Correct Answer - Option 3 : 300%
n = 0.5, V2 = V1/2
\(\begin{array}{l} V{T^n} = C \Rightarrow V{T^{0.5}} = C \Rightarrow V \propto \frac{1}{{\sqrt T }}\\ So,\frac{{{V_2}}}{{{V_1}}} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} \Rightarrow \frac{1}{2} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \frac{1}{4} \Rightarrow {T_2} = 4{T_1} \end{array}\)
% increases in tool life
\(= \frac{{{T_2} - {T_1}}}{{{T_1}}} \times 100 = 300\%\)