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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..........
$$ l = { FL \over AY } = { FL^2 \over (AY) Y } = { FL^2 \over AY } $$ $ 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 ?
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 $
To keep constant time, watches are fitted with balance wheel made of........
Because dimension of invar does not vary with temperature.
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 ?
$ l = {FL \over \pi r^2 y } $ $ \therefore l \alpha { L \over r^2 } $ ( Y and F are constant )
The Coefficient of linear expansion of brass & steel are $ \alpha_1 and \alpha_2 $ . 2 If we take a brass rod of length $ l_1 $ & steel rod of length $l_2 $ at $ 0 ^\circ $ , their difference in length $ ( l_2 - l_1 ) $ will remain the same at a temperature if ........................
$ l_2 = l_2 ( 1 + \alpha_2 \triangle Q) and L_1 = l_1 ( 1 + \alpha_1 \triangle Q ) $ $ \Rightarrow (l_2 - l_1) = ( l_2 -l_1 ) + \triangle Q ( l_2 \alpha_2 - l_1 \alpha _1 ) $ $ now (L_2 - L_1 ) = ( l_2 - l_1 ) $ $ so , l_2 \alpha_2 = l_2 \alpha_1 = 0 $
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 $
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 ) $
$ 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............
$ 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 ?
$ 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 ?
$ Y = { MgL \over Al } $
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