Screw Gauge & Vernier Callipers — Least count, reading
Units & Measurements
10
JEE Qs
8%
Hard
75
min
Master the calculation and application of zero error and least count, as these are the most common traps in JEE questions for this topic.
🧮 Key Formulas
✅ Key Points for JEE
- 1Always calculate the Least Count (LC) first for the given instrument, as it's fundamental to all readings and error calculations.
- 2Correctly identify positive and negative zero errors: Positive zero error occurs when the zero of the Vernier/Circular scale is ahead of the Main scale zero; Negative zero error occurs when it is behind.
- 3Zero Correction is always the negative of the Zero Error. If Zero Error is positive, subtract it; if Zero Error is negative, add its magnitude to the observed reading.
- 4The Main Scale Reading (MSR) for Vernier Callipers is the last reading on the main scale *before* the zero of the Vernier scale. For Screw Gauge, it's the reading on the linear scale visible just before the edge of the circular scale.
- 5The Vernier/Circular Scale Coincidence (VSC/CSC) is the division on the secondary scale that perfectly aligns with *any* division on the main/linear scale.
⚠️ Common Mistakes
- ✕Incorrectly identifying the sign of the zero error (e.g., confusing positive zero error with negative).
- ✕Forgetting to apply zero correction (subtracting zero error) or applying it with the wrong sign.
- ✕Miscalculating the least count, especially confusing pitch with least count for a screw gauge.
- ✕Incorrectly reading the Main Scale Reading (MSR) or the Vernier/Circular Scale Coincidence (VSC/CSC).
- ✕Not understanding that zero correction always subtracts the error, regardless of its sign (e.g., if error is -0.02 cm, correction is +0.02 cm).
📝 Practice Questions
See allQ43.The maximum percentage error in the measurment of density of a wire is [Given, mass of wire = (0.60 ± 0.003)g radius of wire = (0.50 ± 0.01)cm length of wire = (10.00 ± 0.05)cm] (1) 8 (2) 5 (3) 4 (4) 7 for another diatomic molecules, but for rigid molecules and γ2 =
Q43.The energy of a system is given as E(t) = α3e−βt , where t is the time and β = 0.3 s−1 . The errors in the measurement of α and t are 1.2% and 1.6% , respectively. At t = 5 s, maximum percentage error in the energy is : (1) 6% (2) 8.4% (3) 11.6% (4) 4%
Q39.Given below are two statements : Statement I : In a vernier callipers, one vernier scale division is always smaller than one main scale division. Statement II : The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions. In the light of the above statements, choose the correct answer from the options given below. (1) Statement I is true but Statement II is false (2) Statement I is false but Statement II is true (3) Both Statement I and Statement II are false (4) Both Statement I and Statement II are true 2025 (22 Jan Shift 1) JEE Main Previous Year Paper
Q40.Which one of the following is the correct dimensional formula for the capacitance in F ? M, L, T and C stand for unit of mass, length, time and charge, (1) [F] = [C2M−1 L−2 T2] (2) [F] = [C2M−2 L2 T2] (3) [F] = [CM−2 L−2 T−2] (4) [F] = [CM−1 L−2 T2]
Q28.The position of a particle moving on x-axis is given by x(t) = A sin t + B cos2 t + Ct2 + D , where t is time. The dimension of ABC is D (1) L2T −2 (2) L2 (3) L (4) L3 T−2
Q45.For an experimental expression y = 32.3×112527.4 , where all the digits are significant. Then to report the value of y we should write (1) y = 1326.19 (2) y = 1330 (3) y = 1326.186 (4) y = 1326.2
NCERT Chapters
- Class 11 Physics Ch 2: Units and Measurements