Question

A rectangular sheet of paper 35cm×21cm can be transformed into the curved surface of a right cicular cylinder in two ways i.e, either by rolling the paper along its length or along its breadth find the ratio of the volume of the two cylinder thus formed​

Answer 1

Answer:

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Step-by-step explanation:

rolling along length:

height = 30, circumference = 18

2πr = 18

r = 9/π

volume = π(81/π^2)(30) = 2430/π

rolling along breadth

height = 18

circumf = 30

2πr = 30

r = 15/π

Volume = π(225/π^2)(18) = 4050/π

ratio of volume breadth : volume of height

=(4050/π) : 2430/π

= 5 : 3

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Answer 2

Step-by-step explanation:

Given:-

Dimensions of the rectangular sheet are:-

Length (l) = 35 cm

Breath (b) = 21 cm

Curved surface area

= area of the rectangular sheet

=> l × b

=> 35 × 21 cm²

=> 735 cm²

Now as curved surface area = 2πrh

Therefore:-

2 x π × r × h = 735 cm²..............Equation 1

The two ways given in the question are :-

(i) By rolling along its length,

Then height of the cylinder formed(h1)

= 35 cm

Putting this value of h1 in equation 1 and solving we get,

$\sf{2 \times \frac{22}{7} \times r \times 35 = 735 {cm}^{2}}$

$\sf{r \: = \frac{735 \times 7}{2 \times 22 \times 35}}$

$\sf{r = \frac{147}{44}cm}$

(ii) By rolling along its breath,

Then height of the cylinder formed (h2)

= 21 cm

Putting this value of h2 in equation 1 we get,

$\sf{2 \times \frac{22}{7} \times r \times 21 = 735 {cm}^{2}}$

$\sf{r \: = \frac{735 \times 7}{2 \times 22 \times 21} cm}$

$\sf{r = \frac{245}{44} cm}$

Now the volume of the respective cylinders are :-

$\sf{v_{1} = \pi \times {r}^{2} \times h}$

$\sf{v_{2} = \pi \: \times ({ \frac{147}{44} })^{2} \times 35}$

Also:-

$\sf{ \frac{v_{1}}{v_{2}} = \frac{\pi \times { (\frac{147}{44} }^{2}) \times 35 }{\pi \times ( { \frac{245}{44} )}^{2} \times 21} } \\ \\ \sf{ \frac{v_{1}}{v_{2}} = \frac{ \cancel\pi \times { \frac{147}{44} } \times \frac{147}{44} \times 35 }{ \cancel\pi \times { \frac{245}{44} } \times \frac{245}{44} \times 21} } \\ \\ \sf{ \frac{v_{1}}{v_{2}} = \frac{147 \times 147 \times \cancel44 \times \cancel44 \times \cancel35}{\cancel44 \times \cancel44 \times 245 \times 245 \times \cancel21}} \\ \\ \sf{ \frac{v_{1}}{v_{2}} = \frac{ \cancel21 \times \cancel7 \times \cancel21 \times \cancel7 \times 5}{ \cancel35 \times \cancel7 \times \cancel35 \times \cancel7 \times 3} } \\ \\ \sf{ \frac{v_{1}}{v_{2}} = \frac{3 \times \cancel3 \times \cancel5}{5 \times \cancel5 \times \cancel3} } \\ \\ \sf{ \frac{v_{1}}{v_{2}} = \frac{3}{5} }$

Hence:-

The ratio of the volumes of the cylinder is 3:5.

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