Question

A solid sphere weight $40N$ in water and $45N$ in a liquid of relative density $0.75$. Relative density of sphere will be

• A. $3$
• B. $2$
• C. $1$
• D. $4$

Answer 1

The correct option is A $3$

If apparent weight is $w^{\prime }$ in a liquid It means $w^{\prime }$ is the weight of the sphere minus the weight of the liquid the sphere's volume displaces.

So $w^{\prime }={\rho }_s{V}_{s}g-{\rho }_l{V}_{s}g$

for water $w_1={\rho }_s{V}_{s}g-{\rho }_w{V}_{s}g$

$40={\rho }_s{V}_{s}g-1\times V_{s}g$ (1)

for liquid $w_1={\rho }_s{V}_{s}g-{\rho }_w{V}_{s}g$

$45={\rho }_s{V}_{s}g-0.75\times V_{s}g$ (2)

Solving equations we get

${\rho }_s=3$

Hence relative density of sphere is 3.