Question

The diameters of the internal and external surfaces of a hollow spherical shell are 6 cm and 10 cm respectively. if it is melted and recast into a solid cylinder of length 8 3 cm, then the diameter of the cylinder will be –

Answer 1

Answer:

The  Height  of the cylinder is 8/3 cm.

Step-by-step explanation:

Given :  

Internal diameter of a hollow spherical shell =6 cm

Internal radius of a hollow spherical shell ,r = 3 cm

External diameter of a hollow spherical shell = 10 cm

External radius of a hollow spherical shell ,R = 5 cm

Diameter of a cylinder = 14 cm

Radius of the cylinder, r1 = 14/2 = 7 cm

Volume of the hollow spherical shell = 4/3π(R³ − r³)

Volume of the solid cylinder = πr1²×h

Since, the hollow spherical shell is melted and recast into a solid  cylinder , so volume of both are equal

Volume of the hollow spherical shell  = Volume of the solid cylinder

4/3π(R³ − r³) = πr²×h

4/3π(5³ - 3³) = π(7)² × h

4/3 (125 - 27) = 49 h

4/3 × 98  = 49 h

h = (4/3 × 98)/49

h = (4 × 98)/ 3 × 49

h = 8/3

Height  of the cylinder = 8/3  cm

Hence, the  Height  of the cylinder is 8/3 cm.

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Answer 2

SOLUTION

$\textbf{Here,\:we\:have:-}$

$\:$

• Diameter of internal surfaces of a hollow spherical shell = 10 cm

$\:$

• $\sf \:\:\:Radius,\: r = \bf\dfrac{10}{2}\:=\:5\:cm$

• $\sf \:\:External \:radii,\:R= \bf\dfrac{6}{2}\:=\:3\:cm$

$\:\:$

$\:\Large{\star}\:\underline{ \boxed{ \sf\small{{ Volume_ {( hollow\:spherical\:cell)}\:=\:\purple{\bf\dfrac{4}{3}\pi(R^2-r^2)}}}}}$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\dashrightarrow\:v_1 = \bf\dfrac{4}{3}\pi (5^2-3^2)cm^3\:\:\:\:\:\:\:\:\:\:\:\:..(1)$

$\:\:$

${\large{\frak{\pmb{\underline{Also\:given, }}}}}$

$\sf \:\:\:\:\:\:\:\:\:\bullet \:\:Length\:of\:solid\: cylinder,\:h=\bf\dfrac{8}{3}$

$\:\:\:\:\:\:\:\:\:\:──────────────────$

$\:$

${\large{\frak{\pmb{\underline{Now,}}}}}$

$\star\:{\bold { \underline{\small{Let \:radius\:of\:solid\: cylinder\:be\:'r' }}}}$

$\:\:$

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

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\mapsto\:v_2 = \pi r^2 \bigg( \bf\dfrac{8}{3}\bigg)cm^3\:\:\:\:\:\:\:\:\:\:\:...(2)$

$\:\:$

$\huge\mathtt\red{\:\:\:\:\:\:\:\:\:\:\:\:\:v_1\:=\:v_2}$

$\:\:$

${\normalsize{\frak{\pmb{\underline{ Equating\:(1)\:and\:(2)}}}}}$

$\:$

$\sf \:\:\: :\implies\bf\dfrac{4}{3}\pi(25-9)\:=\:\pi r^2 \bigg( \bf\dfrac{8}{3}\bigg)$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\dashrightarrow\: \sf \frac{\dfrac{4}{3}\sf \pi(16)}{\pi \bigg(\dfrac{8}{3}\bigg)}\:=\:r^2$

$\:$

$\sf\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\dashrightarrow\:r^2=49\:cm$

$\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\dashrightarrow\: r=7\:cm$

$\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\;\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\looparrowright \:d=2r=14\:cm$

$\:$

$\:\:\:\:\underline{\boxed{\sf{\therefore \: Diameter\:of\: cylinder=\frak{\orange{14\:cm}}}}}$