An object weighs 60.0N in air, 48.2N in a certain liquid X, and 44.9N in water, Calculate the relative density of X
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Correct Answer: Option D
Explanation:
R.d = Wt in Air-Wt in Air-Wt in water
= \(\frac{60.0 - 48.2}{60.0 - 44.9}\)
= 0.782
R.d = Wt in Air-Wt in Air-Wt in water
= \(\frac{60.0 - 48.2}{60.0 - 44.9}\)
= 0.782