Question

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

Answer 1

\[\text{ Let the original radius of the base of cylinder = r } \]

\[\text{ Let the original height of cylinder = h } \]

\[ \text{ Now, original base area of the cylinder,}  S = \pi r^2 \]

\[\text{ Now, original LSA of cylinder, } A = 2\pi rh . . . . . . . . . . . . \left( 1 \right)\]

\[\text{ When the altitude is increased to 6 times of its initial value and base area is decreased one - ninth of its initial value: } \]

\[\text{ Let the new height of the cylinder } = h'\]

\[\text{ Let the new radius of base of cylinder } = r'\]

\[\text{ Now, new base area of cylinder, } S' = \pi \left( r' \right)^2 \]

\[\text{ Now, it is given that, } \]

\[\text{ new height of cylinder } = 6 \times \text{ original height of cylinder } \]

\[ \Rightarrow h' = 6h\]

\[\text{ Also, new base area of cylinder } = \frac{1}{9}\left( \text{ original base area of the cylinder } \right)\]

\[ \Rightarrow S' = \frac{1}{9}S\]

\[ \Rightarrow \pi \left( r' \right)^2 = \frac{1}{9}\left( \pi r^2 \right)\]

\[ \Rightarrow \left( r' \right)^2 = \left( \frac{r}{3} \right)^2 \]

\[ \Rightarrow r' = \frac{r}{3}\]

\[\text{ Now, new LSA of cylinder } , A' = 2\pi r'h'\]

\[ \Rightarrow A' = 2\pi \times \left( \frac{r}{3} \right) \times \left( 6h \right) = 2 \times \left( 2\pi rh \right)\]

\[ \Rightarrow A' = 2 A \left[ \text{ Using }  \left( 1 \right) \right]\]

\[\text{ Hence, lateral surface area of the cylinder becomes twice of the original }  . \]