The lower surface of a steel plate having a face area of 4 cm^2 and a thickness of 0.5 cm is fixed rigidly. A tangential force of 10 N is applied to the upper surface. Determine the lateral displacement of the upper

Correct answer: B. 1.5 × 10^ -9 m Given, tangential force F = 10 N Face area A = 4 cm^2 = 4 × 10^ -4 m^2 Thickness L = 0.5 cm = 0.5 × 10^ -2 m Rigidity modulus = 8.4 × 10^ 10 N/m^2 The formula for rigidity modulus is given by: = Shear Stress Shear Strain = (F / A)/(Δ x / L) Rearranging for lateral displacement Δ x : Δ x = (F L)/(A ) Substituting the given values: Δ x = 10 × 0.5 × 10^ -2 4 × 10^ -4 × 8.4 × 10^ 10 Δ x = 5 × 10^ -2 33.6 × 10^6 Δ x = (5)/(33.6) × 10^ -8 Δ x 1.488 × 10^ -9 m Δ x 1.5 × 10^ -9 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

The lower surface of a steel plate having a face area of 4 cm^2 and...

The lower surface of a steel plate having a face area of

4 \text{ cm}^2

and a thickness of

0.5 \text{ cm}

is fixed rigidly. A tangential force of

10 \text{ N}

is applied to the upper surface. Determine the lateral displacement of the upper surface relative to the lower surface. The rigidity modulus of steel is

8.4 \times 10^{10} \text{ N/m}^2

Options

A $1.5 \times 10^{-11} \text{ m}$
B $1.5 \times 10^{-9} \text{ m}$
C $1.5 \times 10^{-5} \text{ m}$
D $1.5 \times 10^{-7} \text{ m}$

Correct answer

B. $1.5 \times 10^{-9} \text{ m}$

Step-by-step solution

Given, tangential force

F = 10 \text{ N}

Face area

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

Thickness

L = 0.5 \text{ cm} = 0.5 \times 10^{-2} \text{ m}

Rigidity modulus

\eta = 8.4 \times 10^{10} \text{ N/m}^2

The formula for rigidity modulus is given by:

\eta = \dfrac{\text{Shear Stress}}{\text{Shear Strain}} = \dfrac{F / A}{\Delta x / L}

Rearranging for lateral displacement

\Delta x

\Delta x = \dfrac{F L}{A \eta}

Substituting the given values:

\Delta x = \dfrac{10 \times 0.5 \times 10^{-2}}{4 \times 10^{-4} \times 8.4 \times 10^{10}}

\Delta x = \dfrac{5 \times 10^{-2}}{33.6 \times 10^6}

\Delta x = \dfrac{5}{33.6} \times 10^{-8}

\Delta x \approx 1.488 \times 10^{-9} \text{ m}

\Delta x \approx 1.5 \times 10^{-9} \text{ m}

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