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.



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