A load of 10 kg is suspended by a metal wire that is 3 m long and possesses a cross-sectional area of 4 mm^2 . Determine the elongation. The Young modulus of the metal is 2.0 × 10^ 11 N/m^2 .

Correct answer: C. 3.675 × 10^ -4 m Given: m = 10 kg L = 3 m A = 4 mm^2 = 4 × 10^ -6 m^2 Y = 2.0 × 10^ 11 N/m^2 g = 9.8 m/s^2 The force applied on the wire is F = mg = 10 × 9.8 = 98 N . The elongation Δ L is given by the formula: Δ L = (FL)/(AY) Substituting the values: Δ L = 98 × 3 4 × 10^ -6 × 2.0 × 10^ 11 Δ L = (294)/(8 × 10^5) Δ L = 36.75 × 10^ -5 m = 3.675 × 10^ -4 m

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Some Mechanical Properties of Matter

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Some Mechanical Properties of Matter

A load of 10 kg is suspended by a metal wire that is 3 m long and...

A load of

10 \text{ kg}

is suspended by a metal wire that is

3 \text{ m}

long and possesses a cross-sectional area of

4 \text{ mm}^2

. Determine the elongation. The Young modulus of the metal is

2.0 \times 10^{11} \text{ N/m}^2

Options

A $3.75 \times 10^{-4} \text{ m}$
B $4.17 \times 10^{-5} \text{ m}$
C $3.675 \times 10^{-4} \text{ m}$
D $1.25 \times 10^{-4} \text{ m}$

Correct answer

C. $3.675 \times 10^{-4} \text{ m}$

Step-by-step solution

Given:

m = 10 \text{ kg}

L = 3 \text{ m}

A = 4 \text{ mm}^2 = 4 \times 10^{-6} \text{ m}^2

Y = 2.0 \times 10^{11} \text{ N/m}^2

g = 9.8 \text{ m/s}^2

The force applied on the wire is

F = mg = 10 \times 9.8 = 98 \text{ N}

.

The elongation

\Delta L

is given by the formula:

\Delta L = \dfrac{FL}{AY}

Substituting the values:

\Delta L = \dfrac{98 \times 3}{4 \times 10^{-6} \times 2.0 \times 10^{11}}

\Delta L = \dfrac{294}{8 \times 10^5}

\Delta L = 36.75 \times 10^{-5} \text{ m} = 3.675 \times 10^{-4} \text{ m}

Practice Some Mechanical Properties of Matter on Quantrex Academy →

More from Some Mechanical Properties of Matter