In the figure shown tension in string `AB` always lies between `m_(1)g` and `m_(2)g`. `(m_(1)!=m_(2))` <
Tension in massless string is uniform throughout.

Options

A. If both Accertion and Reason are true and the reason is correct explanation of the Assertion.
B. If both Assertion and Reason are true but reason is not the correct explanation of Assertion.
C. If Assertion is true, but the Reason is false.
D. If Assertion is false but the Reason is true.

Answer

B

Explanation

If `m_(1) gt m_(2)`
`a = ("Net pulling force")/("Total mass") = ((m_(1) - m_(2)) g)/(m_(1) + m_(2))`
FBD of `m_(1)`
`m_(1)g - T = m_(1)a = (m_(1)(m_(1) - m_(2)) g)/(m_(1) + m_(2))`
:. `T = (2m_(1) m_(2))/((m_(1) + m_(2)) g`
`= m_(1)g [(2 m_(2))/(m_(1) + m_(2))] = m_(2) g [(2m_(1))/(m_(1) + m_(2))]`
`= m_(1)g [( m_(2) + m_(2))/(m_(1) + m_(2))] = m_(2) g [(m_(1) + m_(1))/(m_(1) + m_(2))]`
Since `m_(1) gt m_(2)`, `T lt m_(1)g` but `T gt m_(2)g`
Similarly, we can prove that
`T lt m_(2)g`
`T gt m_(1)g`
`m_(2) gt m_(1)`
Therefore under all condition `T` lines between `m_(1)g` and `m_(2)`.