Question

volume of two spheres is 64:27. Find difference of their surface area if sum of their radii is 7cm.

Answer 1

Answer:

SA/sa = ¹⁶/₉

Step-by-step explanation:

Let radii of spheres be 'R' and 'r'.

V/v = ⁶⁴/₂₇

(⁴/₃ π R³) / (⁴/₃ π r³) = ⁶⁴/₂₇

R³ / r³ = ⁶⁴/₂₇

R / r = ⁴/₃

Now, let R = 4k and r = 3k

Given that,

R + r = 7

4k + 3k = 7

7k = 7

or k = 1

∴ R = 4 and r = 3

ATQ.

SA/sa

= (4 π R²) / (4 π r²)

= R² / r²

= (4)² / (3)²

= ¹⁶/₉

Answer 2

$\bf{\underline{\underline{Answer:}}}$

$\therefore \:\:88\:cm^2$

$\bold{\underline {Given:}}$

• Volume of two spheres in the ratio = 64: 27

• The sum of their radii is = 7cm

$\bold{\underline {To\:find:}}$

• The difference of their surface areas = ?

$\bf{\underline{\underline{Step\: by\: step \:explanation:}}}$

Let $r_1$ and $r_2$ be the radii of the two squares and $V_1$ and $V_2$ with their corresponding volumes.

$\because\:\:\:\:\:\:\: \dfrac{V_1}{V_2}=\dfrac{64}{27}\\ \\ \implies \dfrac{\dfrac{4}{3}\pi r_1^3}{\dfrac{4}{3}\pi r^3_2}=\dfrac{64}{27}\\ \\ \implies \left(\dfrac{r_1} {r_2}\right)^3=\left(\dfrac{4}{3}\right)^3\\ \\ \implies \dfrac{r_1}{r_2}=\dfrac{4}{3}$

$\therefore \:\:\:\:\:\:\:r_1=\dfrac{4}{3}×r^2\:\:\:\:\:\:\:\:\:\:$ .......(1)

But, $r_1+r_2=7 \implies \dfrac{4}{3}×r^2+r^2=7$[Using(1)]

$\implies \dfrac{7}{3}×r_2=7\\ \\ \implies r_2=3cm\\ \\ And, r_1=\dfrac{4}{3}×r_2=\dfrac{4}{3}×3=4cm$

Now, surface area of the first sphere $(S_1)=4\pi r_1^2=4×\pi×4×4=64\pi cm^2$

and surface area of the second sphere $(S_2)=4\pi r_2^2=4×\pi×3×3=36\pi cm^2$

•°•$S_1-S_2=(64-36)\pi=28\pi$

$=28× \dfrac{22}{7}=4×22=88cm^2$

Hence, the difference of their surface areas = 88cm²