Physical Quantities and Measurements

Answer:
Parallax error occurs when a scale is viewed from an angle instead of directly in front. The pointer, liquid level or edge of an object then appears to coincide with different scale marks depending on the observers position. This produces a reading larger or smaller than the true reading. It can be avoided by placing the eye directly above the ruler or at the same level as the pointer or meniscus, with the line of sight perpendicular to the scale.

Answer:
Vernier callipers consist of a fixed main scale and a movable Vernier scale. The main scale commonly has millimetre divisions, while the Vernier scale divides a shorter length into equal parts to provide greater precision. The outside jaws measure external dimensions such as thickness or diameter. The inside jaws measure internal diameter, and the projecting depth gauge measures depth. Vernier callipers are used for measurements too small or precise for an ordinary ruler, commonly down to 0.1 mm.

Answer:
The least count of Vernier callipers is the difference between one main-scale division and one Vernier-scale division. If one main-scale division is 1 mm and ten Vernier divisions occupy 9 mm, one Vernier division equals 0.9 mm. Therefore, least count = 1.0 mm - 0.9 mm = 0.1 mm. It may also be calculated by dividing the value of one main-scale division by the total number of Vernier divisions.

Answer:
First close the jaws and check for zero error. Place the object gently between the appropriate jaws. Note the main-scale reading immediately before the zero of the Vernier scale. Find the Vernier division that exactly coincides with a main-scale mark. Multiply this division number by the least count and add it to the main-scale reading. Finally apply zero correction: subtract a positive zero error or add the magnitude of a negative zero error. Record the result with the proper unit.

Answer:
Zero error exists when the main-scale and Vernier-scale zeros do not coincide with the jaws closed. If the Vernier zero lies to the right of the main-scale zero, the instrument reads more than the true value; this is positive zero error and it is subtracted from the observed reading. If the Vernier zero lies to the left, the instrument reads less than the true value; this is negative zero error and its magnitude is added. Applying this correction gives the actual measurement.

Answer:
A micrometer screw gauge contains a frame, anvil, spindle, sleeve or main scale, rotating thimble with circular scale and ratchet. The object is placed between the anvil and spindle. Rotating the thimble moves the spindle because rotational motion is converted into a small linear displacement by the screw. The sleeve gives the main-scale reading and the thimble gives the circular-scale reading. The ratchet prevents excessive tightening and applies uniform pressure. The instrument measures very small lengths such as wire diameter and sheet thickness.

Answer:
Pitch is the distance moved by the spindle during one complete rotation of the thimble. Least count is calculated by dividing the pitch by the total number of circular-scale divisions. If the spindle moves 0.5 mm per rotation and the circular scale has 50 divisions, least count = 0.5/50 = 0.01 mm. A smaller least count allows more precise measurements.

Answer:
First close the anvil and spindle to check zero error. Place the object between them and rotate the ratchet until it clicks. Note the main-scale reading visible just before the thimble edge. Note the circular-scale division aligned with the reference line. Multiply the circular-scale reading by the least count and add it to the main-scale reading. Apply the correct zero correction and record the result with its unit.

Answer:
When the anvil and spindle touch, the zero of the circular scale should align with the reference line. If it lies below the line, the instrument gives a reading greater than the actual value, so the positive zero error is subtracted. If it lies above the line, the instrument gives a smaller reading, so the magnitude of the negative zero error is added. The corrected reading represents the actual dimension.

Answer:
Mass is the amount of matter in a body and its SI unit is kilogram. Weight is the gravitational force acting on the body and its SI unit is newton. A physical balance measures mass by comparison with known standard masses. First level and balance the instrument. Place the object on the left pan and standard masses on the right pan using forceps. Adjust the masses until the pointer remains at zero or oscillates equally on both sides. The sum of the standard masses equals the objects mass.