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Self Induction

Current " I"...

Question

Current " $I$ " flows through an inductive circuit. The work done is equal to $\frac{{L I}^{2}}{2}$ . The dimensions of work done is:

A

{M L}^{2} T^{- 2}

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B

Not expressible in

M, L, T

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C

{M L}^{- 1}

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D

M^{2} L^{2} T^{2}

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Solution

The correct option is A ${M L}^{2} T^{- 2}$
Energy stored by an inductor is given by $\frac{1}{2} {L I}^{2}$ .
Thus the dimension of ${L I}^{2}$ is of energy only. $\Rightarrow M L^{2} T^{- 2}$

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Similar questions

Q. Match the physical quantities given in Column I with suitable dimensions expressed in Column II.

Column I	Column II
a) Angular momentum	g) ${M L}^{2} T^{- 2}$
b) Latent heat	h) ${M L}^{2} Q^{- 2}$
c) Torque	i) ${M L}^{2} T^{- 1}$
d) Capacitane	j) ${M L}^{3} T^{- 1} Q^{- 2}$
e) Inductance	k) $M^{- 1} L^{- 2} T^{2} Q^{2}$
f) Resistivity	l) $L^{2} T^{- 2}$

L

denotes the inductance of an inductor through which a current

i

is flowing, the dimensions of

L i^{2}

Q. Match the physical quantities given in Column I with suitable dimensions expressed in Column II.

Column I	Column II
a) Angular momentum	1) $M^{- 1} L^{3} T^{- 2}$
b) Torque	2) ${M L T}^{- 2}$
c) Gravitational constant	3) ${M L}^{2} T^{- 2}$
d) Tension	4) ${M L}^{2} T^{- 1}$

Q. The dimensions ML⁻¹ T⁻² may correspond to
(a) work done by a force
(b) linear momentum
(c) pressure
(d) energy per unit volume.

C

denotes capacitance and

V

denotes potential difference across the capacitor, the dimensions of

C V^{2}

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