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
A.
B.
C.
D. 2

Answer 1

lateral surface of the new  cylinder is two times the area of existing cylinder

Step-by-step explanation:

Let say Cylinder Dimensions are

radius = r

Height (h) / altitude  = h

lateral surface of the cylinder A  = 2πrh

base Area = πr²

Ne Cylinder

radius = r'

Height (h) / altitude  = h'

The altitude of a circular cylinder is increased six times

=> h' = 6h

the base area is decreased one-ninth of its value.

=>  base area = π(r')² = πr²/9

=> r' = r/3

A' = lateral surface of the new  cylinder = 2πr'h'

= 2π ( r/3) (6h)

= 2 ( 2πrh)

= 2A

lateral surface of the new  cylinder is two times the area of existing cylinder

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Answer 2

Lateral surface area of new circular cylinder is two times lateral surface area of original circular cylinder. Option D is correct.

Step-by-step explanation:

• Given data about original circular cylinder

        Let

        Base radius $\mathbf{=R_{o}}$

        Altitude $\mathbf{=H_{o}}$

         From formula of lateral surface area and base area of cylinder

         $\mathbf{\textrm{Base area of original cylinder}\ (A_{Bo})=\pi R_{o}^{2}}$       ...1)

         $\mathbf{\textrm{Lateral surface area of original cylinder}\ (A_{Lo})=2\pi R_{o}H_{o}}$      ...2)

• Given data about new circular cylinder

        Let

        Base radius $\mathbf{=R_{N}}$

        Altitude $\mathbf{=H_{N}}$

        From formula of lateral surface area and base area of cylinder

        $\mathbf{\textrm{Base area of new cylinder}\ (A_{BN})=\pi R_{N}^{2}}$       ...3)

        $\mathbf{\textrm{Lateral surface area of new cylinder}\ (A_{LN})=2\pi R_{N}H_{N}}$      ...4)

• It is given that he altitude of a new circular cylinder is increased six times original cylinder.

        $\mathbf{\frac{H_{N}}{H_{o}}=\frac{6}{1}}$            ...5)

• It is given that the base area is decreased one-ninth of its value.

        Means

        $\mathbf{\frac{A_{BN}}{A_{Bo}}=\frac{1}{9}}$

        From equation 1) and equation 3)

       $\mathbf{\frac{A_{BN}}{A_{Bo}}=\frac{\pi R_{N}^{2}}{\pi R_{o}^{2}}=\frac{1}{9}}$

       We can also write above

       $\mathbf{\left ( \frac{R_{N}}{R_{o}} \right )^{2}=\left ( \frac{1}{3} \right )^{2}}$

       $\mathbf{ \frac{R_{N}}{R_{o}}= \frac{1}{3} }$            ...6)

• Now ratio of lateral surface area new cylinder and original cylinder

        $\mathbf{\frac{A_{LN}}{A_{Lo}}=\frac{2\pi R_{N}H_{N}}{2\pi R_{o}H_{o}}}$

         It can be written as

         $\mathbf{\frac{A_{LN}}{A_{Lo}}=\frac{R_{N}}{R_{o}}\times \frac{H_{N}}{H_{o}} }$

         $\mathbf{\frac{A_{LN}}{A_{Lo}}=\frac{1}{3}\times \frac{6}{1}=2 }$