A bottle in the shape of a right-circular con | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $R$ be the radius of the base of the cone and $h$ be its height. Let $V$ be the volume of water.Case $1$: Base is on the surface. The empty pa

Vedclass · Accepted Answer

$\frac{1+\sqrt{85}}{8}$

Vedclass · Answer

(B) Let $R$ be the radius of the base of the cone and $h$ be its height. Let $V$ be the volume of water.Case $1$: Base is on the surface. The empty part is a smaller cone at the top with height $a$. By similar triangles,the radius of the water surface is $r = \frac{R}{h}(h-a)$. The volume of water is $V = \frac{1}{3}\pi R^2 h - \frac{1}{3}\pi r^2 a = \frac{1}{3}\pi R^2 h - \frac{1}{3}\pi (\frac{R}{h}(h-a))^2 a = \frac{1}{3}\pi R^2 h [1 - (\frac{h-a}{h})^2 \frac{a}{h}] = \frac{1}{3}\pi R^2 h [1 - (1 - \frac{a}{h})^2 \frac{a}{h}]$.Case $2$: Upside down. The water forms a smaller cone at the bottom with height $h - \frac{a}{4}$. The radius of the water surface is $r_1 = \frac{R}{h}(h - \frac{a}{4})$. The volume of water is $V = \frac{1}{3}\pi r_1^2 (h - \frac{a}{4}) = \frac{1}{3}\pi [\frac{R}{h}(h - \frac{a}{4})]^2 (h - \frac{a}{4}) = \frac{1}{3}\pi R^2 h (\frac{h - a/4}{h})^3$.Equating the volumes: $1 - (1 - \frac{a}{h})^2 \frac{a}{h} = (1 - \frac{a}{4h})^3$. Let $x = \frac{a}{h}$. Then $1 - (1-x)^2 x = (1 - \frac{x}{4})^3$.$1 - x(1 - 2x + x^2) = 1 - 3(\frac{x}{4}) + 3(\frac{x}{4})^2 - (\frac{x}{4})^3$.$1 - x + 2x^2 - x^3 = 1 - \frac{3x}{4} + \frac{3x^2}{16} - \frac{x^3}{64}$.Multiply by $64$: $64 - 64x + 128x^2 - 64x^3 = 64 - 48x + 12x^2 - x^3$.$63x^3 - 116x^2 + 16x = 0$. Since $x 
eq 0$,$63x^2 - 116x + 16 = 0$.Using the quadratic formula for $\frac{h}{a} = \frac{1}{x}$: $x = \frac{116 \pm \sqrt{116^2 - 4(63)(16)}}{2(63)} = \frac{116 \pm \sqrt{13456 - 4032}}{126} = \frac{116 \pm \sqrt{9424}}{126} = \frac{116 \pm 8\sqrt{147.25}}{126}$.Re-evaluating the volume equation: The volume of water in Case $1$ is the total volume minus the volume of the empty cone: $V = \frac{1}{3}\pi R^2 h - \frac{1}{3}\pi (\frac{R}{h}a)^2 a = \frac{1}{3}\pi R^2 h (1 - \frac{a^3}{h^3})$.Equating: $1 - \frac{a^3}{h^3} = (1 - \frac{a}{4h})^3$. Let $k = \frac{h}{a}$. $1 - \frac{1}{k^3} = (1 - \frac{1}{4k})^3 = (\frac{4k-1}{4k})^3$.$\frac{k^3-1}{k^3} = \frac{(4k-1)^3}{64k^3} \Rightarrow 64(k^3-1) = (4k-1)^3 = 64k^3 - 48k^2 + 12k - 1$.$64k^3 - 64 = 64k^3 - 48k^2 + 12k - 1 \Rightarrow 48k^2 - 12k - 63 = 0$.Divide by $3$: $16k^2 - 4k - 21 = 0$. $k = \frac{4 \pm \sqrt{16 - 4(16)(-21)}}{32} = \frac{4 \pm \sqrt{16 + 1344}}{32} = \frac{4 \pm \sqrt{1360}}{32} = \frac{4 \pm 4\sqrt{85}}{32} = \frac{1 \pm \sqrt{85}}{8}$.Since $k > 0$,$k = \frac{1+\sqrt{85}}{8}$.

Similar Questions

If the intercepts made by a variable circle on the $X$-axis and $Y$-axis are $8$ and $6$ units respectively,then the locus of the centre of the circle is

The locus of the center of a circle which touches the circle $x^{2} + y^{2} - 6x - 6y + 14 = 0$ externally and also touches the $y$-axis is:

Let $S = 0$ be the locus of the center of a variable circle which intersects the circle $x^2 + y^2 - 4x - 6y = 0$ orthogonally at the point $(4, 6)$. If $P$ is a variable point on $S = 0$,then the least value of $OP$ is (where $O$ is the origin).

The locus of the centre of a circle which cuts orthogonally the circle $x^2 + y^2 - 20x + 4 = 0$ and which touches the line $x = 2$ is:

If a point $P$ moves such that the distance from $(0, 2)$ to $P$ is $\frac{1}{\sqrt{2}}$ times the distance of $P$ from $(-1, 0)$,then the locus of the point $P$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module