Question

A solid weighs 15 gm in air and 13 gm when completely immersed in a liquid of relative density 0.8 find the volume of solid and density

Answer 1

Given, weight of solid in air = 30 gf and weight of solid in liquid = 26 gf., R.D of liquid = 0.8

Therefore, Density of liquid = 0.8 g cm-3

(i) Let V be the volume of the solid.

Weight of liquid displaced = Volume of liquid displaced xdensity of liquid x g

= V 0.8g dyne

= V 0.8 gf

Loss in the weight of the solid when immersed in the liquid = 30 - 26 = 4gf

But the weight of liquid displaced is equal to the loss in weight of the solid when immersed in liquid.

Therefore, Vx0.8 = 4 or V = = 5 .

(ii) Given, weight of solid = 30gf

Therefore, Mass of solid = 30 g

Density of solid =

Hence relative density of solid = 6

Answer 2

The volume and density of the solid is 2.5 cm³ and 6 gm/cm³

Explanation:

Given that,

Weight of solid in air = 15 gm

Weight of solid in water = 13 gm

Relative density in liquid= 0.8

Let ρ and $\rho_{l}$ are the densities of solid and liquid.  

Weight of the solid in air = 15 g wt

V is the volume of the solid

$V\rho g=15g$

$V\rho=15$...(I)

Let V is the volume of solid

$V\rho g=15g$

$V\rho=15$...(I)

Weight of the solid in water = 13 gm

Reduced weight of the solid = Weight of the solid - Weight of the liquid displaced by the stone

$W_{s}\times g=V\rho\times g-V\rho_{l}\times g$

Put the value into the formula

$13\times g=V\times \rho\times g-V\times0.8\times g$

$13=V\times \rho-V\times0.8$

Put the value from equation (I)

$13=15-V\times 0.8$

$-V=\dfrac{13-15}{0.8}$

$V=2.5\ cm^3$

Put the value of V in equation (I)

$\rho=\dfrac{15}{2.5}$

$\rho=6\ gm/cm^3$

Hence, The volume and density of the solid is 2.5 cm³ and 6 gm/cm³

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Topic : Density

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