Question

) If the percentage errors in measuring mass and velocity of a particle are respectively 2% and 1% percentage error in measuring its kinetic energy i

Answer 1

Answer:

4%

Explanation:

KE = ½mv²

% error in KE =

% error in m + 2× %error in v

=2+ 2×1

=2+2

=4%

Answer 2

Answer:

The percentage error in measuring kinetic energy is 4%

Explanation:

Given that,

The percentage error in measuring mass of a particle is 2%

Therefore,

$\frac{∆m}{m} \times 100 = 2$

Also given the percentage error in measuring velocity is 1%

Hence,

$\frac{∆v}{v} \times 100 = 1$

The kinetic energy is related with mass and velocity as

$K.E = \frac{1}{2} mv {}^{2}$

Applying error derivative on both sides we get

$\frac{∆K.E}{K.E} = \frac{∆m}{m} + 2 \frac{∆v}{v}$

Multiplying both sides by 100 to get the percentage error,we get

$\frac{∆K.E}{K.E} \times 100 \\ = \frac{∆m}{m} \times 100 + 2 \frac{∆v}{v} \times 100$

Substituting the given values,we get

$\frac{∆K.E}{K.E} = 2+ 2(1) = 4 \: percent$

Therefore,The percentage error in measuring kinetic energy is 4%

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