NTA Abhyas JEE Main2020PhysicsUnits and DimensionsPractice
A uniform wire of length l and mass M is stretched between two fixed points, keeping a tension F. A sound of frequency μ is impressed on it. Then the maximum vibrational energy is existing in the wire when μ =
Options
- A1 2 ML F
- BFL M
- C2 × FM L
- D1 2 F ML
Correct answer
D. 1 2 F ML
Step-by-step solution
We should check it dimensionally Let μ = K F a M b L c μ = F a M b L c = MLT 2 a M b L c M 0 L 0 T - 1 = M a + b L a + c T - 2 a a + b = 0 ⇒ b = - a a + c = 0 ⇒ c = - a - 2 a = - 1 ⇒ a = 1 2 b = - 1 2 c = - 1 2 ⇒ μ = K F 1 2 M - 1 2 L - 1 2 μ = K F ML So only option (iv) satisfies it dimensionally.