A) \[2\sqrt{{{P}_{1}}{{P}_{2}}}\]
B) \[\sqrt{P_{1}^{2}\,+P_{2}^{2}}\]
C) \[\frac{\sqrt{P_{1}^{2}\,+P_{2}^{2}}}{2}\]
D) \[\sqrt{\frac{P_{1}^{2}\,+P_{2}^{2}}{2}}\]
Correct Answer: D
Solution :
Let P and F be the two forces, then \[{{P}_{1}}=P+F\] and \[{{P}_{2}}=P-F\,\,\,(\because \,\,P>F)\] \[\Rightarrow \] \[P=\frac{{{P}_{1}}+{{P}_{2}}}{2}\] and \[F=\frac{{{P}_{1}}-{{P}_{2}}}{2}\] The resultant of the forces P and F is given by, \[R=\sqrt{{{P}^{2}}+Fh2+2P\,F\,\cos \theta }\,\,\,\,[\because \,\,\theta ={{90}^{o}}\,(given)]\] \[R=\sqrt{\frac{{{({{P}_{1}}+{{P}_{2}})}^{2}}}{2}+\frac{{{({{P}_{1}}-{{P}_{2}})}^{2}}}{2}+2PF\,\cos \,{{90}^{o}}}\] \[R=\frac{1}{2}\sqrt{{{({{P}_{1}}+{{P}_{2}})}^{2}}+{{({{P}_{1}}-{{P}_{2}})}^{2}}+0}\] \[R=\frac{1}{2}\sqrt{P_{1}^{2}+P_{2}^{2}+P_{1}^{2}+P_{2}^{2}}\] \[=\frac{1}{2}\sqrt{2(P_{1}^{2}+P_{2}^{2}}\] \[R=\sqrt{\frac{P_{1}^{2}+P_{2}^{2}}{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
