A) 1
B) 2
C) 3
D) 5
Correct Answer: C
Solution :
Using relation \[\theta ={{\omega }_{0}}t+\frac{1}{2}a{{t}^{2}}\] \[{{\theta }_{1}}=\frac{1}{2}(\alpha ){{(2)}^{2}}=2\alpha \] ?(i) (As \[{{\omega }_{0}}=0,t=2\,\sec \]) Now using same equation for t = 4 sec, w0 = 0 \[{{\theta }_{1}}+{{\theta }_{2}}=\frac{1}{2}\alpha {{(4)}^{2}}=8\alpha \] ?(ii) From (i) and (ii),\[{{\theta }_{1}}=2\alpha \]and\[{{\theta }_{2}}=6\alpha \]\ \[\frac{{{\theta }_{2}}}{{{\theta }_{1}}}=3\]You need to login to perform this action.
You will be redirected in
3 sec