Q121 / 160 · AP EAMCET 2022 (05 Jul Shift 1) · Physics
The energy of $E$ of a system is function of time $t$ and is given by $E(t)=\alpha t-\beta t^3$, where $\alpha$ and $\beta$ are constants. The dimensions of $\alpha$ and $\beta$ are
- A$\left[\mathrm{ML}^2 T^{-1}\right]$ and $\left[\mathrm{ML}^2 T\right]$
- B$\left[\mathrm{LT}^{-1}\right]$ and $[\mathrm{LT}]$
- Correct answerC$\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right]$ and $\left[\mathrm{ML}^2 \mathrm{~T}^{-5}\right]$
- D$\left[\mathrm{MLT}^{-1}\right]$ and $[\mathrm{MLT}]$
Solution
Energy of the system as a function of time,
$E(t)=\alpha t-\beta t^3$
Where, $\alpha$ and $\beta$ are constant.
According to principle of homogenity, dimension of $\alpha t=$ dimension of energy
i.e
$\Rightarrow \quad[\alpha][t]=[E]$
$\begin{array}{rrr}\Rightarrow & {[\alpha][\mathrm{T}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]} \\ \Rightarrow & {[\alpha]=\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right]}\end{array}$
Again, dimension of $\beta t^3=$ dimension of energy
$\begin{array}{ll}\Rightarrow & {[\beta]\left[t^3\right]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]} \\ \Rightarrow & {[\beta]\left[\mathrm{T}^3\right]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}\end{array}$
$\Rightarrow \quad[\beta]=\left[\mathrm{ML}^2 \mathrm{~T}^{-5}\right]$
Hence, the correct answer is option: 'C'.