Question

A rectangular sheet of paper 30cm×18cm can be transformed into the curved surface of a right circular cylinder in two ways i.e either by rolling the paper along its length or by rolling it along with its breadth. The ratio of the curved surface area of the two cylinders thus formed is
a 1:1
b 2:3
c 3:4
d 4:5 ​

Answer 1

Answer:

Case1: when the sheet is folded along its length :

In this case, it forms a cylinder having height h1=18 cm and the circumference of its base equal to 30 cm.

Let the radius of its base be r1. Then,

2pi r1=30

r1=15/pi.

V1=pi r^2 h=pi*(15/pi)^2 cm3=4050/pi.cm3

Case.2

When the sheet folded along its breadth :

Heights h2=30cm and the circumference of its base equal to 18cm.

Let the radius be r2. Then,

2pi r2=18=(9/pi)

V2=pi r^2 h=(pi*(9/11)^2) cm^3 =2430/pi cm^3.

V1/v2=(4050/pi*pi/2430)=4050/2430=5/3

Ans.5:3.

Step-by-step explanation:

Answer 2

Answer:

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A rectangular sheet of paper 30cm×18cm can be transformed into the curved surface of a right circular cylinder in two ways i.e either by rolling the paper along its length or by rolling it along with its breadth. The ratio of the curved surface area of the two cylinders thus formed is

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Consider r1 as the radius and V1 as the volume

So we get 2πr1 = 30

It can be written as r1 = 30/2 π = 15/π

We know that

V1 = π x r x 12 x h1

By substituting the values

V1 = π × (15/π) 2 × 18

We get -

V1 = (225/ π) × 18 cm³

We know that If the sheet is folded along its breadth it forms a cylinder of height h2 = 30cm and perimeter 18cm

Consider r2 as the radius and V2 as the volume

So we get 2 πr2 = 18

It can be written as r2 = 18/2 π = 9/π

We know that V2 = πr22h2

By substituting the values V2 = π × (9/π) 2 × 3

We get V2 = (81/ π) × 30 cm3

So we get V1/V2 = {(225/ π) × 18}/ {(81/ π) × 30)}

On further calculation V1/V2 = (225 × 18)/ (81 × 30)

We get

V1/V2 = 5/3

We can write it as V1:V2 = 5:3

Therefore, the ratio of the volumes of the two cylinders thus formed is 5:3.