The side of a copper cube is $$1 \mathrm{~m}$$ at $$0^{\circ} \mathrm{C}$$. What will be the change in its volume, when it is heated to $$100^{\circ} \mathrm{C}$$ ? $$\left[\alpha_{\text {copper }}=18 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right]$$

If current '$$I$$' is flowing in the closed circuit with collective resistance '$$R$$', the rate of production of heat energy in the loop as we pull it along with a constant speed '$$\mathrm{V}$$' is ( $$\mathrm{L}=$$ length of conductor, $$\mathrm{B}=$$ magnetic field)

Two coils $$\mathrm{A}$$ and $$\mathrm{B}$$ have mutual inductance 0.008 $$\mathrm{H}$$. The current changes in the coil A, according to the equation $$\mathrm{I}=\mathrm{I}_{\mathrm{m}} \sin \omega \mathrm{t}$$, where $$\mathrm{I}_{\mathrm{m}}=5 \mathrm{~A}$$ and $$\omega=200 \pi ~\mathrm{rad} ~\mathrm{s}^{-1}$$. The maximum value of the e.m.f. induced in the coil $$B$$ in volt is

A thin rod of length $$L$$ has magnetic moment $$M$$ when magnetised. If rod is bent in a semicircular arc what is magnetic moment in new shape?