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Answer 1

Question :-

The internal and external radii of spherical shell are 3 cm and 5 cm respectively. It is melted and recast in to a solid cylinder of diameter 14 cm. Find the height of cylinder. Also, find the total surface area of cylinder.

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

Given that,

• Inner radius of spherical shell, x = 3 cm

• External radius of spherical shell, y = 5 cm.

• Diameter of cylinder = 14 cm

• So, radius of cylinder, r = 7 cm

Let assume that height of cylinder be h cm.

As it is given that, spherical shell is melted and recast into a solid cylinder.

We know, If one object is melted and recast in to another object, then volume of first object is equal to volume of second object.

So, it means,

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

$\rm \: \pi \: {r}^{2}h \: = \: \dfrac{4}{3}\pi( {y}^{3} - {x}^{3})$

$\rm \: {r}^{2}h \: = \: \dfrac{4}{3}( {y}^{3} - {x}^{3})$

On substituting the values, we get

$\rm \: {7}^{2} \times h \: = \: \dfrac{4}{3}( {5}^{3} - {3}^{3})$

$\rm \: 49 h \: = \: \dfrac{4}{3}(125 - 27)$

$\rm \: 49 h \: = \: \dfrac{4}{3}(98)$

$\rm \: h \: = \: \dfrac{4}{3} \times 2$

$\rm\implies \:h \: = \: \dfrac{8}{3} \: cm$

Now, Total Surface Area of cylinder is evaluated as

$\rm \:TSA_{(cylinder)}$

$\rm \: = 2\pi \: r \: (h \: + \: r)$

$\rm \: = \: 2 \times \dfrac{22}{7} \times 7 \times \bigg(\dfrac{8}{3} + 7\bigg)$

$\rm \: = \: 44 \times \bigg(\dfrac{8 + 21}{3}\bigg)$

$\rm \: = \: 44 \times \dfrac{29}{3}$

$\rm \: = \: \dfrac{1276}{3} \: {cm}^{2}$

Hence,

$\rm\implies \:TSA_{(cylinder)}= \: \dfrac{1276}{3} \: {cm}^{2}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

Question :-

The internal and external radii of spherical shell are 3 cm and 5 cm respectively. It is melted and recast in to a solid cylinder of diameter 14 cm. Find the height of cylinder. Also, find the total surface area of cylinder.

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

Solution−

Given that,

Inner radius of spherical shell, x = 3 cm

External radius of spherical shell, y = 5 cm.

Diameter of cylinder = 14 cm

So, radius of cylinder, r = 7 cm

Let assume that height of cylinder be h cm.

As it is given that, spherical shell is melted and recast into a solid cylinder.

We know, If one object is melted and recast in to another object, then volume of first object is equal to volume of second object.

So, it means,

\begin{gathered}\rm \: Volume_{(cylinder)} = Volume_{(spherical\:shell)} \\ \end{gathered}

Volume

(cylinder)

=Volume

(sphericalshell)

\begin{gathered}\rm \: \pi \: {r}^{2}h \: = \: \dfrac{4}{3}\pi( {y}^{3} - {x}^{3}) \\ \end{gathered}

πr

2

h=

3

4

π(y

3

−x

3

)

\begin{gathered}\rm \: {r}^{2}h \: = \: \dfrac{4}{3}( {y}^{3} - {x}^{3}) \\ \end{gathered}

r

2

h=

3

4

(y

3

−x

3

)

On substituting the values, we get

\begin{gathered}\rm \: {7}^{2} \times h \: = \: \dfrac{4}{3}( {5}^{3} - {3}^{3}) \\ \end{gathered}

7

2

×h=

3

4

(5

3

−3

3

)

\begin{gathered}\rm \: 49 h \: = \: \dfrac{4}{3}(125 - 27) \\ \end{gathered}

49h=

3

4

(125−27)

\begin{gathered}\rm \: 49 h \: = \: \dfrac{4}{3}(98) \\ \end{gathered}

49h=

3

4

(98)

\begin{gathered}\rm \: h \: = \: \dfrac{4}{3} \times 2 \\ \end{gathered}

h=

3

4

×2

\begin{gathered}\rm\implies \:h \: = \: \dfrac{8}{3} \: cm \\ \end{gathered}

⟹h=

3

8

cm

Now, Total Surface Area of cylinder is evaluated as

\begin{gathered}\rm \:TSA_{(cylinder)} \\ \end{gathered}

TSA

(cylinder)

\begin{gathered}\rm \: = 2\pi \: r \: (h \: + \: r) \\ \end{gathered}

=2πr(h+r)

\rm \: = \: 2 \times \dfrac{22}{7} \times 7 \times \bigg(\dfrac{8}{3} + 7\bigg)=2×

7

22

×7×(

3

8

+7)

\rm \: = \: 44 \times \bigg(\dfrac{8 + 21}{3}\bigg)=44×(

3

8+21

)

\begin{gathered}\rm \: = \: 44 \times \dfrac{29}{3} \\ \end{gathered}

=44×

3

29

\begin{gathered}\rm \: = \: \dfrac{1276}{3} \: {cm}^{2} \\ \end{gathered}

=

3

1276

cm

2

Hence,

\begin{gathered}\rm\implies \:TSA_{(cylinder)}= \: \dfrac{1276}{3} \: {cm}^{2} \\ \end{gathered}

⟹TSA

(cylinder)

=

3

1276

cm

2

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}

MoreFormulae

MoreFormulae

★CSA

(cylinder)

=2πrh

★Volume

(cylinder)

=πr

2

h

★TSA

(cylinder)

=2πr(r+h)

★CSA

(cone)

=πrl

★TSA

(cone)

=πr(l+r)

★Volume

(sphere)

=

3

4

πr

3

★Volume

(cube)

=(side)

3

★CSA

(cube)

=4(side)

2

★TSA

(cube)

=6(side)

2

★Volume

(cuboid)

=lbh

★CSA

(cuboid)

=2(l+b)h

★TSA

(cuboid)

=2(lb+bh+hl)

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