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# If error in measurement of speed is 10% then find the error in measureent of kinetic energy. AskIITians Expert Hari Shankar IITD
17 Points
12 years ago

Kinetic Energy K = (1/2) m v2

Take log of both sides.

log(K) = log(1/2) + log(m) + 2 log(v)

Now, differentiate both sides.

Differential of log(x) is 1/x dx.

x can be anything.

So, here, we have

(1/K) dK = 0 + (1/m) dm + 2 (1/v) dv    [Because log(1/2) is a constant, and derivative of any constant is zero]

Now, dK is nothing but delta(K).

So, dK is the error in measurement of K.

dK/K is the relative error in measurement of K.

Percentage error is relative error x 100

Hence,

dK/K = dm/m + 2 dv/v

We are given that dv/v = 10% = 0.1

dm/m is not given. So we will assume there is no error in measurement of m.

Therefore, Relative error in measurement of K = dK/K = 0 + 2(0.1) = 0.2

Percentage error = 0.2 x 100 = 20 %

12 years ago

error in KE turns out to be dm/m+2dv/v when we differentiate the expression for KE.

Assuming no error in mass measurement the error in KE should be 2*10 percent=>20%.

please specify doubts specifically and which part of question did you stuck while solving the problem and attempt questions well before you give up.!!!