Q114 / 125 · BITSAT 2016 · Physics

The frequency of vibration of string is given by $\mathrm{v}=\frac{\mathrm{p}}{2 l}\left[\frac{\mathrm{F}}{\mathrm{m}}\right]^{1 / 2}$ Here $\mathrm{p}$ is number of segments in the string and $l$ is the length. The dimensional formula for $\mathrm{m}$ will be

  • A
    $\left[\mathrm{M}^{0} \mathrm{LT}^{-1}\right]$
  • B
    $\left[\mathrm{ML}^{0} \mathrm{~T}^{-1}\right]$
  • Correct answer
    C
    $\left[\mathrm{ML}^{-1} \mathrm{~T}^{0}\right]$
  • D
    $\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$

Solution

$\mathrm{v}=\frac{\mathrm{p}}{2 \ell}\left[\frac{\mathrm{F}}{\mathrm{m}}\right]^{1 / 2}$ $v^{2}=\frac{p}{4 \ell^{2}} \frac{F}{m} \Rightarrow m=\frac{p^{2} F}{4 \ell^{2} v^{2}}$ Now, dimensional formula of R.H.S. $=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^{2}\left(\frac{1}{\mathrm{~T}}\right)^{2}}$ [p will have no dimension as it is an integer] $=\mathrm{ML}^{-1} \mathrm{~T}^{0}$ So, dimensions of $\mathrm{m}$ will be $\mathrm{ML}^{-1} \mathrm{~T}^{0}$.

Hence, the correct answer is option: 'C'.

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