Discussion Forum

In an experiment to determine the acceleration due to  gravity g, the formula used for the  time  period of a periodic motion is 

$T=2\pi\sqrt{\frac{7(R-r)}{5g}}$ . The values of R and r are measured to be (60±1) mm and   (10± 1) mm respectively. In five successive measurements. the time period is found to be 0.52s, 0.56s, 0.57s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01s, Which of the following statement(s) is (are)  true?

Answer:

Explanation:

Mean time period

$=\frac{0.52+0.56+0.57+0.54+0.59}{5}$

 =0.556=0.56 sec as per significant figures

 Error in reading=| T_mean- T₁|=0.04

$| T_{mean}-T_{2}|=0.00\Rightarrow|T_{mean}-T_{3}|=0.01$

$| T_{mean}-T_{4}|=0.02\Rightarrow|T_{mean}-T_{5}|=0.03$

 Mean error=0.1/5=0.02

% of errot in T= $\frac{\triangle T}{T}\times 100=\frac{0.02}{0.56}\times100 =3.57$ %

% error in r= $\frac{0.001 \times 100}{0.010}=10$ %

% error in R=  $\frac{0.010 \times 100}{0.600}=1.67$

% error in $\frac{\triangle g}{g}\times 100$

=$\frac{\triangle (R-r)}{R-r}\times 100+2\times\frac{\triangle T}{T}$

$=\frac{0.002\times 100}{0.05}+2\times 3.57$

= 4%+7%=11%