1.

A solid weight 0.09N in air and0.06N in water what is R.d of the solid

Answer»

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.



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