1
MHT CET 2023 10th May Morning Shift
+1
-0

A square loop of area $$25 \mathrm{~cm}^2$$ has a resistance of $$10 \Omega$$. This loop is placed in a uniform magnetic field of magnitude $$40 \mathrm{~T}$$. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in one second, will be

A
$$1 \times 10^{-4} \mathrm{~J}$$
B
$$1.0 \times 10^{-3} \mathrm{~J}$$
C
$$5 \times 10^{-3} \mathrm{~J}$$
D
$$2.5 \times 10^{-3} \mathrm{~J}$$
2
MHT CET 2023 10th May Morning Shift
+1
-0

Two conducting circular loops of radii '$$R_1$$' and '$$R_2$$' are placed in the same plane with their centres coinciding. If $$R_1>R_2$$, the mutual inductance $$M$$ between them will be directly proportional to

A
$$\frac{R_1}{R_2}$$
B
$$\frac{R_2}{R_1}$$
C
$$\frac{R_1^2}{R_2}$$
D
$$\frac{R_2^2}{R_1}$$
3
MHT CET 2023 9th May Evening Shift
+1
-0

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)

A
$$\frac{\mathrm{BLV}}{\mathrm{R}}$$
B
$$\frac{\mathrm{B}^2 \mathrm{~L}^2 \mathrm{~V}^2}{\mathrm{R}^2}$$
C
$$\frac{B L V}{R^2}$$
D
$$\frac{\mathrm{B}^2 \mathrm{~L}^2 \mathrm{~V}^2}{\mathrm{R}}$$
4
MHT CET 2023 9th May Evening Shift
+1
-0

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
$$4 \pi$$
B
$$8 \pi$$
C
$$10 \pi$$
D
$$16 \pi$$
EXAM MAP
Medical
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