Question

A spherical shell of lead whose external diameter is 18 cm is melted, and the material is used to make a solid right cylinder of height 8 cm and diameter 12 cm. Find the internal diameter of the spherical shell.


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Answer 1

Step-by-step explanation:

Given :

• Diameter of external shell = 18 cm
• Height of cylinder = 8 cm
• Diameter of cylinder = 12 cm

To find :

• Internal diameter of external shell

Solution :

❑ Let's assume the internal radius of the shell be r cm

$\\ {\bold{\rm{\underline{\red{According \: to \: the \: question : }}}}}$

${\bold{\sf{⟼ \: \: \: \: \: Volume_{(material \: of \: the \: shell)} \: = \: \frac{4π}{3}\Big[\Big({\frac{18}{2}\Big)}^{3} - r³\Big]cm³}}}$

❑ We know that

${\bold{⟼ \: \: \: \: \: \: \: \: \: \: }}{\bold{\underline{\boxed{\sf{Volume_{(cylinder)} = πr²h}}}}}$

${\bold{⟼ \: \: \: \: \: \: \: \: \: \: }}{\bold{\sf{Volume_{(cylinder)} \: = \: π{\Big(\frac{12}{2}\Big)}^{2} × 8 \: cm³}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: \frac{4π}{3}(9³ \: - \: r³) \: = \: π \: × \: 6² \: × \: 8}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: 9³ \: - \: r³ \: \: = \: \frac{3}{4} \: × \: 6² \: × \: 8}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: r³ \: = \: 729 \: - \: 216 \: = \: 513}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: log \: r³ \: = \: log \: 513}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: 3 \: log \: r \: = \: 2.7101}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: log r \: = \: 0.9033}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: r \: = \: antilog (0.9033)}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: \: \: \: \: r \: = \: 8.004}}}$

${\bold{⟼ \: \: \: \: \: \: \: \: \: \: }} {\large{\rm{\pink{Internal \: diameter \: \: = \: \: 16 \: cm \: (approx.)}}}}$

Answer 2

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

Given that,

A spherical shell of lead whose external diameter is 18 cm is melted, and the material is used to make a solid right cylinder of height 8 cm and diameter 12 cm.

Case :- 1 Dimensions of Cylinder

Height of cylinder, h = 8 cm

Diameter of cylinder, d = 12 cm

So, Radius of cylinder, r = 6 cm

We know, Volume of cylinder of radius r and height h is

$\red{\boxed{ \sf{ \: Volume_{(cylinder)} = \pi \: {r}^{2} h}}}$

So, on substituting the values, we get

$\rm :\longmapsto\: \: Volume_{(cylinder)} = \pi \times \: {(6)}^{2} \times 8$

$\rm :\longmapsto\: \: Volume_{(cylinder)} = \pi \times \: 36 \times 8$

$\bf\implies \: \: Volume_{(cylinder)} = 288\pi \: {cm}^{3} - - - (1)$

Now,

Case :- 2 Dimensions of Spherical shell

External diameter of spherical shell = 18 cm

So, External radius of spherical shell, x = 9 cm

Let assume that internal radius of spherical shell be y.

We know, Volume of spherical shell of external radius x and internal radius y is

$\purple{\boxed{ \sf{ \: Volume_{(spherical \: shell)} = \dfrac{4}{3} \pi \:( {x}^{3} - {y}^{3})}}}$

So, on substituting the values we get

$\rm :\longmapsto\:Volume_{(spherical \: shell)} = \dfrac{4}{3} \pi \:( {9}^{3} - {y}^{3})$

$\rm :\longmapsto\:Volume_{(spherical \: shell)} = \dfrac{4}{3} \pi \:( 729 - {y}^{3})$

Now, According to statement

A spherical shell of lead whose external diameter is 18 cm is melted, and the material is used to make a solid right cylinder of height 8 cm and diameter 12 cm.

Since, spherical shell is melted and recast in to cylinder.

So,

$\bf\implies \:Volume_{(spherical \: shell)} = Volume_{(cylinder )}$

$\rm :\longmapsto\: \dfrac{4}{3} \pi \:( 729 - {y}^{3}) = 288\pi$

$\rm :\longmapsto\: \dfrac{4}{3} \:( 729 - {y}^{3}) = 288$

$\rm :\longmapsto\: \dfrac{1}{3} \:( 729 - {y}^{3}) = 72$

$\rm :\longmapsto\: 729 - {y}^{3}= 216$

$\rm :\longmapsto\: - {y}^{3}= 216 - 729$

$\rm :\longmapsto\: - {y}^{3}= - 513$

$\rm :\longmapsto\: {y}^{3}= 513$

$\rm :\longmapsto\: {y}^{3}= 3 \times 3 \times 3 \times 19$

$\bf\implies \:y = 3 \sqrt[3]{19}$

So,

$\rm :\longmapsto\:Internal \: diameter \: of \: spherical \: shell \: is$

$\rm \: = \: 2y$

$\rm \: = \: 2 \times 3 \sqrt[3]{19}$

$\rm \: = \: 6 \: \sqrt[3]{19} \: cm$

Additional Information :-

$\boxed{ \sf{ \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2} h}}$

$\boxed{ \sf{ \: Volume_{(cuboid)} = lbh}}$

$\boxed{ \sf{ \: Volume_{(cube)} = {(edge)}^{3} }}$

$\boxed{ \sf{ \: CSA{(cuboid)} = 2(l + b) \times h}}$

$\boxed{ \sf{ \: CSA{(cylinder)} = 2\pi \: rh}}$

$\boxed{ \sf{ \: CSA{(cube)} = 4 \times {(edge)}^{2} }}$

$\boxed{ \sf{ \: CSA{(cone)} = \pi \: rl}}$

$\boxed{ \sf{ \: TSA{(cylinder)} = 2\pi \: r(h + r)}}$

$\boxed{ \sf{ \: TSA{(cone)} = \pi \: r(l + r)}}$