Question

A metal cube of edge $5.0\text{cm}$ and density $9.0{\text{g/cm}}^3$ is suspended by a thread so as to be completely immersed in a liquid of density $1.2{\text{g/cm}}^3$. Find the tension in the thread. (Take g = $10{\text{m/s}}^2$) $\left[6\text{marks}\right]$

Answer 1

Given:

Density of metal cube = $9.0{\text{g/cm}}^3$
Density of liquid = $1.2{\text{g/cm}}^3$

Side of the cube = $5\text{cm}$
Volume of the cube $=5\times 5\times 5=125{\text{cm}}^3$

To find the weight of the cube:
Mass of the cube = volume of the cube x density of the cube
= $125\times 9=1125\text{g}$
∴ Weight of the cube = $1125\text{gf}$ $\left[2\text{marks}\right]$

The weight of the cube acts downwards
Upthrust acting on the cube = weight of the liquid moved
=$\text{volume of the cube}\times \text{density of the liquid}\times g$

= $125\times 1.2\times g=150\text{gf}$

Upthrust on the cube acts in the upward direction
Tension in thread = total force acting in the downward direction
= weight of the cube acting downwards – upthrust acting on the cube
= $1125–150=975\text{gf}\text{(or)}9.75\text{N}$ $\left[4\text{marks}\right]$