The magnetic needle of an oscillation magnetometer makes 10 oscillations per minute under the action of earth's magnetic field alone. When a bar magnet is placed at some distance along the axis of the needle it makes 14 oscillations per minute. If the bar magnet is turned so that its poles interchange their position, then the new frequency of oscillation of the needle is
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a
10 vibrations Per minute
b
14 vibrations per minute
c
4 vibrations per minute
d
2 vibrations per minute
answer is D.
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Detailed Solution
T=2πIMHFor I case: 6010=2πIMH ............(i)For II case: 6014=2πIM(H+R) ............(ii)Equation (ii) / (i),3076=HH+RR=2425H ....(iii)For III case: 60n=2πIM(H-R)=2π x 5IMH-2425H ∵R=2425H=2π x 5 x IMH ........(iv)From equations (i) and (iv), we get60n=30⇒ n = 2 vibrations per minute.