JEE Physics 2023 Question with Solution

Given: The wave speed is related as $v=\lambda^{a} g^{b} \rho^{c}$.

Find: The values of $a$, $b$ and $c$.

Write the dimensions of each quantity:

$$[v]=LT^{-1},\quad [\lambda]=L,\quad [g]=LT^{-2},\quad [\rho]=ML^{-3}$$

Substitute these into the given relation:

$$[v]=[\lambda]^a[g]^b[\rho]^c$$ $$LT^{-1}=(L)^a(LT^{-2})^b(ML^{-3})^c$$

Simplifying the right-hand side,

$$LT^{-1}=M^{c}L^{a+b-3c}T^{-2b}$$

Now equate powers of $M$, $L$ and $T$ separately.

For mass:

$$c=0$$

For time:

$$-2b=-1 \Rightarrow b=\frac{1}{2}$$

For length:

$$a+b-3c=1$$

Substituting $b=\frac{1}{2}$ and $c=0$,

$$a+\frac{1}{2}=1 \Rightarrow a=\frac{1}{2}$$

Therefore,

$$a=\frac{1}{2},\quad b=\frac{1}{2},\quad c=0$$

So the correct option is C.

Given: $v=\lambda^{a} g^{b} \rho^{c}$

Find: The exponents $a$, $b$ and $c$.

Since speed does not contain mass in its dimensions, the power of density must vanish immediately, so $c=0$.

Then the relation reduces to combining only $\lambda$ and $g$:

$$[v]=[\lambda]^a[g]^b=L^a(LT^{-2})^b=L^{a+b}T^{-2b}$$

Compare with

$$[v]=LT^{-1}$$

From time powers,

$$-2b=-1 \Rightarrow b=\frac{1}{2}$$

From length powers,

$$a+b=1 \Rightarrow a=\frac{1}{2}$$

Hence, $\left( a,b,c \right)=\left(\frac{1}{2},\frac{1}{2},0\right)$ and the correct option is C.