Question

A block of steel of size $5cm\times 5cm\times 5cm$ is weighted in water. If the relative density of steel is 7, its apparent weight is :-

• A. $6\times 5\times 5\times 5$ gm wt
• B. $4\times 4\times 4\times 7$ gm wt
• C. $5\times 5\times 5\times 7$ gm wt
• D. $4\times 4\times 4\times 6$ gm wt

Answer 1

The correct option is A $6\times 5\times 5\times 5$ gm wt

Weight of object in vacuum = W ; It's density = $\rho$

Therefore, the volume of the object will be V = $W/\rho$

Now, this object is immersed in a fluid of density ${\rho }_1$

Its apparent loss in weight would be equal to the weight of fluid displaced (as is evident from Archimedes' principle)

Apparent loss = ${\rho }_{1}V$ = ${\rho }_{1}W/\rho$

Therefore, the weight in water = weight in vacuum - the loss in weight

Weight in water = $W-{\rho }_{1}W/\rho$

= $W(1-{\rho }_1/\rho )$

Here, W = $\rho V$ = $7\times 125=875$ gm wt

$\rho =7$ , ${\rho }_1$ = 1

Substituting the values,

$W^{\prime }=875(1-1/7)=875\times 6/7=750$ gm wt