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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`
2 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
```
2 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
```
2 years ago
```							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.
```
one year ago
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