The speed of a longitudinal wave in a metallic bar is 400 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.

The speed of a longitudinal wave in a solid is given by the formula:

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

where $$Y$$ is Young's modulus and $$\rho$$ is the density.

Given initial speed $$v = 400 \, \text{m/s}$$.

Density increases by 0.5%, so new density $$\rho' = \rho \left(1 + \frac{0.5}{100}\right) = 1.005\rho$$.

Young's modulus increases by 1%, so new Young's modulus $$Y' = Y \left(1 + \frac{1}{100}\right) = 1.01Y$$.

The new speed $$v'$$ is:

$$v' = \sqrt{\frac{Y'}{\rho'}} = \sqrt{\frac{1.01Y}{1.005\rho}} = \sqrt{\frac{1.01}{1.005}} \times \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{1.01}{1.005}} \times 400$$

Compute the fraction $$\frac{1.01}{1.005}$$:

$$\frac{1.01}{1.005} = \frac{1010}{1005} = \frac{202}{201} \approx 1.004975$$

Now, $$\sqrt{1.004975}$$ can be approximated using $$\sqrt{1+x} \approx 1 + \frac{x}{2}$$ for small $$x$$, where $$x = 0.004975$$:

$$\sqrt{1.004975} \approx 1 + \frac{0.004975}{2} = 1 + 0.0024875 = 1.0024875$$

Thus,

$$v' \approx 1.0024875 \times 400 = 400.995 \approx 401 \, \text{m/s}$$

Alternatively, using the formula for fractional change:

$$\frac{dv}{v} = \frac{1}{2} \frac{dY}{Y} - \frac{1}{2} \frac{d\rho}{\rho}$$

Given $$\frac{dY}{Y} = 0.01$$ and $$\frac{d\rho}{\rho} = 0.005$$,

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

So, $$v' = v (1 + 0.0025) = 400 \times 1.0025 = 401 \, \text{m/s}$$.

Therefore, the speed changes approximately to $$401 \, \text{m/s}$$.

The correct option is C. 401.