(a) Assuming that the frequency gamma of a vibrating string may depend upon (i) applied force (F) (ii) length

Question 53 of 61

Q. (a) Assuming that the frequency gamma of a vibrating string may depend upon

(i) applied force (F)

(ii) length (l)

(iii) mass per unit length (m), prove that gamma is proportional to (1/l) x sqrt(F/m) using dimensional analysis. OR

(b) State and prove Bernoulli's theorem for a flow of incompressible, non-viscous, and streamlined flow of fluid.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2024Subjective· 5mImportance★★★★★

87% · 53/61 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Dimensional Analysis

The dimensions of a physical quantity record how it is built from the base quantities — how many powers of mass, length, time (and, where needed, electric current, temperature, amount, luminous intensity) it contains. Written in square brackets, force is [M L T⁻²] and energy is [M L² T⁻²]. This single idea — that an equation must be dimensionally consistent — is one of the most powerful cross-checks in physics, and JEE Main mines it heavily.

1 — Dimensional formulae. Every derived quantity has a dimensional formula obtained from its defining relation: velocity [LT⁻¹], acceleration [LT⁻²], force [MLT⁻²], work/energy/torque [ML²T⁻²], power [ML²T⁻³], pressure/stress [ML⁻¹T⁻²], momentum/impulse [MLT⁻¹]. For constants, isolate the constant in its equation and read off its dimensions — from F = Gm₁m₂/r², [G] = [M⁻¹L³T⁻²]; from E = hν, [h] = [ML²T⁻¹]; from PV = nRT, [R] = [ML²T⁻²K⁻¹mol⁻¹]. Electrical quantities carry the base dimension of current [A]: charge [AT], potential [ML²T⁻³A⁻¹], resistance [ML²T⁻³A⁻²].

2 — The principle of homogeneity. In any valid equation, every additive term has the same dimensions. This lets you (a) test whether a given equation can be correct, (b) find a missing exponent by matching the powers of M, L and T on both sides, and (c) reject a proposed formula that is dimensionally inconsistent. A crucial corollary: the argument of any sin, cos, log or exponential — and any exponent — must be dimensionless. So in y = A sin(ωt), ωt is dimensionless, forcing [ω] = [T⁻¹].

3 — Deriving a relation. When a quantity depends on a few others, assume a power-law y = k·x₁ᵃ x₂ᵇ x₃ᶜ, write the dimensions of both sides, and equate the exponents of M, L, T to solve for a, b, c. This recovers the form of many results — the pendulum's T ∝ √(L/g), the speed of a wave on a string v ∝ √(T/μ), Stokes' drag F ∝ ηrv. The dimensionless constant k (like the 2π in the pendulum) is what dimensional analysis cannot supply.

4 — Converting between systems of units. Because a physical quantity is unit-independent, n₁u₁ = n₂u₂. Using the dimensional formula, n₂ = n₁ [M₁/M₂]ᵃ [L₁/L₂]ᵇ [T₁/T₂]ᶜ. This is how 1 N = 10⁵ dyne, 1 J = 10⁷ erg, and how the numerical value of a constant like G changes from SI to CGS. The same machinery lets you express a quantity when a new set of quantities (say force, velocity, time) is chosen as fundamental. …

Start your 14-day free trial to unlock the full solution →

By assuming γ = k F^a l^b m^c and equating powers of M, L, T on both sides, dimensional analysis gives a = 1/2, b = -1, c = -1/2, i.e., γ ∝ (1/l)√(F/m).

Let the frequency γ of a vibrating string depend on the applied force F, the length l, and the mass per unit length m, as:

γ = k F^a l^b m^c ... (1)

where k is a dimensionless constant.

Write the dimensions of each quantity:

[γ] = T^-1 (frequency)

[F] = M L T^-2 (force)

[l] = L (length)

[m] = M L^-1 (mass per unit length)

Substituting into equation (1):

T^-1 = (M L T^-2)^a (L)^b (M L^-1)^c

= M^(a+c) L^(a+b-c) T^(-2a)

Equating the powers of M, L, T on both sides:

Power of M: 0 = a + c ... (i)

Power of L: 0 = a + b - c ... (ii)

Power of T: -1 = -2a → a = 1/2 ... (iii)

From (i): c = -a = -1/2

From (ii): b = c - a = (-1/2) - (1/2) = -1

So a = 1/2, b = -1, c = -1/2.

