Fluid Mechanics Viscosity Question 265
Question: A hemispherical bowl just floats without sinking in a liquid of density$ 1.2\times 10^{3}kg/m^{3} $ . If outer diameter and the density of the bowl are 1 m and $ 2\times 10^{4}kg/m^{3} $ respectively then the inner diameter of the bowl will be
Options:
A) 0.94 m
B) 0.97 m
C) 0.98 m
D) 0.99 m
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Weight of the bowl =mg $ =V\rho g=\frac{4}{3}\pi \left[ {{\left( \frac{D}{2} \right)}^{3}}-{{\left( \frac{d}{2} \right)}^{3}} \right]\rho g $ Where $ D=Outerdiameter $ d= Inner diameter, $ \rho =Densityofbowl $ Weight of the liquid diplaced by the bowl $ =V\sigma g=\frac{4}{3}\pi {{\left( \frac{D}{2} \right)}^{3}}\sigma g $ where $ \sigma $ is the density of the liquid For the floation $ \frac{4}{3}\pi {{\left( \frac{D}{2} \right)}^{3}}\sigma g=\frac{4}{3}\pi \left[ \left( {{\frac{D}{2}}^{3}} \right)-{{\left( \frac{d}{2} \right)}^{3}} \right]pg $ $ \Rightarrow ,{{\left( \frac{1}{2} \right)}^{3}}\times 1.2\times 10^{3}=\left[ {{\left( \frac{1}{2} \right)}^{3}}-{{\left( \frac{d}{2} \right)}^{3}} \right]2\times 10^{4} $ By solving we get $ d=0.98m. $
BETA
