Question

If \(P=Fv\sin\beta t\), where \(F\) is force and \(v\) is velocity, then the dimensions of \(P\) and \(\beta\) are

• A. \(ML^2T^{-3},\ T^{-1}\)
• B. \(MLT^{-2},\ T^{-2}\)
• C. \(ML^2T^{-1},\ T^{-1}\)
• D. \(ML^2T^3,\ T^{-2}\)

Hint:

The angle of \(\sin\), \(\cos\), or \(\tan\) must always be dimensionless. Use this to find unknown dimensions.

Answer 1

The Correct Option is A

Concept: The argument of a trigonometric function must always be dimensionless. Also, force multiplied by velocity gives power.

Step 1: Given: \[ P=Fv\sin\beta t \] Since \(\sin\beta t\) is a trigonometric term, its angle \(\beta t\) must be dimensionless. \[ [\beta t]=1 \] \[ [\beta][t]=1 \] \[ [\beta]=T^{-1} \]

Step 2: Now find the dimension of \(P\).
Since \(\sin\beta t\) is dimensionless, \[ [P]=[F][v] \] Dimension of force: \[ [F]=MLT^{-2} \] Dimension of velocity: \[ [v]=LT^{-1} \] Therefore: \[ [P]=MLT^{-2}\times LT^{-1} \] \[ [P]=ML^2T^{-3} \]

Step 3: Hence: \[ [P]=ML^2T^{-3}, \qquad [\beta]=T^{-1} \] Therefore, \[ \boxed{ML^2T^{-3},\ T^{-1}} \]