Torque on a bar magnet :I=MB‌sin‌θ Here, θ=3o°,I=0.018N−m,B=0.06T ⇒0.018=M×0.06×sin‌30° ⇒0.018=M×0.06×
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⇒M=0.6A−m2 Now v=−MB‌cos‌θ Position of stable equilibrium (θ=0°): ui=−MB Position of unstable equilibrium (θ=180°): uf=MB ⇒ work done :ΔU ⇒W=2MB ⇒W=2×0.6×0.06 ⇒W=7.2×10−2J option (4) is correct