91影视

i) The linear density of the heaviest string, ${\mathrm{渭}}_1=3.0\mathrm{g}/\mathrm{m}$

ii) The linear density of the lightest string, ${\mathrm{渭}}_2=0.29\mathrm{g}/\mathrm{m}$

The mass of the unit length of the wire is known as linear density. By writing the mass as the product of density and volume, in the formula for linear density. Finally, by taking the ratio of the linear densities, we can find the ratio of the diameters.

The volume of a cylinder of heightI ,

$V={\mathrm{蟺诲}}^2\frac{\mathrm{I}}{4}$ (i)

The mass of a body in terms of density,

role="math" $m=蚁V$ (ii)

Here, d is the diameter, I is the length of the cylinder, p is the density of the wire and V is the volume.

Let, the strings are long narrow cylinders, one of diameterand other of diameter$d_2$.

So that, the mass per unit length that is the linear density of a body is given as:

role="math" localid="1660978010280" $$\begin{gathered} \mathrm{渭}=\frac{\mathrm{m}}{\mathrm{I}} \\ =\frac{{\displaystyle \frac{{\mathrm{p蟺诲}}^2\mathrm{I}}{4}}}{\mathrm{I}}\left(\mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{m}\mathrm{using}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\right) \\ =\frac{{\mathrm{蚁蟺诲}}^2}{4}.......................\left(\mathrm{a}\right) \end{gathered}$$

Now, using equation (a), their ratio of linear densities is given by:

$$\begin{gathered} \frac{{渭}_1}{{渭}_2}=\frac{{\displaystyle \frac{p{\mathrm{蟺诲}}_1^2}{4}}}{\frac{p{\mathrm{蟺诲}}_2^2}{4}} \\ \frac{{渭}_1}{{渭}_2}=\frac{{\displaystyle d_1^2}}{d_2^2} \end{gathered}$$

Therefore, the ratio of the diameters is given as:

$$\begin{gathered} \frac{d_1}{d_2}=\sqrt{\frac{{渭}_1}{{渭}_2}} \\ =\sqrt{\frac{3.0}{0.29}} \\ =3.22 \end{gathered}$$

Hence, the value of the ratio is 3.22