In a machining operation, doubling the cutting speed reduces the tool life to $1/8^{th}$ of the original value. The exponent $n$ in Taylor's tool life equation $V{T}^n=C$, is

The correct option is C $\frac{1}{3}$
Taylor's tool life equation,
$V{T}^n=C$ ... (i)
$\text{When cutting speed is doubled and tool life reduces to}1/8^{th},then$
$2V\times {\left(\frac{1}{8}T\right)}^n=C$ ... (ii)
From equation (i) and (ii), we get
$2V{\left(\frac{T}{8}\right)}^n=V{T}^n$
${\left(\frac{T}{8}\right)}^n=\frac{1}{2}$
$3n=1$
$\therefore n=1/3$