A solid weighs 45N and 15N respectively in air and water....
A solid weighs 45N and 15N respectively in air and water. Determine the relative density of the solid
0.33
0.50
1.50
3.00
Correct answer is C
R.d = \(\frac{\text{weight in air}}{\text{apparent loss of weight in water}} = \frac{45}{45 - 15}\)
= \(\frac{45}{30} = 1.50\)
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