CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A solid weighs $$32\ gf$$ in air and $$28.8\ gf$$ in water. Find: $$R.D.$$ of solid.


Solution

Weight of the solid in air, $$W_1=32\ gf$$
weight of the solid immersed completely in water, $$W_2=28.8\ gf$$
density of the liquid $$=0.9\ gcm^{-3}$$
To find the R.D. of the solid
Relative density of the solid $$=\dfrac{W_1}{W_3-W_2}\times $$ relative density of water
$$=\dfrac{32}{32-28.8}\times 1=10$$

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image