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NCERT Class 11 Physics Solutions: Chapter 9 – Mechanical Properties of Solid-Part 7
Question 9.18:
A rod of length having negligible mass is supported at its ends by two wires of steel (wire ) and aluminum (wire ) of equal lengths as shown in Figure. The cross-sectional areas of wires are , respectively. At what point along the rod should a mass m be suspended in order to produce
(a) Equal stresses and
(b) Equal strains in both steel and aluminum wires.
Answer:
Cross-sectional area of wire
Cross-sectional area of wire
Young՚s modulus for steel,
Young՚s modulus for aluminum,
(a) Equal stresses from the steel-wire end
Explanation:
Let a small mass be suspended to the rod at a distance from the end where wire is attached.
Stress in the wire
If the two wires have equal stresses, then:
Where,
Force exerted on the steel wire
Force exerted on the aluminum wire
The situation is shown in the following figure:
Taking torque about the point of suspension, we have:
Using equations, we can write:
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of from the end where wire is attached.
(b) Equal strains in both steel and aluminum wires: from the steel-wire end
Explanation:
Young՚s modulus
If the strain in the two wires is equal, then:
Taking torque about the point where mass , is suspended at a distance from the side where wire attached, we get:
Using equations, we get:
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of from the end where wire is attached.