A solid weight 0.09N in air and0.06N in water what is R.d of the solid
Answers
Answered by
0
Given,
Weight in air (W₁) = 0.09 N
Weight in water (W₂) = 0.06 N
R.D = W₁/(W₁-W₂)
= 0.09 / (0.09-0.06)
= 0.09/0.03
= 3
The Relative Density (R.D) is 3.
I hope this helps.
Similar questions