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A massless rod of length L is suspended ...

A massless rod of length `L` is suspended by two identical string `AB` and `CD` of equal length. A block of mass `m` is suspended from point `O` such that `BO` is equal to 'x'. Further it is observed that the frequency of `1st` harmonic in `AB` is equal to `2nd` harmonic frequency in `CD`. 'x' is

A

`L/5`

B

`(4L)/5`

C

`(3L)/4`

D

`L/4`

Text Solution

Verified by Experts

The correct Answer is:
A
Frequency of the first harmonic of `AB`
`=(1)/(2l)sqrt((T_(AB))/(m))`, When `l` is the length of the two strings.
Frequency of the 2nd harmonic of CD
`=(1)/(l)sqrt((T_(CD))/(m))`
Given that the two frequencies are equal.
`therefore" "(1)/(2l)sqrt((T_( AB))/(m))=(1)/(l)sqrt((T_(CD))/( m))rArr(T_(AB))/(4)=T_(CD)`
`rArr" "T_(AB)=4T_(CD)" "`...(ii)

For rotation equilibrium of masscles rod taking torque about point O.
`T_(AB)xx x=T_(CD)(L-x)" "`...(iii)
Solve to get, `x=(L)/(5)`
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