Question

A metal sphere weighing 50 grms in air is
suspended by a thread in an oil of
specific gravity 0.4. If the tension in the
thread is 40 grm wt. The specific gravity
of the metal sphere is
Nö + 0​

Answer 1

Given :

Weight of the metal sphere = 50 gm

Specific gravity of the oil = 0.4

Tension in the thread = 40 gm W

To Find :

Specific gravity of the metal sphere = ?

Solution :

Since the metal sphere is in vertical equilibrium , so the upward force on the sphere will be equal to downward force on it .

Here the upward forces are force of buoyancy and the tension in the thread .

And the downward force is the weight of the sphere .

So, we have :

$mg =F_b + T$     - (1)

we know , $F_b = V\rho g$  , where V is volume of displaced liquid or vol of sphere    

                                    and $\rho$ is the specific gravity of the liquid .

So, from eq (1) we have :

$mg = V \rho g + T$

Or , $50 \times 10 ^{-3} g = V \times 0.4 g + 40 \times 10 ^{-3} g$

Or, $50 \times 10 ^{-3} = 0.4V + 40 \times 10^{-3}$

Or, $V = \frac{(50- 40 ) \times 10^{-3}}{0.4}$

          = $\frac{10\times 10^{-3}}{0.4}$

          = $2.5 \times 10^{-2} m^3$

Now , density of sphere = $\frac{mass \hspace3 of \hspace3 sphere }{vol \hspace3 of \hspace3 sphere }$

                                         = $\frac{50 \times 10^{-3} kg}{2.5 \times 10^{-2}m^3} = 2 \frac{kg}{m^3}$

So, the specific gravity of sphere = $\frac{2 \frac{kg}{m^3}}{997\frac{kg}{m^3}} = 2 \times 10^{-3}$

Hence, the specific gravity of the metal sphere is 0.002 .