Question

A cylindrical wire of length $0.1\text{m}$ has specific gravity $11$. What is the maximum diameter of the wire that can float on the surface of water, if surface tension of water is $0.07\text{N/m}$. (Take $g=10{\text{m/s}}^2$)

• A. $0.7\text{mm}$
• B. $\frac{0.7}{11}\text{mm}$
• C. $\frac{14}{11}\text{mm}$
• D. $\frac{0.8}{3}\text{mm}$

Answer 1

The correct option is C $\frac{14}{11}\text{mm}$


We know,
Mass $=$ Volume $\times$ density
$\therefore$ Mass of wire $=\pi \times {\left(\frac{d}{2}\right)}^2\times l\times 11\times {10}^3$

Length of wire on which surface tension force is acting $=2\times l$
$\therefore L=2\times l=(2\times 0.1)\text{m}=0.2\text{m}$

Now, we know,
$T=\frac{F}{L}$
$\Rightarrow F=T\times L=T\times \left(2l\right)$
As wire floats on the surface of water,
$\Rightarrow$ Weight of the wire $\left(mg\right)=$ Surface tension force $\left(F\right)$
$\Rightarrow \pi \times {\left(\frac{d}{2}\right)}^2\times l\times 11000\times 10=T\times \left(2l\right)\Rightarrow d=\sqrt{\frac{0.07\times 4\times 2}{\frac{22}{7}\times 10\times 11000}}\Rightarrow d=\frac{14}{11}\times {10}^{-3}\text{m}\Rightarrow d=\frac{14}{11}\text{mm}$