Exams › JEE Main › Physics
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?
- 0.70 mm
- 0.50 mm
- 0.75 mm
- 0.80 mm
Correct answer: 0.80 mm
Solution
Least count = pitch/divisions = 0.5/50 = 0.01 mm. With jaws closed the 45th division coincides and main-scale zero is just visible -> negative zero error = -(50-45)*0.01 = -0.05 mm. Observed reading = 0.5 + 25*0.01 = 0.75 mm; true thickness = 0.75 - (-0.05) = 0.80 mm.
Related JEE Main Physics questions
- A body has a measured mass of 5.00 ± 0.05 kg and a measured volume of 1.00 ± 0.05 m³. What is the maximum possible percentage error in its density?
- A substance has a density of 4 g/cm³ in the CGS system. If a new unit system is used where the unit of length is 10 cm and the unit of mass is 100 g, what is the numerical value of the density in that system?
- Which of the following physical quantities has dimensions that do not match the other three?
- If the percentage uncertainties in measuring mass M, length L, and time T are 1%, 1.5%, and 3% respectively, what is the percentage error in a quantity having dimensions M L⁻¹ T⁻¹?
- If E, m, J and G denote energy, mass, angular momentum and the gravitational constant, respectively, then the dimensional formula of EJ²/(m⁵G²) is identical to that of which quantity?
- A rod was found to have a measured length of 3.50 cm. Which measuring device was used for this reading?
⚔️ Practice JEE Main Physics free + battle 1v1 →