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A wire of cross-section $4 mm^2$ is stretched by 0.1 mm by a certain weight. How far (length) will be wire of same material and length but of area $8 mm^2$ stretched under the action of same force.

A. 0.05 mm B. 0.10 mm C. 0.15 mm D. 0.20 mm

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Explanation

$ l = { FL \over AY } $ $ \therefore l \alpha { 1 \over A} $ ( F , L and Y are constant ) $ { l_2 \over l_1} = { A_1 \over A_2} = {4 \over 8} = {1 \over 2} \Rightarrow l_2 = { l_1 \over 2 } = { 0.1 \over 2 } = 0.05 mm $

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According to Hooke's law of elasticity stress is increased the ratio of stress to i strain.........

A. increases B. Decreases C. becomes zero D. Remains constant

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Explanation

$ Y = { stress \over strain } = constant $ It depends only on nature of material.

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Young's modulus of rubber is $ 10 ^ 4 N /m^2 $ and area of cross section is $ 2 cm^2 $ if force of $2 \times 10^5 dyne $ is applied along its length then its initial length L becomes .............

A. 3L B. 4L C. 2L D. None of the above

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Explanation

$ l = { FL \over AY } $ $ \therefore Final length = initial lengh + increament = 2L $

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A copper wire of length 4 mand area of cross-section $1.2 cm^2$ is stretched with a force of $ 4.8 \times 10^3 N$ . If young's modulus for copper is $ 1.2 \times 10^{11} N/m^2 $ . What will be the length increase of the wire ?

A. 1.33 mm B. 1.33cm C. 2.66mm D. 2.66cm

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Explanation

$ l = { FL \over AY } $

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A wire of length 2 m is made from $10cm^3$ of copper. A force F is applied so that its length increases by 2mm. Another wire of length 8 m is made from the same volume of copper. If the force F is applied to it, its length will increase by............

A. 0.8 cm B. 1.6 cm C. 2.4 cm D. 3.2 cm

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Explanation

$ l = { FL \over Ay } = {FL^2 \over (AL) Y} = {FL^2 \over VY} $ $ \therefore l \alpha L^ 2 If the volume of wire remains constant. $

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In steel the young's modulus and the strain at the breaking point are $ 2 \times 10^{11} N /m^2 $ and 0.15 respectively the stress at the breaking point for steel is therefore ...........

A. $ 1.33 \times 10^{11} N/m^2 $ B. $ 1.33 \times 106{12} N/m^2 $ C. $ 7.5 \times 10^{-13} N/m^2 $ D. $ 3 \times 10^{10 } N/m^2 $

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Explanation

$ Breaking stress = Strain \times young's modulus $

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Which of the following statement is correct

A. Hooke's law is applicable only with in elastic limit B. The adiabatic and isothermal elastic constants of a gas area equal. C. Young's modulus is dimensions less. D. Stress multiplied by strain is equal to stored energy.

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Explanation

In accordance with Hook's law

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The force required to stretch a steel wire of $1cm^2$ cross-section to 1.1 times its length would be $ ( Y = 2 \times 10^{11} N /m^2 ) $

A. $ 2 \times 10^6 N $ B. $ 2 \times 10^3 N$ C. $ 2 \times 10^{-6} N $ D. $ 2 \times 10^{-7} N $

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Explanation

$ F = A \times Y \times Strain $

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Which one of the following quantities does not have unit of force per unit area..........

A. stress B. strain C. Young's modulus of elasticity D. pressure

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Explanation

Because strain is a dimensionless and unitless quantity.

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A copper wire and a steel wire of same diameter and length are connected end to end a force is applied, which stretches their combined length by 1 cm, the two wires will have.......

A. different stresses and strains B. the same stress and strain C. the same strain but different stresses D. the same stress but different strains

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Explanation

stress = force /area In the present case, force applied and area of cross-section of wires are same. Therefore stress has to be the same Strain = stress /Y Since, the young's modulus of steel wire is greater than the copper wire, therefore strain in case of steel wire is less than that of in case of copper wire

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