| \[\underset{t\to x}{\mathop{\lim }}\,\,\frac{\int\limits_{0}^{t}{\sqrt{1-{{\{f(S)\}}^{2}}}\,dS-\int\limits_{0}^{x}{\sqrt{1-{{\{f(S)\}}^{2}}}dx}}}{f(t)-f(x)}\] |
A) \[\left\{ \sqrt{7},\,\,\sqrt{15} \right\}\]
B) \[\left\{ \frac{\sqrt{7}}{2},\,\,\frac{\sqrt{15}}{2} \right\}\]
C) \[\left\{ \frac{\sqrt{7}}{3},\,\,\frac{\sqrt{15}}{3} \right\}\]
D) \[\left\{ \frac{\sqrt{7}}{4},\,\,\frac{\sqrt{15}}{4} \right\}\]
Correct Answer: D
Solution :
Given \[\Rightarrow \] \[v'\frac{v}{\sqrt{2}}\] for all \[E=\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}(2m){{v}^{2}}\] \[=mv{{'}^{2}}+m{{v}^{2}}\] \[\Rightarrow \] By L ?Hospital rule, \[m{{\left( \frac{v}{\sqrt{2}} \right)}^{2}}+m{{v}^{2}}=\frac{3}{2}m{{v}^{2}}\] \[{{T}_{1}}-{{T}_{2}}=12a\] \[{{T}_{2}}=3a\] \[{{T}_{2}}\] \[Eq\] \[{{T}_{1}}=15a\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{15a}{3a}=\frac{5}{1}\] \[\Rightarrow \] \[{{T}_{1}}:{{T}_{2}}=5:1\] \[T=M\,\left( g-\frac{g}{4} \right)=\frac{3Mg}{4}\] \[W=\mathbf{T}\cdot \mathbf{d}\Rightarrow \,W=Td\] \[\Rightarrow \] \[W=-Td=-\frac{3Mgd}{4}\] \[\Sigma mvr=\,({{l}_{system}})\omega \] \[\Rightarrow \] \[mv\frac{l}{2}=\frac{(2m)\,{{(2l)}^{2}}}{12}\omega =\frac{2m(4{{l}^{2}})}{12}\omega \] \[\Rightarrow \]
You need to login to perform this action.
You will be redirected in
3 sec
