A) 1.00
B) 0.75
C) 0.50
D) 0.25
Correct Answer: D
Solution :
\[{{R}_{d}}=\frac{{{v}_{f}}}{2u}(tan{{\beta }_{1}}+tan{{\beta }_{2}})\] \[\frac{{{v}_{f}}}{u}=\frac{1}{2},\] \[tan{{\beta }_{1}}=\tan {{\beta }_{2}}=\frac{1}{2}\] \[\frac{{{v}_{f}}}{u}=\frac{1}{2},\] \[tan{{\beta }_{1}}=\tan {{\beta }_{2}}=\frac{1}{2}\] \[{{R}_{d}}=\frac{1}{4}\,\,\left( \frac{1}{2}+\frac{1}{2} \right)=0.25\]
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