Question

A cube of size 10 cm is floating in equilibrium in a tank of water. When a mass of 10 gm is placed on the cube, the depth of cube inside water increases by $g=10m/s^2$ density of water $=1000kg/m^3)$

• A. 0.1 m
• B. 1 mm
• C. 1 m
• D. 0.31 m

Answer 1

The correct option is B 1 mm

Let $x$ be the initial depth upto which the cube is sinked in water.

Let $d$ be the density of the cube

Then

$x\times 10\times 10\times 1\times g$

$=10\times 10\times 10\times 1\times g\times d$........(1)

$\Rightarrow x=10d$

Let $x^1$ be the new depth, then

$x^1\times 100\times g=1000\times d\times g+10g$........(2)

subtracting (1) and (2) we get

$\Rightarrow 100x^1=10x+10$

$x^2-x=\frac{1}{10cm}=1mm$

Hence,

option $B$ is correct answer.