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In the figure shown tension in string AB...

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.

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.

Text Solution

Verified by Experts

The correct Answer is:
B
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)`.
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