Mechanical Properties of Solids

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) Forces at cross section is F.

Now applying formula . stress = tension/area=F/A

Tension = applied force =F

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Payal Gupta

Contributor-Level 10

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(c), (d) The ultimate tensile strength for material ii is greater hence material ii is elastic over larger region as compared to material (i) for material (ii) fracture point is nearer, hence it is more brittle.

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Payal Gupta

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(d) a mass M is attached at the centre. As the mass is attached to both the rods, both rod will be elongated, but due to different elastic properties of material rubber changes shape also.

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Payal Gupta

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(c) 2Tsin θ -mg=0

2Tsin θ =mg

Total horizontal forces = Tcos θ - T c o s θ = 0

T=mg/2sin θ

As mg is constant T 1 s i n θ

Tmax= mg/sin θ min

Sin θ min=0, θ min= 0

Tmin=mg/2sin θ max

s i n θ max= 1, θ =900

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar
 
(a)

 BO+OC- (BD+DC)

= 2BO -2BD

= 2 (BO-BD)

=2 [ (x2+L2)1/2-L]

=2L [ (1+ x 2 L 2 )1/2-L]

= 2L [ (1+ 1 2 x 2 L 2 - 1 ]= x 2 L

Strain = ? L 2 L = x 2 L 2 L = x 2 2 L 2

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Payal Gupta

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Y= s t r e s s s t r a i n = F A ? L L = F π ( D / 2 ) 2 * L ? L

D2= 4 F L π ? L Y

D= 4 F L π ? L Y

D c o p p e r D i r o n = Y i r o n Y c o p p e r

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Payal Gupta

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(c) Consider the diagram where a spring is stretched by applying a load to its free end. Clearly the length and shape of the spring changes. The change in length corresponds to longitudinal strain and change in shape corresponds to shearing strain.

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Payal Gupta

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(d) Y young's modulus is inversely proportional to temperature, so if we increase temperature , young's modulus decreases.

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Payal Gupta

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(d) breaking stress = breaking force/area of cross section

Breaking force will not depend upon length.

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Payal Gupta

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(b) No frictional force exists in case of ideal fluid. Hence tangential forces are zero.

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