JEE Physics 2026 Question with Solution

Given: The speed of the longitudinal wave is $400 \, \text{m/s}$. The Young’s modulus increases by $1\%$ and the density increases by $0.5\%$.

Find: The new speed of the wave.

Principle: For a longitudinal wave in a rod,

$$v=\sqrt{\frac{Y}{\rho}}$$

Using relative change,

$$\frac{\Delta v}{v}=\frac{1}{2}\left(\frac{\Delta Y}{Y}-\frac{\Delta \rho}{\rho}\right)$$

Substitute the given percentage changes:

$$\frac{\Delta Y}{Y}=0.01, \qquad \frac{\Delta \rho}{\rho}=0.005$$

Therefore,

$$\frac{\Delta v}{v}=\frac{1}{2}(0.01-0.005)=0.0025$$

Now,

$$\Delta v=400\times 0.0025=1 \, \text{m/s}$$

So the new speed is

$$v_{\text{new}}=400+1=401 \, \text{m/s}$$

Therefore, the speed changes approximately to $401 \, \text{m/s}$. The correct option is D.

Given: $v \propto \sqrt{\frac{Y}{\rho}}$.

Find: The approximate new speed.

Since the speed depends on a square root, the net percentage change in the bracket is first found and then halved.

Increase in $Y$ gives $+1\%$, while increase in $\rho$ gives $-0.5\%$ effect on $v$. So net percentage effect before square root is

$$1\%-0.5\%=0.5\%$$

After the square root, the percentage change in speed is half of this:

$$\frac{0.5\%}{2}=0.25\%$$

Now $0.25\%$ of $400$ is

$$400\times \frac{0.25}{100}=1$$

Hence the new speed is $401 \, \text{m/s}$. The correct option is D.