JEE Physics 2023 Question with Solution

Given: The new fundamental quantities are $c$, $G$ and $h$.

Find: The dimensions of mass in terms of $h$, $c$ and $G$.

Let

$$M = h^x c^y G^z$$

Now use the dimensional formulas:

• $[h] = [M L^2 T^{-1}]$
• $[c] = [L T^{-1}]$
• $[G] = [M^{-1} L^3 T^{-2}]$

So,

$$[M] = [M L^2 T^{-1}]^x [L T^{-1}]^y [M^{-1} L^3 T^{-2}]^z$$ $$[M^1 L^0 T^0] = [M^{x-z} L^{2x+y+3z} T^{-x-y-2z}]$$

Equating powers of $M$, $L$ and $T$,

$$x - z = 1$$ $$2x + y + 3z = 0$$ $$-x - y - 2z = 0$$

Adding the second and third equations,

$$x + z = 0$$

Together with $x - z = 1$, we get

$$2x = 1$$

Hence,

$$x = \frac{1}{2}, \qquad z = -\frac{1}{2}$$

Using $-x - y - 2z = 0$,

$$-\frac{1}{2} - y - 2\left(-\frac{1}{2}\right) = 0$$ $$-\frac{1}{2} - y + 1 = 0$$ $$y = \frac{1}{2}$$

Therefore,

$$M = h^{\frac{1}{2}} c^{\frac{1}{2}} G^{-\frac{1}{2}}$$

So the correct option is D.

The solution states option C and the answer key states option (4), but the worked dimensional analysis clearly gives $h^{\frac{1}{2}} c^{\frac{1}{2}} G^{-\frac{1}{2}}$, which matches option D.

Given: $M = h^x c^y G^z$.

Find: Values of $x$, $y$ and $z$.

Write each dimension separately:

$$[h^x] = [M^x L^{2x} T^{-x}]$$ $$[c^y] = [L^y T^{-y}]$$ $$[G^z] = [M^{-z} L^{3z} T^{-2z}]$$

Multiplying,

$$[M] = [M^{x-z} L^{2x+y+3z} T^{-x-y-2z}]$$

For pure mass, the powers must be:

• power of $M$ is $1$
• power of $L$ is $0$
• power of $T$ is $0$

Hence,

$$x-z=1, \qquad 2x+y+3z=0, \qquad -x-y-2z=0$$

Now add the last two equations:

$$(2x+y+3z)+(-x-y-2z)=0$$ $$x+z=0$$

Solve with $x-z=1$:

$$2x=1 \Rightarrow x=\frac{1}{2}$$ $$z=-\frac{1}{2}$$

Substitute into $-x-y-2z=0$:

$$-\frac{1}{2}-y+1=0$$ $$y=\frac{1}{2}$$

Thus,

$$[M] = [h^{\frac{1}{2}} c^{\frac{1}{2}} G^{-\frac{1}{2}}]$$

Therefore, the correct option is D.