A circular wire of diameter 10,cm is cut and | vedclass

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

Question

(C) The length of the wire is equal to the circumference of the circular wire with diameter $d_1 = 10\,cm$.Length of wire $(l)$ = $\pi 	imes d_1 = 10

Vedclass · Accepted Answer

$\frac{\pi}{5}\,	ext{radian}$

Vedclass · Answer

(C) The length of the wire is equal to the circumference of the circular wire with diameter $d_1 = 10\,cm$.Length of wire $(l)$ = $\pi 	imes d_1 = 10\pi\,cm$.The wire is placed along the circumference of a circle with diameter $d_2 = 1\,m = 100\,cm$.The radius of this circle $(r)$ = $\frac{d_2}{2} = \frac{100}{2} = 50\,cm$.The angle $(	heta)$ subtended by an arc of length $l$ at the centre of a circle of radius $r$ is given by $	heta = \frac{l}{r}$.Substituting the values,$	heta = \frac{10\pi}{50} = \frac{\pi}{5}\,	ext{radian}$.

$A$ circular wire of diameter $10\,cm$ is cut and placed along the circumference of a circle of diameter $1\,m$. The angle subtended by the wire at the centre of the circle is equal to

Similar Questions

The value of the expression $2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1$ is

$\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7} =$

If $y = 3\,\sin\,x + 4\,\cos\,x$,then find the maximum value of $y$.

$ABC$ is a triangle. If $\sin \left(\frac{A+B}{2}\right) = \frac{\sqrt{3}}{2}$,then the value of $\sin \frac{C}{2}$ is

If $\theta$ is in the first quadrant and $\tan \theta = \frac{3}{4},$ then $\frac{\tan \left(\frac{\pi}{2}-\theta\right)-\sin (\pi-\theta)}{\sin \left(\frac{3 \pi}{2}+\theta\right)-\cot (2 \pi-\theta)} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module