NTA Abhyas JEE Main2020PhysicsWaves and SoundPractice
A massless rod of length l is hung from the ceiling with the help of two identical wires attached at its ends. A block is hung on the case, the frequency of the 1 st harmonic of the wire on the left end is equal to the frequency of the 2 nd harmonic of the wire on the right. The value of X is
Options
- Al 2
- Bl 3
- Cl 4
- Dl 5
Correct answer
D. l 5
Step-by-step solution
v = λ f T 1 μ T 2 μ = 2 l f l f T 1 T 2 = 4 T 1 + T 2 = M g ; T 2 = M g 5 Σ t B = 0 ⇒ M g × X = M g 5 × l ; X = l 5