Question

A metal ball of density $7800kg/m^3$ is suspected to have a large number of liner cavities. It weighs $9.8kg$ when weighed directly on a balance and $1.5kg$ less when immersed in water. The fraction by volume of the cavities in the metal bell is approrximately:

• A. 20%
• B. 30%
• C. 16%
• D. 11%

Answer 1

The correct option is D 11%

$\begin{array}{c}dvg=9.8kg.........\left(1\right)\\ dvg-5v^{\prime }g=8.3kg..............\left(2\right)\\ \Rightarrow \rho v^{\prime }g=1.5kg\\ 1000.v^{\prime }\times 10=1.5kg\\ v^{\prime }=\frac{1.5}{{10}^4}\\ also,7800\times v\times 10=9.8\\ v=\frac{9.8}{78\times {10}^3}\\ \therefore \frac{v^{\prime }}{v}=\frac{1.5\times 78\times {10}^3}{{10}^4\times 9.8}\\ \frac{v^{\prime }}{v}\times 100=\frac{15\times 78}{98}\times 100\end{array}$

$\begin{array}{c}dvg=9.8kg.........\left(1\right)\\ dvg-5v^{\prime }g=8.3kg..............\left(2\right)\\ \Rightarrow \rho v^{\prime }g=1.5kg\\ 1000.v^{\prime }\times 10=1.5kg\\ v^{\prime }=\frac{1.5}{{10}^4}\\ also,7800\times v\times 10=9.8\\ v=\frac{9.8}{78\times {10}^3}\\ \therefore \frac{v^{\prime }}{v}=\frac{1.5\times 78\times {10}^3}{{10}^4\times 9.8}\\ \frac{v^{\prime }}{v}\times 100=\frac{15\times 78}{98}\times 100\end{array}$

$\approx 11$%

$\therefore$ Option $D$ is correct answer.