Mensuration
- The volume of a right circular cone is 1232 cm3 and its vertical height is 24 cm. Its curved surface area is
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1 πr²h = 1232 3 ⇒ 1 × 22 × r² × 24 = 1232 3 7 ⇒ r²= 1232 × 3 × 7 = 49 22 × 24
⇒ r = √49 = 7 cm.
∴ Slant height ( l ) = √h² + r²
= √24² + 7² = √625 = 25 cm.
∴ Curved surface of cone = πrl= 22 × 7 × 25 = 550 cm² 7 Correct Option: B
1 πr²h = 1232 3 ⇒ 1 × 22 × r² × 24 = 1232 3 7 ⇒ r²= 1232 × 3 × 7 = 49 22 × 24
⇒ r = √49 = 7 cm.
∴ Slant height ( l ) = √h² + r²
= √24² + 7² = √625 = 25 cm.
∴ Curved surface of cone = πrl= 22 × 7 × 25 = 550 cm² 7
- If h, c, v are respectively the height, curved surface area and volume of a right circular cone, then the value of 3πvh³ – c²h² + 9v² is
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Radius of the base of cone = r units
∴ Volume (v) = 1 πr²h 3
Curved surface area = πr√h² + r²
∴ 3πvh³ - c²h² + 9v²= 3π × 1 πr²h × h³ 3 - π²r²(h² + r²)h² + 9 × 1 π²r4h² 9
= π²r²h4 – π²r²h4 – π²r4h² + π²r4h² = 0Correct Option: D
Radius of the base of cone = r units
∴ Volume (v) = 1 πr²h 3
Curved surface area = πr√h² + r²
∴ 3πvh³ - c²h² + 9v²= 3π × 1 πr²h × h³ 3 - π²r²(h² + r²)h² + 9 × 1 π²r4h² 9
= π²r²h4 – π²r²h4 – π²r4h² + π²r4h² = 0
- If the radius of a sphere is increased by 2 cm. its surface area increased by 352 cm². The radius of sphere before change is : use (π = 22/7)
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Let the radius of first sphere be r cm
and the radius of second sphere = (r + 2) cm
Now, Difference between surface area = 352
⇒ 4π {(r + 2)² – r²} = 352
or,4 × 22 {(r + 2 -r) + (r + 2 + r)} = 352 7 ⇒ 2 × 2(r + 1) = 352 × 7 4 × 22 ⇒ r + 1 = 352 × 7 4 × 4 × 22
⇒ r + 1 = 7
∴ r = 7 – 1 = 6 cmCorrect Option: D
Let the radius of first sphere be r cm
and the radius of second sphere = (r + 2) cm
Now, Difference between surface area = 352
⇒ 4π {(r + 2)² – r²} = 352
or,4 × 22 {(r + 2 -r) + (r + 2 + r)} = 352 7 ⇒ 2 × 2(r + 1) = 352 × 7 4 × 22 ⇒ r + 1 = 352 × 7 4 × 4 × 22
⇒ r + 1 = 7
∴ r = 7 – 1 = 6 cm
- Spheres A and B have their radii 40 cm and 10 cm respectively. Ratio of surface area of A to the surface area of B is :
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Surface area of A Surface area of B = 4πr1² = r1² 4πr2² r2²
Where r1 and r2 are radii of spheres A and B respectively.= 40 × 40 = 16 10 × 10 1
⇒16 : 1Correct Option: D
Surface area of A Surface area of B = 4πr1² = r1² 4πr2² r2²
Where r1 and r2 are radii of spheres A and B respectively.= 40 × 40 = 16 10 × 10 1
⇒16 : 1
- The volume of a sphere is
88 × (14)³ cm³ 21
. The curved surface of the sphere is (Take π = 22/7)
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Volume of the sphere = 4 πr³ 3
or= 4 πr³ = 88 × (14)³ 3 21 or 4 × 22 × r³ = 4 × 22 × (14)³ 3 7 3 7
or r = 14
The curved surface of the sphere = 4rπ²= 4 × 22 × 14 × 14 = = 2464 cm². 7 Correct Option: D
Volume of the sphere = 4 πr³ 3
or= 4 πr³ = 88 × (14)³ 3 21 or 4 × 22 × r³ = 4 × 22 × (14)³ 3 7 3 7
or r = 14
The curved surface of the sphere = 4rπ²= 4 × 22 × 14 × 14 = = 2464 cm². 7
