IIT JEE Entrance Exam> -value-of-y...
Let y = l^2-l^3/z where l = 2.0 + 0.1, z = 1.0 + 0.1 then the value of y is given by
(a) + 2 + 0.8
(b) –4 + 1.6
(c) –4 + 0.8
(d) + 2 + 1.6
1 Answers(b)
y = l^2-l^3/z
dy = 2ldl-((z.3l^2 dl-l^3 dz)/z^2 )=(2l-(3l^2)/z)dl+l^3/z^2 dz
= (2×2-(3×2^2)/1)(±0.1)+8/1 (±0.1)
= – 8 (+0.1) +8 (+0.01) = + 1.6
y = l^2-l^3/z=2^2-2^3/1 = 4 – 8 = –4
=> y = – 4 + 1.6
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