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In Melde's experiment in parallel positi...

In Melde's experiment in parallel position the mass of the pan is `M_(0)`. When a mass `m_(1)` is kept in the pan, the number of loops formed is `p_(1)`. For the mass `m_(2)`, the number of loops, formed is `p_(2)`. Then the mass of the pan `M_(0)`, in terms of `m_(1) , m_(2), p_(1) " and " p_(2)` is given by

A

`m_(0)=(p_(1)^(2)-p_(2)^(2))/(m_(2)p_(2)^(2)-m_(1)p_(1)^(2))`

B

`m_(0)=(m_(2)p_(2)^(2)-m_(1)p_(1)^(2))/(p_(1)^(2)-p_(2)^(2))`

C

`m_(0)=(m_(2)p_(2)^(2)+m_(1)p_(1)^(2))/(p_(1)^(2)-p_(2)^(2))`

D

`m_(0)=(m_(2)p_(2)^(2)-m_(1)p_(1)^(2))/(p_(1)^(2)+p_(2)^(2))`

Text Solution

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The correct Answer is:
B
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