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The dimensions of four wires of the same material are given below, in which wire the increase in length will be maximum when the same strain is applied.

A. Length 100 cm, Diameter 1 mm B. Length 200 cm, Diameter 2 mm C. Length 300 cm, Diameter 3 mm D. Length 50 cm, Diameter 0.5 mm

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Explanation

$ Y = { F \over A} { L \over l} \Rightarrow l \alpha { l \over A } \alpha {L \over \pi d^2 } $ $ \therefore l \alpha { L \over d^2 } ( As F and Y are constant ) $ The ratio of $ { L \over d^2 } $ is maximum for case CD

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On increasing the length by 0.5 mm in a steel wire of length 2 mand area of cross-section $2 mm^2$ the force required is..............$ [ Y for steel = 2.2 \times 10^{11} N/m ] $

A. $ 1.1 \times 10^ 5 N $ B. $ 1.1 \times 10^ 4 N $ C. $ 1.1 \times 10^ 3 N $ D. $ 1.1 \times 10^2 N $

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Explanation

$ F = { YAI \over L } $

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A stress of $ 3.8 \times 10^8 N m^2 $ is applied to steel rod of length 1 m along its length. Its young's modulus is $ 2 \times 10^ {11} N / m^2 0$. Then what is the elongation produced in the rod in mm ?

A. 3.18 B. 6.36 C. 5.18 D. 1.59

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Explanation

$ = { { F \over A} \over {\triangle l \over l} } $ $ given stress 3.18 \times 10^8 N/m^2 $ $ \therefore \triangle l ={ \triangle F /A \over Y } $

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A force F is needed to break a copper wire having radius R, The force needed to break a copper wire of radius 2R will be........

A. F/2 B. 2F C. 4F D. F/4

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Explanation

Breaking force $ \alpha $ area of crossection of $( \pi r^2 ) $ wire If radius of wire is doubled then breaking force will become four times.

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A rubber cord 10m long is suspended vertically. How much does it stretch under its own weight. $( Density of rubber is1500 kg /m^3 , Y = 5 \times 10^ 8 N /m^2 , g = 10 m/s ) $

A. $ 15 \times 10^ {-4} m $ B. $ 7.5 \times 10^ {-4} m $ C. $ 12 \times 10^ {-4} m $ D. $ 25 \times 10^ {-4} m $

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Explanation

$ l = { L^2 \over 2 Y } $

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If x, longitudinal strain is produced in a wire of young's modulus y then energy stored in the material of the wire per unit volume is..........

A. $ yx ^2 $ B. $ 2 yx^2 $ C. $ { 1 \over 2 } y^2 x $ D. $ {1 \over 2} yx^2 $

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Explanation

Energy stored per unit volume $ = {1 \over 2 } \times stress \times strain $

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A steel wire of cross-sectional area $ 3 \times 10^ {-6} m^2 $ can with stand a maximum strain of $ 10 ^ {-3} $ Young's modulus of steel is $ 2 \times 10 ^ {11} N /m^2 $ . The maximum mass the wire can hold is ........$ ( g = 10 m/s^2 ) $

A. 40 kg B. 60 kg C. 80 kg D. 100 kg

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Explanation

$ Y = { stress \over strain } \Rightarrow Max.strain = { Max.stress \over V}$ $ Max. strain = { mg /A \over Y } $

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The young's modulus of a rubber string 8 cm long and density $ 1.5 kg /m^3 $ is $ 5 \times 10^8 N /m^2 $ .What will be the length increase due to its own weight?

A. $ 9.6 \times 10^ {-5} m $ B. $ 9.6 \times 10^ {-11} m $ C. $ 9.6 \times 10^ {-3} m $ D. 9.6 m

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Explanation

$ l = { L^2 dg \over 2 Y } = { (8 \times 10^{-2} )^2 \times 1.5 \times 9.8 \over 2 \times 5 \times 10^8 } = 9.6 \times 10^{-11} m $

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A and B are two wires. The radius of A is twice that of B. They are stretched by the same load. Then what is the stress on B ?

A. Equal to that on A B. Four times that on A C. Two times that on A D. Half that on A

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Explanation

$ stress = { force \over area } \Rightarrow \therefore stress \alpha { 1 \over \pi r^2 } $ $ \therefore { S_B \over S_A} = \left( rA \over rB \right) ^2 = (2)^2 \Rightarrow S_B = $ S_A $

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If the length of wire is reduced to half then it can hold the .........load.

A. Half B. Same C. Double D. One fourth

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Explanation

Breaking force $ \alpha $ area of cross section of wire. i.e. load hold by the wire does not depend upon the length of the wire.

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