if percentage errors in measuring mass and speed of a body are 2% and 3% than what is the maximum percentage error in measuring the kinetice energy

Dear Vinayak,
KE= 1/2.m.v^2
log KE =  log 1/2 + log m + 2logv
differentiating,
 delta(KE)/KE =  delta (m)/m  +  2.delta v/v
but given.... [delta m/m=2/100] and [delta v/v=3/100]
putting the value we get.
delta KE/KE= 8/100
% error(max)= [ delta KE/KE ].100
            = 8% (ANS)
Regards.

percentage error in measuring mass=2%

percentage error in measuring speed=3%

kinetic energy =mv²/2

for multiplication relative errors are added to get the relative error and power is multiplied

%error in kinetic energy=2%+2*3%=8%

about 0.0018

8%

ΔK/K = ΔM/M + 2ΔV/V

where ΔK/K is % error in kinetic energy and so on....

8%