The drawing shows a frictionless incline and pulley. The two blocks are connected by a wire (mass per unit length, $\mu =25\text{\hspace{0.17em}g}/\text{m}$) 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  $m_2$ be increased by 1%, the speed (in m/s) of the transverse wave relative to string will be

 

  $\begin{array}{l}Initially, m_{2}g=T----\left(1\right)\text{}\\ T=m_{1}g\sin 30---\left(2\right)\text{}\\ from \left(1\right) \& \left(2\right)\text{}\\ \Rightarrow m_{2}g=\frac{m_1}{2}g\text{}\end{array}$

$\Rightarrow m_1=2m_2$
$v=\sqrt{\frac{T}{\mu }}$

$\Rightarrow \frac{\text{d}v}{v}=\frac{1}{2}\frac{\text{d}T}{T}$
where  is velocity of the wave, T is tension and $\mu$ is the mass for unit length of the string.
Now, after increasing value of $m_2$ , tension in string also increases.

New mass of 2^nd block : $m_2\text{'}=1.01m_2$

Let say new value of tension is $T\text{'}$

Applying Newton’s second law for both blocks, 
$m_2\text{'}g-T\text{'}=m_2\text{'}a$

and

$T\text{'}-m_{1}g\sin {30}^0=m_{1}a$

$\text{After solving, }$

$T\text{'}=\frac{3m_1{m}_2\text{'}g}{2\left(m_1+m_2\text{'}\right)}$

$\Rightarrow T\text{'}=\frac{3\left(2m_2\right)\left(1.01m_2\right)g}{2\left(2m_2+1.01m_2\right)}=\frac{6.06}{6.02}{m}_{2}g \left(\because m_1=2m_2\right)$

$\frac{∆T}{T}=\frac{T\text{'}-T}{T}=\frac{6.06}{6.02}-1=\frac{0.04}{6.02}$

Now, 
$\Rightarrow \frac{\text{d}v}{v}=\frac{1}{2}\frac{\text{d}T}{T}=\frac{1}{2}\times \frac{0.04}{6.02}$

$\Rightarrow \frac{\text{d}v}{60}=\frac{1}{2}\times \frac{0.04}{6.02}$

$\Rightarrow \text{d}v=60\times \frac{1}{2}\times \frac{0.04}{6.02}$

$\Rightarrow \text{d}v=0.2$
Therefore, the correct answer is  60.20