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Gervais Mills
5 years ago
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1 Measurements and Errors A capillary tube is attached horizontally to a constant heat arrangement. If the radius of the capillary tube is increased by 10%, then the rate of flow of liquid will change nearly by a) +10% b) +6% c) -10% d) -0%. If momentum is increased by 0%, then kinetic energy increases by a) 8% b) % c) 0% d) 6%. If increase in linear momentum of a body is 50%, then change in its kinetic energy is a) 5% b) 15% c) 150% d) 50%. Choose the incorrect statement out of the following. a) Every measurement by any measuring instrument has some errors. b) Every calculated physical quantity that is based on measured values has some error. c) A measurement can have more accuracy but less precision and vice versa. d) The percentage error is different from relative error By what percentage should the pressure of a given mass of a gas be increased, so as to decrease its volume by 10% at a constant temperature? a) 5% b) 7.% c) 1.5% d) 11.1% 6. Percentage error in the measurement of mass and speed are % and % respectively. The error in the estimation of kinetic energy obtained by measuring mass and speed will be a) 1% b) 10% c) % d) 8%.

2 If the error in the measurement of radius of a sphere is %, then the error in the determination of volume of the sphere will be a) $ b) 6% c) 8% d) % 8. Assertion: The error in the measurement of radius of the sphere is 0.%. The permissible error in its surface area is 0.6%. ΔA Δr Reason: The permissible error is calculated by the formula =. A r a) Both assertion and reason are true reason is the correct explanation of assertion. b) Both assertion and reason are true but reason is not the correct explanation of assertion. c) Assertion is true but reason is false. d) Both assertion and reason are false. 9. The period of oscillation of a simple pendulum in the experiment is recorded as.6s, s,.s,.71s and.80s respectively. The average absolute error is a) 0.1s b) 0.11s c) 0.01s d) 1.0s 10. A wire has a mass (0.± 0.00) g, radius (0.5 ± 0.005) mm and length (6 ± 0.006) cm. The maximum percentage error in the measurement of its density a) 1 b) c) d) 11. If voltage V = (100 ± 5) volt and current I = (10 ± 0.) A, the percentage error in 005 resistance R is a) 5.% b) 5% c) 7% d) 10% 1. If radius of the sphere is (5. ± 0.1)cm. Then percentage error in its volume will be a) b) c) d)

3 1. The radius of sphere is measured to be (.1± 0.5) cm. Calculate its surface area with error limits a) (55. ± 6.)cm b) (55. ± 0.0)cm c) (55. ±.6)cm d) (55. ± 0.6)cm 1. If the length of rod A is.5 ± 0.01cm and that of B is.19 ± 0.01cm. Then the rod B is 00 longer than rod A by a) 0.9 ± 0.00cm b) 0.9 ± 0.01cm c) 0.9 ± 0.0cm d) 0.9 ± 0.005cm 15. A force F is applied on a square plate of side L. If percentage error in determination 00 of L is % and that in F is %, permissible error in pressure is a) % b) % c) 6% d) 8% 16. The value of two resistors are R1 = (6 ± 0.) kω and R = (10 ± 0.) kω. The percentage error in the equivalent resistance when they are connected in parallel is a) 5.15% b) % c) 10.15% d) 7% 17. The difference in the lengths of a mean solar day and a sidereal day is about a) 1 min b) min c) 15 min d) 56 min 1) b ) b Key ) b ) d 5) d 6) d 7) b 8) c 9) b 10) d 11) c 1) c 1) a 1) c 15) d 16) c 17) b

4 Hints 1. Volume of liquid coming out of the tube of per second V = V pπ r 8η l r = V1 r1 110 V = V V = V = V = 1 (1.1) 1.61 volt ΔV V V V = = = 6% V V V p. The kinetic energy is given by KE = m pδp pδp So, Δ KE = = m m KE Δp = KE p Thus, the final momentum becomes 1.p final KE initial KE So, percentage change in KE = 100 initial KE 1.( p / m) ( p / m) ( / ) = = % p m. We know that linear momentum p = mk Now we have p1 = p, p = p1+ 50% of p1 = 1.5 p1 K p p = K = K =.5K K p p So change in KE =.5 1 = 15%

5 5. When T is constant, pv = constant. When volume is decreased by 10% that is volume becomes 90, the pressure must become 100/99. Thus percentage increase in pressure 100 = (100 90) = 11.1% 6. Kinetic energy 1 K = mv Fractional error in kinetic energy ΔK Δm Δv = + K m v Percentage error in kinetic energy is Δm Δv = m v Δ m Δ v As we know, 100 = % and 100 = % m v So, percentage error in kinetic energy = + x = + 6 = 8% 7. Volume of a sphere Or V = π R ( ) = π radius Taking logarithm on both sides, we have logv = log π + log R Differentiating, we get Δ R Accordingly, = % R ΔV = % = 6% V ΔV ΔR = 0 + V R

6 9. Average value = =.6s Now, Δ T1 =.6.6 = 0.01 Δ T =.6.56 = 0.06 Δ T =.6. = 0.0 Δ T =.71.6 = 0.09 Δ T 5 =.80.6 = 0.18 Δ T = Δ T + Δ T + Δ T + Δ T + ΔT = = 0.11s 5 m 10. Density ρ = π rl 1 5 Δρ Δm Δr ΔL 100 = 100 ρ + + m r L After substituting the values, we get the maximum percentage error in density = % 11. Given, voltage V = (100 ± 5) volt, Current I = (10 ± 0.) A From ohm s law V = IR Resistance V R = I Maximum percentage error in resistance ΔR ΔV ΔI 100 = R V I 5 0. = = 5 + = 7%

7 1. Volume of sphere ( V) = π r % error in volume Δr = 100 r 0.1 = Surface area S = π r = (.1) = 55. = 55.cm 7 ΔS Δr Further, =. S r Δr Or Δ S = ( S) r = = 6.8 = 6.cm.1 S = (55. ± 6.) cm 1. Length of rod A is L A =.5 ± 0.01 And that of B is L =.19 ± 0.01 B Then, the rod B is longer than rod A by a length Δ l = L L B A Δ l = (.19 ± 0.01) (.5 ± 0.01) Δ l = (0.9 ± 0.0) cm 15. Pressure = force = F area L dp df dl 100 = p F L = + x = 8% R1R 16. Rparallel = R + R 1

8 ΔRp ΔR1 ΔR Δ ( R1+ R) = + + R R R R + R p 1 1 Δ Rp = + + R p ΔRp 100 = 10.15% R p 17. Sidereal day is about min shorter than our normal solar h or one day. It is about min 56s.

Honor Physics Final Exam Review. What is the difference between series, parallel, and combination circuits?

Name Period Date Honor Physics Final Exam Review Circuits You should be able to: Calculate the total (net) resistance of a circuit. Calculate current in individual resistors and the total circuit current.