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Grade 11IIT JEE Entrance Exam

if the error in the measurement of momentum of a particle is 10% and mass is known exactly,the permissible error in the determination of kinetic energy is

Profile image of Bidisha Baruah
11 Years agoGrade 11
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6 Answers

Profile image of mohit vashishtha
11 Years ago
2*10= 20%
Profile image of Bidisha Baruah
11 Years ago
are u sure??the answer in my book is 21%
Profile image of Nicho priyatham
11 Years ago
yes 21%
u can apply calucus only when error is very small or less then 1 
here error is 10% so by calucus u get 2*10 =20 which is wrong 
to do such problems first take P1=100 and calulate KE1
now put p2=110 and calculate KEthen find the error
PLEASE APPROVE
Profile image of Bidisha Baruah
11 Years ago
i did not get it i am extemely sorry.i mean why should we put p=110
Profile image of Ramgopal tiwari
9 Years ago
Here , K.E- 1/2mv^2Mass is known exactly but there is always a minimum error in any calculation so, Here we let the minimum in mass =1% Now, ∆K/K=∆ M/M +2∆V/V = 1% + 2(10%) = 1% + 20% = 21%
Profile image of Aan
8 Years ago
P=mv
Ke = 1/2 mv^2
Ke = P^2/2m
Here error is 10%
Thus
P'= 10% of P
P'=0.1P
Ke'=P'/2m
=1.21×Ke
Percentage error = (Final - Initial)×100/initial
=(1.21Ke - Ke)×100/Ke
= 21%