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The unit of 1//lsqrt((T//mu)) is the sam...

The unit of `1//lsqrt((T//mu)`) is the same as that of (`l`-length, `T`-tension and `mu`-mass/unit length)

A

wave-length

B

wave number

C

frequency

D

time period

Text Solution

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The correct Answer is:
To solve the problem of finding the unit of \( \frac{1}{L} \sqrt{\frac{T}{\mu}} \), where \( L \) is length, \( T \) is tension, and \( \mu \) is mass per unit length, we can follow these steps: ### Step 1: Identify the dimensions of each variable - **Length (L)**: The dimension of length is \( [L] \). - **Tension (T)**: Tension is a force, and its dimension is given by \( [T] = [M][L][T^{-2}] \) (mass times acceleration). - **Mass per unit length (\(\mu\))**: The dimension of \(\mu\) is \( [\mu] = [M][L^{-1}] \). ### Step 2: Substitute the dimensions into the expression We need to find the dimension of \( \sqrt{\frac{T}{\mu}} \): \[ \frac{T}{\mu} = \frac{[M][L][T^{-2}]}{[M][L^{-1}]} = \frac{[M][L][T^{-2}]}{[M]} \cdot [L] = [L^2][T^{-2}] \] ### Step 3: Take the square root of the expression Now, we take the square root of \( [L^2][T^{-2}] \): \[ \sqrt{\frac{T}{\mu}} = \sqrt{[L^2][T^{-2}]} = [L][T^{-1}] \] ### Step 4: Combine with \( \frac{1}{L} \) Now, we need to multiply this result by \( \frac{1}{L} \): \[ \frac{1}{L} \sqrt{\frac{T}{\mu}} = \frac{1}{[L]} \cdot [L][T^{-1}] = [T^{-1}] \] ### Step 5: Identify the unit The dimension \( [T^{-1}] \) corresponds to frequency, which is measured in hertz (Hz) or \( \text{s}^{-1} \). ### Conclusion Thus, the unit of \( \frac{1}{L} \sqrt{\frac{T}{\mu}} \) is the same as that of frequency. ---
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