Mechanical Properties of Solids MCQs for NEET — Physics Questions with Answers

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A fixed volume of iron is drawn into a wire of length L. The extension x produced in this wire be a constant force F is propotional to..........

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Explanation

$$ l = { FL \over AY } = { FL^2 \over (AY) Y } = { FL^2 \over AY } $$ $ If volume is fixed then l \alpha L^ 2 $

On applying a stress of $ 20 \times 10 ^8 N/m^2 $ the length of a perfect elastic wire is doubled. What will be its Young's modulus ?

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Explanation

Young's modulus = stress /strain As the length of wire get doubled therefore strain = 1 $ \therefore Y = strain = 20 \times 10^8 N/m^2 $

A wire is stretched by 0.01 m by a certain force F. Another wire of same material whose diameter and length are double to the original wire is stretched by the same force ? Then what will be its elongation ?

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Explanation

$ l = {FL \over \pi r^2 y } $ $ \therefore l \alpha { L \over r^2 } $ ( Y and F are constant )

A rod is fixed between two points at $ 20 ^\circ $ The Coefficient of linear expansion of material of rad is $ 1.1 \times 10^{-5} / ^\circ C $ and Young's modulus is $ 1.2 \times 10 ^{11} N/m^2 $ . Find the stress developed in the rod if temperature of rod becomes $ 10 ^\circ C $

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Explanation

Thermal stress = $ y \alpha \triangle Q $

How much force is required to produce an increase of 0.2% in the length of a bross wire of diameter 0.6 mm $ (Young's modulus for brass = 0.9 \times 10^{11} N /m^2 ) $

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Explanation

$ F = { YA l \over L } $

A 5m long aluminium wire $ ( Y = 7 \times 10 ^{10 } N /m^2 ) $ of diameter 3mmsupports a 40 kg mass. In order to have the same elongation in a copper wire $ Y = 12 \times 10 ^ {10 } N /m^2 $ of the same length under the same weight, the diameter should now be in mm............

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Explanation

$ l= { FL \over \pi r^2 y } \Rightarrow r^2 \alpha {1 \over Y } $ ( F ,L and l are constant )

Two similar wires under the same load yield elongation of 0.1 mmand 0.05 mmrespectively. If the area of Cross - section of the first wire is 4mm2. Then what is the area of cross - section of the second wire ?

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Explanation

$ l = {FL \over \pi r^2 y } $ $ \therefore l \alpha {1 \over A} ( F , L .Y are constant ) $ $ {A_2 \over A_1 } = {l_1 \over l_2 } $

An iron rod of length 2m and cross-section area of $50 mm^2$ stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. What is young's modulus of the iron rod ?

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Explanation

$ Y = { MgL \over Al } $

A load W produces an extension of 1mm in a thread of radius r. Now if the load is made 4 w and radius is made 2r all other things remaining same the extension will becomes..............

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Explanation

$ l = { FL \over AY} $ $ \therefore l \alpha {F \over r^2 } $

A steel wire of 1m long and $1mm^2$ cross sectional area is hung from rigid end when weight of 1 kg is hung from it then change in length will be ……$ ( Y = 2 \times 10 ^{11} N/m^2 ) $

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Explanation

$ L = { mgL \over AY } $

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