Question

The radii of two spheres are in the ratio 5 : 7. Find the ratio between their surface areas.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

The radii of two spheres are in the ratio 5 : 7

Let assume that

Radius of first sphere be 5r

Radius of second sphere be 7r

We know,

$\boxed{\sf{  \:\rm \: Surface \: Area_{(sphere)} = 4\pi \: {r}^{2} \: \: }}$

where,

r is the radius of sphere.

So, using this result, surface area of first sphere is

$\rm \: Surface \: Area_{(sphere_{1})} \: = \: 4 \: \pi \: {(5r)}^{2}$

$\rm \: =  \: 4\pi \: (25 {r}^{2})$

$\rm \: =  \: 100\pi \: {r}^{2}$

$\rm\implies \:Surface \: Area_{(sphere_{1})} = 100\pi \: {r}^{2} - - - (1)$

Now, surface area of second sphere of radius 7r is

$\rm \: Surface \: Area_{(sphere_{2})} \: = \: 4 \: \pi \: {(7r)}^{2}$

$\rm \: =  \: 4\pi \: (49 {r}^{2})$

$\rm \: =  \: 196\pi \: {r}^{2}$

$\rm\implies \:Surface \: Area_{(sphere_{2})} = 196\pi \: {r}^{2} - - - (2)$

Now,

$\rm \: Surface \: Area_{(sphere_{1})} : Surface \: Area_{(sphere_{2})}$

$\rm \: =  \: 100\pi \: {r}^{2} : 196\pi \: {r}^{2}$

$\rm \: =  \: 100 : 196$

On dividing both by 4, we get

$\rm \: =  \: 25 : 49$

Hence,

$\rm \: Surface \: Area_{(sphere_{1})} : Surface \: Area_{(sphere_{2})} = 25 : 49$

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Additional Information :-

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²

Answer 2

Answer:

8:34

Step-by-step explanation:

Given the ratio of radii of two sphere is 2:3

Then the ratio of the surface area of two spheres =4π(2r)

⇒4:9

⇒4:9Then ratio of volume of two spheres= 34

3: 34

3: 34 π(3r) 3

⇒8:27

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