A) 0.94 m
B) 0.97 m
C) 0.98 m
D) 0.99 m
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 \[D=Outer\text{ }diameter\] d= Inner diameter, \[\rho =Density\text{ }of\text{ }bowl\] 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 \[d=0.98m.\]
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