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 $W_1=8.00\pm 0.05$N and the weight in water is $W_2=6.00\pm 0.05$N, then the relative density ${\rho }_r=\frac{W_1}{W_1-W_2}$ with the maximum permissible error is?

• A. $4.00\pm 0.62\%$
• B. $4.00\pm 0.82\%$
• C. $4.00+3.2\%$
• D. $4.00\pm 5.62\%$

Answer 1

The correct option is D $4.00\pm 5.62\%$

Weight in air $\left(W_1\right)=8.00\pm 0.05$ N

Weight in water $\left(W_2\right)=6.00\pm 0.05$N

Loss of weight in water $(W_1-W_2)=2.00\pm 0.1$N

Now, Relative Density $(R.D.)=\frac{W_1}{W_1-W_2}$

$R.D.=\frac{8.00\pm 0.05}{2.00\pm 0.1}$

Relative density with maximum permissible error is

$=\frac{8}{2}\pm (\frac{0.05}{8}+\frac{0.1}{2})\times 100$

$=4\pm (0.62+5)$%

$=4.00\pm 5.62$ %