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Units and Measurements question

2025 · Shift 1 · Q36
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Units and Measurements question

2025 · Shift 1 · Q36

JEE AdvancedPhysicsUnits and MeasurementsMCQ+3 / −1
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter D of a tube. The measured value of D is: JEE Advanced 2025 Paper 1 Online Physics - Units & Measurements Question 2 English
  1. A
    0.12 cm
  2. B
    0.11 cm
  3. C
    0.13 cm
  4. D
    0.14 cm
View written solutionFree

Correct answer: C

The solution involves reading the Vernier calipers, accounting for the zero error, to find the correct measurement. This is done in four steps:

Step 1: Determine the Least Count (LC) of the Vernier Calipers.

The main scale is graduated in millimeters (mm). Thus, 1 Main Scale Division (MSD) = 1 mm = 0.1 cm. The Vernier scale has 10 divisions. From the figures, it's clear that 10 divisions on the Vernier scale coincide with 9 divisions on the main scale. Therefore, 10 Vernier Scale Divisions (VSD) = 9 Main Scale Divisions (MSD). This gives us 1 VSD = (9/10) MSD = 0.9 MSD = 0.9 mm. The least count (LC) is the difference between one MSD and one VSD. LC=1 MSD−1 VSDLC = 1 \text{ MSD} - 1 \text{ VSD}LC=1 MSD−1 VSD LC=1 mm−0.9 mm=0.1 mmLC = 1 \text{ mm} - 0.9 \text{ mm} = 0.1 \text{ mm}LC=1 mm−0.9 mm=0.1 mm Converting to centimeters: LC=0.01 cmLC = 0.01 \text{ cm}LC=0.01 cm

Step 2: Determine the Zero Error from Figure 1.

Figure 1 shows the reading when the jaws of the Vernier calipers are closed. The zero mark of the Vernier scale is to the right of the zero mark of the main scale, which indicates a positive zero error. To find the magnitude of the error, we identify the Vernier scale division that coincides perfectly with any main scale division. Looking closely at Figure 1, the 5th division of the Vernier scale aligns with a division on the main scale. The zero error is calculated as: Zero Error=+(Coinciding Vernier Division)×LC\text{Zero Error} = + (\text{Coinciding Vernier Division}) \times LCZero Error=+(Coinciding Vernier Division)×LC Zero Error=+5×0.01 cm=+0.05 cm\text{Zero Error} = + 5 \times 0.01 \text{ cm} = +0.05 \text{ cm}Zero Error=+5×0.01 cm=+0.05 cm

Step 3: Determine the Observed Reading from Figure 2.

Figure 2 shows the instrument measuring the diameter D.

  1. Main Scale Reading (MSR): This is the reading on the main scale immediately to the left of the zero mark of the Vernier scale. The zero of the Vernier scale is just past the 1 mm mark. MSR=1 mm=0.1 cm\text{MSR} = 1 \text{ mm} = 0.1 \text{ cm}MSR=1 mm=0.1 cm
  2. Vernier Scale Coincidence (VSC): This is the division on the Vernier scale that aligns perfectly with a division on the main scale. Observing Figure 2, the 8th division of the Vernier scale coincides with a main scale division. VSC=8\text{VSC} = 8VSC=8
  3. Observed Reading: The total observed reading is given by the formula: Observed Reading=MSR+(VSC×LC)\text{Observed Reading} = \text{MSR} + (\text{VSC} \times \text{LC})Observed Reading=MSR+(VSC×LC) Observed Reading=0.1 cm+(8×0.01 cm)\text{Observed Reading} = 0.1 \text{ cm} + (8 \times 0.01 \text{ cm})Observed Reading=0.1 cm+(8×0.01 cm) Observed Reading=0.1 cm+0.08 cm=0.18 cm\text{Observed Reading} = 0.1 \text{ cm} + 0.08 \text{ cm} = 0.18 \text{ cm}Observed Reading=0.1 cm+0.08 cm=0.18 cm

Step 4: Calculate the Corrected Reading (True Value).

The true value of the diameter D is found by subtracting the zero error from the observed reading. Corrected Reading=Observed Reading−Zero Error\text{Corrected Reading} = \text{Observed Reading} - \text{Zero Error}Corrected Reading=Observed Reading−Zero Error D=0.18 cm−(+0.05 cm)D = 0.18 \text{ cm} - (+0.05 \text{ cm})D=0.18 cm−(+0.05 cm) D=0.13 cmD = 0.13 \text{ cm}D=0.13 cm

The measured value of the diameter D is 0.13 cm. This matches option C.

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