question_answer

A hemispherical bowl is made of steel of \[0.25\text{ }cm\]thickness. The inner radius of the bowl is 5 cm. The volume of steel used is ___. (Use\[\pi =3.141\])                

A)  \[42.15\,c{{m}^{3}}\]        

B) \[41.52\,c{{m}^{3}}\]     

C)                     \[41.25\,c{{m}^{3}}\]    

D) \[40\,c{{m}^{3}}\]                          

Correct Answer: C

Solution :

       Let r and R be inner and outer radius of bowl respectively. Then, \[r=5\text{ }cm,\text{ }R=5+0.25=5.25\text{ }cm.\] Volume of steel used = Outer volume - Inner volume \[=\frac{2}{3}\pi {{(5.25)}^{3}}-\frac{2\pi }{3}{{(5)}^{3}}=41.25c{{m}^{3}}\]