Question

A cone used as an oil funnel has a radius of 5 in. and a height of 10 in. What volume of oil could it hold if the opening at the bottom is plugged?

Answer 1

Answer:

part =

2

18

=9 cm

Radius (r) of lower circular end of frustum part = Radius of circular end of cylindrical part =

2

8

=4 cm

Height (h

1

) of frustum part =22−10=12 cm

Height (h

2

) of cylindrical part =10 cm

Slant height (l) of frustum part =

(R−r)

2

+h

1

2

=

(9−4)

2

+12

2

=

25+144

=

169

=13 cm

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part

=π(R+r)l+2πrh

2

=

7

22

×(9+4)×13+2×

7

22

×4×10

=

7

22

[169+80]=

7

22×249

=782

7

4

cm

2