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_1p_2^2-m_2p_1^2)/(p_1^2-p_2^2)`
B. `(m_2p_2^2-m_1p_1^2)/(p_1^2-p_2^2_`
C. `(p_1^2-p_2^2)/(m_2p_2^2-m_1p_1^2)`
D. `(p_2^2-p_1^2)/(m_1p_2^2-m_1p_1^2)`

Answer

B

Explanation

`P_1^2T_1 =P_2^2T_2`
`P_1^2(m_1+m_0) = P_2^2(m_2+m_0)`
`P_1^2m_1+P_2^2m_0 = P_2^2m_2+P_2^2m_0`
`m_0 (P_1^2-P_2^2) = m_2P_2^2 -m_1P_2^2`
`m_0 = (m_2P_2^2-m_1P_1^2)/(P_1^2-P_2^2)`