JEE PYQ: Units & Measurements - Question ID 90a065751921 (JEE Main 2021)

ID: 90a065751921JEE Main 2021Single Correct MCQ
The force is given in terms of time t and displacement x by the equation

F = A cos Bx + C sin Dt

The dimensional formula of ADB{{AD} \over B} is :

Select Option

Step-by-step Explanation

Core Formula & Concept:

In dimensional analysis, every physical quantity can be expressed in terms of the fundamental dimensions: mass (MM), length (LL), and time (TT). The key principle is that the arguments of trigonometric functions (like cos⁡\cos and sin⁡\sin) must be dimensionless. This ensures that the functions yield pure numbers, which is essential for mathematical consistency.

Given the force equation: F=Acos⁡(Bx)+Csin⁡(Dt)F = A \cos(Bx) + C \sin(Dt)

  • The term Acos⁡(Bx)A \cos(Bx) implies that BxBx must be dimensionless.
  • The term Csin⁡(Dt)C \sin(Dt) implies that DtDt must be dimensionless.
Force (FF) has the dimensional formula [M1L1T−2][M^1 L^1 T^{-2}]. The coefficients AA and CC must also have the same dimensions as force since they multiply dimensionless trigonometric functions.

Step-by-Step Derivation:

Step 1: Determine the dimensions of BB
The argument BxBx must be dimensionless. Since xx is displacement, its dimensions are [L1][L^1]. Thus: [B]⋅[L1]=[M0L0T0]  ⟹  [B]=[L−1][B] \cdot [L^1] = [M^0 L^0 T^0] \implies [B] = [L^{-1}] Step 2: Determine the dimensions of DD
The argument DtDt must be dimensionless. Since tt is time, its dimensions are [T1][T^1]. Thus: [D]⋅[T1]=[M0L0T0]  ⟹  [D]=[T−1][D] \cdot [T^1] = [M^0 L^0 T^0] \implies [D] = [T^{-1}] Step 3: Determine the dimensions of AA and CC
Since F=Acos⁡(Bx)+Csin⁡(Dt)F = A \cos(Bx) + C \sin(Dt) and cos⁡(Bx)\cos(Bx) and sin⁡(Dt)\sin(Dt) are dimensionless, the dimensions of AA and CC must match those of force: [A]=[C]=[M1L1T−2][A] = [C] = [M^1 L^1 T^{-2}] Step 4: Compute the dimensions of ADB{{AD} \over B}
Substitute the dimensions of AA, DD, and BB: [ADB]=[A]⋅[D][B]=[M1L1T−2]⋅[T−1][L−1]\left[ \frac{AD}{B} \right] = \frac{[A] \cdot [D]}{[B]} = \frac{[M^1 L^1 T^{-2}] \cdot [T^{-1}]}{[L^{-1}]} Simplify the expression: =[M1L1T−3][L−1]=[M1L1+1T−3]=[M1L2T−3]= \frac{[M^1 L^1 T^{-3}]}{[L^{-1}]} = [M^1 L^{1+1} T^{-3}] = [M^1 L^2 T^{-3}] Step 5: Match with the given options
The derived dimensional formula is [M1L2T−3][M^1 L^2 T^{-3}], which corresponds to Option B.

Common Traps & Exam Tip:

  1. Ignoring the dimensionless nature of trigonometric arguments: Students often forget that cos⁡(Bx)\cos(Bx) and sin⁡(Dt)\sin(Dt) must be dimensionless, leading to incorrect dimensions for BB and DD.
  2. Miscalculating the dimensions of AA and CC: Some assume AA and CC are dimensionless or have different dimensions than force, which is incorrect.
  3. Algebraic errors in dimensional simplification: When computing ADB{{AD} \over B}, students may incorrectly add or subtract exponents, especially for length (LL). For example, dividing by [L−1][L^{-1}] is equivalent to multiplying by [L1][L^1], not subtracting.
  4. Confusing the order of operations: Ensure that the dimensions of AA, DD, and BB are substituted correctly before simplifying. A common mistake is to misplace the exponents during multiplication or division.
Exam Tip: Always verify that the arguments of trigonometric, exponential, or logarithmic functions are dimensionless. This is a golden rule in dimensional analysis and will help you avoid critical errors.

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