A body of volume $100 c m^{3}$ weights $1 k g f$ in air. Find its relative density.

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

100

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $10$
Volume of body $= 100 {c m}^{3}$

Actual Mass of body $= 1 k g = 1000 g$

Since the density of air is very low, the buoyant force experienced by the object in air is negligible. Therefore, the weight of the object as measured in air is equal to its actual weight.

Density of body $= \frac{M a s s}{V o l u m e} = \frac{1000}{100} = 10 g / {c m}^{3}$

Relative Density $= \frac{{D e n s i t y}_{B o d y}}{{D e n s i t y}_{W a t e r}} = \frac{10}{1} = 10$

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.

100 c m^{3}

1 k g f

X g f

\frac{X}{100}

100 c m^{3}

10 N

g = 10 m s^{- 2}

= 1000 k g m^{- 3}

100 c m^{3}

5 k g f

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

Join BYJU'S Learning Program