A glass stopper weighs $0.250 k g f$ in air, $0.150 k g f$ in water and $0.125 k g f$ in liquid. Calculate the:
relative density of solid, and
relative density of the liquid.

Open in App

Solution

Step 1: Given data,
Weight of the glass stopper in air = $0.250 k g f$
Weight of the glass stopper in water = $0.150 k g f$
Weight of the glass stopper in a liquid = $0.125 k g f$
We know, the density of water at $4^{\circ} C$ is = $1000 k g / m^{3}$
Part (a):
Step 2: Finding the relative density of the solid (glass stopper),
$r e l a t i v e d e n s i t y = \frac{w e i g h t o f t h e s o l i d i n a i r}{w e i g h t o f s o l i d i n a i r - w e i g h t o f s o l i d i n w a t e r} = \frac{0.250 k g f}{0.250 k g f - 0.150 k g f} = 2.5$
Hence, the relative density of the solid is 2.5
Step 3: Finding the density of the solid,
$r e l a t i v e d e n s i t y o f t h e s o l i d = \frac{d e n s i t y o f t h e s o l i d}{d e n s i t y o f w a t e r a t 4^{\circ} C} S o, d e n s i t y o f t h e s o l i d = r e l a t i v e d e n s i t y o f t h e s o l i d \times d e n s i t y o f w a t e r a t 4^{\circ} C = 2.5 \times 1000 k g / m^{3} = 2500 k g / m^{3}$
Step 4: Finding the relative density of the solid w.r.t. (with respect to) the liquid,
$r e l a t i v e d e n s i t y = \frac{w e i g h t o f t h e s o l i d i n a i r}{w e i g h t o f s o l i d i n a i r - w e i g h t o f s o l i d i n l i q u i d} = \frac{0.250 k g f}{0.250 k g f - 0.125 k g f} = 2$
Part (b):
Step 5: Finding the density of the liquid,
$a l s o, r e l a t i v e d e n s i t y o f t h e s o l i d (w . r . t t h e l i q u i d) = \frac{d e n s i t y o f t h e s o l i d}{d e n s i t y o f t h e l i q u i d} S o, d e n s i t y o f t h e l i q u i d = \frac{d e n s i t y o f t h e s o l i d}{r e l a t i v e d e n s i t y} = \frac{2500 k g / m^{3}}{2} = 1250 k g / m^{3}$
Step 6: Finding the relative density of the liquid,
$r e l a t i v e d e n s i t y o f t h e l i q u i d = \frac{d e n s i t y o f t h e l i q u i d}{d e n s i t y o f w a t e r a t 4^{\circ} C} = \frac{1250 k g / m^{2}}{1000 k g / m^{3}} = 1.25$
Hence the relative density of the liquid is 1.25

Suggest Corrections

Similar questions

A body of volume 100 $c m^{3}$ weighs 1 kgf in air.

Find : (i) its weight in water and (ii) its relative density.

A glass ball weighs $0.188 k g f$ in air, $0.116 k g f$ in water and $0.125 k g f$ in turpentine. Calculate the density of glass and relative density of turpentine.

Q. A solid weighs

1.5 k g f

in air and

0.9 k g f

in a liquid of density

1.2 \times 10^{3} k g m^{- 3}

. Calculate relative density of solid.

Q. A solid weighs

1.5 kgf

in air and

0.9 kgf

in a liquid of density

1.2 \times 10^{3} {kg/m}^{3}

. Calculate relative density of the solid.

[3 marks]

Q. A solid weighs

200 kgf

in air and

186 kgf

when immersed completely in water. What is the relative density of this solid?

Join BYJU'S Learning Program

A glass stopper weighs 0.250kgf in air, 0.150kgf in water and 0.125kgf in liquid. Calculate the:relative density of solid, andrelative density of the liquid.

A glass stopper weighs $0.250 k g f$ in air, $0.150 k g f$ in water and $0.125 k g f$ in liquid. Calculate the:
relative density of solid, and
relative density of the liquid.