Question

MCQ


If a frustum,having radio of it's circular ends as x and 2x and its height as 3x,melted to form a cube,then the side of cube is of length

$1. \sqrt[3]{7} x \\ \\ 2. \sqrt[3]{3} \pi \: x \\ \\ 3. \sqrt[3]{7} \pi \: x \\ \\ 4. \sqrt[3]{2\pi \: x}$
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Answer 1

Hey there!

Given,

Radius of lower end (R)= 2x

Radius of upper end (r) = x

Height (h) = 3x

We know,

$\textsf{Volume of frustum} = \dfrac{\pi h}{3} (R^2 + Rr + r^2)\\\\\textsf{Volume of frustum} = \dfrac{\pi \ 3x}{3} [(2x)^2 + 2x \times x + x^2)\\\\\textsf{Volume of frustum} = \pi x }(4x^2 + 2x^2 + x^2)\\\\\textsf{Volume of frustum} =\pi x(7x^2)\\\\\boxed{\textsf{Volume of frustum} = 7 \pi x^3}$

Now,

Volume of cube = Volume of frustum

$\sf a^3 = \sqrt[3]{7\pi x^3}\\\\\boxed{a = x \sqrt[3]{7\pi}}$

$\textsf{Hence, length of side of cube is option }3. \sqrt[3]{7\pi}}\ x.$

Hope it Helps You!

Answer 2

Yo

Let R and r be the radii of the frustum

Volume = π × 3x/3 [(2x)² + x² + 2x(x)]

» V = π × x × 7x²

» V = 7πx³

Volume of Cube = Vol of Frustum

» s³ = 7πx³

» s = x 3√7x

Option (3) is correct

TGA