A) 67.5
B) 32.5
C) 45.3
D) 63.1
Correct Answer: A
Solution :
\[{{P}_{1}}=\frac{60}{4\pi \times {{4}^{2}}}\times \frac{\pi \times {{(2\times {{10}^{-3}})}^{2}}}{4}\] \[=9.375\times {{10}^{-7}}\,J/s\] \[n\] Now, \[1\times \cos \,{{60}^{o}}+2\,\cos \,{{60}^{o}}+x-4\,\cos \,{{60}^{o}}=0\] \[\therefore \] or \[x=0.5\,\,N\] \[\frac{M}{\pi {{R}^{2}}}\times \pi {{\left\{ \frac{R}{2} \right\}}^{2}}=\frac{M}{4}\] \[=M-\frac{M}{4}=\frac{3M}{4}\] \[{{X}_{CM}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}\] \[{{X}_{CM}}\] Volume % of \[{{m}_{2}}{{x}_{2}}=-{{m}_{1}}{{x}_{1}}\]You need to login to perform this action.
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