A) \[\frac{P}{2}\]
B) \[\frac{P}{3}\]
C) \[\frac{P}{1.088}\]
D) \[\frac{P}{2.088}\]
Correct Answer: C
Solution :
\[{{G}_{1}}={{G}_{2}},\] \[{{d}_{1}}={{d}_{2}}\] \[{{D}_{wl}}=180\,mm,\,D\] \[\frac{8{{W}_{1}}D_{m2}^{3}{{n}_{1}}}{{{G}_{1}}d_{1}^{4}}=\frac{8{{W}_{2}}D_{m2}^{3}{{n}_{2}}}{{{G}_{2}}d_{2}^{4}}\] \[\frac{{{W}_{1}}}{{{W}_{2}}}={{\left( \frac{D{{m}_{2}}}{D{{m}_{1}}} \right)}^{3}}={{\left( \frac{80}{180} \right)}^{3}}=\frac{1}{11.39}\] \[{{W}_{1}}+{{W}_{2}}=P\] \[{{W}_{1}}(1+11.39)=P\] \[{{W}_{1}}=\frac{1139}{1.088}\times P=\frac{P}{1.088}\]
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