Question

If a block of iron of density $5g/c{m}^3$ and size $5cm\times 5cm\times 5cm$ was weighed whilst completely submerged in water, what would be the apparent weight in $g-f(gram-force)?$

Answer 1

Density of block $=5gm/c{m}^2$

Volume of block $V=(5\times 5\times 5)c{m}^3$

$=125c{m}^3$

Mass of block.M=density $\times$ volume

$=5\times 125$

$=325gm=0625kg$

Density of water, $\rho =1gm/c{m}^3$

On submerging the iron block completely into water the apparent weight of the block is given from the box of the container

Drawing free body diagram of the block

Now as the block is stationery upwards and downward form on the block should balance each other

Thus. $N+F_B=Mg$

Now, $F_B$= density of water $\times$ volume of block $\times g$

$=\left(1gm/c{m}^3\right)\times \left(125c{m}^3\right)\times \left(98m/s^2\right)$
$=\left(125gm\right)\times \left(98m/s^2\right)$
$=(0.125\times 9.8)kgm/s^2$
$=(0.125\times 9.8)N(1N=1kgm/s^2)$
Putting values is $\left(1\right)$

$N=(0.625\times 9.8)-(0.125\times 9.8)$ Newton

$\Rightarrow$ Apparent weight $=(0.5\times 9.8)$ Newton

($\because$ Apparent weight = normal force)

Now, $1gf(gram-force)=9.8\times {10}^{-3}N$

$\Rightarrow 1N=\frac{1}{9.8}\times {10}^{3}gf$

Thus, Apparent weight $+(0.5\times 9.8)\times (\frac{1}{9.8}\times {10}^3)$

$=500gf$

Hence apparent weight of the block completely submerged in water is $500$ gram- force