A) \[{{P}_{1}}={{P}_{2}}\]
B) \[{{P}_{1}}>{{P}_{2}}\]
C) \[{{P}_{1}}<{{P}_{2}}\]
D) \[{{P}_{1}}={{P}_{2}}=0\]
Correct Answer: B
Solution :
For \[\text{X, }{{\text{F}}_{1}}\text{=100 N, }{{\text{A}}_{1}}\text{=100 c}{{\text{m}}^{\text{2}}}\] \[\therefore {{P}_{1}}=\frac{{{F}_{1}}}{{{A}_{1}}}=\frac{100}{100}=1Nc{{m}^{-2}}\] For \[Y,{{F}_{2}}=100\,\,N,\,\,{{A}_{2}}=250c{{m}^{2}}\] \[\therefore {{P}_{2}}=\frac{{{F}_{2}}}{{{A}_{2}}}=\frac{100}{250}=0.4Nc{{m}^{-2}},\,\,Hence,\,\,{{P}_{1}}>{{P}_{2}}\]You need to login to perform this action.
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