Q141 / 200 · Manipal (MET) 2014 · Physics

The equation $\left(p+\frac{a}{V^2}\right)(V-b)=$ constant. The unit of $a$ is

  • A
    Dyne $\times \mathrm{cm}^5$
  • Correct answer
    B
    Dyne $\times \mathrm{cm}^4$
  • C
    Dyne $/ \mathrm{cm}^3$
  • D
    Dyne $/ \mathrm{cm}^2$

Solution

According to the principle of dimensional homogenity $[p]=\left[\frac{a}{v^2}\right]$ $\left.\Rightarrow \quad[a]=[p] V^2\right]=\left[M L^{-1} T^{-2}\right]\left[L^6\right]=\left[M L^5 T^{-2}\right]$
Hence, the units of $a=\mathrm{gm} \times \mathrm{cm}^5 \times \mathrm{s}^{-2}$ $=\text { Dyne } \times \mathrm{cm}^4$

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

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