A) \[{{T}_{1}}{{T}_{2}}g\]
B) \[\frac{{{T}_{1}}{{T}_{2}}g}{2}\]
C) \[\left( T_{1}^{2}+T_{2}^{2} \right)g\]
D) \[\frac{\left( T_{1}^{2}+T_{2}^{2} \right)}{2}g\]
Correct Answer: B
Solution :
In case first, \[{{T}_{1}}=\frac{2u\,\sin \theta }{g}\]and in case second \[{{T}_{2}}=\frac{2u\,\cos \theta }{g}\] Since one of the angle is \[\theta \] and other is \[\left( \frac{\pi }{2}-\theta \right)\] While, \[R=\frac{{{u}^{2}}\sin 2\theta }{g}=\frac{2{{u}^{2}}\cos \theta \sin \theta }{g}\] \[\Rightarrow \] \[=\frac{g}{2}\times \left( \frac{2u\,\sin \theta }{g} \right)\,\left( \frac{2u\,\cos \theta }{g} \right)\] \[\Rightarrow \] \[R=\frac{{{T}_{1}}{{T}_{2}}g}{2}\]
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