KVPY 2020-21 SA Exam Previous Papers

Question : 64 of 80

Marks: +1, -0

A bottle in the shape of a right-circular cone with height h contains some water. When its base is placed on a flat surface, the height of the vertex from the water level is a units. When it is kept upside down, the height of the base from the water level is

a/4

units. Then the ratio

h/a

${1+\√{85}}/{4}$
${1+\√{85}}/{8}$
${1+\√{65}}/{4}$
${1+\√{65}}/{8}$

Validate

Solution:

Volume of water in both the cases will be equal i.e.,

{1}/{3} \πr_{1}^{2}(h-{a}/{4})={1}/{3} \πR^{2} h-{1}/{3} \πr^{2} a

....…(1)

{r}/{a}={R}/{h}

.....…(2)

r_{1} h=R(h-{a}/{4})

....…(3)
By

(1)

and

(2)

and

(3)

, we get

16({h}/{a})^{2}-4({h}/{a})-21=0 \

⇒{h}/{a}={1+\√{85}}/{8}

Go to Question:

Prev Question

Next Question