A) 0.94 m
B) 0.97 m
C) 0.98 m
D) 0.99 m
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 bowl Weight 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 \[\frac{4}{3}\pi {{\left( \frac{D}{2} \right)}^{3}}\sigma g=\frac{4}{3}\pi \left[ {{\left( \frac{D}{2} \right)}^{3}}-{{\left( \frac{d}{2} \right)}^{3}} \right]\rho g\] Þ \[{{\left( \frac{1}{2} \right)}^{3}}\times 1.2\times {{10}^{3}}=\left[ {{\left( \frac{1}{2} \right)}^{3}}-{{\left( \frac{d}{2} \right)}^{3}} \right]\,2\times {{10}^{4}}\] By solving we get d = 0.98 m.
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