HSC Science (Computer Science) 12th Board ExamMaharashtra State Board
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The Area of the Upper Face of a Rectangular Block is 0.5 M ' 0.5 M and the Lower Face is Fixed. the Height of the Block is 1 Cm. a Shearing Force Applied at the Top Face Produces a Displacement of 0.015 Mm. Find the Strain and Shearing Force. - HSC Science (Computer Science) 12th Board Exam - Physics

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Question

The area of the upper face of a rectangular block is 0.5 m by 0.5 m and the lower face is fixed. The height of the block is 1 cm. A shearing force applied at the top face produces a displacement of 0.015 mm. Find the strain and shearing force.

(Modulus of rigidity: η = 4.5 x 1010 N/m2)

Solution

Given:

Area under shear, A = 0.5m × 0.5m = 0.25m2

Height of the block, h = 1cm = 1 × 10-2 m,

Displacement of top face = x = 0.015 mm = 15 × 10-6 m,

Modulus of rigidity, η = 4.5 × 1010 N/m2

To Find:

Strain, θ = ?

Shearing force, F = ?

Formulae:

θ = x/h

F = ηAθ

Solution:

θ = x/h

`theta = (15 xx 10^-6)/10^-2`

`theta = 1.5 xx 10^-3`

Shearing strain (θ) is 1.5 × 10-3

F = ηAθ

F = 4.5 × 1010 × 0.25 × 1.5 × 10-3

F = 1.688 × 107 N

Shearing force (F) is 1.688 × 107 N

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APPEARS IN

 2013-2014 (October) (with solutions)
Question 4.1 | 7.00 marks
Solution The Area of the Upper Face of a Rectangular Block is 0.5 M ' 0.5 M and the Lower Face is Fixed. the Height of the Block is 1 Cm. a Shearing Force Applied at the Top Face Produces a Displacement of 0.015 Mm. Find the Strain and Shearing Force. Concept: Definition of Stress and Strain.
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