Question

The reading of a spring balance when a body is suspended in air is $60N$. When the body is immersed in water, the reading is $40N$ and when the body is immersed in a liquid, the reading is $48N$. The specific gravity of liquid is

• A. $0.8$
• B. $0.6$
• C. $0.2$
• D. $0.4$

Answer 1

The correct option is B $0.6$

Given, weight of body in air,
$W_0=60N$

Weight of body when it completely immersed in water,
$W_a=40N$

Weight of body when it immersed in liquid, $=48N$

Specific gravity or relative density of body

$RD=\frac{W_0}{W_0-W_a}=\frac{60}{60-40}=3$

When the body is immersed in water,
Weight of body = buoyancy force

$W_0=V_{p}g$

$V\times 3\times 10=60$

$V=2m^3$

Again, $2\times p_l\times 10=60-48$

$\Rightarrow p_l=\frac{12}{20}=0.6$