A transverse wave is represented by:
 $ y = \dfrac{{10}}{\pi }\sin \left( {\dfrac{{2\pi }}{T}t - \dfrac{{2\pi }}{\lambda }x} \right) $
For what value of the wavelength is the wave velocity twice the maximum particle velocity
(A) 40cm
(B) 20cm
(C) 10cm
(D) 60cm

Hint
Particle velocity is perpendicular to the wave velocity. This implies that it propagates on $ y $ axis while the wave velocity propagates on $ x $.
Formula used: $ v = \dfrac{\lambda }{T} $ where $ v $ is the wave velocity, $ \lambda $ is the wavelength of the wave and $ T $ is the period.

Complete step by step answer
Comparing the above equation with the equation of a travelling transverse wave
 $\Rightarrow y = {y_{\max }}\sin \left( {\dfrac{{2\pi }}{T}t - \dfrac{{2\pi }}{\lambda }x} \right) $. $ y $ is the vertical position of the particle at a time $ t $ and horizontal position $ x $, $ {y_{\max }} $ is the maximum displacement of the particle from its equilibrium position.
Therefore,
 $\Rightarrow {y_{\max }} = \dfrac{{10}}{\pi } $
The particle velocity is derived by differentiating the particle position as shown below
 $\Rightarrow \dfrac{{dy}}{{dt}} = \left( {\dfrac{{2\pi }}{T}} \right){y_{\max }}\cos \left( {\dfrac{{2\pi }}{T}t - \dfrac{{2\pi }}{\lambda }x} \right) $
 $\Rightarrow {\left( {\dfrac{{dy}}{{dt}}} \right)_{\max }} = \left( {\dfrac{{2\pi }}{T}} \right){y_{\max }} $
Therefore, particle velocity equals to two times wave velocity implies that
 $\Rightarrow \left( {\dfrac{{2\pi }}{T}} \right){y_{\max }} = 2v $
But wave velocity is given by $ v = \dfrac{\lambda }{T} $.
Thus, $ \left( {\dfrac{{2\pi }}{T}} \right){y_{\max }} = 2\dfrac{\lambda }{T} $
Making $ \lambda $ subject of the formula, and cancelling out $ T $ and 2 we have
 $\Rightarrow \lambda = \pi {y_{\max }} = \pi \times \dfrac{{10}}{\pi } $
 $ \therefore \lambda = 10cm $
Hence, the answer is C.

Additional Information
Transverse wave is a type of wave whose particles vibrate perpendicularly to the propagation of the wave itself. This implies that wave velocity is always perpendicular to particle velocity. This is why we have two positions in the wave equation, one vertical and the other horizontal.

Note
How we identify the maximum velocity [and in some cases acceleration (not in the question)], may have eluded you. But the maximum of any of these quantities is simply the value of the quantities when the phase angle is equal to one i.e. $ \sin \left( {\dfrac{{2\pi }}{T}t - \dfrac{{2\pi }}{\lambda }x} \right) = 1 $ (for the position) and $ \cos \left( {\dfrac{{2\pi }}{T}t - \dfrac{{2\pi }}{\lambda }x} \right) = 1 $ (for velocity) which always leaves us with the first term on our right hand side of the equation.