Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: 2010 · Marks: · Difficulty:
A solid is in the form of a right circular cylinder with hemispherical ends. The total length of the solid is 35 cm . The diameter of the cylinder is $\frac{1}{4}$ of its height. The surface area of the solid is
(a)$462 \mathrm{cm}^{2}$
(b)$693 \mathrm{cm}^{2}$
(c)$750 \mathrm{cm}^{2}$
(d)$770 \mathrm{cm}^{2}$
Answer
Answer (as printed): D
Explanation
Let the radius of the cylinder and the hemi-sphere be $r \mathrm{cm}$. Diameter of the cylinder $=(2 r) \mathrm{cm}$. Height of the cylinder $=(4 \times 2 r) \mathrm{cm}=(8 r) \mathrm{cm}$. Total length of the solid $=(8 r+r+r) \mathrm{cm}=(10 r) \mathrm{cm}$. $10 r=35 \Rightarrow r=3.5 \mathrm{cm}$. ∴ Surface area of the solid = Curved surface area of the cylinder + 2 × (curved surface area of the hemisphere) $=\left(2 \times \frac{22}{7} \times 3.5 \times 28+2 \times 2 \times \frac{22}{7} \times 3.5 \times 3.5\right) \mathrm{cm}^{2}$ $=(616+154) \mathrm{cm}^{2}=770 \mathrm{cm}^{2}$.