Question

Please help it's urgent
tomorrow is my exam
please give full explanation
chapter - Surface area and volume
Don't post irrelevant or unrelated answers to the question otherwise 10 answers will be deleted
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Answer 1

Answer :-

$\\\;\large{\underbrace{\underline{Question's\;Analysis\;:-}}}$

Here the concept Volume of Cylinders has been used. We are given that radii of both are in ratio and their heights as well are in ratio. So there will be a constant value for both by which they are multiplied to make their values original.

Let's do it !!

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★ Equations Used :-

$\\\;\boxed{\sf{Volume\;of\;Cylinder\;=\;\bf{\pi r^{2} h}}}$

$\\\;\boxed{\sf{Ratio\;of\;Volumes\;=\;\bf{\dfrac{Volume\;of\;First\;Cylinder}{Volume\;of\;Second\;Cylinder}}}}$

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★ Solution :-

Given,

» Ratio of radii of two cylinders = 2 : 3

» Ratio of height of two cylinders = 5 : 3

Now let us assume that first ratio part denotes first Cylinder and second ratio part denotes second Cylinder.

Then,

• Let x be the constant by which both the radii should be multiplied.

• Let k be the constant by which both the heights should be multiplied.

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~ For Measurements of different part of both Cylinders :-

✒ Radii of first cylinder = 2x

✒ Height of first cylinder = 5k

✒ Radii of second Cylinder = 3x

✒ Height of second cylinder = 3k

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~ For the Volume of First Cylinder :-

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Cylinder_{(First)}\;=\;\bf{\pi r^{2} h}}$

Note :- Here we are not going to apply the value of π since we are going to cancel it at end.

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Cylinder_{(First)}\;=\;\bf{\pi\;\times\;(2x)^{2}\;\times\;5k}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Cylinder_{(First)}\;=\;\bf{\pi\;\times\;4x^{2}\;\times\;5k}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Cylinder_{(First)}\;=\;\bf{20\:\pi\:x^{2}\:k}}$

$\\\;\underline{\boxed{\tt{Volume\;\;of\;\;first\;\;Cylinder\;\;=\;\bf{20\:\pi\:x^{2}\:k}}}}$

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~ For the Volume of Second Cylinder :-

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Cylinder_{(Second)}\;=\;\bf{\pi r^{2} h}}$

Note :- Here we are not going to apply the value of π since we are going to cancel it at the end.

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Cylinder_{(Second)}\;=\;\bf{\pi\;\times\;(3x)^{2}\;\times\;3k}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Cylinder_{(Second)}\;=\;\bf{\pi\;\times\;9x^{2}\;\times\;3k}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Cylinder_{(First)}\;=\;\bf{27\:\pi\:x^{2}\:k}}$

$\\\;\underline{\boxed{\tt{Volume\;\;of\;\;first\;\;Cylinder\;\;=\;\bf{27\:\pi\:x^{2}\:k}}}}$

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~ For the Ratio of Volumes of Two Cylinders :-

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Ratio\;of\;Volumes\;=\;\bf{\dfrac{Volume\;of\;First\;Cylinder}{Volume\;of\;Second\;Cylinder}}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Ratio\;of\;Volumes\;=\;\bf{\dfrac{20\:\pi\:x^{2}\:k}{27\:\pi\:x^{2}\:k}}}$

Cancelling πx²k from numerator snd denominator, we get,

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Ratio\;of\;Volumes\;=\;\bf{\dfrac{20}{27}}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Ratio\;of\;Volumes\;=\;\bf{20\;:\;27}}$

Hence, Option b.) 20 : 27 is the correct option.

$\\\;\large{\underline{\underline{\rm{Thus,\;ratio\;of\;Volumes\;of\;two\;cylinders\;is\;\;\boxed{\bf{20\;:\;27}}}}}}$

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★ MorE FormulaS to KnoW :-

$\\\;\sf{\leadsto\;\;\;TSA\;of\;Cylinder\;=\;2\pi rh\;+\;2\pi r^{2}}$

$\\\;\sf{\leadsto\;\;\;CSA\;of\;Cylinder\;=\;2\pi rh}$

$\\\;\sf{\leadsto\;\;\;TSA\;of\;Cone\;=\;\pi rL\;+\;\pi r^{2}}$

$\\\;\sf{\leadsto\;\;\;CSA\;of\;Cone\;=\;\pi rL}$

$\\\;\sf{\leadsto\;\;\;Volume\;of\;Cube\;=\;(Side)^{3}}$

$\\\;\sf{\leadsto\;\;\;Volume\;of\;Cuboid\;=\;Length\;\times\;Breadth\;\times\;Height}$

$\\\;\sf{\leadsto\;\;\;Volume\;of\;Cone\;=\;\dfrac{1}{3}\;\times\;\pi r^{2}h}$

$\\\;\sf{\leadsto\;\;\;Volume\;of\;Hemisphere\;=\;\dfrac{2}{3}\;\times\;\pi r^{3}}$

$\\\;\sf{\leadsto\;\;\;Volume\;of\;Sphere\;=\;\dfrac{4}{3}\;\times\;\pi r^{3}}$

$\\\;\sf{\leadsto\;\;\;Volume\;of\;Hollow\;Sphere\;=\;\pi (R^{2}\;-\;r^{2})h}$

Answer 2

We know,

Volume of Cylinder = πr²h.

Given that, ratio of radius = 2 : 3.

Ratio of heights = 5 : 3.

$\therefore$ Ratio of Volumes = (V${}_1$)/(V${}_2$)

$\longrightarrow$ (πr²h)/(πR²H)

$\longrightarrow$ r²h ÷ R²H

$\longrightarrow$ (2² × 5)/(3² × 3)

$\longrightarrow$ 20/27

$\longrightarrow$ 20 : 27 (b).