Properties Of Solids And Liquids Question 263
Question: A hemispherical bowl just floats without sinking in a liquid ofdensity 1.2 × 103kg/m3. If outer diameter and the density of the bowl are 1 m and 2 × 104 kg/m3 respectively, then the inner diameter of the bowl will be [SCRA 1998]
Options:
A)0.94 m
B)0.97 m
C)0.98 m
D)0.99 m
Show Answer
Answer:
Correct Answer: C
Solution:
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 = Outer diameter ,
d = Inner diameter $ \rho $ = Density of bowlWeight of the liquid displaced 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 flotation $ \frac{4}{3}\pi {{\left( \frac{D}{2} \right)}^{3}}\sigma g=\frac{4}{3}\pi \left[ {{\left( \frac{D}{2} \right)}^{3}}-{{\left( \frac{d}{2} \right)}^{3}} \right]\rho g $
therefore $ {{\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 = 0.98 m.
