Q. The relative density of material of a body is found by weighing it first in air and then in water. If the weight in air is ($5.00 \pm 0.05$) $newton$ and weight in water is $(4.00 \pm 0.05)\ newton\ .$Then the relative density along with the maximum permissible percentage error is
A) $5.0 \pm 11\%$
B) $5.0 \pm 1\%$
C) $5.0 \pm 6\%$
D) $1.25 \pm 5\%$
Explanation
Weight in air =$\ (5.00 \pm 0.05)N$ Weight in water =$(4.00\ \pm 0.05)N$ Loss of weight in water =$(1.00 \pm 0.1)N$ Now relative density=$\frac{weight\ in\ air}{weight\ loss\ in\ water}$ $i.e.\ R.D = \frac{5.00 \pm 0.05}{1.00 \pm 0.1}$ Now relative density with max permissible error $= \frac{5.00}{1.00} \pm \left( \begin{array}{r} \begin{array}{r} \frac{0.05}{5.00} \\ \ \end{array} + \frac{0.1}{1.00} \\ \ \end{array} \right) \times 100 = 5.0 \pm (1 + 10)\%$ $= 5.0 \pm 11\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.