The linear momentum of a body is increased by$100\%$.The kinetic energy increases by

$$\begin{gathered} p_1=p,p_2=p_1+100\% p_1=P+\frac{100}{100}P=2P \\ {K}_1=K,K_2=? \\ {K}_1=\frac{p^2}{2m}\Rightarrow K_2=\frac{{p_2}^2}{2m}=\frac{{\left(2p\right)}^2}{2m}=4K_1=4(100\%)=400\% \\ K.E increased to=400\% \\ K.E increased by =300\% \end{gathered}$$