\Delta T _{ m }}{ T } 	imes 100=\frac{0.02}{0.56} 	imes 100=3.57 \%$

Error in the measurement of $r : \frac{\Delta r }{ r } 	imes 100=\frac{1}{10} 	imes 100=10 \%$

From the equation given, we get: $g =\frac{28 \pi^2( R - r )}{5 T^2}$

$	herefore$ Error in the measurement of $g$ :

$\frac{\Delta g }{ g } 	imes 100=\frac{\Delta R +\Delta r }{( R - r )} 	imes 100+2 \frac{\Delta T _{ m }}{ T } 	imes 100$

$\Rightarrow \frac{\Delta g }{ g } 	imes 100=\frac{1+1}{(60-10)} 	imes 100+2(3.57) \%=11.14 \%$

Thus options $A, B$ and $D$ are correct.

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