A) 1900
B) 2000
C) 2100
D) 2200
Correct Answer: C
Solution :
\[v'\frac{v}{\sqrt{2}}\] \[E=\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}(2m){{v}^{2}}\] By binomial theorem, \[=mv{{'}^{2}}+m{{v}^{2}}\] \[\Rightarrow \] \[m{{\left( \frac{v}{\sqrt{2}} \right)}^{2}}+m{{v}^{2}}=\frac{3}{2}m{{v}^{2}}\] \[{{T}_{1}}-{{T}_{2}}=12a\] \[{{T}_{2}}=3a\] \[{{T}_{2}}\]\[Eq\] \[{{T}_{1}}=15a\] Similarly, \[\underset{^{\theta \to \frac{{{\pi }^{-}}}{2}}}{\mathop{\lim }}\,\,\frac{{{(1+\tan \,\theta )}^{10}}}{{{\tan }^{9}}\theta }=20+\,\underset{\theta \to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\tan \theta \], \[\underset{^{\theta \to \frac{{{\pi }^{-}}}{2}}}{\mathop{\lim }}\,\,\frac{{{(3+\tan \,\theta )}^{10}}}{{{\tan }^{9}}\theta }=30+\,\underset{\theta \to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\tan \theta \] ?????????????????????????????????????????????????????? \[\underset{\theta \to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\frac{{{(20+\tan \theta )}^{10}}}{{{\tan }^{9}}\theta }=200+\,\,\underset{\theta \to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\,\,\tan \theta \] On adding all these terms; we get \[\underset{\theta \to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,f(\theta )=(10+20+...+200)+20\,\,\underset{\theta \to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\,\,\tan \theta \] \[-\underset{\theta \to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,20\,d\,\tan \,\theta \] \[\Rightarrow \] \[\underset{\theta -\frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,f(\theta )=10\,(1+2+3+...+20)\] \[=\frac{10\times 20\times 21}{2}=2100\]
You need to login to perform this action.
You will be redirected in
3 sec
