Q.26. The dimensional formula of ( = permittivity of vacuum and E = electric field) is . The value of =

Sol. The expression $\frac{1}{2} \epsilon_0 E^2$ represents energy density (energy per unit volume). Dimensional formula of Energy = $[ML^2T^{-2}]$ Dimensional formula of Volume = $[L^3]$ So, Dimensional formula of Energy Density = $\frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$ Comparing this with $M^a L^b T^c$, we get: $a=1$ $b=-1$ $c=-2$ Now, we need to find the value of $2a-b+c$: $2a-b+c = 2(1) - (-1) + (-2) = 2 + 1 - 2 = 1$.

What is the correct answer for JEE Main Physics Q26?

The correct answer is B. Sol. The expression $\frac{1}{2} \epsilon_0 E^2$ represents energy density (energy per unit volume). Dimensional formula of Energy = $[ML^2T^{-2}]$ Dimensional formula of Volume = $[L^3]$ So, Dimensional formula of Energy Density = $\frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$ Comparing this with $M^a L^b T^c$, we get: $a=1$ $b=-1$ $c=-2$ Now, we need to find the value of $2a-b+c$: $2a-b+c = 2(

Q.26. The dimensional formula of ( = permittivity of vacuum and E = electric field) is . The value of =

The correct answer is B. Sol. The expression $\frac{1}{2} \epsilon_0 E^2$ represents energy density (energy per unit volume). Dimensional formula of Energy = $[ML^2T^{-2}]$ Dimensional formula of Volume = $[L^3]$ So, Dimensional formula of Energy Density = $\frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$ Comparing this with $M^a L^b T^c$, we get: $a=1$ $b=-1$ $c=-2$ Now, we need to find the value of $2a-b+c$: $2a-b+c = 2(

JEE Main 2026 Physics PYQ · With Answer · Free

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Q.26. The dimensional formula of $\frac{1}{2} \epsilon_0 E^2$ ($\epsilon_0$ = permittivity of vacuum and E = electric field) is $M^a L^b T^c$. The value of $2a-b+c$ =

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