Question

In Searle’s apparatus for measurement of Young’s modulus, a steel wire of length 2m and
area of cross-section 6 2 2.5 10^-6 m^2 − × is suspended from a torsion head. When a weight of
5 kg is suspended at its free-end, its length increases. Calculate the work done on the
wire. Take Young’s modulus of steel as 2 10^11 Nm .

Answer 1

Answer on Question #45597-Physics-Mechanics

In Searle's apparatus for measurement of Young's modulus, a steel wire of length $l = 2m$ and area of cross-section $A = 2.5 \cdot 10^{-6} \, m^2$ is suspended from a torsion head. When a weight of $m = 5 \, kg$ is suspended at its free-end, its length increases. Calculate the work done on the wire. Take Young's modulus of steel as $Y = 2 \cdot 10^{11} \, Nm$.

Solution

The work done on the wire is

$$W = \frac{1}{2} F e,$$

where $F = mg$ is the applied force, $e$ is extension obtained at force $F$.

$$e = \frac{F l}{Y A},$$

where $A$ is the area of the cross section of the object and $l$ is the length of the object, $Y$ is Young's modulus.

Therefore

$$W = \frac{1}{2} F \frac{F l}{Y A} = \frac{1}{2} \frac{F^2 l}{Y A} = \frac{1}{2} \frac{(m g)^2 l}{Y A} = \frac{1}{2} \frac{(5 \cdot 9.8)^2 \cdot 2}{2 \cdot 10^{11} \cdot 2.5 \cdot 10^{-6}} = 4.8 \cdot 10^{-3} J = 4.8 \, mJ.$$

Answer: $4.8 \cdot 10^{-3} J$

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