Fluid Mechanics Viscosity Question 200
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 waterIf 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 ) $ Þ$ \frac{36}{9}(9-1)=\frac{48}{{\rho_{2}}}({\rho_{2}}-1) $ By solving we get $ {\rho_{2}}=3 $ .
BETA
