A magnet makes 12 oscillation per minute at a place where horizontal component of earth's field is 6.4 × 10–3 T. It is found to require 8 seconds per oscillation at another place X. The vertical component of earth's field at X, where resultant field makes angle 600 with horizontal is ---×10–4 T
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a
253
b
3
c
253
d
25
answer is C.
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Detailed Solution
T∝1BH ; T2T1=B1B285=B1B2 ; B2=2.5 x 10-3B cos 60º = 2.5 x 10-3Bv=B sin 600=2.5 x 10-3cos 600 x sin 600Bv=2.5 x 10-3 x 32 x 12=253 x 10-4 T