A solid wooden toy is in the from of a hemisphere surmounted by a cone of same radius.the radius of hemisphere
Answers
Plz complete the question.
Q.A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. The radius of hemisphere is 3.5cm and the total wood used in the making of toy is 166 5/6cm³. Find the height of the toy. Also,find the cost of painting the hemi-spherical part of the toy at the rate of ₹ 10 per cm².(use π = 22/7 ).
SOLUTION:
Given: Radius of hemisphere = radius of cone = r = 3.5 cm
Volume of total wood used in making a toy=166⅚=1001/6 cm³
Let h be height of a cone.
Volume of total wood used in making a toy = volume of hemisphere + volume of cone
=(⅔) × πr³ + (⅓)π²h
= (⅓)πr²(2r + h)
1001/6 = (⅓) ×( 22/7) × (3.5)² × (h +2×3.5)
1001/6 = (⅓) ×( 22/7) × 3.5 ×3.5 ×( h+7)
1001/6 = (⅓) × 22 × .5 ×3.5 × (h +7)
1001 × 3 = 6 × 22 × .5 ×3.5 × (h+7)
h +7 = 1001 × 3 / (6 × 22 × .5 ×3.5)
h +7 = 1001 × 3 / 132 × 1.75
h +7 = 1001 × 3 × 100 / 132 × 175
h +7 = 91 × 3 × 4 / 12 × 7
h +7 = 91 × 12 / 12×7
h +7 = 13
h + 7 = 13 -7 = 6
Height of a cone(h) = 6
Height of the toy = Height of a cone + Height of a hemisphere
Height of the toy=6+3.5=9.5 cm
Curved surface area of hemisphere = 2πr²
CSA of hemisphere = 2 ×( 22/7) × 3.5 × 3.5
= 2 × 22 × .5 × 3.5 = 44 × 1.75 = 77 cm²
Rate of painting the hemispherical part of the toy = ₹ 10 per m².
Cost of painting the hemispherical part of the toy = 77 × 10 = ₹ 770 .
Hence, the Height of the toy is 9.5 m & Cost of painting the hemispherical part of the toy is ₹ 770.
HOPE THIS WILL HELP YOU....