Units And Measurement

Both the options A and B are Correct.

Step 1: Identify the least count (L.C) of the Vernier calipers
The least count is the smallest measurement that can be accurately read using the instrument. It is given as:
L.C = 0.01 cm
This means the Vernier calipers can measure up to one-hundredth of a centimeter accurately.

Step 2: Account for the systematic error
Sometimes, the instrument may have a zero error or a systematic error due to misalignment or calibration issues. In this case, it is given that there is an error equivalent to 4 least counts:
Error = 4 × L.C = 4 × 0.01 cm = 0.04 cm
This error needs to be added to the observed measurement to obtain the true value.

Step 3: Determine the observed reading
The observed measurement from the Vernier calipers consists of two parts:
- Main scale reading = 3.1 cm
- Vernier scale reading = 1 division = 1 × 0.01 cm = 0.01 cm
So, Observed reading = 3.1 cm + 0.01 cm = 3.11 cm

Step 4: Add the error to get the corrected value
To get the correct diameter of the sphere, the error is added to the observed value:
Corrected diameter = 3.11 cm + 0.04 cm = 3.15 cm

Final Answer:
Therefore, the correct diameter of the sphere is 3.15 cm

Correct option: Option (C)

LC=1000.1​=0.001mm
Zero error =4×0.001=0.004mm
Reading 1=0.5×4+20×0.001−0.004=2.16mm
Reading 2=0.5×4+16×0.001−0.004=2.12mm
Mean value =2.14mm
Mean absolute error =
Diameter =
Area =
Therefore correct answer is

• Length: 10.5 cm
• Breadth: 0.05 mm = 0.05 / 10 cm = 0.005 cm
• Thickness: 6.0 m = 6.0 / m = 6.0 / cm = 0.0006 cm

• Volume = Length Breadth Thickness
• Volume = 10.5 cm 0.005 cm 0.0006 cm = 3.15 cm

• Length: 10.5 cm (3 significant figures)
• Breadth: 0.005 cm (1 significant figure)
• Thickness: 0.0006 cm (1 significant figure)
• When multiplying, the result should have the same number of significant figures as the measurement with the \textit{least} number of significant figures. In this case, that's 1.

• 3.15 cm rounded to 1 significant figure is 3 cm