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. Shopkeeper use a weight W to weigh and article by false balance. in two weighing by using alternating pans let `W_(1)` and `W_(2)` are the weights of the article given to the customer i.e., customer gets `(W_(1)+W_(2))` weight of article. Loss of gain to the shopkeeper is:

Options

A. `(W(l_(2)-l_(1))^(2))/(l_(1)l_(2))` gain
B. `(W(l_(2)-l_(1))^(2))/(l_(1)l_(2))` loss
C. No gain, no loss
D. Data insufficient

Answer

B

Explanation

Since `P_(1)l_(1)=P_(2)l_(2)` and `P_(2)ltP_(1)impliesl_(2)gtl_(1)`
`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))`