Physics - Unit & Dimension Question with Solution | TestHub

Key Idea for checking the correctness of an equation dimensional analysis is done using the principle of homogeneity.
The dimensions of given variables of SHM are as
Displacement,$\left[x\right]=\left[M^{0}L{T}^0\right]$
Velocity,$\left[v\right]=\left[M^{0}L{T}^{-1}\right]$
Acceleration,$\left[a\right]=\left[M^{0}L{T}^{-2}\right]$
and time period,$\left[T\right]=\left[M^0{L}^{0}T\right]$
Now, checking each option for these values for option (a)
$\frac{\left[a\right]\left[T\right]}{\left[x\right]}=\frac{\left[M^{0}L{T}^{-2}\right]}{\left[M^{0}L{T}^0\right]}=\left[M^0{L}^0{T}^{-1}\right]$
As it depends on time, so change with it.
For option (b),
$\left[a\right]\left[T\right]+2\pi \left[v\right]=\left[M^{0}L{T}^{-2}\right]\left[M^0{L}^{0}T\right]+\left[M^0{L}^0{T}^{-1}\right]$
=$\left[M^{0}L{T}^{-1}\right]$
It is also dependent on time and hence changes with it.
For option$\left(c\right)$,
$\frac{\left[a\right]\left[T\right]}{\left[v\right]}=\frac{\left[M^{0}L{T}^{-2}\right]\left[M^0{L}^{0}T\right]}{\left[M^{0}L{T}^{-1}\right]}=\left[M^0{L}^0{T}^0\right]$
As it is a constant having no dimension, so it does not change with time.
For option (d),
$\left[a\right]\left[t\right]+4{\pi }^2{\left[d\right]}^2=\left[M^{0}L{T}^{-2}\right]\left[M^0{L}^{0}T\right]+{\left[M^{0}L{T}^{-1}\right]}^2$
$=\left[L{T}^{-1}\right]+\left[L^2{T}^{-2}\right]$
As, the terms is dependent on time, so changes with it.
Also, it is dimensionally incorrect.
Hence, option$\left(c\right)$is correct.