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Question: A short bar magnet allowed it to fall along the axis of the horizontal metallic ring. Starting from ...

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. 4.0m4.0m
B. 5.0m5.0m
C. 6.0m6.0m
D. 7.0m7.0m

Explanation

Solution

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+12gt2s = ut + \dfrac{1}{2}g{t^2}
ss is the distance,
uu is the initial speed,
tt is the time and
gg is acceleration due to gravity.

Complete answer:
Applying Newton's second law of motion.
s=ut+12gt2\because s = ut + \dfrac{1}{2}g{t^2}
According to the question,
Time, t=1sect = 1\sec
Initial velocity, u=0u = 0
Not substituting the values in above equation,
s=ut+12gt2 s=0(1)+12×10×(1)2 s=5m  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 5m5m .
Hence the correct option is BB .

Note:
The distance falling by the magnet in one second is less than 5m5m 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 gg.