Mensuration
- The radius of the base and height of a right circular cone are in the ratio 5 : 12. If the volume of the cone is
314 2 cm³ 7
, the slant height (in cm) of the cone will be
-
View Hint View Answer Discuss in Forum
Let the radius of the base of the cone be 5x cm and its height be 12x cm.
∴ V = 1 πr²h 3 = 314 2 = 1 × 22 × 5x × 5x × 12x 7 3 7 ⇒ x³ = 2200 × 3 × 7 = 1 7 × 22 × 25 × 12
⇒ x = 1
∴ Slant height of the cone = √5² + 12² = √25 + 144 = √169 = 13 cm.
[Note : For a right circular cone, 5² + 12² = 13²]
Correct Option: B
Let the radius of the base of the cone be 5x cm and its height be 12x cm.
∴ V = 1 πr²h 3 = 314 2 = 1 × 22 × 5x × 5x × 12x 7 3 7 ⇒ x³ = 2200 × 3 × 7 = 1 7 × 22 × 25 × 12
⇒ x = 1
∴ Slant height of the cone = √5² + 12² = √25 + 144 = √169 = 13 cm.
[Note : For a right circular cone, 5² + 12² = 13²]
- Two solid right cones of equal height and of radii r1 and r2 are melted and made to form a solid sphere of radius R. Then the height of the cone is
-
View Hint View Answer Discuss in Forum
Let the height be h units.
∴ 1 πh(r1² + r2²) = 4 π R³ 3 3
⇒ h(r1² + r2²) = 4R³∴ V = 1 πr²h 3 Correct Option: C
Let the height be h units.
∴ 1 πh(r1² + r2²) = 4 π R³ 3 3
⇒ h(r1² + r2²) = 4R³∴ V = 1 πr²h 3
- The ratio of the volume of two cones is 2 : 3 and the ratio ofradii of their base is 1 : 2. The ratio of their height is
-
View Hint View Answer Discuss in Forum
Let the height be h1 and h2 and radii be r and 2r respectively.
∴ = 1 πr²h1 V1 3 V2 1 π(2r)² × h2 3 ⇒ r² × h1 = 2 4r² × h2 3 ⇒ h1 = 2 × 4 = 8 = 8 : 3 h2 3 1 3 Correct Option: B
Let the height be h1 and h2 and radii be r and 2r respectively.
∴ = 1 πr²h1 V1 3 V2 1 π(2r)² × h2 3 ⇒ r² × h1 = 2 4r² × h2 3 ⇒ h1 = 2 × 4 = 8 = 8 : 3 h2 3 1 3
- If a right circular cone is separated into solids of volumes V1, V2 , V3 by two planes parallel to the base, which also trisect the altitude, then V1 : V2 : V3 is
-
View Hint View Answer Discuss in Forum
Let FQ = r1, DP = r2 and BO = r3
Also, AQ = QP = PO = h/3
From ∆AFQ and ∆ADP,
FQ = AQ BO AO ⇒ r1 = 1 ⇒ r3 = 3r1 r3 3 ∴ V1 : V2 : V3 = 1 πr1² × h : 1 π 1 (r1² + r2² + r1r2) : 1 π 1 (r2² + r3² + r2r3) 3 3 3 3 3 3
= r1² : (r1² + 4r1² + 2r1²) : (4r1² + 9r1² + 6r1²)
= r1² : 7r1² : 19r1²
= 1 : 7 : 19Correct Option: D
Let FQ = r1, DP = r2 and BO = r3
Also, AQ = QP = PO = h/3
From ∆AFQ and ∆ADP,FQ = AQ BO AO ⇒ r1 = 1 ⇒ r3 = 3r1 r3 3 ∴ V1 : V2 : V3 = 1 πr1² × h : 1 π 1 (r1² + r2² + r1r2) : 1 π 1 (r2² + r3² + r2r3) 3 3 3 3 3 3
= r1² : (r1² + 4r1² + 2r1²) : (4r1² + 9r1² + 6r1²)
= r1² : 7r1² : 19r1²
= 1 : 7 : 19
- If the radii of the circular ends of a truncated conical bucket which is 45cm high be 28 cm and 7 cm, then the capacity of the bucket in cubic centimetre is ( use π = 22/7)
-
View Hint View Answer Discuss in Forum
Volume of bucket = 1 πh (r1² + r2² + r1r2) 3 = 1 × 1 × 45( 28² + 7² + 28 × 7) 3 7 = 1 × 22 × 45( 784 + 49 + 196) 3 7 = 1 × 22 × 45 × 1029 = 48510 cu.cm. 3 7 Correct Option: A
Volume of bucket = 1 πh (r1² + r2² + r1r2) 3 = 1 × 1 × 45( 28² + 7² + 28 × 7) 3 7 = 1 × 22 × 45( 784 + 49 + 196) 3 7 = 1 × 22 × 45 × 1029 = 48510 cu.cm. 3 7
