Question

A short bar magnet allowed it to fall along the axis of the horizontal metallic ring. Starting from rest, the distance falls by the magnet in one second may be:
A. $4.0m$
B. $5.0m$
C. $6.0m$
D. $7.0m$

Answer 1

Hint: Here we know that whenever a short bar magnet falls along the axis of a horizontal metallic ring, then it falls under the influence of gravity. So, we will apply the second equation of motion and solve to get the required answer.

Formula used:
$s = ut + \dfrac{1}{2}g{t^2}$
$s$ is the distance,
$u$ is the initial speed,
$t$ is the time and
$g$ is acceleration due to gravity.

Complete answer:
Applying Newton's second law of motion.
$\because s = ut + \dfrac{1}{2}g{t^2}$
According to the question,
Time, $t = 1\sec $
Initial velocity, $u = 0$
Not substituting the values in above equation,
$
  s = ut + \dfrac{1}{2}g{t^2} \\
   \Rightarrow s = 0(1) + \dfrac{1}{2} \times 10 \times {(1)^2} \\
   \Rightarrow s = 5m \\
 $
So, the distance to the magnet in one second is $5m$ .
Hence the correct option is $B$ .

Note:
The distance falling by the magnet in one second is less than $5m$ because as the magnet falls freely in the ring, flux gets increased and a field is induced whose direction is opposite to that of the bar magnet. the magnet experiences a repulsive force and hence acceleration of the magnet becomes less than $g$.