Q101 / 150 · MHT CET 2026 (20 April Shift 1) · Physics
The percentage error in the measurement of mass of a body and its speed are $0.73\%$ and $1.84\%$ respectively. The percentage error in the measurement of its momentum and kinetic energy are respectively
- A$2.57\%$ , $2.75\%$
- Correct answerB$2.57\%$ , $4.41\%$
- C$1.46\%$ , $1.84\%$
- D$1.84\%$ , $3.70\%$
Solution
Given percentage error in mass, $\dfrac{\Delta m}{m} \times 100 = 0.73\%$
Given percentage error in speed, $\dfrac{\Delta v}{v} \times 100 = 1.84\%$
Momentum is given by $p = mv$
Percentage error in momentum is $\dfrac{\Delta p}{p} \times 100 = \dfrac{\Delta m}{m} \times 100 + \dfrac{\Delta v}{v} \times 100$
$\dfrac{\Delta p}{p} \times 100 = 0.73\% + 1.84\% = 2.57\%$
Kinetic energy is given by $K = \dfrac{1}{2}mv^2$
Percentage error in kinetic energy is $\dfrac{\Delta K}{K} \times 100 = \dfrac{\Delta m}{m} \times 100 + 2\dfrac{\Delta v}{v} \times 100$
$\dfrac{\Delta K}{K} \times 100 = 0.73\% + 2(1.84\%) = 0.73\% + 3.68\% = 4.41\%$
Answer: $2.57\%$ , $4.41\%$
Given percentage error in speed, $\dfrac{\Delta v}{v} \times 100 = 1.84\%$
Momentum is given by $p = mv$
Percentage error in momentum is $\dfrac{\Delta p}{p} \times 100 = \dfrac{\Delta m}{m} \times 100 + \dfrac{\Delta v}{v} \times 100$
$\dfrac{\Delta p}{p} \times 100 = 0.73\% + 1.84\% = 2.57\%$
Kinetic energy is given by $K = \dfrac{1}{2}mv^2$
Percentage error in kinetic energy is $\dfrac{\Delta K}{K} \times 100 = \dfrac{\Delta m}{m} \times 100 + 2\dfrac{\Delta v}{v} \times 100$
$\dfrac{\Delta K}{K} \times 100 = 0.73\% + 2(1.84\%) = 0.73\% + 3.68\% = 4.41\%$
Answer: $2.57\%$ , $4.41\%$
Hence, the correct answer is option: 'B'.