If the velocity of light (c), gravitational constant (G) and Planck's constant (h) are chosen as fundamental units, then the dimensions of mass in new system is

The correct option is c⁰.⁵G⁻⁰.⁵h⁰.⁵.

To find the dimensions of mass in the new system of units, we express mass ($M$) as a product of powers of the new fundamental units: velocity of light ($c$), gravitational constant ($G$), and Planck's constant ($h$).

Let the relationship be:
$M\propto c^x{G}^y{h}^z$
where $x$, $y$, and $z$ are the dimensions to be determined.

First, we write down the dimensional formulas of the given physical quantities in the standard Mass-Length-Time (M-L-T) system:
1. Mass:
$\left[M\right]=M^1{L}^0{T}^0$
2. Velocity of light:
$\left[c\right]=L{T}^{-1}$
3. Gravitational constant: From Newton's law of gravitation, $F=\frac{G{m}_1{m}_2}{r^2}$, which gives:
$\left[G\right]=M^{-1}{L}^3{T}^{-2}$
4. Planck's constant: From the energy relationship, $E=h\nu$, which gives:
$\left[h\right]=M^1{L}^2{T}^{-1}$

Substituting these dimensions into our initial equation:
$M^1{L}^0{T}^0={\left(L{T}^{-1}\right)}^x$
Combining the terms on the right-hand side, we get:
$M^1{L}^0{T}^0=M^{-y+z}{L}^{x+3y+2z}{T}^{-x-2y-z}$

Equating the exponents of $M$, $L$, and $T$ from both sides:
For $M$:
$-y+z=1$ (Equation 1)
For $L$:
$x+3y+2z=0$ (Equation 2)
For $T$:
$-x-2y-z=0\Rightarrow x+2y+z=0$ (Equation 3)

Adding Equation 2 and Equation 3 eliminates $x$:
$(x+3y+2z)-(x+2y+z)=0$
$y+z=0\Rightarrow z=-y$ (Equation 4)

Substituting Equation 4 ($z=-y$) into Equation 1:
$-y-y=1$
$-2y=1\Rightarrow y=-0.5$

Using $z=-y$, we find:
$z=-(-0.5)=0.5$

Substituting the values of $y$ and $z$ into Equation 3:
$x+2(-0.5)+0.5=0$
$x-1+0.5=0$
$x-0.5=0\Rightarrow x=0.5$

Hence, the dimensions of mass in the new system of units are:
$\left[M\right]=c^{0.5}{G}^{-0.5}{h}^{0.5}$