Units and Measurement
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Plane angle and solid angle are dimensionless but have units.
The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are:
The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are:
As the factors controlling temperature and voltage supply are beyond prediction and control so the error occurred due to unpredictable fluctuations of temperature and voltage would be random errors.
The radius of a sphere is measured to be (7.50±0.85) cm. Suppose the percentage error in its volume is x. The value of x, to the nearest integer, is.
The radius of a sphere is measured to be (7.50±0.85) cm. Suppose the percentage error in its volume is x. The value of x, to the nearest integer, is.
V = (4/3)πr³
⇒ ΔV/V = 3 (Δr/r) ⇒ Relative error
% error in volume = (ΔV/V) × 100 = 3 (Δr/r) × 100 = 3 × (0.85/7.50) × 100 = 33.999 ≈ 34
The time period of a simple pendulum is given by T = 2π√(l/g). The measured value of the length of pendulum is 10cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to the nearest integer is,
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The time period of a simple pendulum is given by T = 2π√(l/g). The measured value of the length of pendulum is 10cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to the nearest integer is,
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l = 10.0 ± 0.1cm.
T = (100 ± 1) / 200 = 0.5 ± 0.005
The vernier-scale used for measurement has a positive zero error of 0.2 mm. If while taking a measurement it was noted that '0' on the vernier-scale lies between 8.5 cm and 8.6 cm, vernier coincidence is 6, then the correct value of measurement is …………… cm. (least count = 0.01 cm)
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The vernier-scale used for measurement has a positive zero error of 0.2 mm. If while taking a measurement it was noted that '0' on the vernier-scale lies between 8.5 cm and 8.6 cm, vernier coincidence is 6, then the correct value of measurement is …………… cm. (least count = 0.01 cm)
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Positive zero error = 0.02 cm
Reading = 8.5 + 6 * 0.01 = 8.56 cm
Actual reading = Reading - Positive zero error
Actual reading = 8.56 - 0.02 = 8.54 cm
A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm; 5.65 mm. The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as:
A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm; 5.65 mm. The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as:
Since significant figures show the degree of correctness of any measurement, so in any mathematical calculation we cannot increase the number of significant digits. Because the four reading has 3 significant digits so the answer should also have 3 significant digits only.
A quantity X is given by (IFv2MLL4) in terms of moment of inertia I, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of:
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A quantity X is given by (IFv2MLL4) in terms of moment of inertia I, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of:
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x = (IFV²)/ (WL? )
I = [ML²]
F = [MLT? ²]
V² = [L² T? ²]
W = [ML² T? ²]
Q? = [L? ]
X = [ML? ¹ T? ²]
Dimensional formula for thermal conductivity is (here K denotes the temperature):
Dimensional formula for thermal conductivity is (here K denotes the temperature):
K = (Q?)Δx / AΔT
⇒ (ML²T?²)(L) / (L²)(θ)(T)
⇒ M¹L¹T?³θ?¹
