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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 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

Grade:11

4 Answers

tulika singla
24 Points
5 years ago
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
jitender
114 Points
5 years ago
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
harshita joshi
11 Points
5 years ago
let the original lenght be L and A is area
stress=youngs modulas*strain
so 8/A=Y*(X-L)\L----------------(1)
10\A=Y*(y-L)\L-------------------(2)
18\A=Y*(Z-L)\L--------------------(3)
Operating eq 1\2
On solving we get L =5X-4y
Oprating eq 3\2
we get   9y-9L=5Z-5L
now put the value of L as solved above
on solving the equation we get
Z=5y-4X
this is the required ans.
 
 
 
Yash Chourasiya
askIITians Faculty 256 Points
3 years ago
Dear Student

Let, original length of the spring is L metre and, Y = F.L/A.l

Now, 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.L​m/A.(y−L)........(II)
From equation (I) and (II) we get,
8(y−L) = 10(x−L)
or, 4y − 4L = 5x − 5L
or, L = 5x − 4y
When, 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 − 4x

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

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