# 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 - Mathematics

MCQ

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

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

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

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

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

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

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

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

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

$\text{ Now, it is given that, }$

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

$\Rightarrow h' = 6h$

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

$\Rightarrow S' = \frac{1}{9}S$

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

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

$\Rightarrow r' = \frac{r}{3}$

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

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

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

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

Concept: Surface Area of Cylinder
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 19 Surface Areas and Volume of a Circular Cylinder
Exercise 19.4 | Q 20 | Page 30

Share