The drawing shows a frictionless incline and pulley. The two blocks are connected by a wire (mass per unit length, μ=25 g/m) and remain stationary. A transverse wave on the wire has a speed of  60m/s relative to it. Neglect the weight of the wire relative to the tension in the wire. If the mass  m2 be increased by 1%, the speed (in m/s) of the transverse wave relative to string will be

Initially, m2g=T−−−−1​T=m1gsin30−−−2​from 1 & 2​⇒m2g=m12g​⇒m1=2m2v=Tμ⇒dvv=12dTTwhere  is velocity of the wave, T is tension and μ is the mass for unit length of the string.Now, after increasing value of m2 , tension in string also increases.New mass of 2nd block : m2'=1.01m2Let say new value of tension is T'Applying Newton’s second law for both blocks, m2'g−T'=m2'aandT'-m1gsin300=m1aAfter solving, T'=3m1m2'g2m1+m2'⇒T'=32m21.01m2g22m2+1.01m2=6.066.02m2g        ∵m1=2m2∆TT=T'-TT=6.066.02-1=0.046.02Now, ⇒dvv=12dTT=12×0.046.02⇒dv60=12×0.046.02⇒dv=60×12×0.046.02⇒dv=0.2Therefore, the correct answer is  60.20