Question

A solid is composed of a cylinder with hemisphercal ends. If the whole length of the solid is 104 cm and the radius of each of it's hemisphercal ends is 7 cm , find the cost of polishing it's surface at the rate of rupees 10 per dm².

Answer 1

Radius of each hemisphercal end = 7 cm.

Height of each hemisphercal part = It's radius = 7 cm


Height of the cylindrical part = ( 104 - 2 × 7 ) cm = 90 cm.

Area of surface to be polished = 2 ( Curved surface area of the hemisphere ) + ( Curved surface area of the cylinder ).


=> [ 2 ( 2πR² ) + 2πRH ] cm².




=> [ ( 4 × 22/7 × 7 × 7 ) + ( 2 × 22/7 × 7 × 90 ) ] cm².



=> ( 616 + 3960 ) cm².


=> 4576 cm².



=> ( 4576 / 10 × 10 ) dm².



=> 45.76 dm². [ 10 cm = 1 dm ]




Therefore,

Cost of polishing the surface of the solid = Rs ( 45.76 × 10) = Rs 457.60

Answer 2

Answer:

Hemispherical part: r=7cm

Cylindrical part: r=7cm,h=104−(7+7)=90cm

Total surface area of the solid= CSA of cylindricall part+2×CSA of hemispherical part

=2πrh+2×2πr

2

=2πr(h+2r)

=2×

7

22

×7[90+2(7)]cm

2

=44×104cm

2

=4576cm

2

∴ TSA of the solid=4.576cm

2

Cost of polishing at the rate of Rs.4 per 100cm

2

=4576×

100

Rs.4

=Rs.183.04

∴ Cost of polishing the solid=Rs.183.04.