SSC JE Civil Engineering 29-Oct-2020 Shift 1 Paper

Given:
$\frac{\mathrm{P}}{\mathrm{W}}=\frac{1}{4}$
$\mathit{g}=10\mathrm{m}∕{\mathrm{s}}^2$ , $\mathrm{V}=20\mathrm{m}∕\mathrm{s}=72\mathrm{kmph}$
Acceleration $\mathit{a}=0.25\mathrm{m}∕{\mathrm{s}}^2$
Rate of change of centrifugal Acceleration $\left(\mathrm{C}\right)=\frac{80}{\left(75+\mathrm{V}\right)}$
where V is in kmph
i.e. $\left(\mathrm{C}\right)=\frac{80}{\left(75+\mathrm{V}\right)}$ $=\frac{80}{\left(75+72\right)}$ $=0.544\mathrm{m}∕{\sec }^3$
We know that $\frac{\mathrm{P}}{\mathrm{W}}=\frac{{\mathrm{V}}^2}{\mathit{g}\mathrm{R}}$
$\frac{1}{4}=\frac{{\left(20\right)}^2}{10\times \mathrm{R}}\Rightarrow \mathrm{R}=160\mathrm{m}$
Also,
$\mathit{L}=\frac{{\mathrm{V}}^3}{\mathrm{RC}}$ $=\frac{{\left(20\right)}^3}{160\times 0.54}=91.91\mathrm{m}$
∴ R = 160 m & L = 92 m is most appropriate answer.