A solid of density 7600 kg/m³ is found to weigh 0.950 kgf in air. If 4/5 volume of solid is completely immersed in a solution of density 900 kg/m³, find the apparent weight of solid in the liquid.

Open in App

Solution

Step 1. Given data
Density of solid, $ρ = 7600 k g / m^{3}$
Weight of solid, $w_{s} = 0.950 k g f$
Mass of solid, $m_{s} = 0.95 k g$
Volume of solid $= v$
Volume of solid immersed in the solution, $v_{1} = \frac{4}{5} v$
Density of the solution, $ρ_{s} = 900 k g / m^{3}$
Apparent weight of the solid in the liquid, $w_{a} = ?$
Step 2. Calculation of the apparent weight of the solid in liquid
Volume of solid, $v = \frac{m_{s}}{ρ_{s}}$
$v = \frac{0.95 k g}{7600 k g / m^{3}} = 0.000125 m^{3}$
Submerged volume, $v_{1} = \frac{4}{5} v_{}$
$v_{1} = \frac{4}{5} \times 0.000125 m^{3} = 0.0001 m^{3}$
Mass of the solution displaced, $m_{d} = v_{1} \times ρ_{w_{}}$
$m_{d} = 0.0001 m^{3} \times 900 k g / m^{3}$
$m_{d} = 0.09 k g$
Weight of the solution displaced, $w = m_{d} \times g$
$w = 0.09 k g \times 10 m / s^{2} = 0.9 N$
According to the law of floatation. “upthrust is equal to the weight of the liquid displaced” and therefore
Upthrust, $f_{u p} = w = 0.9 N$
$f_{u p} = 0.9 N \times 10^{- 1} = 0.09 k g f$
Apparent weight of the solid in the liquid, $w_{a} = w_{s} - f_{u p}$
$w_{a} = 0.950 k g f - 0.09 k g f$
$w_{a} = 0.86 k g f$
Hence, the apparent weight of the stone in the liquid is equal to 0.86 kgf.

Suggest Corrections

Similar questions

A solid of density $ρ = 5000 k g m^{- 3}$ weighs $= 0.5 k g f i n a i r .$ . It is completely immersed in a liquid of density $1000 k g m^{- 3}$ . Calculate the apparent weight of the solid in the liquid

. A solid of density 5000kg/m3weights 0.5 kg in air. It is completely immersed in water of density1000kg/m3

(a) Calculate the apparent weight of solid in water.(ans. 0.4 kg)

(b) What will be its apparent weight if water is replaced by a liquid of density 8000kg/m3

A solid of density 5000 $\frac{k g}{m^{3}}$ weighs 0.5 kg in air .It is completely immersed in a liquid of density 8000 $\frac{k g}{m^{3}}$ .Calculate the apparent weight of the solid in the liquid.

Q. A body of volume 100 c

m^{3}

weighs 5 kg in air. It is completely immersed in a liquid of density 1.8 x

10^{3}

kg/

m^{3}

. Find the apparent weight of the body in the liquid.

Q. A solid block of material density 6000

k g / m^{3}

weighs 10 kg wt. in air. When it is immersed completely into a liquid, it weighs 6 kg wt. What is the density of the liquid?

Join BYJU'S Learning Program

A solid of density 7600 kg/m3 is found to weigh 0.950 kgf in air. If 4/5 volume of solid is completely immersed in a solution of density 900 kg/m3, find the apparent weight of solid in the liquid.

A solid of density 7600 kg/m³ is found to weigh 0.950 kgf in air. If 4/5 volume of solid is completely immersed in a solution of density 900 kg/m³, find the apparent weight of solid in the liquid.