. The height of the cone is 20 cm. A small cone is cut off from the top by a plane parallel to base. If its volume be 1 / 125 of the volume of the original cone determine at what height above the base the section is made.
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Solution:-
Given: Height of the cone = 20 cm
Let the small cone is cut off at a height 'h' from the top.
Let the radius of the big cone be 'R' and of small cone be 'r'
Let the volume of the big cone be V1 and of small cone be V2
Volume of the big cone = 1/3R²h
= 1/3πR²*20
= 20/3πR² cu cm
Volume of small cone = 1/3πr²h
⇒ V2 = 1/125th of the volume of big cone.
⇒ V2/V1 = 1/125
⇒ (1/3πr²h)/(20/3πR²) = 1/125
⇒ r²h/20R² = 1/125
⇒ r²/R² × h/20 = 1/125 ...(1)
⇒ From the figure. Δ ACD ~ Δ AOB (By AA similarity criterion)
⇒ r/R = h/20
Putting this value of r/R = h/20 in equation (1), we get.
(h/20)² × (h/20) = 1/125
(h/20)³ = 1/125
(h/20)³ = (1/5)³
h/20 = 1/5
h = 20/5
h = 4 cm
So, the height above the base where the section is made is 20 - 4 = 16 cm
therefore answer = 16
Given: Height of the cone = 20 cm
Let the small cone is cut off at a height 'h' from the top.
Let the radius of the big cone be 'R' and of small cone be 'r'
Let the volume of the big cone be V1 and of small cone be V2
Volume of the big cone = 1/3R²h
= 1/3πR²*20
= 20/3πR² cu cm
Volume of small cone = 1/3πr²h
⇒ V2 = 1/125th of the volume of big cone.
⇒ V2/V1 = 1/125
⇒ (1/3πr²h)/(20/3πR²) = 1/125
⇒ r²h/20R² = 1/125
⇒ r²/R² × h/20 = 1/125 ...(1)
⇒ From the figure. Δ ACD ~ Δ AOB (By AA similarity criterion)
⇒ r/R = h/20
Putting this value of r/R = h/20 in equation (1), we get.
(h/20)² × (h/20) = 1/125
(h/20)³ = 1/125
(h/20)³ = (1/5)³
h/20 = 1/5
h = 20/5
h = 4 cm
So, the height above the base where the section is made is 20 - 4 = 16 cm
therefore answer = 16
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