A) \[\frac{{{I}_{m}}}{9}(4+5\cos \phi )\]
B) \[\frac{{{I}_{m}}}{3}\left( 1+2{{\cos }^{2}}\frac{\phi }{2} \right)\]
C) \[\frac{{{I}_{m}}}{5}\left( 1+4{{\cos }^{2}}\frac{\phi }{2} \right)\]
D) \[\frac{{{I}_{m}}}{9}\left( 1+8{{\cos }^{2}}\frac{\phi }{2} \right)\]
Correct Answer: D
Solution :
| Let \[{{a}_{1}}=a,\,\,{{I}_{1}}={{a}_{1}}^{2}={{a}^{2}}\] |
| \[{{a}_{2}}=2a,\,{{I}_{2}}={{a}_{2}}^{2}=4{{a}^{2}}\] |
| \[{{I}_{2}}=4{{I}_{1}}\] |
| \[{{I}_{r}}={{a}_{1}}^{2}+{{a}_{2}}^{2}+2{{a}_{1}}{{a}_{2}}\cos \phi \]\[={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi \] |
| \[{{I}_{r}}={{I}_{1}}+4{{I}_{1}}+2\sqrt{4{{I}_{1}}^{2}}\cos \phi \] |
| \[\Rightarrow \,\,{{I}_{r}}=5{{I}_{1}}+4{{I}_{1}}\cos \phi \] ?(1) |
| Now' \[{{I}_{\max }}={{({{a}_{1}}+{{a}_{2}})}^{2}}={{(a+2a)}^{2}}=9{{a}^{2}}\] |
| \[{{I}_{\max }}=9{{I}_{1}}\Rightarrow \,{{I}_{1}}=\frac{{{I}_{\max }}}{9}\] |
| Substituting in equation (1) |
| \[{{I}_{r}}=\frac{5{{I}_{\max }}}{9}+\frac{4{{I}_{\max }}}{9}\cos \phi \] |
| \[{{I}_{r}}=\frac{{{I}_{\max }}}{9}[5+4\cos \phi ]\] |
| \[{{I}_{r}}=\frac{{{I}_{\max }}}{9}\left[ 5+8{{\cos }^{2}}\frac{\phi }{2}-4 \right]\] |
| \[{{I}_{r}}=\frac{{{I}_{\max }}}{9}\left[ 1+8{{\cos }^{2}}\frac{\phi }{2} \right]\] |
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