Explanation

(c) The frequency of sonometer wire is
$v = \frac{1}{2l}\sqrt{\frac{T}{m}}$
Taking logarithm and differentiating, we get
$\frac{\mathrm{\Delta}v}{v} = \frac{\mathrm{\Delta}l}{l} + \frac{1}{2}\frac{\mathrm{\Delta}T}{T} + \frac{1}{2}\frac{\mathrm{\Delta}m}{m}$
$\therefore\frac{\mathrm{\Delta}v}{v} = \frac{\mathrm{\Delta}l}{l} + 1 + 1 = 1\%$
Hence, frequency will increase by 1%