Question

Two spheres's volume is in ratio of 64:27and the sum both radius is 21cm what is the radii of both the spheres ?

Answer 1

Answer:

• Radius of first sphere (r₁) = 12 cm
• Radius of second sphere (r₂) = 9 cm

Step-by-step explanation:

Given:

• Volume of 2 sphere in ratio of 64:27.
• Sum of both radius = 21 cm

To Find:

• Radius of both spheres.

Now, we know that,

$\tt{\implies Volume\;of\;sphere=\dfrac{4}{3}\pi r^{2}}$

Let radius of first sphere be r₁ and radius of second sphere be r₂.

Now, Put the values in the formula,

$\tt{\implies \dfrac{\dfrac{4}{3}\pi r_{1}^{3}}{\dfrac{4}{3}\pi r_{2}^{3}}=\dfrac{64}{27}}$

$\tt{\implies \dfrac{r_{1}^{3}}{r_{2}^{3}}=\dfrac{64}{27}}$

$\tt{\implies \dfrac{r_{1}}{r_{2}}=\Bigg(\dfrac{64}{27}\Bigg)^{\frac{1}{3}}}$

$\tt{\implies \dfrac{r_{1}}{r_{2}}=\dfrac{4}{3}}$

$\tt{\implies r_{1}=\dfrac{4}{3}r_{2}\;\;\;\;\;\;............(1)}$

As it is given that the sum of both radius = 21 cm.

So, r₁ + r₂ = 21      ...........(2)

$\tt{Now,\;replace\;r_{1}\;by\;\dfrac{4}{3}r_{2}.}$

$\tt{\implies \dfrac{4}{3}r_{2}+r_{2}=21}$

$\tt{\implies \dfrac{4r_{2}+3r_{2}}{3}=21}$

$\tt{\implies \dfrac{7r_{2}}{3}=21}$

$\tt{\implies 7r_{2}=63}$

$\tt{\implies r_{2}=\dfrac{63}{7}}$

$\tt{\implies r_{2}=9\;cm}$

Now, put the value of r₂ in equation (2).

⇒ r₁ + r₂ = 21

⇒ r₁ + 9 = 21

⇒ r₁ = 21 - 9

⇒ r₁ = 12 cm.

Hence,

• Radius of first sphere (r₁) = 12 cm
• Radius of second sphere (r₂) = 9 cm

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