Question

A cylindrical object floats in water such that 3/4th of its volume is immersed in water what is its density in kgm

Answer 1

Answer:

Given:

Cylindrical object floats in water

3/4th of its volume is immersed in water

To Find:

Density of the Object ($\rm d_o$)

Solution:

We know that,

Density of water is 1000 kg/m³

Given that,

Cylindrical object floats in water such that 3/4th of its volume is immersed in water

Let us take,

Density of Water ($\rm d_w$) = 1000 kg/m³

Density of Object ($\rm d_o$) is what we have to find.

So,

Using the formula,

$\rm \implies \dfrac{d_o}{d_w} = \dfrac{3}{4}$

Substituting, $\rm d _w= 1000\;kg/m^3$

We get,

$\rm \implies d_o=\dfrac{3}{4} \times 1000\;\;kg/m^3$

$\rm \implies d_o=0.75 \times 1000\;\;kg/m^3$

$\rm \implies d_o=750\;\;kg/m^3$

$\rm \implies d_o=750\;\;kgm^-3$

Hence,

The Density of Object ($\bold{\rm d_o}$) = 750 kgm⁻³