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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Mathematics in Physics

JEE Main · Physics · Mathematics in Physics

The momentum ( p ) and mass ( m ) of a moving body are measured in...

The momentum (pp) and mass (mm) of a moving body are measured in a laboratory as

p=(50±1)p = (50 \pm 1) kg m/s
m=(10±0.5)m = (10 \pm 0.5) kg

The percentage error in the calculated kinetic energy of the body is:

Options

  1. A9%9\%
  2. B7%7\%
  3. C6%6\%
  4. D1%1\%

Correct answer

A. 9%9\%

Step-by-step solution

The kinetic energy KK of a body is related to its momentum pp and mass mm by the formula:
K=p22mK = \dfrac{p^2}{2m}

The maximum fractional error in kinetic energy is given by:
ΔKK=2Δpp+Δmm\dfrac{\Delta K}{K} = 2\dfrac{\Delta p}{p} + \dfrac{\Delta m}{m}

Multiplying by 100 to get the percentage error:
ΔKK×100=2(Δpp×100)+(Δmm×100)\dfrac{\Delta K}{K} \times 100 = 2 \left( \dfrac{\Delta p}{p} \times 100 \right) + \left( \dfrac{\Delta m}{m} \times 100 \right)

Given values:
Δpp×100=150×100=2%\dfrac{\Delta p}{p} \times 100 = \dfrac{1}{50} \times 100 = 2\%
Δmm×100=0.510×100=5%\dfrac{\Delta m}{m} \times 100 = \dfrac{0.5}{10} \times 100 = 5\%

Substituting these values into the error formula:
Percentage error in K=2(2%)+5%=4%+5%=9%K = 2(2\%) + 5\% = 4\% + 5\% = 9\%

Answer: 9%9\%
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