Question

weight of the metal block is 120g in air and 60g in liquid.with relative density of the liquid 0.9.then what will be the relative density of the metal?
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Answer 1

Given :

Weight of the metal block in air = 120 g

Weight of this block in liquid = 60 g

Relative density of the liquid = 0.9

To Find :

What is the relative density of this metal block ?

Solution :

The free body diagram is shown in the attached fig .

Now the net weight of the metal block = $mg - F_b$

Here , $F_b$ is force of buoyancy .

So, $\frac{60}{1000}kg \times 10 = \frac{120}{1000}\times 10kg - F_b$

Or, $F_b = \frac{1200}{1000}- \frac{600}{1000}$

          =$\frac{600}{1000}$

          = $\frac{6}{10}$

Or, $F_b = \frac{3}{5}$

Now , $F_b = v\rho g$

Here , v is volume of block and $\rho$ is density of liquid .

So, $\frac{3}{5} = \frac{120\times 0.9 \times 10}{1000 \times relative \hspace3 density \hspace3 of \hspace3 block }$

Or, relative density of block = $\frac{120 \times 0.9 \times 10 \times 5}{1000\times 3}$

                                               = 1.8

Hence , the relative density of metal block is 1.8