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

Step 1: Understanding the Question:
The question asks for the dimensional formulas of a variable \(P\) and a constant \(\beta\) given a physical relation.
The principle of homogeneity states that dimensions on both sides of an equation must be identical.
Arguments of trigonometric functions (like \(\sin \theta\)) must be dimensionless.
Step 2: Key Formula or Approach:
Dimension of Force \(F = [MLT^{-2}]\).
Dimension of Velocity \(v = [LT^{-1}]\).
Argument of \(\sin(\beta t)\) must be \([M^0L^0T^0]\).
Since \(\sin(\beta t)\) is dimensionless, the dimension of \(P\) is simply the product of dimensions of \(F\) and \(v\).
Step 3: Detailed Explanation:

Finding dimensions of \(\beta\):
The argument inside the sine function is \(\beta t\). For it to be dimensionless:
\[ [\beta] [t] = [1] \]
\[ [\beta] [T] = [T^0] \implies [\beta] = [T^{-1}] \]

Finding dimensions of \(P\):
From the given equation \(P = F \cdot v \sin(\beta t)\), and since \(\sin(\beta t)\) is a numeric ratio with no dimensions:
\[ [P] = [F] \cdot [v] \]
\[ [P] = [MLT^{-2}] \cdot [LT^{-1}] \]
\[ [P] = [ML^2T^{-3}] \]

Note: The product of Force and Velocity is Power, which indeed has dimensions \(ML^2T^{-3}\).

Step 4: Final Answer:
The dimensional formula for \(P\) is \(ML^2T^{-3}\) and for \(\beta\) is \(T^{-1}\).