Free Measurements MCQs with Answers

211. 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.

212. 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.

213. 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

214. 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.

215. 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.

216. 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.

217. A student makes measurements from which she calculated the speed of sound as 321 ms-1 . She estimates that the result is accurate to +3% or -3% . Which of the following give results expressed to the appropriate number of significant figures?

• A. 327 ms-1
• B. 328 ms-1
• C. 330 ms-1
• D. 300 ms-1

Explanation: a) 327 m/s:This option provides a result with three significant figures. However, the given measurement is accurate to +3% or -3%, which means the uncertainty in the measurement is ±3%. Therefore, the appropriate number of significant figures should consider the uncertainty, and in this case, it should be expressed as 321 ± 3 m/s. Rounded to three significant figures with the appropriate uncertainty, it would be written as 321 ± 3 m/s.

218. The density of a steel ball was determined by measuring its mass and diameter. The mass was measured within 1% and the diameter with 3%. The error in the calculated density of the steel ball is at most:

• A. 10%
• B. 4%
• C. 3%
• D. 2%

Explanation: b) 4%:This option suggests that the maximum error in the calculated density of the steel ball is 4%. While this seems reasonable, let's examine the errors in measuring the mass and diameter to confirm if this is accurate.

219. Five energies are listed.5 KJ5mJ 5MJ 5nJStarting with the smallest first, what is the order of increasing magnitude of these energies?

• A. 5nJ -> 5 mJ -> 5 kJ -> 5 MJ
• B. 5 nJ -> 5 kJ -> 5 MJ -> 5 mJ
• C. 5 kJ -> 5 mJ -> 5 MJ -> 5 nJ
• D. 5 mJ -> 5 nJ -> 5 kJ -> 5 MJ

Explanation: In order to solve this you must know the value and symbol of the number prefix. (Refer to the table below)5 kJ = 5 x10^3J5 mJ = 5 x10^-3 J5 MJ = 5 x10^6 J5 nJ = 5 x10^-9 JThe order of increasing magnitude will be 5 x10^-9 J < 5 x10^-3 J < 5 x10^3J < 5 x 10^6 J5 nJ < 5 mJ < 5 kJ < 5 MJ

220. Michelson measured the length of standard metre in terms of the wavelength of:

• A. Green cadmium light
• B. Violet cadmium light
• C. Red cadmium light
• D. Blue cadmium light

Explanation: Michelson measured the length of the standard meter using the interference fringes produced by red cadmium light. He used a device called an interferometer to measure the wavelength of the light and used it to determine the length of the standard meter.The interferometer creates interference patterns by splitting a beam of light into two paths, allowing them to recombine and create an interference pattern. By measuring the fringe pattern and knowing the wavelength of the light source, the length of the standard meter can be determined.