Question

One end of a metal wire is fixed to a ceiling and a load of 2 kg hangs from the other end. A similar wire in attached to the bottom of the load and another load of 1 kg hangs from this lower wire. Find the longitudinal strain in both the wires. Area of cross section of each wire is $0.005c{m}^2$ and Young modulus of the metal is $2.0\times {10}^{11}N{m}^{-2}.$ Take $g=10m{s}^{-2}.$

• A. ${10}^{-5},5\times {10}^{-5}$
• B. ${10}^{-3},2\times {10}^{-3}$
• C. ${10}^{-4},3\times {10}^{-4}$
• D. None of the above

Answer 1

The correct option is C ${10}^{-4},3\times {10}^{-4}$

The situation is described in figure (14-W2). As the 1 kg mass is in equilibrium, the tension in the lower wire equal the weight of the load.

Thus $T_1=10N$
Stress $=10N/0.005c{m}^2$
$=2\times {10}^{7}N{m}_2.$

Longitudinal strain $=\frac{stress}{Y}=\frac{2\times {10}^{7}N{m}^{-2}}{2\times {10}^{11}N{m}^{-2}}={10}^{-4}.$

Considering the equilibrium of the upper block, we can write,

$T_2=20N+T_1,or,T_2=30N.$

Stress $=30N/0.005c{m}^2$

$=6\times {10}^{7}N{m}^{-2}.$

Longitudinal strain $=\frac{6\times {10}^{7}N{m}^{-2}}{2\times {10}^{11}N{m}^{-2}}=3\times {10}^{-4}.$