Free Physical Quantities and Units MCQs with Answers

1. Supplementary S.I units of radian and steradian were established to measure

• A. Geometrical quantities
• B. Luminous intensity
• C. Electric current
• D. Temperature

Explanation: his is correct. The radian and steradian are indeed used to measure specific geometrical concepts:Radian: measures plane angles.Steradian: measures solid angles

2. The option that shows three base quantities' units in System International (SI) is

• A. Gram, joule, and second.
• B. Gram, centimetre, and dyne.
• C. Kilogram, metre, and second.
• D. Kilogram, newton, and second.

Explanation: The correct answer is Option C: 'Kilogram, metre, and second'. These are the base units in the SI system for the fundamental quantities of mass, length, and time. Option D is incorrect because 'newton' is a derived unit of force. Options A and B list units that are either derived or belong to the CGS system, which are not considered SI base units.

3. The least count of a metre scale is

• A. 0.001 cm.
• B. 0.1 mm.
• C. 1 mm.
• D. 1 cm

Explanation: The least count of a measuring instrument is the smallest measurement that can be accurately read. For a standard meter scale, the smallest division is 1 millimeter (mm), making the least count 1 mm. Option A, 0.001 cm, suggests an unrealistic precision not possible with a regular meter scale. Option B, 0.1 mm, is also not achievable with a typical meter scale. Option D, 1 cm, is too large to be the least count, as meter scales have finer divisions.

4. The option that shows three base quantities' units in System International (S.I.) is

• A. Gram, joule and second.
• B. Gram, centimetre and dyne.
• C. Kilogram, metre and second
• D. Kilogram, newton and second,

Explanation: The correct answer is Option C:kilogram, metre and second. These are the base units in the SI system for mass, length, and time, respectively. Option D is incorrect because although kilogram and second are base units, newton is not; it is a derived unit. Options A and B list units from the CGS system and include joules and dynes, which are also derived units.

5. A bag contains 50 metallic spheres of different diameters. A task is given to students to classify these spheres in terms of their diameters and find the number of spheres that liebetween the range of 0.3 cm and 1.5 cm out of 50 metallic spheres.For classifying as accurately as possible, the MOST appropriate instrument the students should pick from the science lab is

• A. spherometre.
• B. screw gauge.
• C. measuring tape.
• D. Vernier callipers.

Explanation: The correct answer is the spherometre because it is designed to measure spherical objects accurately, making it the most appropriate instrument for classifying metallic spheres based on their diameters. Screw gauge, measuring tape, and Vernier callipers are not as suitable for measuring spherical objects and may not provide the accuracy needed for this task.

6. The given image shows a measuring instrument.On the main scale of this instrument, the purpose of markings above and below the zero mark is to measure theI. depth.II. height.III. thickness.

• A. I only
• B. III only
• C. I and II
• D. II and III

Explanation: A spherometer is a precision instrument primarily used to measure: * The radius of curvature of spherical surfaces, such as lenses and mirrors. * The thickness of small plates or objects. * The depth of depressions in a surface.The purpose of the markings above and below the zero mark on the main scale of a spherometer is to measure the vertical displacement of the central screw relative to the plane formed by its three outer legs.

7. The question is given below:

• A. A
• B. B
• C. C
• D. D

Explanation: L = n * (λ/4)where 'n' represents the harmonic number (n = 1, 3, 5, ... for the first, second, third, etc., resonances).Let's analyze each length shown in the image: * For n = 1 (First Resonance - L1): The image correctly shows L1 as the first resonance, where a node is at the water surface and an antinode is at the open end. L1 = 1 * (λ/4) = λ/4 This matches the value given in the diagram. * For n = 2 (Second Resonance - L2): The "second resonance" in this context refers to the second possible resonant length for the closed-end tube, which corresponds to the third harmonic (n=3). Here, there will be two nodes and two antinodes, or generally (2n-1) nodes and antinodes. L2 = 3 * (λ/4) = 3λ/4 * For n = 3 (Third Resonance - L3): The "third resonance" refers to the third possible resonant length, which corresponds to the fifth harmonic (n=5). L3 = 5 * (λ/4) = 5λ/4

8. The given diagram shows two different scales on a Vernier callipers.The reading on the main scale is

• A. 6 cm
• B. 6.5 cm
• C. 7 cm
• D. 7.4 cm

Explanation: To determine the reading on the main scale of the Vernier caliper shown in the diagram, we need to observe the position of the zero mark of the Vernier scale relative to the main scale.The main scale is marked in centimeters (cm) and millimeters (mm). Each major division represents 1 cm, and the smaller divisions represent 1 mm (or 0.1 cm).Upon close inspection of the image: * The zero mark of the Vernier scale is located past the 6 cm mark. * It has also passed the 6.1 cm, 6.2 cm, 6.3 cm, and 6.4 cm marks. * However, the zero mark has not yet reached the 6.5 cm mark.In standard Vernier caliper operation, the main scale reading is the value on the main scale immediately to the left of the zero mark of the Vernier scale. Based on this, the main scale reading would be 6.4 cm.

9. The given figure shows two different scales on a screw gauge

• A. 35
• B. 38
• C. 40
• D. 45

Explanation: To find the measurement using a screw gauge, you need to add the reading from the main scale to the reading from the circular scale. If the main scale shows 30 and the circular scale shows 8, the total measurement is 38. Option B is correct because it accurately combines these readings. Options A, C, and D are incorrect due to errors in reading or combining the scales.

10. In daily life, the relative change in mass, length and time of an object cannot be observed because the

• A. size of the object is very small.
• B. mass of the object is very large.
• C. speed of the object is equal to the speed of light
• D. speed of the object is much smaller than that of light.

Explanation: The relative change in mass, length, and time of an object can be observed when the speed of the object is much smaller than that of light. This is because at such speeds, relativistic effects are insignificant, allowing for the direct observation of these changes. The other options are incorrect because they do not address the impact of the object's speed on observing these relative changes.