JEE Main 2026 — Physics Mechanics

The dimensional formulas for the given physical quantities are:

Magnetic field intensity $[H] = [M^0 L^{-1} T^0 A^1]$[H]=[M0L−1T0A1]

Distance $[x] = [M^0 L^1 T^0 A^0]$[x]=[M0L1T0A0]

Permittivity $[\epsilon] = [M^{-1} L^{-3} T^4 A^2]$[ϵ]=[M−1L−3T4A2]

Electric field $[E] = [M^1 L^1 T^{-3} A^{-1}]$[E]=[M1L1T−3A−1]

Time $[t] = [M^0 L^0 T^1 A^0]$[t]=[M0L0T1A0]

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

$$[M^0 L^{-1} T^0 A^1] = [L]^p [M^{-1} L^{-3} T^4 A^2]^q [M L T^{-3} A^{-1}]^r [T]^{-s}$$[M0L−1T0A1]=[L]p[M−1L−3T4A2]q[MLT−3A−1]r[T]−s

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

$$[M^0 L^{-1} T^0 A^1] = [M^{-q+r} L^{p-3q+r} T^{4q-3r-s} A^{2q-r}]$$[M0L−1T0A1]=[M−q+rLp−3q+rT4q−3r−sA2q−r]

Equating the powers of corresponding fundamental quantities from both sides:

For $M$M: $0 = -q + r \implies q = r$0=−q+r⟹q=r

For $A$A: $1 = 2q - r$1=2q−r

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

$$1 = 2q - q \implies q = 1$$1=2q−q⟹q=1

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

For $L$L: $-1 = p - 3q + r$−1=p−3q+r

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

$$-1 = p - 3(1) + 1 \implies -1 = p - 2 \implies p = 1$$−1=p−3(1)+1⟹−1=p−2⟹p=1

For $T$T: $0 = 4q - 3r - s$0=4q−3r−s

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

$$0 = 4(1) - 3(1) - s \implies 0 = 1 - s \implies s = 1$$0=4(1)−3(1)−s⟹0=1−s⟹s=1

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

Answer: $1, 1, 1, 1$1,1,1,1