A solid weight 45N and 15N respectively in air ... - WAEC Physics 2002 Question
A solid weight 45N and 15N respectively in air and water. Determine the relative density of the solid
A
0.33
B
0.50
C
1.50
D
3.00
correct option: c
R.d = \(\frac{\text{weight in air}}{\text{apparent loss of weight in water}} = \frac{45}{45 - 15}\)
= \(\frac{45}{30} = 1.50\)
= \(\frac{45}{30} = 1.50\)
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