Measurement Question

Document Title

1. The diagram below shows the measurement of a diameter of a wire.

                                    

However, its actual diameter is 2.45 mm. What is the zero error?

(A) –0.14 mm
(B) +0.14 mm
(C) –0.36 mm
(D) +0.36 mm

Analysis:
From Micrometer Screw Gauge, the reading is 2.5 + 0.09 = 2.59mm
However, the actual reading is only 2.45mm. Therefore, there is a positive zero error of 2.59 - 2.45 = +0.14mm. So, answer is B.

2. Diagram 1 shows the scale when the vernier calipers is closed.
Diagram 2 shows the scale when it is used to measure the diameter of a steel ball bearing.

                  

What is the radius of the ball bearing?

(A) 2.01 cm
(B) 2.03 cm
(C) 4.01 cm
(D) 4.05 cm

Analysis:
From diagram 1, there is a negative zero error as the zero mark on vernier scale is slightly to the left of the zero mark on main scale. Recall that negative zero error has to count from 10 on vernier scale. So, negative zero error = -0.02cm
From diagram 2, the reading is 4.03cm.
So, actual reading for diameter of steel ball is 4.03 - (-0.02) = 4.05cm
However, look at the question carefully. It asks for radius of the steel ball. So radius = 4.05 / 2 = 2.03cm, which is B
Comment: Many students didnt look at the question carefully, and choose D as the answer!

3. Which of the following will cause the period of a pendulum to be doubled?
(A) Doubling the mass.
(B) Doubling the length.
(C) Doubling the acceleration of free fall.
(D) Swinging it at double the angle.

Analysis:
This question is a recall question. If you manage to remember the formula for the period of pendulum as T = sqrt(2∏ L/g), you should know the answer is B.

Excel@Physics