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A wire of length L is hanging from a fix...

A wire of length L is hanging from a fixed support. The length changes to `L_(1) and L_(2)` when masses `M_(1)and M_(2)` are suspended respectively from its free end. Then L is equal to

A

`(L_(1) + L_(2))/(2)`

B

`sqrt(L_(1)L_(2))`

C

`(L_(1) M_(2) + L_(2) M_(1))/(M_(1) + M_(2))`

D

`(L_(1) M_(2)- L_(2) M_(1))/(M_(2)- M_(1))`

Text Solution

Verified by Experts

The correct Answer is:
D
`Delta l_(1)= L_(1)- L " " F_(1)= M_(1)g`
`Delta l_(2)= L_(2)- L " " F_(2) = M_(2)g`
`Y= (F_(1)L_(1))/(A Delta l_(1))= (F_(2) L_(2))/(A Delta l_(2))`
`rArr (M_(1)g)/(L_(1)- L) = (M_(2)g)/(L_(2) - L) rArr L= (L_(1) M_(1) - L_(2) M_(1))/(M_(2)- M_(1))`
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