Question

the height of the cone is 120cm a small cone is cut off at the top of the plane parallel to the base and its volume is 1/64th. find the volume of original cone and the height​

Answer 1

Answer ⇒ Height of the remaining part from the base = 90 cm.

Step-by-step explanation ⇒  

Height of the Cone = 120 cm.

Volume of the Cutted Cone = 1/64 of V original.

Let the radius of the small cone cutted be x cm.

Therefore, Volume of the Cutted cone = 1/3 πx²h, where h is the Height of small cone.

Volume of the Original cone = 1/3 πR²H, H is the height of the Original cone = 120 cm.

∴ 1/3 πx²h = 1/64 × 1/3πR²(120)

x²h = R² × 120/64

(x/R)²h = 120/64  ----eq(i)

Now, Refer to the attachment,

From the figure,

h/120 = x/R

x/R = h/120

Putting this in the eq(i),

(h/120)² × h = 120/64

∴ h³ = (120 × 120 × 120)/64

∴ h³ = (120/4)³

On Comparing,

h = 120/4

∴ h = 30 cm.

Therefore, Height of the remaining part from the base = H - h

= 120 - 30

= 90 cm.

Hope it helps.

Answer 2

answer : 90cm from the base of which the section is made.

height of the cone , h = 120cm

Let a small cone is cut off at the top by a plane parallel to the base.

length of small cone is l and radius of small cone is x.

a/c to question,

volume of small cone = 1/64 × volume of original cone

or, 1/3 πx²l = 1/64 × πR²h

or, x²l = 1/64 × R² × 120

or, x²l = 15R²/8 ..........(1)

see figure, from ∆ABC and ∆AED

$\angle{ABC}=\angle{AED}$ [corresponding angles ]

$\angle{ACB}=\angle{ADE}$ [ corresponding angles ]

so, $\Delta ABC\sim\Delta AED$

$\therefore\frac{AC}{AD}=\frac{CB}{DE}$

$\frac{l}{h}=\frac{x}{R}$

or, lR = hx = 120x

or, l = 120x/R .......(2)

from equations (1) and (2),

x² × (120x/R) = 15R²/8

or, 120x³ = 15R³/8

or, 64x³ = R³

or, x/R = 1/4

hence, x = R/4 , put it in equation (2),

l = 120 × (R/4)/R = 30cm

hence, (120cm - 30cm) = 90cm from the base of which the section is made.