The dimensional formula of current density is

[M^{0} L^{- 2} T^{- 1} Q]

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[M^{0} L^{2} T^{1} Q^{- 1}]

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[M L T^{- 1} Q]

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[M L^{- 2} T^{- 1} Q^{2}]

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Solution

The correct option is A $[M^{0} L^{- 2} T^{- 1} Q]$
Current density $= \frac{C u r r e n t}{a r e a} = \frac{Q}{a r e a \times t} = [M^{0} L^{- 2} T^{- 1} Q]$

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Similar questions

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

is the dimensional formula for

Q. Match the Column I with Column II

	Column I (Units)		Column II (Dimensional formulae)
(A)	Pa s	(p)	$[M^{0} L^{2} T^{- 2} K^{- 1}]$
(B)	N m K $^{- 1}$	(q)	$[M L T^{- 3} K^{- 1}]$
(C)	J kg $^{- 1}$ K $^{- 1}$	(r)	$[M L^{- 1} T^{- 1}]$
(D)	W m $^{- 1}$ K $^{- 1}$	(s)	$[M L^{2} T^{- 2} K^{- 1}]$

Q. Tick (✓) the correct answer:
pq² + q(p − 1) − 1 = ?
(a) (pq + 1)(q − 1)
(b) p(q + 1)(q − 1)
(c) q(p − 1)(q + 1)
(d) (pq − 1) (q + 1)

Capillary tube is a very thin tube. If it is dipped in water, water rises upto height $h = \frac{2 T}{r ρ g},$ where r is the radius of tube, $ρ$ is density, and g is acceleration due to gravity. Find dimensional formula of [T]

$p q^{2} + q (p - 1) - 1 = ? (a) (p q + 1) (q - 1) (b) p (q + 1) (p - 1) (c) q (p - 1) (q + 1) (d) (p q - 1) (q + 1)$

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The dimensional formula of current density is

The correct option is A [M0L−2T−1Q]Current density =Currentarea=Qarea×t=[M0L−2T−1Q]

The correct option is A $[M^{0} L^{- 2} T^{- 1} Q]$
Current density $= \frac{C u r r e n t}{a r e a} = \frac{Q}{a r e a \times t} = [M^{0} L^{- 2} T^{- 1} Q]$