Question

The internal and external diameters of a hollow cylinder measured with vernier calipers are (5.73 \(\pm\) 0.01) cm and (6.01 \(\pm\) 0.01) cm respectively. Then the thickness of the cylinder wall is

• A. (0.28 \(\pm\) 0.01) cm
• B. (0.28 \(\pm\) 0.02) cm
• C. (0.14 \(\pm\) 0.02) cm
• D. (0.14 \(\pm\) 0.01) cm \bigskip

Hint:

The thickness of a hollow cylinder is found by subtracting the internal diameter from the external diameter and dividing by 2. When dealing with errors, always add the absolute uncertainties before division. This ensures the correct final uncertainty.

Answer 1

The Correct Option is C

The thickness of the wall is calculated as the difference between the external and internal radii. The internal diameter \( d_1 = 5.73 \) cm and external diameter \( d_2 = 6.01 \) cm, so the internal radius is \( r_1 = \frac{d_1}{2} = 2.865 \) cm and the external radius is \( r_2 = \frac{d_2}{2} = 3.005 \) cm. The thickness of the wall is: \[ t = r_2 - r_1 = 3.005 - 2.865 = 0.14 \, \text{cm}. \] The uncertainties are added, so the total uncertainty in thickness is \( \pm 0.02 \) cm. Thus, the thickness is \( 0.14 \pm 0.02 \) cm.