P=a21b2c3d−4
(pΔp)max=21aΔa+2bΔb+3cΔc+4dΔd
100×PΔP=(21aΔa+2bΔb+3cΔc+4dΔd)×100
=32
JEE Main 2017 — Physics Mechanics
A physical quantity P is described by the relation P=a21b2c3d−4. If the relative errors in the measurement of a, b, c and d respectively, are 2, 1, 3 and 5. Then the relative error in P will be:
Held on 9 Apr 2017 · Verified 6 Jul 2026.
12
8
25
32
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