1

IIT-JEE 2001 Screening

A quantity X is given by ${\varepsilon _0}L{{\Delta V} \over {\Delta t}}$ where ${\varepsilon _0}$ is the permittivity of the free space, L is a length, ${\Delta V}$ is a potential difference and ${\Delta t}$ is a time interval. The dimentional formula for X is the same as that of
A
resistance
B
charge
C
voltage
D
current

Explanation

Given that $X = {\varepsilon _0}L\left( {{{\Delta V} \over {\Delta t}}} \right)$

Therefore, dimensional formula for X is

$[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}][L]{{[M{L^2}{T^{ - 3}}{I^{ - 1}}]} \over {[T]}} = [I]$

Thus, dimensional formula for X is same as that of current.

2

IIT-JEE 2000 Screening

The dimension of $\left( {{1 \over 2}} \right){\varepsilon _0}{E^2}$
( ${\varepsilon _0}$ : permittivity of free space, E electric field )
A
MLT-1
B
ML2T-2
C
ML-1T-2
D
ML2T-1

Explanation

Dimension of $\varepsilon$0 = M$-$1L$-$3T4A2

Dimension of electric field E = MLT$-$3A$-$1

Now, dimension of

${1 \over 2}$$\varepsilon$0E2 = (M$-$1L$-$3T4A2)(MLT$-$3A$-$1)2

= M$-$1L$-$3T4A2M2L2T$-$6A$-$2 = ML$-$1T$-$2