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

Correct answer: C. [ML^ -1 ~T^ 0 ] v= p 2 [ F m ]^ 1 / 2 v^ 2 = p 4 ^ 2 (F)/(m) m= p^ 2 F 4 ^ 2 v^ 2 Now, dimensional formula of R.H.S. = MLT^ -2 L^ 2 ( 1 T )^ 2 [p will have no dimension as it is an integer] =ML^ -1 T^ 0 So, dimensions of m will be ML^ -1 T^ 0 .

BITSAT 2016

Exam: BITSAT
Year: 2016
Subject: Physics
Chapter: Units and Dimensions

BITSAT 2016 · Physics · Units and Dimensions

The frequency of vibration of string is given by v= p 2 l [ F m ]^...

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

Options

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

Correct answer

C. $\left[\mathrm{ML}^{-1} \mathrm{~T}^{0}\right]$

Step-by-step 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}

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