Q143 / 160 · TS EAMCET 2015 · Physics

Force $F$ is given by the equation $F=\frac{X}{\text { Linear density }}$. Then dimensions of $X$ are

  • Correct answer
    A
    $\left[\mathrm{M}^2 \mathrm{~L}^0 \mathrm{~T}^{-2}\right]$
  • B
    $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$
  • C
    $\left[\mathrm{L}^2 \mathrm{~T}^{-2}\right]$
  • D
    $\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$

Solution

$F=\frac{X}{\text { Linear density }}$ Dimension of $F=\left[\mathrm{MLT}^{-2}\right]$ Dimension of linear density $=\left[\mathrm{ML}^{-1}\right]$ $\begin{aligned}<br />\therefore X & =F \text { (linear density) } \\<br />& =\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{ML}^{-1}\right]=\left[\mathrm{M}^2 \mathrm{~L}^0 \mathrm{~T}^{-2}\right]<br />\end{aligned}$

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

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