a
Let the length of the spring is \(l\). When the system is whirled round in a horizontal circle the centripetal force is given by

\(F=\frac{m{{v}^{2}}}{r}=\frac{m{{(r\omega )}^{2}}}{r}=mr{{\omega }^{2}}\)

then \(r=l+\) elongation Given: elongation \(=1 \;cm\) (in the first case)

or angular velocity \(\omega\) the force required is

\({{F}_{1}}=m(l+1){{\omega }^{2}}=kx=k\times 1=k\)

\(k=m(l+1){{\omega }^{2}}\) .......(i)

\({{F}_{2}}=m(l+5){{(2\omega )}^{2}}=kx=k\times 5=5k\)

\(5k=4m(l+5){{\omega }^{2}}\).......(ii)

Now, dividing Eq. \((i)\) by Eq.\((ii)\), we get

\(\frac{k}{5k}=\frac{m(l+1){{\omega }^{2}}}{4m(l+5){{\omega }^{2}}}\)

\(5(l+1)=4(l+5)\)

\(l=20-5=15\,cm\)