Percentage error in measurement of displacement= 5/200x100 Percentage error in measurement of time = 0.2/20x100 Maximum permissible error = 2.5 + 1 = 3.5%
Percentage error in measurement of displacement= 5/200×100
Percentage error in measurement of time = 0.2/20×100
Maximum permissible error = 2.5 + 1 = 3.5%
When the planet Jupiter is at a distance of 824.7 million km from the earth, its angular diameter is measured to be 35.72′′ of arc. Calculate diameter of Jupiter.
θ = 35.72′′ 1′′ = 4.85 × 10⁻⁶ radian ⇒ 35.72′′ = 35.72 × 4.85 ×10⁻⁶ rad. d = DQ = 824.7 × 35.72 × 4.85 × 10⁻⁶ = 1.4287 × 10⁵ km
θ = 35.72′′
See less1′′ = 4.85 × 10⁻⁶ radian ⇒ 35.72′′ = 35.72 × 4.85 ×10⁻⁶ rad.
d = DQ = 824.7 × 35.72 × 4.85 × 10⁻⁶
= 1.4287 × 10⁵ km
A physical quantity is measured as Q = (2.1 ± 0.5) units. Calculate the percentage error in (1) Q2 (2) 2Q.
P = Q2 ∆p/p= 2∆Q/Q = 2(0.5/2.1) = 1.0/2.1 = 0.476 ∆p/p x 100% = 47.6% = 48% R=2Q ∆R/R=∆Q/Q ⇒ 0.5/2.1 = 0.238 ∆R/R x 100% = 24%
P = Q2
See less∆p/p= 2∆Q/Q = 2(0.5/2.1) = 1.0/2.1 = 0.476
∆p/p x 100% = 47.6% = 48%
R=2Q
∆R/R=∆Q/Q ⇒ 0.5/2.1 = 0.238
∆R/R x 100% = 24%
The length of a rod as measured in an experiment was found to be 2.48m, 2.46m, 2.49m, 2.50m and 2.48m. Find the average length, absolute error and percentage error. Express the result with error limit.
Average length= 2.48+2.46+2.49+2.50+2.48/5 = 12.41/5 = 2.48m Mean absolute error = 0.00+0.02+0.01+0.02+0.00/5 = 0.05 = 0.013m Percentage error = 0.01/2.48x100%=0.04x100% = 0.40% Correct length = (2.48 ± 0.01)m Correct length = (2.48m ± 0.40%)
Average length= 2.48+2.46+2.49+2.50+2.48/5 = 12.41/5 = 2.48m
See lessMean absolute error
= 0.00+0.02+0.01+0.02+0.00/5 = 0.05 = 0.013m
Percentage error = 0.01/2.48×100%=0.04×100%
= 0.40%
Correct length = (2.48 ± 0.01)m
Correct length = (2.48m ± 0.40%)
If displacement of a body s = (200 ± 5) m and time taken by it t = (20 + 0.2) s, then find the percentage error in the calculation of velocity.
Percentage error in measurement of displacement= 5/200x100 Percentage error in measurement of time = 0.2/20x100 Maximum permissible error = 2.5 + 1 = 3.5%
Percentage error in measurement of displacement= 5/200×100
See lessPercentage error in measurement of time = 0.2/20×100
Maximum permissible error = 2.5 + 1 = 3.5%
5.74 g of a substance occupies 1.2 cm³. Express its density to correct significant figures.
Density = Mass/Volume = 5.74/1.2 = 4.783g/cm³ Here least significant figure is 2, so density = 4.8 g/cm³.
Density = Mass/Volume = 5.74/1.2 = 4.783g/cm³
See lessHere least significant figure is 2, so density = 4.8 g/cm³.