Dimensional Analysis & Its Applications (NCERT) - NEET Physics Class 11

Only the dimensional accuracy can be checked with the help of principle of homogeneity of dimensions. This is limitation of principle of homogeneity of dimensions.

Taking dimensions on both sides, we get

∴ LHS = RHS
Now, T² = 4π²r²
Taking dimensions on both sides
[T]² = [L]² 
∴ LHS ≠ RHS

As, we know, E = mc² and energy released can be given as:
E = 931MeV
for, m = 1u
1u = 931.5MeV/c² - Dimensionally correct [Dimension of mass]
1u = 1.67 × 10⁻²⁷ kg - Dimensionally correct [Dimension of mass]
but, 931.5 MeV is having the dimension of energy so it can't be equated to dimension of mass or dimension of 1u.
Hence, dimensionally correct data is given in option(A)

Dimensions on RHS must be displacement [L].
Arguments of sine and cosine are dimensionless. Hence, Option (c) is not correct. 

y = Asin(ωt − kx)
As (ωt − kx) represents an angle which is dimensionless, therefore

Let l ∝ c^xh^y G^z ;l = kc^x h^y G^z
where k is a dimensionless constant and x, y and z are the exponents.
Equating dimensions on both sides, we get
[M⁰ LT⁰] =[LT⁻¹] x [ML² T⁻¹] y [M⁻¹ L³ T⁻²]^z
= [M^y−z L^x+2y+3z T ^{−x−y−2z}]
Applying the principle of homogeneity of dimensions, we get
y − z = 0             ....(i)
x + 2y + 3z = 1     ....(ii)
− x − y − 2z = 0     ...(iii)
On solving Eqs. (i), (ii) and (iii), we get

Joule is a unit of energy.

Dimensional formula of energy is [ML² T⁻²].
Comparing with [M^a L^b T^c], we get a = 1, b = 2,c = -2

E = G^p h^q c^r…(i)
[M¹L²T⁻²] = [M⁻¹L³T⁻²]p[ML²T⁻¹]q[LT⁻¹]r
= [M^−p+q L^3p+2q+r T^{−2p−q−r}]
Applying principle of homogeneity of dimensions, we get
−p + q = 1…(ii)
3p + 2q + r = 2…(iii)
−2p − q − r = −2…(iv)
Adding (iii) and (iv),
we get p + q = 0…(v)
Adding (ii) and (v), we get q = 1/2
From (ii), we get p = q − 1

Dimension of a/V³ = Dimensions of P
∴ Dimensions of a = dimensions of PV³ 

Dimensions of b²  = dimensions of V
∴ [b] = [V]^1/2 = [L³]^1/2 or [b] = [L^3/2]
 

v = at + bt²
[v] = [bt²] or [LT⁻¹] = [bT²] or [b] = [LT⁻³]