Question 235: Properties of Matter

Question

One end of a metal wire is fixed to a ceiling and a load of $2 \mathrm{~kg}$ hangs from the other end. A similar wire is attached to the bottom of the load and another load of $1 \mathrm{~kg}$ hangs from this lower wire. Then the ratio of longitudinal strain of upper wire to that of the lower wire will be ________. [Area of cross section of wire $=0.005 \mathrm{~cm}^2, \mathrm{Y}=2 	imes 10^{11} \mathrm{Nm}^{-2}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ]

Accepted Answer

To solve this problem, we use the fact that the longitudinal strain in a wire is given by the formula:
$$	ext{Strain} = \frac{\Delta L}{L} = \frac{F}{AY}$$
where $\Delta L$ is the change in length, $L$ is the original length, $F$ is the force applied, $A$ is the area of cross-section of the wire, and $Y$ is the Young's modulus of the material of the wire.
The force applied by each load due to gravity is calculated using $F = mg$, where $m$ is the mass of the load and $g$ is the acceleration due to gravity ($10 	ext{ m/s}^2$ in this case).
For the upper wire, the total force applied is the weight of both masses (2 kg and 1 kg):
$$F_1 = (2 	ext{ kg} + 1 	ext{ kg}) 	imes 10 	ext{ m/s}^2 = 30 	ext{ N}$$
For the lower wire, the force applied is just the weight of the 1 kg mass:
$$F_2 = 1 	ext{ kg} 	imes 10 	ext{ m/s}^2 = 10 	ext{ N}$$
Since the area $A$ and Young's modulus $Y$ are the same for both wires, these values cancel out when we calculate the ratio of the strains. Therefore, the ratio of the strains is directly proportional to the ratio of the forces:
$$\frac{	ext{Strain of upper wire}}{	ext{Strain of lower wire}} = \frac{F_1}{F_2} = \frac{30 	ext{ N}}{10 	ext{ N}} = 3$$
Hence, the ratio of longitudinal strain of the upper wire to that of the lower wire is 3.

Properties of Matter

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