A) \[\frac{3}{4}\text{m}/\text{s}\]
B) \[\frac{1}{3}\text{m}/\text{s}\]
C) \[\frac{3}{2}\text{m}/\text{s}\]
D) \[\frac{2}{3}\text{m}/\text{s}\]
Correct Answer: D
Solution :
Given,\[=2.12\times {{10}^{8}}m/s\] \[X=r\tan \theta \] \[V=\frac{dX}{dt}\] \[=r{{\cos }^{2}}\theta \left( \frac{d\theta }{dt} \right)\] \[\sqrt{1+1+2\cos 2\alpha }<1\] \[\frac{d\theta }{dt}=\omega \] \[\sqrt{2(1+\cos 2\alpha )}<1\] \[v=\omega r{{\sec }^{2}}\theta \] \[\phi ={{45}^{o}}\,\,so,\,\,\theta ={{45}^{o}}\] \[\therefore \] \[v=0.1\times 3\times {{\sec }^{2}}{{45}^{o}}\] \[=0.1\times 3\times 2=0.6m/s\] \[{{\left( \text{CD} \right)}^{\text{2}}}={{\left( \text{SC} \right)}^{\text{2}}}-{{\left( \text{SD} \right)}^{\text{2}}}\] \[{{X}^{2}}={{(L)}^{2}}-{{\left( \frac{L}{2} \right)}^{2}}\] \[{{X}^{2}}={{\frac{3L}{4}}^{2}}\]You need to login to perform this action.
You will be redirected in
3 sec