Physics And Measurement Question 180
Question: In an experiment, the following observation’s were recorded : L = 2.820 m, M = 3.00 kg, l = 0.087 cm, Diameter D = 0.041 cm Taking g = 9.81 $ m/s^{2} $ using the formula , Y= $ \frac{4MgL}{\pi D^{2}l} $ , the maximum permissible error in Y is
Options:
A) 7.96%
B) 4.56%
C) 6.50%
D) 8.42%
Show Answer
Answer:
Correct Answer: C
Solution:
$ Y=\frac{4MgL}{\pi D^{2}l} $ so maximum permissible error in Y = $ \frac{\Delta Y}{Y}\times 100=( \frac{\Delta M}{M}+\frac{\Delta g}{g}+\frac{\Delta L}{L}+\frac{2\Delta D}{D}+\frac{\Delta l}{l} )\times 100 $
$ =( \frac{1}{300}+\frac{1}{981}+\frac{1}{2820}+2\times \frac{1}{41}+\frac{1}{87} )\times 100 $
$ =0.065\times 100=6.5$ %
