Physics - Unit & Dimension Question with Solution | TestHub

PhysicsUnit & DimensionMiscellaneousEasy2 minPYQ_2020

PhysicsEasynumerical

Let $ε_{0}$ denote the dimensional formula of the permittivity of vacuum. If $M =$ mass, $L =$ length, $T =$ time and $A =$ electric current, then dimensions of permittivity is given as $[M^{p} L^{q} T^{r} A^{s}]$ . Find the value of $\frac{p - q + r}{s}$

Answer:

3.00

Solution:

From Coulomb's law,
$F = \frac{q_{1} q_{2}}{4 π ε_{0} r^{2}}$
$∴ ε_{0} = \frac{q_{1} q_{2}}{4 π F r^{2}} = \frac{(A^{1} T^{1}) (A^{1} T^{1})}{[M^{1} L^{1} T^{- 2}] [L^{2}]} = [M^{- 1} L^{- 3} T^{4} A^{2}]$
$∴ p = - 1, q = - 3, r = 4, s = 2$
$∴ \frac{p - q + r}{s} = \frac{- 1 - (- 3) + 4}{2} = \frac{6}{2} = 3$

Stream:NTA_ABHYASSubject:PhysicsTopic:Unit & DimensionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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