NCERT Class 11 Physics Solutions: Chapter 9 – Mechanical Properties of Solid-Part 7 (For CBSE, ICSE, IAS, NET, NRA 2022)

Glide to success with Doorsteptutor material for CBSE/Class-7 : get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-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.

A Rod of Length 1.05 M Having Negligible Mass is Supported a …

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:

Two Equal Stress Steel and Aluminium Wire Showing in 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.

Developed by: