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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$$T=2$$$$π$$IMHFor I case: $$6010=2$$$$π$$IMH   ............(i)For$$ II case: $$6014=2$$$$π$$$$IM(H+R)   ............(ii)Equation$$ $$(ii) $$/$$ (i),$$3076=HH+RR=2425H$$    ....(iii)For$$ III case: $$60n=2$$$$π$$$$IM(H-R)=2$$π x $$5IMH-2425H$$     ∵$$R=2425H=2$$π x $$5$$ x IMH   ........$$(iv)From$$ equations $$(i) and (iv), we $$get60n=30$$⇒ n $$=$$ $$2$$ vibrations per minute.

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