Substituting back into equation (1): …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 50 on this concept.

CBSE 2026Set 55/2/11 markMCQ
$\phi$
›Reveal solutionSolution
The key idea is dimensional consistency: each term in $\phi = 5At^2 + Bt - 2C$ must have the same unit as $\phi$ (Wb). This forces $A$ to have units $\text{Wb s}^{-2}$ and $B$ to have units $\text{Wb s}^{-1}$ , so the correct option is (C).

The problem gives you a relation between magnetic flux $\phi$ (in webers) and time $t$ (in seconds):
$\phi = 5At^2 + Bt - 2C.$
You’re asked for the SI units of $A$ and $B$ . The constants $5$ and $2$ are pure numbers — they have no units. So the only way this equation makes physical sense is if every term on the right-hand side has the same unit as $\phi$ , which is the weber (Wb). This is the principle of dimensional homogeneity, and it’s the entire foundation of the solution.

Let’s apply it term by term.

First term: $5At^2$ Since $5$ is dimensionless, the unit of $5At^2$ is the unit of $A$ multiplied by the unit of $t^2$ . Time $t$ is in seconds, so $t^2$ has unit $\text{s}^2$ . For this term to equal a flux in Wb, we need:

$\text{unit of }A \times \text{s}^2 = \text{Wb}.$
Therefore:
$\text{unit of }A = \frac{\text{Wb}}{\text{s}^2} = \text{Wb s}^{-2}.$

Second term: $Bt$ Here $B$ is multiplied by $t$ (unit s). So:

$\text{unit of }B \times \text{s} = \text{Wb}.$
Hence:
$\text{unit of }B = \frac{\text{Wb}}{\text{s}} = \text{Wb s}^{-1}.$

Third term: $-2C$ …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following physical quantity has no dimensions?$
(1) Angular Velocity
(2) Angular Acceleration
(3) Angular Displacement
(4) Stress
›Reveal solutionSolution
Angular displacement is dimensionless because it is defined as the ratio of two lengths (arc / radius).

Check each option's dimensional formula:

Angular velocity ω = dθ/dt → dimension [T^-1] (angle is dimensionless, so only 1/time survives).

Angular acceleration α = dω/dt → dimension [T^-2].

Angular displacement θ = arc length / radius = a ratio of two lengths → the length dimensions cancel, leaving a pure number (dimensionless), measured in radians. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The dimensional formula of work done is the same as the dimensional formula of$
(a) Momentum
(b) Power
(c) Energy
(d) Torque
›Reveal solutionSolution
Work done and energy share the exact same dimensional formula [ML^2T^-2], because work is defined as a mode of energy transfer.

Work done W = Force x displacement = [MLT^-2] x [L] = [ML^2T^-2].

Check each option:

Momentum p = mv = [M][LT^-1] = [MLT^-1] -- different.

Power P = Work/time = [ML^2T^-2]/[T] = [ML^2T^-3] -- different.

Energy (kinetic or potential) = (1/2)mv^2 or mgh = [ML^2T^-2] -- matches exactly. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. What is the dimensional formula of angular momentum?$
(a) [ML^2T^-1]
(b) [MLT^-2]
(c) [MLT^-1]
(d) [M^-1L^3T^2]
›Reveal solutionSolution
Angular momentum L = mvr, so its dimensions are mass x velocity x radius = [ML^2T^-1].

Angular momentum is defined as L = r x p = mvr (magnitude, for a point mass moving with linear velocity v at perpendicular distance r from the axis).

Dimensions:

Mass, m = [M] …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following physical quantities has the same dimensions as impulse?$
(a) Force
(b) Momentum
(c) Work
(d) Power
›Reveal solutionSolution
Impulse J = F x delta t has dimensions [MLT^-1], identical to momentum p = mv.

Impulse J = Force x time = [MLT^-2] x [T] = [MLT^-1].

Compare:

Force = [MLT^-2] -- different.

Momentum = mv = [M][LT^-1] = [MLT^-1] -- matches exactly.

Work = [ML^2T^-2] -- different.

