Question

a solid weighs 50gf in air and 34gf when completely immersed in a liquid of relative density 0.8, find the volume of the solid and the relative density of the solid​

Answer 1

Answer:

Volume of the solid = 0.02 m³

Relative Density of the Solid = 3.125

Explanation:

Mass of the solid in Air : 50 g = 0.05 kg

⇒ Weight of the solid in Air ($W_{a}$) = 0.05 × 10 = 0.5 N

(∵ W = mg)

Mass of the solid in a liquid = 34 g = 0.034 kg

⇒ Weight of the solid in the liquid ($W_{l}$) = 0.034 × 10 = 0.34 N

Buoyant Force = $W_{a} - W_{l}$

⇒ B = 0.50 - 0.34

⇒ B = 0.16 N

0.16 = $V_{b}* d_{l}* g$

Where :

$V_{b}$ = Volume of the body (solid)

$d_{l}$ = Density of liquid

⇒ 0.16 = $V_{b}$ · 0.8 · 10

⇒ 0.16 = $V_{b}$ · 8

∴ $V_{b}$ = 0.02 m³ (OR) 2 × 10⁴ cm³

Density of the solid = mass/volume

⇒ $d_{s}$ = 0.05/0.02 = 2.5 kg/m³

Relative density of the solid = $\frac{d_{b}}{d_{l}}$

Where :

$d_{b}$ = Density of the body (solid)

$d_{l}$ = Density of the liquid

⇒ Relative density of the solid = 2.5/0.8 = 3.125

Thanks !