AP EAPCET 24-Aug-2021 Shift 2 Paper

Given that, mass of body, $m=4.9\ \mathrm{kg}$ Time period of oscillation of spring. $T=0.5\ \mathrm{s}$ Acceleration due to gravity, $g=10\ \mathrm{m/s}^2$ and $\pi^{2}=10$ We know that, time period of spring $T=2\pi\sqrt{\frac{m}{k}}$ By squaring on both sides, $T^{2}=4\pi^{2}\;\frac{m}{k}$ $\Rightarrow\frac{m}{k}=\frac{T^{2}}{4\pi^{2}}$ $=\frac{(0.5)^{2}}{4\times10}=\frac{0.25}{40}=000625$ After removal of mass, length of spring decrease is equal to extension produced in spring. By using equilibrium condition, $F=mg$ $\Rightarrow kx=mg$ $x=\frac{m}{k}g$ Substituting the values, we get $x=0.00625\times10$ $=0.0625\ \mathrm{m}$ $=6.25\ \mathrm{cm}$ Hence, the spring is shortened by $6.25\ \mathrm{cm}$ after removal of mass.