Q. 755.0( 1 Vote )
Match the following columns:
a—(q), b—(s), c—(p), d—(r)
a) Given: The radii of the circular ends of a bucket are 20 cm and 10 cm respectively.
Height of the bucket is 30cm
Bucket is in the shape of frustum.
Let V be the Volume of the Bucket(Frustum)
Volume of the frustum is given by: × h × (R2 + r2 + Rr) (here r and R are the radii of smaller and larger circular ends respectively)
∴ V = × h × (R2 + r2 + R × r)
⇒ V = × 30 × (202 + 102 + 20 × 10)
⇒ V = × 30 × (400 + 100 + 200) = × 30 × (700)
⇒ V = × 30 × (700) = 22000 cm3
∴ The capacity of the bucket is: 22000 cm3
b) Given: Height of the frustum of a cone: 15 cm
radii of the Circular ends: 28cm and 20 cm.
Here slant height h can be found by using Pythagoras theorem.
∴ s2 = h2 + (R-r)2 (here R is 28cm and r is 20cm)
⇒ s2 = 152 + (28—20)2
⇒ s2 = 152 + (8)2
⇒ s2 = 225 + 64
⇒ s2 = 289
⇒ s = √289 = 17
∴ Slant height 0f the Frustum is 17cm
c) Given: The radii of the end of a bucket are 33 cm and 27 cm and its slant height is 10 cm
TSA of a frustum of a cone = πl(r1 + r2) + πr12 + πr22 (here l , r1, r2 are the slant height, radii of the frustum)
Let S be the TSA of the Frustum.
∴ S = πl(r1 + r2) + π(r2)2 + π(r1)2
⇒ S = π × 10 × (33 + 27) + π(33)2 + π(27)2
⇒ S = π(600 + 1089 + 729) = 2418π
∴ TSA of frustum is 2418π.
d) Given: Three solid metallic sphere of radii 3 cm, 4 cm and 5 cm.
Volume of the Solid sphere is: πr3 (here r is the radius of the sphere).
Let V1 be the volume of the sphere with radius 3cm.
Let V2 be the volume of the sphere with radius 4cm.
Let V3 be the volume of the sphere with radius 5cm.
∴ V1 = πr3 = π(3)3
∴ V2 = πr3 = π(4)3
∴ V3 = πr3 = π(5)3
Let V be the Volume of the sphere that is formed by melting the spheres with volumes V1 , V2 , V3 .
∴ V = V1 + V2 + V3 = π(3)3 + π(3)3 + π(3)3
⇒ V = π(33 + 43 + 53) = π(27 + 64+ 125) = π(216)
⇒ V = π(216) = π(6)3
Here, we can see that radius of the Sphere is 6cm, ∴ diameter = 2 × 6 = 12cm
∴ Diameter of sphere formed by melting spheres with volumes V1, V2 , V3 is 12cm
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