Question

the outer and inner diameters of a hemispherical bowl are 17cm and 15cm. Find the cost of polishing it over at the rate of 25 pise per sq. ​

Answer 1

Answer:

• Cost of polishing the bowl will be 21.45 Rs

Explanation:

Given:-

• The Outer diameter of the hemispherical bowl = 17 cm; Outer radius, R = 17/2  cm
• The Inner radius of the hemispherical bowl = 15 cm; Inner radius, r = 15/2  cm
• Rate of polishing it = 25 paise per cm sq.

To find:-

• Cost of polishing the bowl

Formula required:-

• The formula for CSA of the hemisphere  

        CSA of hemisphere = 2 π (radius)²

• Area of circle

       Area of circle = π (radius)²

Solution:-

→ CSA of the bowl with the outer radius

= 2 π R² = 2 × 22/7 × 17/2 × 17/2 = 3179/7 cm²

→ CSA of the bowl with the inner radius

= 2 π r² = 2 × 22/7 × 15/2 × 15/2 = 2475/7  cm²

→ Area of the circular ring (brim of the bowl)

= π R² - π r² = π (R² - r²) = 22/7 × [(17/2)² - (15/2)²] = 22/7 × ( 289/4 - 225/4 ) = 22/7 × 64/4 = 352/7  cm²

Now,

→ Total area to be polished

= (3179/7) + (2475/7) + (352/7) = 6006/7 = 858 cm²

Rate of polishing the bowl is 25 paise per cm sq. therefore.

→ Cost of polishing the bowl = 25 paise/cm²  × 858 cm² = 21450 Paise = 21.45 Rs

Hence,

• Cost of polishing the bowl will be 21.45 Rs

Answer 2

$\large{\underline{\underline{\pink{\textbf{Given:-}}}}}$

✒ Outer Diameter of Hemispherical Bowl = 17cm

✒ Inner Diameter of Hemispherical Bowl = 15cm

✒ Rate of polishing = 25paise/cm²

$\large{\underline{\underline{\pink{\textbf{Find:-}}}}}$

✑ Cost of polishing it.

$\large{\underline{\underline{\pink{\textbf{Solution:-}}}}}$

❍ Outer Curved Surface Area:

➮ R = D/2

➮ R = 17/2cm

Now, using

➠ Curved Surface area of Hemisphere = 2πr²

✏ Outer Curved Surface area = 2πR²

where,

• π = 22/7
• R = 17/2cm

✗ Substituting these values ✗

⟹ Outer Curved Surface area = 2πR²

⟹ Outer Curved Surface area = 2×22/7×(17/2)²

⟹ Outer Curved Surface area = 44/7×289/4

⟹ Outer Curved Surface area = 12716/28

⟹ Outer Surface area = 454.142(approx.) cm²

⟹ Outer Curved Surface area = 454.14 cm²

❍ Inner Curved Surface Area:

➮ r = d/2

➮ r = 15/2cm

Now, again using

➠ Curved Surface area of Hemisphere = 2πr²

✎ Inner Curved Surface area = 2πr²

where,

• π = 22/7
• r = 15/2cm

✁ Substituting these values ✁

↬Inner Curved Surface area = 2πr²

↬Inner Curved Surface area = 2×22/7×(15/2)²

↬Inner Curved Surface area = 44/7×225/4

↬Inner Curved Surface area = 9900/28

↬Inner Curved Surface area = 353.571(approx.) cm²

↬Inner Curved Surface area = 454.14 cm²

Now, using

❏ Area of brim = πR² - πr²

where,

• π = 22/7
• R = 17/2cm
• r = 15/2cm

• Substituting these values •

❐ Area of brim = πR² - πr²

❐ Area of brim = π(R² - r²)

❐ Area of brim = 22/7{(17/2)² - (15/2)²}

❐ Area of brim = 22/7(289/4 - 225/4)

❐ Area of brim = 22/7(289-225/4)

❐ Area of brim = 22/7(64/4)

❐ Area of brim = 1408/28

❐ Area of brim = 50.285(approx.) cm²

❐ Area of brim = 50.28 cm²

Now,

➺ Area to be polished = outer C.S.A+Inner C.S.A+Area of brim

➺ Area to be polished = 454.14+454.14+50.28

➺ Area to be polished = 958.56 cm²

Cost of polishing:

⟶ Cost of polishing = 25 × 958.56

⟶ Cost of polishing = 23964paise

Hence, Cost of polishing the Hemispherical Bowl is 23964paise