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An error of 1% is made in calculating the length and breadth of a rectangle. Let the error in measuring the length be L be dL and the error in measuring the breadth W be dW.
The error in the area A which is the product A = W*L is `(dA)/A = sqrt(((dL)/L)^2 + ((dW)/W)^2)`
=> `(dA)/A = sqrt(0.01^2 + 0.01^2)`
=> `(dA)/A = sqrt(2*10^-4)`
=> `(dA)/A = sqrt 2*10^-2`
=> `(dA)/A = sqrt 2 %`
The error in the area is `sqrt 2` % of the area estimated.
I appreciate your effort. But this ques is MCQ. Options are-
And correct answer is b. 2
I needed solution.
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