Question

An object floats in water with half of its volume immersed in water. What will be its relative density?​

Answer 1

Answer:

relative density=1/2

Weight of body = weight of liquid displaced

Vdg =(V/2)*density of water*g

So, density of body /density of water =1/2

Answer 2

$\huge\red{\boxed{\mathcal{SOLUTION}}}$

When a body floats in water, the upthrust it experiences is equal to the weight of the body.

Hence, upthrust (U) = Weight of the body

U = mg

(Where, m = mass of the body and g = acceleration due to gravity)

Mass (m) = Density (d) × Volume (V)

Therefore,

U = d × V × g ...(1)

Upthrust is also equal to the weight of water displaced.

Volume of water displaced is 

$\frac{v}{2}$

V being the volume of the object.

Therefore,

U = Mass of water displaced × g

$\frac{1}{2}v \times {d}^{w} \times g$

(where dw = density of water) ...(2)

Since the weight of the object is equal to the weight of the displaced water,

$d \times v \times g = \frac{1}{2}v \times {d}^{w} \times g$

$\frac{dw}{d} = \frac{1}{2}$

Therefore, the relative density of the object is

$\frac{1}{2}$