Question

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tomorrow is my exam
Chapter Surface area and volume
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otherwise 30 answers will be deleted.
I know the answer please give full explanation​

Answer 1

Answer:

The increase in surface are is 300% when the radius of sphere is doubled . Therefore it is included that if the radius of the sphere is doubled then the ratio of area surface is 4 times the older area . The ratio of the surface area of two sphere is 3:5.

Answer 2

Answer :-

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of Total Surface Areas of Sphere has been used. We need to find the ratio of initial Sphere and final sphere whose radius is doubled than first. So first we shall assume the radius of initial sphere. Then we will find its Total Surface Area. Since, CSA and TSA of Sphere are equal so we can find ratio of Surface areas of any using any formula.

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★ Equations Used :-

$\\\;\boxed{\sf{Total\;Surface\;Area\;of\;Sphere\;=\;\bf{4\pi (radius)^{2}}}}$

$\\\;\boxed{\sf{Radius\;of\;Final\;Sphere\;=\;\bf{2\;\times\;Radius\;of\;Initial\;Sphere}}}$

$\\\;\boxed{\sf{Ratio\;of\;Surface\;Areas\;=\;\bf{\dfrac{Surface\;Area\;of\;Initial\;Sphere}{Surface\;Area\;of\;Final\;Sphere}}}}$

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★ Solution :-

Let us assume the radius of initial sphere to be 'r'.

And the radius of final sphere be 'R'.

Then,

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~ For the Volume of Initial Sphere :-

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Total\;Surface\;Area\;of\;Sphere\;=\;\bf{4\pi (radius)^{2}}}$

Note :- Here we won't substitute the value of π since it is going to be cancelled at end.

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Total\;Surface\;Area\;of\;Sphere_{(Initial)}\;=\;\bf{4\:\times\:\pi\:\times\:(r)^{2}}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Total\;Surface\;Area\;of\;Sphere_{(Initial)}\;=\;\underline{\underline{\bf{4\:\pi\:r^{2}}}}}$

$\\\;\underline{\boxed{\tt{Surface\;\;Area\;\;of\;\;Initial\;\;Sphere\;=\;\bf{4\:\pi\:r^{2}}}}}$

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~ For the Radius of Final Sphere :-

Clearly, its given that the radius of sphere is doubled.

So,

$\\\;\;\;\;\;\sf{:\Rightarrow\;\;\;Radius\;of\;Final\;Sphere,\;R\;=\;\bf{2\;\times\;Radius\;of\;Initial\;Sphere}}$

$\\\;\;\;\;\;\sf{:\Rightarrow\;\;\;Radius\;of\;Final\;Sphere,\;R\;=\;\bf{2\;\times\;r}}$

$\\\;\;\;\;\;\sf{:\Rightarrow\;\;\;Radius\;of\;Final\;Sphere,\;R\;=\;\underline{\underline{\bf{2r}}}}$

$\\\;\underline{\boxed{\tt{Radius\;\;of\;\;Final\;\;Sphere\;=\;\bf{2r}}}}$

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~ For the Volume of Final Sphere :-

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Total\;Surface\;Area\;of\;Sphere\;=\;\bf{4\pi (radius)^{2}}}$

Note :- Here we are not going to substitute the value of π since it is going to be cancelled at the end.

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Total\;Surface\;Area\;of\;Sphere_{(Final)}\;=\;\bf{4\:\times\:\pi\:\times\:(R)^{2}}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Total\;Surface\;Area\;of\;Sphere_{(Initial)}\;=\;\underline{\underline{\bf{4\:\pi\:R^{2}}}}}$

$\\\;\underline{\boxed{\tt{Surface\;\;Area\;\;of\;\;Initial\;\;Sphere\;=\;\bf{4\:\pi\:R^{2}}}}}$

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~ For the Ratio of Surface Areas of Two Spheres :-

• Area of Initial Sphere = 4πr²

• Area of Final Sphere = 4πR²

Then,

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Ratio\;of\;Surface\;Areas\;=\;\bf{\dfrac{Surface\;Area\;of\;Initial\;Sphere}{Surface\;Area\;of\;Final\;Sphere}}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Ratio\;of\;Surface\;Areas\;=\;\bf{\dfrac{4\:\pi\:r^{2}}{4\:\pi\:R^{2}}}}$

Cancelling 4π from numerator and denominator, we get,

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Ratio\;of\;Surface\;Areas\;=\;\bf{\dfrac{r^{2}}{R^{2}}}}$

Also, from above results, R = 2r. Then,

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Ratio\;of\;Surface\;Areas\;=\;\bf{\dfrac{r^{2}}{(2r)^{2}}}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Ratio\;of\;Surface\;Areas\;=\;\bf{\dfrac{r^{2}}{4\;r^{2}}}}$

Cancelling r² from numerator and denominator, we get,

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Ratio\;of\;Surface\;Areas\;=\;\underline{\underline{\bf{\dfrac{1}{4}}}}}$

Hence, Option C.) 1 : 4 is the correct option.

$\\\;\large{\underline{\underline{\rm{Thus,\;ratio\;of\;Surface\;Areas\;of\;Two\;Spheres\;is\;\;\boxed{\bf{1\;:\;4}}}}}}$

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★ More to know :-

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

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

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

$\\\:\sf{\leadsto\;\;\;Volume\;of\;Cylinder\;=\;\pi r^{2}h}$

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

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

$\\\:\sf{\leadsto\;\;\;CSA\;of\;Hemisphere\;=\;2\:\times\:\pi r^{2}}$

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

$\\\:\sf{\leadsto\;\;\;Volume\;of\;Hollow\;Sphere\;=\;\dfrac{4}{3}\:\times\:\pi (R^{3}\;-\;r^{3})}$

$\\\:\sf{\leadsto\;\;\;Volume\;of\;Hollow\;Hemisphere\;=\;\dfrac{2}{3}\:\times\:\pi (R^{3}\;-\;r^{3})}$