Physics · Units And Measurements · NEET
Assume the target quantity equals a constant times each given quantity raised to an unknown power. For a pendulum's time period T depending on length l, mass m and gravity g, write T = k · l^a · m^b · g^c. Here k is a dimensionless constant and a, b, c are the powers you will find. Then replace every symbol by its dimensions and compare M, L, T powers on both sides.
Because 2π (and any pure number) has no dimensions — it is [M^0 L^0 T^0]. Matching powers of M, L and T only fixes the exponents a, b, c, never a dimensionless multiplier. That is why the pendulum comes out as T = k·√(l/g), and only experiment or theory tells us k = 2π. NEET expects you to state that the numerical constant cannot be found.
When you match powers of M, only mass m carries M (its dimension is M^1). Since T = 2π√(l/g) has no mass on the left side, the M equation is b = 0. So the period does not depend on the bob's mass at all — a result you can prove by dimensions alone. This is a favourite NEET point.
Usually no. Dimensional analysis gives only three equations (one each for M, L, T). If the quantity depends on more than three unknowns, you have more unknowns than equations and cannot solve them uniquely. NCERT limits the method to a product of up to three quantities. If you see four, extra physical information must reduce the unknowns.
Dimensional analysis cannot derive formulas with two or more added terms, because it only builds single product-type relations. It works only when the quantity is a pure product of powers. For sums you can only check consistency (each term must have the same dimension), not derive the relation.
A balloon of surface tension S with an outlet of small area A is filled with gas of density ρ and forms a sphere of radius R. As gas flows out, the radius r goes from R to 0 in time T. If the exit speed v(r) depends on r as r^a and T ∝ S^α A^β ρ^γ R^δ, then:
Planck's constant (h), speed of light (c) and gravitational constant (G) are three fundamental constants. Which combination has the dimensions of length?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
1) Assume the quantity is a product of the others with unknown powers and a constant k. 2) Write the dimensional formula of every quantity. 3) Substitute and equate the powers of M, L and T on both sides. 4) Solve the equations for the powers. 5) Write the formula, leaving k unknown.
Assuming T = k·l^a·m^b·g^c and matching M, L, T gives b = 0, a = 1/2, c = -1/2. So T = k·√(l/g). Experiment fixes k = 2π, giving T = 2π√(l/g).
No. It works only for product-type relations with up to three unknown quantities and it cannot find dimensionless constants. It fails for formulas containing sums, trigonometric, exponential or logarithmic functions.
Only mass carries the M dimension, and the period has no M on its left side, so the M-matching equation forces the mass power to zero. Dimensions alone prove the period is independent of mass.
k is a pure number (dimensionless), such as 2π or 1/2. Since it has no M, L or T, matching dimensions can never produce it, so it is left as an unknown until experiment or full theory supplies its value.
