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

Options

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))`

Answer

B

Explanation

In any position , `p_(1)^(2)T_(1) = P_(2)^(2)T_(2)`
`T_(1) = (M_(0) +m_(1))g " and " T_(2) = (M_(0) + m_(2))g`
` :. (M_(0) + m_(1))gp_(1)^(2) = (M_(0)+m_(2)) gp_(2)^(2)`
` :. M_(0)p_(1)^(2) + m_(1)p_(1)^(2) = M_(0)p_(2)^(2) + m_(2)p_(2)^(2)`
` :. M_(0)(p_(1)^(2) - p_(2)^(2)) = m_(2)p_(2)^(2) - m_(1)p_(1)^(2)`
`:. M_(0) = (m_(2)p_(2)^(2) - m_(1)p_(1)^(2))/(p_(1)^(2) - p_(2)^(2))`