Test: Dimensional Analysis and Its Applications (April 8) - NEET MCQ

Taking dimensions on both sides, we get

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

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

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)

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

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]
 

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

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

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

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

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.