Skip to content
Numerical · Q17

Q.In a workshop a worker measures the length of a steel plate with a Vernier callipers having a least count 0.01 cm. Four such measurements of the length yielded the following values: 3.11 cm, 3.13 cm, 3.14 cm, 3.14 cm. Find the mean length, the mean absolute error and the percentage error in the measured value of the length.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
63% · 17/27 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1. Mean =3.11+3.13+3.14+3.144=12.524=3.13 cm= \dfrac{3.11+3.13+3.14+3.14}{4} = \dfrac{12.52}{4} = 3.13\ \text{cm}.

Step 2. Absolute errors: ∣3.11−3.13∣=0.02|3.11-3.13|=0.02, ∣3.13−3.13∣=0|3.13-3.13|=0, ∣3.14−3.13∣=0.01|3.14-3.13|=0.01, ∣3.14−3.13∣=0.01|3.14-3.13|=0.01 (all in cm).

Step 3. Mean absolute error =0.02+0+0.01+0.014=0.044=0.01 cm= \dfrac{0.02+0+0.01+0.01}{4} = \dfrac{0.04}{4} = 0.01\ \text{cm}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.