A) \[{{30}^{o}}\]
B) \[{{76}^{o}}\]
C) \[{{53}^{o}}\]
D) \[{{90}^{o}}\]
Correct Answer: B
Solution :
In projectile motion the maximum length is given by, \[{{H}_{\max }}=\frac{{{u}^{2}}{{\sin }^{2}}\alpha }{2g}\] ?..(i) and \[R=\frac{{{u}^{2}}{{\sin }^{2}}\alpha }{g}\] .....(ii) Given that, \[{{H}_{\max }}=R\] \[\therefore \] \[\frac{{{u}^{2}}{{\sin }^{2}}\alpha }{2g}=\frac{{{u}^{2}}\sin 2\alpha }{g}\] Using equations (i) and (ii) \[\Rightarrow \] \[{{\sin }^{2}}\alpha =4\sin \alpha \,\cos \alpha \] \[(\because \,\,\sin 2\alpha =2\sin \alpha \cos \alpha )\] \[\tan \alpha =4\] or \[\alpha ={{\tan }^{-1}}4\] \[\left( \because \,\frac{\sin \alpha }{\cos \alpha }=\tan \alpha \right)\] \[\Rightarrow \] \[\alpha ={{76}^{o}}\] (Using trignoment table tan \[\tan \,\,4={{76}^{o}}\]) For ideal gas at constant pressure, we have, \[\frac{{{V}_{1}}}{{{T}_{1}}}=\frac{{{V}_{2}}}{{{T}_{2}}}\] \[\Rightarrow \] \[{{T}_{2}}=\frac{{{V}_{2}}}{{{V}_{1}}}\times {{T}_{1}}\] ?..(I) Here \[{{V}_{1}}-V,\,\,{{V}_{2}}=3V\] \[{{T}_{1}}={{0}^{o}}C=273\,K.\] Putting values in equation (i) we get, \[{{T}_{2}}=\frac{3V}{V}\times 273.\] \[\Rightarrow \] \[{{T}_{2}}=3\times 273\,K=819\,K\] \[{{T}_{2}}=819-273={{546}^{o}}C\]
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