Power = [ML^2T^-3] -- different. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The pair of physical quantities not having same dimension is:$
(a) Torque and Energy
(b) Surface Tension and Impulse
(c) Angular momentum and Planck's constant
(d) None of the above
›Reveal solutionSolution
Surface tension ( $MT^{-2}$ ) and impulse ( $MLT^{-1}$ ) have different dimensions; the other two pairs match.

Work out each pair's dimensional formula:

Torque $= r \times F$ , dimension $ML^2T^{-2}$ . Energy also has dimension $ML^2T^{-2}$ . Same.

Surface tension $=$ force/length, dimension $MLT^{-2}/L = MT^{-2}$ . Impulse $=$ force $\times$ time, dimension $MLT^{-2}\cdot T = MLT^{-1}$ . Different — an $L$ appears in impulse but not in surface tension. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: Dimensional formula of ____ is same as the dimensional formula of angular momentum.$
›Reveal solutionSolution

Planck's constant has the same dimensional formula, [M L² T⁻¹], as angular momentum.

Angular momentum L = mvr has dimensions [M][LT⁻¹][L] = [M L² T⁻¹].

Planck's constant is defined through E = hν, so h = E/ν. Energy E has dimensions [M L² T⁻²] and frequency ν has dimensions [T⁻¹], so:

h = [M L² T⁻²] / [T⁻¹] = [M L² T⁻¹]

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following has dimensional formula [ML^2T^-2] ?$
(a) Acceleration
(b) Force
(c) Work
(d) Linear momentum
›Reveal solutionSolution
[ML^2T^-2] is the dimension of energy/work, so the answer is (C) Work.

We find the dimensions of each option:

Acceleration = velocity/time = [LT^-2]

Force = mass x acceleration = [MLT^-2]

Work = force x displacement = [MLT^-2] x [L] = [ML^2T^-2]

Linear momentum = mass x velocity = [MLT^-1] …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following couples of quantities have same dimension? (A) Work and energy (B) Force and momentum (C) Force and power (D) Momentum and energy$
›Reveal solutionSolution
Work and energy share the same dimension $[ML^2T^{-2}]$ , both measured in joules.

Check each pair:

Work $W = Fd$ , dimension $[MLT^{-2}][L] = [ML^2T^{-2}]$ . Kinetic energy $= \tfrac{1}{2}mv^2$ , dimension $[M][LT^{-1}]^2 = [ML^2T^{-2}]$ . Same dimension.

Force $[MLT^{-2}]$ vs momentum $[MLT^{-1}]$ — different. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The velocity of a particle$
(v) at an instant t is given by v = at + bt^2. The dimension of b is
(a) [L]
(b) [LT^-1]
(c) [LT^-2]
(d) [LT^-3]
›Reveal solutionSolution

By the principle of dimensional homogeneity, every term added to v must have the dimension of velocity; this forces [b] = [LT^-3].

Given v = at + bt^2, where v is velocity, [v] = [LT^-1].

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANN1 markMCQ
$Q. Find out the fundamental quantity from among the physical quantities given below:$
(a) velocity
(b) temperature
(c) force
(d) density
›Reveal solutionSolution
Temperature is the fundamental quantity here; velocity, force and density are all derived from base quantities.

The seven SI base (fundamental) quantities are length, mass, time, electric current, thermodynamic temperature, amount of substance and luminous intensity. A derived quantity is built from these using multiplication or division.

Checking each option:

velocity = displacement / time = length / time -> derived. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The physical quantity having the dimensions [M¹T⁻²] is$
(a) (A) Density
(b) (B) Tension
(c) (C) Surface tension
(d) (D) Viscosity
›Reveal solutionSolution
[!TLDR]

(C) Surface tension

Why

Surface tension = force/length = (MLT⁻²)/L = MT⁻², exactly matching the g …
Start your 14-day free trial to see the full solution →

←Previous questionQ17 — What are fundamental and derived quantities? Give examples.→Next questionQ19 — Which of the following has the dimension of (mu0 epsilon0)^(-1/2)? (a)…

Q.(a) Assuming that the frequency gamma of a vibrating string may depend upon

🎓Unlock everything free for 14 days

Unlock everything free for 14 days