Question

A phonograph turntable rotates at ${\text{50rev}}\,{\min ^{ - 1}}$. Its angular speed in ${\text{rad}}\,{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$ is?
$\left( A \right){\text{8}}{\text{.164rad}}\,{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$
$\left( B \right){\text{5}}{\text{.23rad}}\,{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$
$\left( C \right){\text{52}}{\text{.3rad}}\,{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$
$\left( D \right){\text{81}}{\text{.64rad}}\,{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$

Answer 1

Here we know the height of the object, height of the image and distance of the object is known to us. Now using the magnification formula of the concave mirror we can find the relation between the heights of the object, height of the image, image distance and the object distance. From that we can find the value of the image distance from the concave mirror. Now using the mirror formula we can calculate the focal length of the mirror.

Complete step by step solution:
As per the problem we have a phonograph turntable rotates at ${\text{50rev}}\,{\min ^{ - 1}}$.
Now we need to find the angular speed of the phonograph in terms of ${\text{rad}}\,{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$.
As we know that,
$1$ revolution = $6.28$ radian
$1$ minute = $60$ seconds
Now converting the above value in the required terms we will get,
${\text{50rev}}\,{\min ^{ - 1}} \to \dfrac{{{\text{50}} \times {\text{6}}{\text{.28rad}}}}{{60\sec }}$
Now on solving further we can write,
${\text{50rev}}\,{\min ^{ - 1}} \to {\text{5}}{\text{.28rad}}\,{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$
Hence the angular speed of the phonograph is ${\text{5}}{\text{.28rad}}\,{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$.

Note:
Remember that the angular speed of a rotating body is the rate of rotation around an axis usually expressed in radians or revolutions per second or per minute. We can calculate it as the ratio of total distance traveled by total time taken by the rotating object.