Question

The weight of solid in a liquid of density $0.9gc{m}^{-3}$ is $Xgf$. Find [$\frac{X}{7}$]. {Where, [] is a step function}

• A. $3$
• B. $6$
• C. $5$
• D. $4$

Answer 1

Answer: (D)

$4$

Let us consider that $D_l$ and $D$ be the density of liquid of density $0.09g/c{m}^3$ and solid respectively.

The actual weight of the solid is $32gf$.

Apparent weight of the body = Actual weight - Buoyant force ($V\times D_l\times g$).

Let V be the volume of the solid.

Density of water $D_w=1g{cm}^3$.

In this case, $28.8=mg-V\times D_w\times g$

That is, $28.8=32-V\times 1$

Therefore, $V=3.2{cm}^3$.

Now, the Force of buoyancy or upthrust of the liquid is given as =$Volumeoftheliquiddisplaced\times Densityoftheliquid\times g$.

That is, upthrust = $3.2\times 0.9\times g=2.88gf.$

Apparent weight of body in the liquid = Actual weight - buoyant force (upthrust of liquid)

$=32gf-2.88gf=29.12gf=X$

Hence, the answer $X/7=4.16$