If the tension and diameter of a sonometer wire of fundamental frequency $n$ are doubled and density is halved, then its fundamental frequency will become

The correct option is C $n$
Frequency of a sonometer wire $n=\frac{1}{2l}\sqrt{\frac{T}{\pi r^2\rho }}$
As the tension and diameter of a sonometer wire are doubled and density is halved
${\rho }^{\prime }=\frac{\rho }{2};T^{\prime }=2T\text{and}{D}^{\prime }=2D\Rightarrow r^{\prime }=2r$

$n^{\prime }=\frac{1}{2l}\sqrt{\frac{2T}{\pi (2r{)}^2\frac{\rho }{2}}}=\frac{1}{2l}\sqrt{\frac{T}{\pi r^2\rho }}=n.$
Hence, the correct answer is option (c)