Question

If the semi vertical angle of a cone is $45^{\circ }$ and its height is $20.025cm$, then the approximate value of its curved surface area (in sq.cm) is

• A. $401\pi \sqrt{2}$
• B. $\frac{401\sqrt{2}}{\pi }$
• C. $401\pi \sqrt{3}$
• D. (401.2) $\pi$

Answer 1

Answer: (A)

Let $r$ be the radius, $h$ be the slant height and $\bar{l}$ be the slant height of a cone of semi vertical angle $45^{\circ }$, then, $r=h$ and $l=\sqrt{2}h$
Let $h=20$ and $h+\Delta h=20.025$
$\Rightarrow \Delta h=0.025$
Let $S$ be surface area, then
$S=\sqrt{2}\pi h^2[\because S=\pi rl\Rightarrow S=\pi h\times \sqrt{2}h]$
$\Rightarrow \frac{dS}{dh}=2\sqrt{2}\pi h$
$\Rightarrow {\left(\frac{ds}{dh}\right)}_{h=20}=40\sqrt{2}\pi$
$\therefore \Delta S=\frac{ds}{dh}\Delta h$
$\Rightarrow \Delta s=40\sqrt{2}\pi \times 0.025=\sqrt{2}\pi$
Also, at $h=20$
We have, $S=400\sqrt{2}\pi$
$\therefore S+\Delta S=400\sqrt{2}\pi +\sqrt{2}\pi =401\sqrt{2}\pi$