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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Mechanical Properties of Solids

NTA Abhyas 2020 · Physics · Mechanical Properties of Solids

One of the square faces of a metal slab of side 50 cm and thickness...

One of the square faces of a metal slab of side 5050 cm and thickness 2020 cm is rigidly fixed on a horizontal surface. If a tangential force of 30N30N is applied to the top face and it is known that the shear modulus of the material is 4×1010N/m24 \times {10}^{10}N / m^{2}, then the displacement (in mm) of the top face isOne of the square faces of a metal slab of side 50 cm and thickness 20 cm is rigidly fixed on a horizontal surface. IfOne of the square faces of a metal slab of side 50 cm and thickness 20 cm is rigidly fixed on a horizontal surface. If

Options

  1. A4×10−124 \times {10}^{- 12}
  2. B4×10−104 \times {10}^{- 10}
  3. C6×10−106 \times {10}^{- 10}
  4. D8×10−108 \times {10}^{- 10}

Correct answer

C. 6×10−106 \times {10}^{- 10}

Step-by-step solution

One of the square faces of a metal slab of side 50 cm and thickness 20 cm is rigidly fixed on a horizontal surface. If

One of the square faces of a metal slab of side 50 cm and thickness 20 cm is rigidly fixed on a horizontal surface. If

Shear strain, ε=tanθ=θ=xlε = tan \theta = \theta = \dfrac{x}{l}

η=σεη = \dfrac{σ}{ε}

Or x=F⋅lA⋅ηx = \dfrac{F ⋅ l}{A ⋅ η}

Given F=30NF = 30N

A=50×50×10−4A = 50 \times 50 \times {10}^{- 4}

η=4×1010η = 4 \times {10}^{10}

x=30×20×10−250×50×10−4×4×1010x = \dfrac{30 \times 20 \times {10}^{- 2}}{50 \times 50 \times {10}^{- 4} \times 4 \times {10}^{10}}

=625×4×10−2×1010= \dfrac{6}{25 \times 4 \times {10}^{- 2} \times {10}^{10}}

=6×10−10m= 6 \times {10}^{- 10}m
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