Properties Of Solids And Liquids Question 281
Question: Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density is 9 g / cm3. If the mass of the other is 48 g, its density in g / cm3 is [CBSE PMT 1994]
Options:
A)$ \frac{4}{3} $
B)$ \frac{3}{2} $
C)3
D)5
Show Answer
Answer:
Correct Answer: C
Solution:
Apparent weight $ =V(\rho -\sigma )g=\frac{m}{\rho }(\rho -\sigma )g $
where $ m= $ mass of the body,$ \rho = $ density of the body
$ \sigma = $ density of water
If two bodies are in equilibrium then their apparent weight must be equal.
$ \therefore $ $ \frac{m _{1}}{{\rho _{1}}}({\rho _{1}}-\sigma )=\frac{m _{2}}{{\rho _{2}}}({\rho _{2}}-\sigma ) $
therefore $ \frac{36}{9}(9-1)=\frac{48}{{\rho _{2}}}({\rho _{2}}-1) $ By solving we get $ {\rho _{2}}=3 $ .
