$\begin {array} {1 1} (a)\;2 \times 10^7\: m/s & \quad (b)\;2 \times 10^6\: m/s \\ (c)\;8 \times 10^6\: m/s & \quad (d)\;8 \times 10^5\: m/s \end {array}$

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Ans : (c)

$ \large\frac{1}{2} mv^2_{max} = hv - hv_{\circ}$

$ \large\frac{1}{2} m \times 4 \times 10^6 \times 4 \times 10^6 = 2v_{\circ} – hv_{\circ}$

Again, $ \large\frac{1}{2} m \times v^2_{max} = 5hv_{\circ} – hv_{\circ} $

Dividing, $\large\frac{v^2_{max} }{(4 \times 10^6 )^2} = \large\frac{4hv_{\circ}}{hv_{\circ}} = 4$

$ \Rightarrow \large\frac{v_{max}}{4 \times 10^6} = 2$

$ \Rightarrow v_{max}\: = 8 \times 10^6 m/s$

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