Question

a solid composed of a cylinder with hemispherical ends.If the whole length of the solid is 104 cm.and radius of each hemispherical end is 7cm. Then find the cost of polishing it's surface at the rate of rupees 2per SQM
​

Answer 1

Answer:

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

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.