Physics And Measurement Question 197
Question: The relative density of a material of a body is found by weighing it first in air and then in water. If the weight of the body in air is $ W _{1}^{{}} $ = $ 8.00\pm 0.05N $ and the weight in water is $ W _{2}^{{}} $ = $ 6.00\pm 0.05 $ N, then the relative density $ \rho _{r}^{{}}=W _{1}^{{}}/(W _{1}^{{}}-W _{2}^{{}}) $ With the maximum permissible error is
Options:
A) $ 4.00\pm 0.62 $ %
B) $ 4.00\pm 0.82 $ %
C) $ 4.00\pm 3.2 $ %
D) $ 4.00\pm 5.62 $ %
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Relative density $ {\rho _{r}}=\frac{W _{1}}{W _{1}-W _{2}}=\frac{8.00}{8.00-6.00}=4.00 $
$ \frac{\Delta {\rho _{r}}}{{\rho _{r}}}\times 100=\frac{\Delta W _{1}}{W _{1}}\times 100+\frac{\Delta (W _{1}-W _{2})}{W _{1}-W _{2}}\times 100 $
$ =\frac{0.05}{8.00}\times 100+\frac{0.05+0.05}{2}\times 100=5.62 $ %
$ \therefore {\rho _{r}}=4.00\pm 5.62 $ %
