A) \[\frac{2}{\sqrt{3}}{{\left( \frac{\tau }{B.i} \right)}^{\frac{1}{2}}}\]
B) \[2{{\left( \frac{\tau }{\sqrt{3}B.i} \right)}^{\frac{1}{2}}}\]
C) \[\frac{2}{\sqrt{3}}\left( \frac{\tau }{B.i} \right)\]
D) \[\frac{1}{\sqrt{3}}\frac{\tau }{B.i}\]
Correct Answer: B
Solution :
[b] \[\tau =MB\,\sin \,\theta ,\,\tau =iAB\,\sin \,{{90}^{o}}\] \[\therefore \,\,\,\,A=\frac{\tau }{iB}=1/2(BC)(AD)\] But \[\frac{1}{2}\,(BC)(AD)\] \[=\frac{1}{2}(l)\sqrt{{{l}^{2}}-{{\left( \frac{l}{2} \right)}^{2}}}=\frac{\sqrt{3}}{4}{{l}^{2}}\] \[\Rightarrow \,\,\,\frac{\sqrt{3}}{4}{{(l)}^{2}}=\frac{\tau }{Bi}\Rightarrow \,\therefore \,\,\,l=2{{\left( \frac{\tau }{\sqrt{3}B.i} \right)}^{\frac{1}{2}}}\]You need to login to perform this action.
You will be redirected in
3 sec