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The percentage error in the measurement of mass and speed of a body are 2% and 3% respectively. What will bek the maximum percentage error in the estimation of kinetic energy of the body? |
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Answer» Solution :`K.E.=(1)/(2)mv^(2)` `THEREFORE (DELTAK)/(k)=(Deltam)/(m)+(2Deltav)/(v)implies(Deltak)/(k)xx100=(Deltam)/(m)xx100+2((Deltav)/(v))xx100` `therefore` Percentage ERROR in `K.E. = 3% + 2 × 2% = 7%` |
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