Exams › JEE Main › Physics
In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:
- 0.0430 cm
- 0.3150 cm
- 0.4300 cm
- 0.2150 cm
Correct answer: 0.2150 cm
Solution
The thickness of the wire can be calculated using the formula: thickness = (main scale reading + circular scale reading) × least count. The least count is determined by dividing the total movement of the screw (0.25 cm) by the number of circular scale divisions (100), which gives 0.0025 cm per division. Thus, the total thickness is (4 × 0.025 cm) + (30 × 0.0025 cm) = 0.2150 cm.
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 →