The ratio of the radii of two spheres whose volume are in the ratio 64:27 is ?

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30 Qs. 30 Marks 30 Mins

Given:

The ratio of the volume of the two spheres = 64 : 27

Formula used:

The volume of the sphere = (4/3) × π × R3

The surface area of the sphere = 4 × π × R2     Where R = The radius of the sphere

Calculation:

Let us assume the ratio of the surface area of the sphere be X : Y and the radius of the spheres be R1 and R2 respectively

⇒ The volume of the first sphere = [(4/3) × π × R13]     ----(1)

⇒ The volume of the second cylinder = [(4/3) × π × R23]     ----(2)

⇒ According to the question equation (1) ÷ (2) = 64 : 27

⇒ (R1/R2)3 = 64/27

⇒ R1/R2 = ∛(64/27)

⇒ R1/R2 = 4/3

⇒ Let us assume the radius of the first sphere = 4x and the second sphere = 3x

⇒ The surface area of the first sphere = 4 × π × (4x)2 = 64πx2     ----(3)

⇒ The surface area of the second sphere = 4 × π × (3x)2 = 36πx2     ----(4)

⇒ The ratio of the surface of the spheres = (64πx2)/(36πx2)

⇒ The ratio of the surface area of the spheres = 16/9

⇒ The ratio of their surface area X : Y = 16 : 9

∴ The required result will be 16 : 9.

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The volumes of the two spheres are in the ratio 64:27. Find the ratio of their surface areas.

Solution

Let the radius of two spheres be $r_1$ and $r_2$.
Given, the ratio of the volume of two spheres $=64:27$

$\frac{V_1}{V_2}=\frac{64}{27}$

$\Rightarrow \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3}=\frac{64}{27}$ $[\because$ volume of sphere $=\frac{4}{3}\pi r^3]$

$\Rightarrow {\left(\frac{r_1}{r_2}\right)}^3={\left(\frac{4}{3}\right)}^3$

$\Rightarrow \frac{r_1}{r_2}=\frac{4}{3}$

Let the surface areas of the two spheres be $S_1$ and $S_2$.

$\therefore \frac{S_1}{S_2}=\frac{4\pi r_1^2}{4\pi r_2^2}={\left(\frac{r_1}{r_2}\right)}^2$ [$\because$ Surface area of sphere $=4\pi r^2$]

$\Rightarrow S_1:S_2={\left(\frac{4}{3}\right)}^2=\frac{16}{9}$

$\Rightarrow S_1:S_2=16:9$

Hence, the ratio of their surface areas is $16:9.$

Mathematics

NCERT Exemplar

Standard IX