JEE Main 2026 — Units and Dimensions Question with Solution

The dimensional formulas for the given physical quantities are:

Magnetic field intensity $[H]=[M^0{L}^{-1}{T}^0{A}^1]$

Distance $[x]=[M^0{L}^1{T}^0{A}^0]$

Permittivity $[ϵ]=[M^{-1}{L}^{-3}{T}^4{A}^2]$

Electric field $[E]=[M^1{L}^1{T}^{-3}{A}^{-1}]$

Time $[t]=[M^0{L}^0{T}^1{A}^0]$

Given the equation $H=\frac{x^p{ϵ}^q{E}^r}{t^s}$, we can write it in terms of dimensions as:

$[M^0{L}^{-1}{T}^0{A}^1]=[L{]}^p[M^{-1}{L}^{-3}{T}^4{A}^2{]}^q[ML{T}^{-3}{A}^{-1}{]}^r[T{]}^{-s}$

Combining the powers of $M,L,T$, and $A$ on the right side:

$[M^0{L}^{-1}{T}^0{A}^1]=[M^{-q+r}{L}^{p-3q+r}{T}^{4q-3r-s}{A}^{2q-r}]$

Equating the powers of corresponding fundamental quantities from both sides:

For $M$: $0=-q+r⟹q=r$

For $A$: $1=2q-r$

Substituting $r=q$ into the equation for $A$:

$1=2q-q⟹q=1$

Since $q=r$, we get $r=1$.

For $L$: $-1=p-3q+r$

Substituting $q=1$ and $r=1$:

$-1=p-3(1)+1⟹-1=p-2⟹p=1$

For $T$: $0=4q-3r-s$

Substituting $q=1$ and $r=1$:

$0=4(1)-3(1)-s⟹0=1-s⟹s=1$

Thus, the values are $p=1$, $q=1$, $r=1$, and $s=1$.

Answer: $1,1,1,1$