Question

Find the dimensions of the quantity \( x \) in the equation \( T = 2\pi \left( \frac{ML^3}{3XY} \right)^{\frac{1}{2}} \), where \( T \) is the time period, \( M \) is mass, \( L \) is length, and \( Y \) is the Young’s modulus.

• A. \( [L^3] \)
• B. \( [L^6] \)
• C. \( [L^2] \)
• D. \( [L^4] \)

Hint:

To find dimensions in equations, substitute the dimensional formula of each quantity and simplify. This helps in solving for unknown dimensions.

Answer 1

The Correct Option is D

Step 1: Analyze the given equation.
The time period \( T \) is related to the equation \( T = 2\pi \left( \frac{ML^3}{3XY} \right)^{\frac{1}{2}} \). We are asked to find the dimensions of \( X \) in this equation. Step 2: Dimension analysis.
The dimension of the time period \( T \) is \( [T] = [M^0 L^0 T^1] \). The dimension of mass \( M \) is \( [M] = [M^1 L^0 T^0] \), the dimension of length \( L \) is \( [L] = [M^0 L^1 T^0] \), and the dimension of Young’s modulus \( Y \) is \( [Y] = [M^1 L^{-1} T^{-2}] \). Now, substituting the dimensions into the equation: \[ T = 2\pi \left( \frac{ML^3}{3XY} \right)^{\frac{1}{2}} \] Taking the dimension of both sides and equating, we get: \[ [M^0 L^0 T^1] = \left( \frac{[M^1 L^3 T^0]}{[X][M^1 L^{-1} T^{-2}]} \right)^{\frac{1}{2}} \] Simplifying the expression, we find that the dimensions of \( X \) are \( [L^4] \). Step 3: Conclusion.
The correct dimensions of \( X \) are \( [L^4] \), which corresponds to option (D).