Q34 (2026)
The speed of a longitudinal wave in a metallic bar is $400 \text{ m/s}$. If the density and Young's modulus of the bar material are increased by 0.5% and 1% respectively then the speed of the wave is changed approximately to ____ m/s.
- $399$
- $398$
- $402$
- $401$
### Related Formula
$$V_{\text{sound}} = \sqrt{\frac{Y}{\rho}}$$
where $Y$ is Young's modulus and $\rho$ is density.
### Core Logic
Taking natural logarithm on both sides:
$$\ln V = \frac{1}{2} \ln Y - \frac{1}{2} \ln \rho$$
Differentiating to find fractional errors/changes:
$$\frac{\Delta V}{V} = \frac{1}{2} \frac{\Delta Y}{Y} - \frac{1}{2} \frac{\Delta \rho}{\rho}$$
### Step 1: Calculate Percentage Change
Multiply by 100 to get percentages:
$$\frac{\Delta V}{V} \times 100 = \frac{1}{2} \left( \frac{\Delta Y}{Y} \times 100 \right) - \frac{1}{2} \left( \frac{\Delta \rho}{\rho} \times 100 \right)$$
Given $\frac{\Delta Y}{Y} \times 100 = 1\%$ and $\frac{\Delta \rho}{\rho} \times 100 = 0.5\%$.
$$\frac{\Delta V}{V} \times 100 = \frac{1}{2}(1\%) - \frac{1}{2}(0.5\%) = 0.5\% - 0.25\% = 0.25\%$$
### Step 2: Calculate New Speed
$$\frac{\Delta V}{V} = \frac{0.25}{100} = \frac{1}{400}$$
$$\Delta V = \frac{1}{400} \times 400 = 1 \text{ m/s}$$
The final speed is:
$$V_{\text{final}} = V + \Delta V = 400 + 1 = 401 \text{ m/s}$$
### Pattern Recognition
For products and quotients of the form $A^a / B^b$, the small percentage change is $\pm a(\%\Delta A) \mp b(\%\Delta B)$. Here, velocity scales with $Y^{1/2} \rho^{-1/2}$, so change is $\frac{1}{2}(\%\Delta Y) - \frac{1}{2}(\%\Delta \rho)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Class 11 Physics: Waves
