A uniform rope of length ' L ' and mass ' m₁ ' hangs vertically from a rigid support. A block of mass ' m₂ ' is attached to the free end of the rope. A transverse wave of wavelength ' λ₁ ' is produced at the lower end of the rope. The wavelength of the wave when it reaches the top of the rope is ' λ₂ '. The ratio (λ₁)/(λ₂) is (a) [(m₂)/(m₁+m₂)]1/2 (b) [(m₁+m₂)/(m₂)]1/2 (c) [(m₁)/(m₁+m₂)]1/2 (d) [(m₂)/(m₁-m₂)]1/2

Answer: (a). Frequency unchanged, so λ ∝ √(T). Bottom tension m₂ g, top tension (m₁+m₂)g. (λ₁)/(λ₂) = [(m₂)/(m₁+m₂)]1/2 So the answer is (a).

A uniform rope of length ' L ' and mass ' m₁ ' hangs vertically from… | MHT CET 2023 (10 May, Shift 2)

Q20MHT CET 2023 · 10 MAY · SHIFT 2Difficult

Speed of Waves

A uniform rope of length ' $L$ ' and mass ' $m_1$ ' hangs vertically from a rigid support. A block of mass ' $m_2$ ' is attached to the free end of the rope. A transverse wave of wavelength ' $\lambda_1$ ' is produced at the lower end of the rope. The wavelength of the wave when it reaches the top of the rope is ' $\lambda_2$ '. The ratio $\frac{\lambda_1}{\lambda_2}$ is

Show answer and solution

Answer: (a)

Frequency unchanged, so $\lambda \propto \sqrt{T}$ .
Bottom tension $m_2 g$ , top tension $(m_1+m_2)g$ .
$\dfrac{\lambda_1}{\lambda_2} = \left[\dfrac{m_2}{m_1+m_2}\right]^{1/2}$
So the answer is (a).

Report a mistake

Speed of Waves question from MHT CET 2023 (10 May, Shift 2)

More questions on Speed of Waves