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A true balance is one whose pans are of equal masses and arms are of equal lengths. When this happens, the net moment of forces about point of suspension is zero and beam remains horizontal without any weight i.e., for the true balance P_(1)=P_(2) and l_(1)=l_(2) also P_(1)l_(1)=P_(2)l_(2) (mass of beam is neglibigle). A shopkeeper uses a false balance to weigh articles. Both arms and pans of this false balance are different, but beam become horizontal without any weight. (P_(1)neP_(2) and l_(1)nel_(2) but P_(1)l_(1)=P_(2)l_(2)). Q. Choose the correct options(s) |
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Answer» If MANN of pan `P_(2)` is less than mass of pan `P_(1)` and shopkeeper puts weight W of P_(1), and article on the pan `P_(2)` in this weight GAIN the shopkeeper is W `((l_(2)-l_(1))/(l_(2)))` `Wl_(1)=W_(2)l_(2)impliesW_(2)=(Wl_(1))/(l_(2))` since `(l_(1))/(l_(2))lt1impliesW_(2)ltW` `implies` there will be gain for shopkeeper `DeltaW_(2)=W-W_(2)=W((l_(2)-l_(1))/(l_(2)))` `W_(1)` be the weight of article when article PUT in `P_(1)` `W_(1)l_(1)=Wl_(2)impliesW_(1)=(Wl_(2))/(l_(1))` since `(l_(1))/(l_(2))lt` `impliesW_(1)GTW` `implies` there will be loss for shopkeeper `DeltaW_(1)=W_(1)-W=W((l_(2)-l_(1))/(l_(1)))` Since weight loss is more so net loss `=W((l_(2)l_(1))/(l_(1)))-W((l_(2)-l_(1))/(l_(2)))=W((l_(2)-l_(1))^(2))/(l_(1)l_(2))` |
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