JEE PYQ: Units & Measurements - Question ID 8b42665875c5 (JEE Main 2020)

ID: 8b42665875c5JEE Main 2020Single Correct MCQ
If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is

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Step-by-step Explanation

Core Formula & Concept:

In dimensional analysis, we express any physical quantity in terms of fundamental quantities (mass, length, time, etc.). Here, the problem redefines the fundamental quantities as momentum (PP), area (AA), and time (TT). Our goal is to express energy in terms of these new fundamental dimensions.

The key formulas we use are:

  • Momentum: P=mvP = mv (mass × velocity)
  • Area: A=L2A = L^2 (length squared)
  • Energy (kinetic): E=12mv2E = \frac{1}{2}mv^2

We need to express mass (MM), length (LL), and velocity (vv) in terms of PP, AA, and TT to rewrite energy.

--- Step-by-Step Derivation:

Step 1: Express mass (MM) in terms of PP, AA, and TT

From momentum: P=mv    m=PvP = mv \implies m = \frac{P}{v}. But we need mm in terms of PP, AA, and TT only. So, we need to express vv in terms of AA and TT.

Step 2: Express velocity (vv) in terms of AA and TT

Velocity is distance over time: v=LTv = \frac{L}{T}. But A=L2    L=A1/2A = L^2 \implies L = A^{1/2}. Thus, v=A1/2Tv = \frac{A^{1/2}}{T}.

Step 3: Substitute vv back into the expression for mass

m=Pv=PA1/2/T=PA1/2Tm = \frac{P}{v} = \frac{P}{A^{1/2}/T} = P A^{-1/2} T.

Step 4: Express energy (EE) in terms of PP, AA, and TT

Kinetic energy: E=12mv2E = \frac{1}{2}mv^2. Substitute m=PA1/2Tm = P A^{-1/2} T and v=A1/2T1v = A^{1/2} T^{-1}:

E=12(PA1/2T)(A1/2T1)2E = \frac{1}{2} \left(P A^{-1/2} T\right) \left(A^{1/2} T^{-1}\right)^2

Simplify the expression:

E=12(PA1/2T)(A1T2)=12PA1/2+1T12=12PA1/2T1E = \frac{1}{2} \left(P A^{-1/2} T\right) \left(A^{1} T^{-2}\right) = \frac{1}{2} P A^{-1/2 + 1} T^{1 - 2} = \frac{1}{2} P A^{1/2} T^{-1}

The constant 12\frac{1}{2} is dimensionless and can be ignored in dimensional analysis. Thus, the dimensional formula for energy is:

[E]=[PA1/2T1][E] = \left[ P A^{1/2} T^{-1} \right]

Step 5: Match with the given options

The derived formula matches option C: [PA1/2T1]\left[ P A^{1/2} T^{-1} \right].

--- Common Traps & Exam Tip:

Students often make the following mistakes:

  • Incorrect substitution of velocity: Forgetting that v=LTv = \frac{L}{T} and L=A1/2L = A^{1/2} leads to wrong expressions for mass and energy.
  • Ignoring dimensional consistency: Some students directly equate energy to P2P^2 or AA without proper derivation, leading to incorrect options like A or B.
  • Sign errors in exponents: Misapplying the rules of exponents (e.g., (A1/2)2=A1(A^{1/2})^2 = A^1, not AA) can result in wrong dimensional formulas.

Exam Tip: Always express all intermediate quantities (like mass and velocity) in terms of the given fundamental quantities before substituting into the target formula. Double-check exponent arithmetic to avoid sign errors.

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