Free Measurements MCQs with Answers

1. The second equation of motion of a uniformly accelerated ball along a straight line is stated ass = vft +1/2 at2The dimension of this equation is

• A. [L]
• B. [T]
• C. [L][T-1]
• D. [L][T-2]

Explanation: The second equation of motion is s = vft + 1/2 at2. The dimension of the first term vft is [L][T-1][T] = [L]. The dimension of the second term 1/2 at2 is [L][T-2][T2] = [L]. Since both terms have the same dimensions, the overall dimension of the equation is [L]. Option D is therefore the correct answer. Options A, B, and C are incorrect as they do not represent the dimensional analysis of the entire equation.

2. Which of the following shows the dimensions of force and acceleration respectively?

• A. MLT-2 and LT-2
• B. ML2 T and LT-2
• C. MLT-2 and LT2
• D. MLT-2 and LT-1

Explanation: Force (F) is defined by Newton's second law as mass (m) multiplied by acceleration (a) (F = ma).* The dimension of mass is [M] (Mass).* The dimension of acceleration is [L][T]^-2 (Length per time squared).Therefore, the dimension of Force is [M][L][T]^-2.Dimensions of Acceleration:Acceleration (a) is the rate of change of velocity, which is velocity per unit time. Velocity is defined as displacement per unit time.* The dimension of displacement (length) is [L].* The dimension of time is [T].* So, the dimension of velocity is [L][T]^-1.* Therefore, the dimension of acceleration is [L][T]^-1 / [T] = [L][T]^-2.

3. The second equation of motion of a uniformly accelerated ball along a straight line is stated asS=vit+ ½ at2The dimension of each term of this equation is

• A. [L]
• B. [T]
• C. [L-1]
• D. [T-1]

Explanation: The second equation of motion is given by S = vit + ½ at2, where S is the displacement, vi is the initial velocity, a is the acceleration, and t is the time. The dimension of displacement (S) is length [L]. Both terms vit and ½ at2 in the equation must also have the dimension of length to ensure dimensional consistency. Therefore, the correct dimension for S is [L]. The other options are incorrect because they do not correspond to the dimension of displacement.

4. The dimensions of momentum are represented as

• A. [MLT-1]
• B. [ML2T-1]
• C. [MLT-2]
• D. [ML-1T]

Explanation: Momentum is defined as the product of mass and velocity, so its dimensions are given by multiplying the dimensions of mass (M) and velocity (LT-1). Therefore, the dimension of momentum is [MLT-1]. Option A correctly represents these dimensions. Option B [ML2T-1] could refer to a different physical concept like angular momentum, Option C [MLT-2] represents force, and Option D [ML-1T] does not correspond to a recognized physical quantity.

5. The length of a strip is measured by a metre scale as 3.4 cm. The reading measured on the Vernier callipers for the length of the strip will be

• A. 0.34 cm.
• B. 3.4 cm.
• C. 3.40 cm.
• D. 3.400 cm.

Explanation: The correct answer is 3.400 cm. Vernier calipers provide more precise measurements than a standard metre scale, typically offering precision to the thousandths of a centimeter. While a metre scale measures to one decimal place (3.4 cm), a Vernier caliper extends this precision, recording the measurement as 3.400 cm to reflect the additional accuracy. The other options either reduce the precision or fail to extend it to the level provided by Vernier calipers.

6. Random errors can be reduced byI. taking zero correctionII. taking mean of several measurementsIII. comparing the instrument with a more accurate one

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

Explanation: Option II is correct because taking the mean of several measurements helps to minimize the effect of random errors by averaging out the variations that occur in repeated measurements.Option I is incorrect because zero correction is used to address systematic errors, which are consistent and predictable, rather than random errors.Option III is incorrect because comparing the instrument with a more accurate one helps identify and correct systematic errors, not random errors.Therefore, only taking the mean of several measurements (Option II) is effective in reducing random errors.

7. The number of significant figures in the sum of 1.21 x 10^5 and 3.26 x 10^3 is

• A. 3
• B. 5
• C. 6
• D. 8

Explanation: To find the number of significant figures in the sum of 1.21 × 105 and 3.26 × 103, first, convert both numbers to the same power of ten: 1.21 × 105 = 121000 and 3.26 × 103 = 3260. Add them to get 124260. When written in scientific notation, this becomes 1.24260 × 105, which has 8 significant figures. Thus, Option D is correct, while the other options underestimate the number of significant figures due to incorrect alignment or rounding.

8. A systematic error can be removed by

• A. Taking mean reading.
• B. Repeating the reading.
• C. Changing the surrounding.
• D. Recalibrating the instrument.

Explanation: Systematic errors are consistent biases in measurement that arise from flaws in the measurement system, such as miscalibrated instruments. To remove systematic errors, recalibration of the instrument is necessary, ensuring that it provides accurate readings. Taking mean readings or repeating measurements do not correct systematic errors because these methods only address random errors. Changing the surroundings may impact measurements but doesn't target the root cause of systematic errors.

9. Uncertainty in measurement can occur due to theI. limitation of instrumentsII. natural variationsIII. frequent errors

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

Explanation: Uncertainty in measurement can arise from multiple sources. Natural variations (II) refer to the inherent variability in the system being measured, which can lead to different results even if the procedure is repeated exactly. Frequent errors (III) refer to mistakes that occur during the measurement process, which can be due to human error or instrument calibration issues. Although the limitation of instruments (I) can also cause uncertainty, the correct combination of factors listed in the options that lead to uncertainty are II and III. Therefore, the correct answer is Option D.

10. Random errors can be reduced byI. taking zero correctionII. taking mean of several measurementsIII. comparing the instrument with another more accurate one

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

Explanation: The correct answer is Option A:II only.Random errors can be reduced by taking the mean of several measurements, as this averages out the unpredictable variations inherent in each measurement. Option III addresses systematic errors, as comparing with a more accurate instrument helps identify consistent deviations. Option I, zero correction, is also a method to correct systematic errors, not random errors.