# NCERT Class 11 Physics Solutions: Chapter 9 – Mechanical Properties of Solid-Part 7

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**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.