Question

A solid has a mass 39kg it loses 1/3 rd of its weight when immersed in water its volume will be


Please give explanation

Answer 1

Answer:

Assume the volume of the body = v cm³ and the density of the body = d gcm⁻³.

The weight of the body in air, Wair = vdg dyne, where g is the acceleration due to gravity.

The buoyant force on the body when it is completely immersed is the weight of displaced water, Wbouyant = v x 1 x g dyne.

The weight of the body in water, Wwater = Wair - Wbuoyant = vg(d−1) dyne

According to the question, Wwater = 1/3 Wair

⇒vg(d−1) = 1/3 vdg

⇒d−1 = 1/3 d

⇒d−d/3 = 1

⇒2d/3 = 1

⇒d=3/2 = 1.5 gcm⁻³.

Answer 2

Answer:

sorry bro snsnikrjfkfugnffiyxskzjsjsisk