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Question

​A defective balance has equal arms. A body weighs w1 when placed in one pan and w2 when placed in the other pan. the weight of the pan is?

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Solution

Dear Student,

Let W be the weight of the pan
Let the Weight of the Body be Wb

Now Balancing moment of Torque across the pivot( given both arms are equal= L )

Left hand moment= right hand moment

(W+Wb)L=W1L or W+Wb=W1....(i)
similarly W2=W+Wb...........(ii)

substracting (ii) from (i)

W-W2+Wb=W1-W-Wb
​2W=W1+W2-2Wb

W=((W1+W2)/2)-Wb ...(Ans)

Regards.

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