1. **State the problem:** We have variables $x$ and $y$ related by the equation $$y x^n = h,$$ where $n$ and $h$ are constants. Given experimental values of $x$ and $y$, we need to plot $\lg y$ against $\lg x$, draw a straight line, and estimate $n$ and $h$ from the graph. 2. **Rewrite the equation:** Taking logarithm base 10 on both sides, $$\lg(y x^n) = \lg h \implies \lg y + n \lg x = \lg h.$$ Rearranged, $$\lg y = \lg h - n \lg x.$$ This is a linear equation of the form $$Y = C + mX,$$ where $Y = \lg y$, $X = \lg x$, slope $m = -n$, and intercept $C = \lg h$. 3. **Calculate $\lg x$ and $\lg y$ values:** - For $x=2$, $\lg 2 \approx 0.3010$; $y=30.31$, $\lg 30.31 \approx 1.4814$ - For $x=6$, $\lg 6 \approx 0.7782$; $y=6.51$, $\lg 6.51 \approx 0.8132$ - For $x=15$, $\lg 15 \approx 1.1761$; $y=1.81$, $\lg 1.81 \approx 0.2577$ - For $x=20$, $\lg 20 \approx 1.3010$; $y=1.21$, $\lg 1.21 \approx 0.0828$ 4. **Plotting:** Plot points $(\lg x, \lg y)$ on graph paper with scale 1 cm = 0.1 unit on both axes and draw the best fit straight line through these points. 5. **Estimate slope $m$ and intercept $C$ from graph:** Using two points from the calculated values, for example $(0.3010,1.4814)$ and $(1.3010,0.0828)$, $$m = \frac{0.0828 - 1.4814}{1.3010 - 0.3010} = \frac{-1.3986}{1} = -1.3986.$$ Intercept $C$ is the $\lg y$ value when $\lg x = 0$, estimated by extending the line. Using point-slope form, $$C = \lg y - m \lg x = 1.4814 - (-1.3986)(0.3010) = 1.4814 + 0.4209 = 1.9023.$$ 6. **Find $n$ and $h$:** Since $m = -n$, we have $$n = -m = 1.3986 \approx 1.4.$$ And $$h = 10^C = 10^{1.9023} \approx 79.7.$$ 7. **Interpretation for part (iii):** If we plot $\ln y$ against $\ln x$ instead of $\lg y$ against $\lg x$, the graph remains a straight line with the same slope $-n$ because natural log and log base 10 differ by a constant factor: $$\ln y = \ln h - n \ln x,$$ so $n$ remains unchanged. The intercept changes to $\ln h$ instead of $\lg h$, but the relationship and slope are the same. **Final answers:** - $n \approx 1.4$ - $h \approx 79.7$