Question 1:
easyA screw gauge gives the following reading while measuring diameter of a wire. Main Scale Reading = \(7\text{ mm}\), Circular Scale Reading = \(67\). Given that \(1\text{ mm}\) on main scale corresponds to \(100\text{ divisions}\) on circular scale. The diameter of the wire is:
Least count is \(LC = \frac{1\text{ mm}}{100} = 0.01\text{ mm}\). Total Reading is \(MSR + CSR \times LC = 7\text{ mm} + 67 \times 0.01\text{ mm} = 7.67\text{ mm}\).
Question 2:
easyA screw gauge gives the following readings when used to measure the diameter of a wire:
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1 mm on main scale corresponds to 100 divisions on the circular scale.
The diameter of the wire from the above data is
Least Count \(\frac{1\text{ mm}}{100} = 0.01\text{ mm} = 0.001\text{ cm}\. Diameter = MSR + (CSR times LC) = 0\text{ mm} + 52 \times 0.001\text{ cm} = 0.052\text{ cm}\.
Question 3:
easyThe pitch of a screw gauge is 1 mm and there are 100 divisions on circular scale. While measuring the thickness of a sheet, the main scale reads 1 mm and \(52^{\text{nd}}\) division on circular scale coincide with the reference line. The thickness of the sheet is
Least count \(LC = \frac{\text{Pitch}}{\text{Number of circular divisions}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}\). Thickness \(= \text{MSR} + \text{CSR} \times LC = 1 \text{ mm} + 52 \times 0.01 \text{ mm} = 1.52 \text{ mm} = 0.152 \text{ cm}\).
Question 4:
easyDashrath measures the length of a wire using a meter scale with a least count of 1 mm and finds it to be L = 75.0 cm. He also measures diameter of thin wire using a screw gauge with a least count of 0.01 mm and finds it to be d = 0.500 cm. He uses these measurements to calculate the volume of wire. The maximum percentage error in volume of wire is nearly
Volume of wire is \(V = \pi \left(\frac{d}{2}\right)^2 L ⇒ \frac{\Delta V}{V} = 2\frac{\Delta d}{d} + \frac{\Delta L}{L} = 2\left(\frac{0.001}{0.500}\right) + \frac{0.1}{75.0} \approx 0.53%\).
Question 5:
easyDashrath measures the length of a wire using a meter scale with a least count of \( 1\text{ mm} \) and finds it to be \( L = 75.0\text{ cm} \). He also measures diameter of thin wire using a screw gauge with a least count of \( 0.01\text{ mm} \) and finds it to be \( d = 0.500\text{ cm} \). He uses these measurements to calculate the volume of wire. The maximum percentage error in volume of wire is nearly
Volume \( V = \frac{\pi d^2 L}{4} ⇒ \frac{\Delta V}{V} = 2 \frac{\Delta d}{d} + \frac{\Delta L}{L} \). Here, \( \Delta d = 0.001\text{ cm} \) and \( \Delta L = 0.1\text{ cm} \). Thus, \( \frac{\Delta V}{V} = 2\left(\frac{0.001}{0.500}\right) + \frac{0.1}{75} = 0.4% + 0.13% = 0.53% \).
Question 6:
easyA screw gauge gives the following readings when used to measure the diameter of a wire
Main scale reading: \(0 \text{mm}\)
Circular scale reading: \(52 \text{divisions}\)
Given that \(1 \text{mm}\) on main scale corresponds to \(100 \text{divisions}\) on the circular scale. The diameter of the wire from the above data is:
[NEET 2020]
Pitch = \(1 \text{mm}\), Number of divisions = \(100\). Least Count (LC) = Pitch / Number of divisions = \(1 \text{mm} / 100 = 0.01 \text{mm}\). Total reading = MSR + (CSR \(\times\) LC) = \(0 \text{mm} + (52 \times 0.01 \text{mm}) = 0.52 \text{mm}\). Convert to cm: \(0.52 \text{mm} = 0.052 \text{cm}\).
Question 7:
easyA screw gauge has least count of \(0.01 \text{mm}\) and there are \(50 \text{divisions}\) in its circular scale. The pitch of the screw gauge is:
[NEET 2020]
Least Count (LC) = \(0.01 \text{mm}\), Number of divisions on circular scale = \(50\). The formula for LC is: LC = Pitch / Number of divisions. Therefore, Pitch = LC \(\times\) Number of divisions = \(0.01 \text{mm} \times 50 = 0.5 \text{mm}\).
Question 8:
easyA student measured the diameter of a small steel ball using a screw gauge of least count \(0.001 \text{cm}\). The main scale reading is \(5 \text{mm}\) and zero of circular scale division coincides with \(25 \text{divisions}\) above the reference level. If screw gauge has a zero error of \(-0.004 \text{cm}\), the correct diameter of the ball is :
[NEET 2018]
Least Count (LC) = \(0.001 \text{cm}\). Main Scale Reading (MSR) = \(5 \text{mm} = 0.5 \text{cm}\). Circular Scale Reading (CSR) = \(25 \text{divisions}\). Observed Reading = MSR + (CSR \(\times\) LC) = \(0.5 \text{cm} + (25 \times 0.001 \text{cm}) = 0.525 \text{cm}\). Correct Reading = Observed Reading - Zero Error = \(0.525 \text{cm} - (-0.004 \text{cm}) = 0.525 \text{cm} + 0.004 \text{cm} = 0.529 \text{cm}\).