question_answer

A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the for of a right circular cone mounted on a hemisphere is immersed into the tub. If the radius of the hemisphere is \[3.5\text{ }cm\]and the height of the cone outside the hemisphere is 5 cm, find the volume of water left in the tub. (Take\[\pi =22/7\])

A)  \[616\,c{{m}^{3}}\]           

B)  \[600\,\,c{{m}^{3}}\]                     

C)  \[535\,\,c{{m}^{3}}\]     

D)         \[716\,c{{m}^{3}}\]                       

Correct Answer: A

Solution :

Volume of water in the cylinder tub = Volume of the tub \[=\pi {{r}^{2}}h=\left( \frac{22}{7}\times 5\times 5\times 9.8 \right)c{{m}^{3}}=770\,c{{m}^{3}}\] Volume of the solid immersed in the tub = Volume of the hemisphere + Volume of the cone             \[=\left[ \left( \frac{2}{3}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times \frac{7}{2} \right)+\left( \frac{1}{3}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times 5 \right) \right]c{{m}^{3}}\] \[=\left( \frac{539}{6}+\frac{385}{6} \right)\,c{{m}^{3}}=\left( \frac{924}{6} \right)c{{m}^{3}}=154c{{m}^{3}}\] Volume of water left in = Volume of the tub - Volume of solid immersed \[=(770-154)c{{m}^{3}}=616\,c{{m}^{3}}\]