The frequency \((n)\) of vibration of a string is given as \(n=\frac{1}{2 l} \sqrt{\frac{T}{m}}\), where…

The frequency \((n)\) of vibration of a string is given as \(n=\frac{1}{2 l} \sqrt{\frac{T}{m}}\), where \(T\) is tension and \(l\) is the length of vibrating string. Then, the dimensional formula for \(m\) is
  1. \(\left[M^{0} L^{1} T^{\prime}\right]\)
  2. \(\left[M^{0} L^{0} T^{0}\right]\)
  3. \(\left[M^{1} L^{-1} T^{0}\right]\)
  4. \(\left[ M L^{0} T^{0}\right]\)

Solution

\(n=\frac{1}{2 l} \sqrt{\frac{T}{m}} \Rightarrow n^{2}=\frac{1}{4 l^{2}} \frac{T}{m}\)
\(n=\frac{T}{4 l^{2} n^{2}}=\left[\frac{M L T^{-2}}{L^{2} \times T^{-2}}\right]=\left[M L^{-1}\right]\)

Asked in: JEE Mains - Units and Dimensions - Chapter Test

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