Question

Outer and inner diameters of hemispherical bowl made by silver sheet are 8 cm and 4 cm respectively at its opening. The bowl is melted and a solid cone of diameter of base 8 cm is made. What is the height of the cone?

Answer 1

ɢɪᴠᴇɴ :

Outer diameter of hemispherical bowl is 8 cm and inner diameter is 4 cm. so,

• Outer radius of hemispherical bowl, r₁ = 4 cm
• Inner radius of hemispherical bowl, r₂ = 2 cm
• bowl is melted to form a Cone
• Base diameter of cone is 8 cm therefore; base radius of cone, r = 4 cm

ᴛᴏ ғɪɴᴅ :

• Height of the cone formed, h = ?

ᴋɴᴏᴡʟᴇᴅɢᴇ ʀᴇǫᴜɪʀᴇᴅ :

• Formula to calculate volume of Hemisphere

        V of hemisphere = 2/3 π R³

• Formula for volume of Cone  

       V of cone = 1/3 π r² h

[ where R is radius of hemisphere, r is radius of Cone and h is height of cone ]

sᴏʟᴜᴛɪᴏɴ :

Calculating volume of hemispherical bowl

➠ V of hemispherical bowl = (2/3 π r₁³) - (2/3 π r₂³)

➠ V of hemispherical bowl = 2/3 π ( r₁³ - r₂³ )

➠ V of hemispherical bowl = 2/3 π × ( (4)³ - (2)³ )

➠ V of hemispherical bowl = 2/3 π × 56

➠ V of hemispherical bowl = 112 π/3   cm³

Calculating volume of Cone formed on melting

➠ V of cone = 1/3 π r² h

➠ V of cone = (π/3) (4)² h  cm³

Calculating height of Cone by equating volume of Hemispherical bowl and cone formed

➠ V of hemispherical bowl = V of cone formed

➠ 112 π / 3 = (π / 3) (4)² h

➠ 112 = (4)² h

➠ 112 = 16 h

➠ h = 112 / 16

➠ h = 7 cm

therefore,

• Height of the Cone will be 7 cm.

Answer 2

$\mathcal{\huge{\underline{\underline{\red{Answer:-}}}}}$

➢ Height of the Cone is 56 cm.

$\mathcal{\huge{\underline{\underline{\green{Solution:-}}}}}$

Given :-

• Outer and inner diameters of hemispherical bowl made by silver sheet are 8 cm and 4 cm at its opening.

• The bowl is melted and a solid cone of diameter of base 8 cm is made.

To Find :-

• The height of the cone.

Calculation :-

According to the question,

➝ Volume of the hemispherical bowl =

($\dfrac{2}{3}$ π r₁³) - ($\dfrac{2}{3}$ π r₂³)

➝ $\dfrac{2}{3}$ π ( r₁³ - r₂³ )

➝ $\dfrac{2}{3}$ π × ( (8/2)³ - (4/2)³ )

$\dfrac{2}{3}$ π × ( (4)³ - (2)³ )

➝ $\dfrac{2}{3}$ π × 64-8

$\dfrac{2}{3}$ π × 56

➝ 112 π/3 cm³

Hence, The Volume of the hemispherical bowl is 112 π/3 cm³ .

★ Volume of cone = 1/3 π r² h

➝ $\dfrac{π}{3}$(4)² h cm³

➝ $\dfrac{16 π}{3}$ h cm³

So, Volume of cone is $\</u></strong><strong><u>d</u></strong><strong><u>frac{16 π}{3}$ h cm³.

✦ The bowl is melted and a solid cone is made.

➝ $\dfrac{112 π}{3}$ = $\dfrac{16 π h}{3}$

➝ 112 = 16 h

➝ Height of the Cone = $\cancel{\dfrac{112 }{16}}$

➝ Height of the Cone = 7 cm

Hence, Height of the Cone is 7 cm.

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