F.3 Maths (4)

1)
Let h cm be the height of the cone cut off.
h/(h + 8) = 3/6
2h = h+8
h=9

Volume of frustum
= [1/3 x Pi x 6 x 6 x (8 + 8) - 1/3 x Pi x 3 x 3 x 8] cm^3
= (192 Pi - 24 Pi ) cm^3
= 168 Pi cm^3

Volume of the semi-sphere
= (1/2 x 4/3 x Pi x 6^3) cm^3
= 144 Pi cm^3

So,volume of X
= (168 Pi + 144 Pi) cm^3
= 312 Pi cm^3, Ans.

Volume of X: Volume of Y
= [sqrt(Surface Area of X: Surface Area of Y)]^3
= (4 : 9) ^ (2/3)
= 8 : 27

So, Volume of Y
= [312 Pi / ( 8 / 27) ]cm^3
= 1053 cm^3, Ans.

2a)Let r cm be the radius of the water surface.
r / 18 = 8 / 24
r = 6

So, Volume of water contained in the vessel
= (1/3 x Pi x 6 x 6 x 8) cm^3
= 96 Pi cm^3, Ans.

2bi) Slant height of the water in the vessel
= sqrt(6^2 + 8^2) cm　　(Pyth. theorem)
=10 cm

So, area of the wet curved suface
= (Pi x 6 x 10 )cm^2
= 60 Pi cm^2, Ans.

ii)Volume of water in the bigger vessel: that in the smaller vessel = 1 : 1 (given)

So, area of wet curved surface of the bigger vessel: that of the smaller vessel = 1 : 1

Hence, area of wet curved surface of the bigger vessel
= Area of the wet curved suface of the smaller vessel
= 60 Pi cm^2, Ans.

1) 1.let h be the height of the small vessel
(h+8)/h =6/3
h+8=2h
h=8
2.volume of solid X=volume of frustum + volume of semi-sphere
=6^2*(8+8)π/3 -3^2*8π//3 + 1/2 x4π//3 x6^3
=312π
(^2=square root.^3=cubic root)
3.let yπ be the volume of solid Y
y/312 =9/4
y=702
so,the volume of solid Y=702π cm^3