Answer:
Number of revolution of smaller pulley of 8 inches is 1200 revolution per minute.
Solution:
Given that number of revolution of two pulleys is inversely proportional to their diameter.
Let say diameter of two pulley be [tex]\mathrm{d}_{1} \text { and } \mathrm{d}_{2}[/tex]
And revolution of two pulleys be [tex]\mathrm{r}_{1} \text { and } \mathrm{r}_{2}[/tex]
From given information [tex]\mathrm{d}_{1}[/tex] is inversely proportional to [tex]\mathrm{r}_{1}[/tex] and [tex]\mathrm{d}_{2}[/tex] is inversely proportional to [tex]\mathbf{r}_{2}[/tex]
Assuming k be constant of proportionality we get
[tex]\mathrm{d}_{1}=\frac{k}{r_{1}} \text { and } d_{2}=\frac{k}{r_{2}}[/tex]
so we get
[tex]\frac{d_{1}}{d_{2}}=\frac{r_{2}}{r_{1}}[/tex]
Given that [tex]{d}_{1}[/tex] = 24 inches, [tex]{r}_{1}[/tex] = 400 revolution per minute , [tex]{d}_{2}[/tex] = 8 inches. we need to calculate [tex]{r}_{2}[/tex]
[tex]\frac{24}{8}=\frac{r_{2}}{400}[/tex]
[tex]{r}_{2} = 24 \times \frac{400}{8}[/tex] = 1200 revolutions per minute.
Hence number of revolution of smaller pulley of 8 inches is 1200 revolution per minute.
