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 [M1L1T2][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]=[L1][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]=[T1][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]=[M1L1T2][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]=[M1L1T2][T1][L1]\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: =[M1L1T3][L1]=[M1L1+1T3]=[M1L2T3]= \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 [M1L2T3][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 [L1][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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