Measurements and Experimentation Quiz - ICSE Class 9 Physics
Free practice quiz: Measurements and Experimentation (ICSE Class 9 Physics). 53 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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A screw gauge has a pitch of 0.5 mm and a circular scale with 100 divisions. What is the least count of the screw gauge?
- 0.005 mm
- 0.05 mm
- 0.0005 mm
- 0.5 mm
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Answer: A. 0.005 mm
The least count of a screw gauge is calculated by dividing the pitch by the number of divisions on the circular scale. Least Count = Pitch / Number of divisions on circular scale = 0.5 mm / 100 = 0.005 mm.
The pitch of a screw gauge is 0.5 mm. If there are 100 divisions on the circular scale, what is the smallest distance that can be measured accurately?
- 0.05 mm
- 0.005 mm
- 0.5 mm
- 0.1 mm
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Answer: B. 0.005 mm
The smallest distance that can be measured accurately is the least count. Least Count = Pitch / Number of divisions on circular scale = 0.5 mm / 100 = 0.005 mm.
Which instrument is most suitable for measuring the diameter of a thin wire with high precision?
- Meter scale
- Measuring tape
- Screw gauge
- Spring balance
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Answer: C. Screw gauge
A screw gauge is specifically designed to measure small lengths, like the diameter of a thin wire or the thickness of a sheet, with a high degree of precision (typically up to 0.01 mm). A meter scale or measuring tape has much lower precision, and a spring balance measures force.
When measuring the length of a pencil, if the zero of the vernier scale is to the right of the zero of the main scale, the zero error is:
- Negative
- Positive
- Zero
- Either positive or negative
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Answer: B. Positive
If the zero of the vernier scale is to the right of the zero of the main scale when the jaws are closed, it means the instrument reads more than the actual value. This is a positive zero error.
A vernier caliper has 100 divisions on its vernier scale, and it is designed to measure up to 10 cm with a least count of 0.01 cm. How many divisions are there on its main scale for 1 cm?
- 10 divisions
- 100 divisions
- 50 divisions
- 20 divisions
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Answer: A. 10 divisions
Least Count = Value of 1 MSD - Value of 1 VSD. We know LC = 0.01 cm. Also, 100 VSD = 99 MSD (standard configuration for this LC and range). Value of 1 MSD = 1 cm / (number of divisions on main scale for 1 cm). Let this be 'm'. Then Value of 1 VSD = (99/100) * m. So, 0.01 cm = m - (99/100) * m = m * (1 - 99/100) = m * (1/100). Therefore, m = 0.01 cm * 100 = 1 cm. This means there is 1 division on the main scale for every 1 cm. Thus, for 1 cm of main scale length, there is 1 division on the main scale.