Question

three metallic balls of the radii 4cm, 5cm and 6cm are mettal and formed a large matalic ball that find the TSA of grater ball
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Answer 1

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Answer 2

Answer:

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• Three metallic balls of the radii 4cm, 5cm and 6cm are melted and formed a large metallic ball. Find TSA of greater ball.

★ANSWER★

Given :-

• Three metallic balls of radius 4 cm, 5 cm and 6 cm. They are melted and recast in to other metallic ball.

To find :-

• Total Surface Area of largest ball

Formula used :-

$\small \bf Volume \: of \: sphere = \dfrac{4}{3}\pi \: {r}^{3}$

$\small \bf Surface \: area \: of \: sphere \: = 4\pi \: {r}^{2}$

Solution:-

Let radius of three balls be represented as

x = 4 cm

y = 5 cm

z = 6 cm

Let radius of resultant metallic ball be 'r' cm.

According to condition,

$\dfrac{4}{3}\pi \: {x}^{3} + \dfrac{4}{3}\pi \: {y}^{3} + \dfrac{4}{3}\pi \: {z}^{3} = \dfrac{4}{3}\pi \: {r}^{3}$

$\bf \implies \: {x}^{3} + {y}^{3} + {z}^{3} = {r}^{3}$

Put the value of x, y and z, we get

$\bf \implies \: {4}^{3} + {5}^{3} + {6}^{3} = {r}^{3}$

$\bf \implies \: {r}^{3} = 64 + 125 + 216$

$\bf \implies \: {r}^{3} = 405$

$\bf \implies \: r = \sqrt[3]{405} = \sqrt[3]{5 \times 3 \times 3 \times 3 \times 3}$

$\bf \implies \:r = 3 \sqrt[3]{15}$

So, total surface area of greater ball is

$\small \bf = 4\pi \: {r}^{2}$

$\bf \implies \: 4 \times \dfrac{22}{7} \times {(3 \sqrt[3]{15} )}^{2}$

$\bf \implies \: \dfrac{88}{7} \times9 \times {( \sqrt[3]{15} )}^{2}$

$\bf \implies \:\dfrac{792}{7} \times {( \sqrt[3]{15} )}^{2} {cm}^{2}$

$\bf \implies \:Surface \: area \: of \: sphere = \dfrac{792}{7} {( \sqrt[3]{15}) }^{2} \: {cm}^{2}$

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