JEE Mains 2 April 2026 shift 1 Physics

# Q1 of 25

The dimension of 1/2 ε0E2 is MaLbTc, then the value of a−2b+c is:

Options :
A.

1

B.

2

C.

3

D.

4

Show Answer
Answer :

1

Solution :

The expression 1/2 ε0E2 represents energy density, which is defined as:


Energy density = Energy /Volume


Dimensions used:

• Energy = ML2T−2

• Volume = L3


Step 1: Find the dimensions of energy density.


Energy density = ML2T−2 /L3


Step 2: Compare with given form MaLbTc. =

a =1, b=−1, c=−2


Step 3: Compute required expression. a −2b+c=1−2(−1)+(−2)=1+2−2=1


Energy density problems can be solved quickly by remembering:


Energy density = ML−1T−2


No need to derive each time in exams.

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