The altitude of a circular cylinder is increased six times and the base area is decreased one-ninth of its value. The factor by which the lateral surface of the cylinder increases, is
Options
- \[\frac{2}{3}\]
- \[\frac{1}{2}\]
- \[\frac{3}{2}\]
2
Solution
\[\text{ Let the original radius of the base of cylinder = r } \]
\[\text{ Let the original height of cylinder = h } \]
\[ \text{ Now, original base area of the cylinder,} S = \pi r^2 \]
\[\text{ Now, original LSA of cylinder, } A = 2\pi rh . . . . . . . . . . . . \left( 1 \right)\]
\[\text{ When the altitude is increased to 6 times of its initial value and base area is decreased one - ninth of its initial value: } \]
\[\text{ Let the new height of the cylinder } = h'\]
\[\text{ Let the new radius of base of cylinder } = r'\]
\[\text{ Now, new base area of cylinder, } S' = \pi \left( r' \right)^2 \]
\[\text{ Now, it is given that, } \]
\[\text{ new height of cylinder } = 6 \times \text{ original height of cylinder } \]
\[ \Rightarrow h' = 6h\]
\[\text{ Also, new base area of cylinder } = \frac{1}{9}\left( \text{ original base area of the cylinder } \right)\]
\[ \Rightarrow S' = \frac{1}{9}S\]
\[ \Rightarrow \pi \left( r' \right)^2 = \frac{1}{9}\left( \pi r^2 \right)\]
\[ \Rightarrow \left( r' \right)^2 = \left( \frac{r}{3} \right)^2 \]
\[ \Rightarrow r' = \frac{r}{3}\]
\[\text{ Now, new LSA of cylinder } , A' = 2\pi r'h'\]
\[ \Rightarrow A' = 2\pi \times \left( \frac{r}{3} \right) \times \left( 6h \right) = 2 \times \left( 2\pi rh \right)\]
\[ \Rightarrow A' = 2 A \left[ \text{ Using } \left( 1 \right) \right]\]
\[\text{ Hence, lateral surface area of the cylinder becomes twice of the original } . \]
