A pan with a set of weights is attached to a light spring. The...

Given, $T=0.5 \mathrm{~s}$, We know that, $T=2 \pi \sqrt{\frac{m}{k}} \Rightarrow 0.5=2 \pi \sqrt{\frac{m}{k}} \frac{1}{4 \pi}=\sqrt{\frac{m}{k}}$ On squaring both sides, we get $\frac{m}{k}=\frac{1}{16 \pi^2}$...(i) $$\begin{aligned} & \Rightarrow \quad T+0.1=2 \pi \sqrt{\frac{m+m^{\prime}}{k}} \\ & \Rightarrow \quad 0.5+0.1=2 \pi \sqrt{\frac{m+m^{\prime}}{k}} \\ & \Rightarrow \quad\left(\frac{0.6}{2 \pi}\right)^2=\frac{m+m^{\prime}}{k}=\frac{m}{k}+\frac{m^{\prime}}{k} \\ & \end{aligned}$$\Rightarrow \quad\left(\frac{0.6}{2 \pi}\right)^2=\frac{1}{16 \pi^2}+\frac{m^{\prime}}{k} \quad$[using Eq. (i)]$\Rightarrow \quad \frac{(0.6)^2}{4 \pi^2}-\frac{1}{16 \pi^2}=\frac{m^{\prime}}{k} \Rightarrow \quad \frac{m^{\prime}}{k}=\frac{1.44-1}{16 \pi^2} \Rightarrow \frac{m^{\prime}}{k}=\frac{0.44}{16 \pi^2}$....(ii)$$\begin{aligned} \because & m^{\prime} g & =k x \\ \therefore & x & =\frac{m^{\prime}}{k} g\end{aligned}$$=\frac{0.44}{16 \pi^2}$ \times 10 \quad$[using Eq. (ii)]$$\begin{aligned} & =0.0279 \\ & =2.79 \times 10^{-2} \mathrm{~m} \\ & =2.7 \mathrm{~cm}\end{aligned}$$