Question

The magnitude of the PE per unit mass of the object at the surface of earth is E. Then the escape velocity of the object is,
a) root of 2E
b) root of 4E^2
c) root of E
d) root of E/2

Answer 1

$$\begin{gathered} \mathrm{Potential}\mathrm{energy}\mathrm{per}\mathrm{unit}\mathrm{mass}=\mathrm{E} \\ \mathrm{Total}\mathrm{potential}\mathrm{energy}=\mathrm{mE} \\ \mathrm{Equate}\mathrm{potential}\mathrm{energy}\mathrm{to}\mathrm{kinetic}\mathrm{energy} \\ \frac{1}{2}{{\mathrm{mv}}_{\mathrm{e}}}^2=\mathrm{mE} \\ \frac{1}{2}{{\mathrm{v}}_{\mathrm{e}}}^2=\mathrm{E} \\ \mathrm{Hence}, \\ {\mathrm{v}}_{\mathrm{e}}=\sqrt{2\mathrm{E}} \\ \mathrm{Ans}:\mathrm{Option}-\mathrm{a} \end{gathered}$$