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The length of an elastic string is X m when the tension is 8 N and Y m when its tension is 10 N . The length in metres when the tension is 18 N isi)4X-5Yii)5Y-4Xiii)9X-4Yiv)4Y-9Y
The answer is (ii)...First let the initial length is `a` now we know force=Y`lA/a where `l` the is change in length so we can write X=a+8a/Y`ASimilarily Y=a+10a/Y`ASolving them we get 5X-4Y=a and Y`A=8a/(x-a)Now for 18N force we can writeP=a+18a/Y`A put the value of Y`A and get the answer
T=kxNow ◇T=k◇x10-8=k (y-x) ...1st18-8=k (z-x) .....2nd2nd÷1st10/2=(z-x)/(y-x)5y-5x=z-xZ=5y-4xSo length of string is 5y-4x when T=18
let the original lenght be L and A is areastress=youngs modulas*strainso 8/A=Y*(X-L)\L----------------(1)10\A=Y*(y-L)\L-------------------(2)18\A=Y*(Z-L)\L--------------------(3)Operating eq 1\2On solving we get L =5X-4yOprating eq 3\2we get 9y-9L=5Z-5Lnow put the value of L as solved aboveon solving the equation we getZ=5y-4Xthis is the required ans.
Dear StudentLet, original length of the spring is L metre and, Y = F.L/A.lNow, when F = 8N, and l = (x - l)m then,Y = 8.L/A.(x−L)..........(I)and when F = 10N, and l = (y - l)m then, Y = 10.Lm/A.(y−L)........(II)From equation (I) and (II) we get,8(y−L) = 10(x−L)or, 4y − 4L = 5x − 5Lor, L = 5x − 4yWhen, F = 18N,Let, length of the wire will be Z metre.∴Y = 18.L/ {A.(Z−L)} ….........(III)From equation (I) and (III) we get,9(x−L) = 4(Z−L)or,4Z = 9x − 9L + 4L = 9x − 5L = 9x − 25x + 20y [putting value of L]or,Z = 5y − 4xI hope this answer will help you.Thanks & RegardsYash Chourasiya
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