Dimensions of electrical resistance areleft -ML-2- T-3- A-1-right -left -ML-2- T-3- A-2-right -left -ML-3- T-3- A-2-right -left -ML-1- L-3- T-3- A-2-right -

Question

Toppr Admin · Accepted Answer

-R-VI-Wql-Work-V-xD7-q-x21D2-V-Wq-R-ML2T-x2212-2AT-A-ML2A-x2212-2T-x2212-3

	Column I		Column II
A.	Electrical resistance	$1.$	$[M L^{3} T^{- 3} A^{- 2}]$
B.	Electrical potential	$2.$	$[M L^{2} T^{- 3} A^{- 2}]$
C.	Specific resistance	$3.$	$[M L^{2} T^{- 3} A^{- 1}]$
D.	Specific conductance	$4.$	None of these

	Column I (Physical quantity)		Column II (Dimensional formula)
(A)	Premittivity of free space	(p)	$[M^{0} L^{0} T^{- 1}]$
(B)	Radiant flux	(q)	$[M L^{3} T^{- 3} A^{- 2}]$
(C)	Resistivity	(r)	$[M L^{2} T^{- 3}]$
(D)	Hubble constant	(s)	$[M^{- 1} L^{- 3} T^{4} A^{2}]$

Dimensions of electrical resistance are
$[M L^{2} T^{- 3} A^{- 1}]$
$[M L^{2} T^{- 3} A^{- 2}]$
$[M L^{3} T^{- 3} A^{- 2}]$
$[M L^{- 1} L^{3} T^{3} A^{2}]$

$R = \frac{V}{I} = \frac{W}{q l} (W o r k = V \times q \Rightarrow V = \frac{W}{q})$

$R = \frac{M L^{2} T^{- 2}}{A T . A} = M L^{2} A^{- 2} T^{- 3}$

Dimensions of electrical resistance are[ML2T−3A−1][ML2T−3A−2][ML3T−3A−2][ML−1L3T3A2]

R=VI=Wql(Work=V×q⇒V=Wq) R=ML2T−2AT.A=ML2A−2T−3

Dimensions of electrical resistance are
$[M L^{2} T^{- 3} A^{- 1}]$
$[M L^{2} T^{- 3} A^{- 2}]$
$[M L^{3} T^{- 3} A^{- 2}]$
$[M L^{- 1} L^{3} T^{3} A^{2}]$

$R = \frac{V}{I} = \frac{W}{q l} (W o r k = V \times q \Rightarrow V = \frac{W}{q})$

$R = \frac{M L^{2} T^{- 2}}{A T . A} = M L^{2} A^{- 2} T^{- 3}$