Dimensional Analysis

Answer: D. D is correct. Dimensional homogeneity is a necessary condition — if dimensions don't match, the equation cannot be correct (NCERT Class 11 Physics Chapter 1, page 8). However, dimensional correctness is not sufficient for numerical correctness.

Why A is wrong: A is wrong because dimensional correctness is necessary but not sufficient. For example, F = 2ma is dimensionally correct but numerically wrong.

Why B is wrong: B is wrong because dimensionless constants (like 2π or ½) cannot be determined by dimensional analysis — it only tracks powers of base dimensions.

Why C is wrong: C is wrong because sin(x) and cos(x) are both dimensionless functions. Dimensional analysis cannot distinguish between them.

Answer: B. B is correct. Match dimensions on both sides: [T] = [L]ᵃ [M]ᵇ [L T⁻²]ᶜ. Mass appears only in mᵇ, so b = 0. Time: 1 = −2c, so c = −½. Length: 0 = a + c, so a = ½. Hence T ∝ √(l/g), which is the form of the pendulum formula; dimensional analysis cannot supply the 2π.

Why A is wrong: A is wrong because a = 1, c = −1 gives T ∝ l/g, which has dimensions [L]/[L T⁻²] = [T²], not [T]. The exponents must make the time dimension come out to the first power.

Why C is wrong: C is wrong because b = ½ gives the period a dependence on mass that nothing else in the expression can cancel. Mass appears in only one factor, so its exponent must be zero for the dimensions to balance.

Why D is wrong: D is wrong because a = −½, c = ½ gives T ∝ √(g/l), which has dimensions of [T⁻¹] — a frequency, not a period. The signs are inverted.

Answer: A. A is correct. Plane angle (radian = arc/radius) and solid angle (steradian = area/r²) are ratios of like quantities, making them dimensionless. They carry unit names for clarity, not because they have dimensions (NCERT Class 11 Physics Chapter 1, page 8).

Why B is wrong: B is wrong because solid angle also has a unit name: steradian (sr). Both radian and steradian are named supplementary SI units.

Why C is wrong: C is wrong because neither plane angle nor solid angle has dimensions. They are both ratios — arc length ÷ radius and surface area ÷ r² — and ratios of like quantities are dimensionless.

Why D is wrong: D is wrong because having a unit name does not imply having dimensions. Radian and steradian are defined as ratios, which are inherently dimensionless.

Answer: B. B is correct. For dimensional homogeneity, [αt²] must have dimensions of force: [α] × [T²] = [M L T⁻²]. Therefore [α] = [M L T⁻²] / [T²] = [M L T⁻⁴].

Why A is wrong: A gives [M L T⁻³], which would be the dimensions of β (since [βt] = [M L T⁻²] gives [β] = [M L T⁻³]). Confusing α with β is a common mix-up when both terms are present.

Why C is wrong: C gives [M L T⁻²], which is the dimensions of force itself. This error comes from ignoring the t² factor entirely — α is not force, it is force divided by time squared.

Why D is wrong: D gives [M L T⁻¹], which results from multiplying the force dimensions by T instead of dividing by T² — moving the time factor to the wrong side of the equation.

Answer: C. C is correct. Pressure = Force / Area = [M L T⁻²] / [L²] = [M L⁻¹ T⁻²] (NCERT Class 11 Physics Chapter 1, page 7).

Why A is wrong: A is wrong. Force has dimensions [M L T⁻²]. The L exponent is +1, not −1. Confusing force and pressure is a common error when reading dimensional formulas quickly.

Why B is wrong: B is wrong. Energy has dimensions [M L² T⁻²]. Both the L and T exponents differ from [M L⁻¹ T⁻²].

Why D is wrong: D is wrong. Power has dimensions [M L² T⁻³]. The L exponent is +2 and the T exponent is −3, neither of which matches the given formula.

Answer: A. A is correct. From F = ma, mass = F/a = F/(v/T) = FT/v = [F T v⁻¹]. Checking dimensions in SI: [M L T⁻²] × [T] × [L⁻¹ T] = [M L T⁻² × T × L⁻¹ T] = [M]. Confirmed.

Why B is wrong: B gives FT⁻¹v⁻¹. In SI: [M L T⁻²][L⁻¹ T][T⁻¹] = [M T⁻²], which is not mass. The error comes from dividing by T instead of multiplying.

Why C is wrong: C gives FT²v⁻¹. In SI: [M L T⁻²][T²][L⁻¹ T] = [M T], which has an extra time dimension. This results from using a = v/T² instead of a = v/T.

Why D is wrong: D gives FvT⁻¹. In SI: [M L T⁻²][L T⁻¹][T⁻¹] = [M L² T⁻⁴], which is not mass. This error multiplies by v instead of dividing.

Answer: C. C is correct. The term at² has dimensions [L T⁻²][T²] = [L], which is length — not velocity [L T⁻¹]. The equation is dimensionally inconsistent and therefore certainly wrong. The correct kinematic equation is v = u + at.

Why A is wrong: A is wrong because at² does not have the dimensions of velocity. [acceleration × time²] = [L T⁻²][T²] = [L], which is displacement, not velocity. The equation fails dimensional homogeneity.

Why B is wrong: B is wrong because the issue here is not about dimensionless constants — it is about a genuine dimensional mismatch. The term at² has dimensions of length, making the equation dimensionally inconsistent regardless of constants.

Why D is wrong: D is wrong on two counts: the equation is NOT dimensionally consistent (at² gives [L], not [L T⁻¹]), and even if it were, dimensional consistency would only be necessary, not sufficient, for correctness.

Answer: D. D is correct. Writing dimensions: [L T⁻¹] = [M L T⁻²]ᵃ [M L⁻¹]ᵇ = [Mᵃ⁺ᵇ Lᵃ⁻ᵇ T⁻²ᵃ]. Matching exponents — M: a + b = 0, T: −2a = −1 so a = ½, then b = −½. L: a − b = ½ − (−½) = 1. All three check out. Therefore v = k √(T/μ).

Why A is wrong: A has a = −½, b = ½. Checking T exponent: −2(−½) = +1 ≠ −1. The time exponent has the wrong sign. This inverts the roles of T and μ.

Why B is wrong: B has b = +½ instead of −½. This violates the M exponent equation: a + b = ½ + ½ = 1 ≠ 0. The error comes from dropping the negative sign when solving b = −a.

Why C is wrong: C has a = 1, b = −1. Checking T exponent: −2(1) = −2 ≠ −1. This fails the time dimension. The error results from setting −2a = −2 (confusing velocity's T⁻¹ with T⁻²).