Measurement Error & Experiment Mock Test

MEASUREMENT ERROR & EXPERIMENT MOCK TEST SECTION - I : STRAIGHT OBJECTIVE TYPE 1. The number of circular divisions on the shown screw gauge is 50. It moves 0.5 mm on main scale for one complete rotation and main scale has 1/2 mm marks . The diameter of the ball is : 0 (A) 2.25 mm 5 (B) 2.20 mm 2 25 (C) 1.20 mm (D) 1.25 mm 2. If a tuning fork of frequency (f0) 340Hz tolerance  1% is used in resonance column method [v = 2f0 (2 – 1)], the first and the second resonance are measured at 1 = 24.0 cm and 2 = 74.0 cm. The max. permissible error in speed of sound is : (A) 1.4 % (B) 1.8 % (C) 1% (D) 0.8 % 3. To find the value of ‘g’ using simple pendulum. T = 2.00 sec; = 50.0 cm was measured. The maximum permissible error in ‘g’ is : (A) 1.4 % (B) 1.1 % (C) 1.5 % (D) 1.2% 4. The values of kinetic energy ‘K’ and potential energy ‘U’ are measured as follows : K = 100.0 ± 2.0 J U = 200.0 ± 1.0 J Then the percentage error in the measurement of mechanical energy is : (A) 2.5% (B) 1% (C) 0.5% (D) 1.5% 5. Let y = 2 – (A) + 2 ± 0.8 6. 3 z where  = 2.0 ± 0.1, z = 1.0 ± 0.1 then the value of y is given by : (B) – 4 ± 1.6 (C) – 4 ± 0.8 (D) none of these Two masses MA & MB (MA < MB) are weighed using same weighing machine. Absolute error and relative error in two measurements are [Assume only systematic errors are involved] (A) Absolute error same for both, relative error greater for MA and lesser for MB. (B) Absolute error same for both, relative error greater for MB and lesser for MA. (C) Relative error same for both, absolute error greater for MA and lesser for MB. (D) Relative error same for both, absolute error greater for MB and lesser for MA. SECTION - II : COMPREHENSION TYPE Comprehension # 1 Determining the value of ‘g’ using a simple pendulum //////// //////// r In this experiment. a small spherical bob is hanged with a cotton thread. This arrangement is called sample pendulum. The bob is displaced slightly and allowed to oscillate. To find time period, time taken for 50 oscillations is noted using a stop watch. L Theoretically T = 2 g  g = 42 L ..............(1) T2 where L = Equivalent length of pendulum = length of thread () + radius (R) of bob, T = time period of the simple pendulum So ‘g’ can be easily determined by equation ...(1). Graphical method to find ‘g’ :  42    T =  g L   * Find T for different values of L. ............(2) 2 So, T2  L  4 2    * Plot T v/s L curve. From equation (2), it should be a straight line, with slope =  g  .   2  4    Find slope of T2 v/s L graph and equate it to  g  and get ‘g’ .   2 7. In certain observation we got  = 23.2 cm, r = 1.32 cm and time taken for 10 oscillation was 10.0 sec. Estimate the value of ‘g’ in proper significant figure. (take 2 = 10) (A) 9.8 m/s2 (B) 9.80 m/s2 (C) 9.800 m/s2 (D) none of these 8. For different values of L, we get different values of ‘T’. The curve between L v/s T 2 is shown. Estimate ‘g’ from this curve. (take 2 = 10) L (meter) 0.49 2 (A) 9.7 m/s2 (B) 9.6 m/s2 T2 (sec2) (C) 9.8 m/s2 (D) 10 m/s2 MEASUREMENT ERROR & EXPERIMENT Comprehension # 2 Working : Resonance tube is a 100 cm tube. Initially it is filled with water. To increase the length of air column in the tube, water level is lowered. The air column is forced with a tuning fork of frequency f0. Let at length 1, we get a first resonance (loud voice) then V eq = 4 f 1 0 V  1 +  = 4 f 0 .........(i) where  is end correction If we further lower the water level, the noise becomes moderate. But at resonance) then 2. We, again get a loud noise (second 3V eq = 4 f 2 0 3V  2 +  = 4 f 0 .........(ii) From (i) and (ii) V = 2f0 (2 – 1) Observation table : Room temperature is 27ºC P o siti o n o f w a te r le v e l (c m ) F re q . o f tu n in g fo rk i n (H z ) (f 0 ) 330 Hz 9. 10. 11. 12. W a te r le ve l is risin g Mean re so n a n t le n g th R e so n a n c e W a te r le ve l is fa l li n g 1 s t R es o na nc e 2 3.9 24 . 1 l 1 = .. ... ... .. 2 nd R e s o na nc e 73 .9 7 4. 1 l 2 = .. ... ... .. Speed of sound calculated is roughly (A) 340 m/sec (B) 380 m/sec Spe e d of so u n d V = 2 f 0 (l 2 – l 1 ) (C) 430 m/sec (D) 330 m/s In the previous question, speed of sound at 0ºC is roughly (A) 324 m/sec (B) 380 m/sec (C) 430 m/sec (D) 314 m/s V = . .... ... .. What should be minimum length of tube, so that third resonance can also be heard. (A) 3 = 421 cm (B) 3 = 214 cm (C) 3 = 124 cm (D) None of these From equation (i) and (ii) end correction can be calcualted. Estimate the diameter of the tube using imparical formula ( 0.3d) (A) 2.5 cm (B) 3.3 cm (C) 5.2 cm (D) None of these SECTION - III : INTEGER TYPE 13. In some observations, value of ‘g’ are coming as 9.81, 9.80, 9.82, 9.79, 9.78, 9.84, 9.79, 9.78, 9.79 and 9.80 m/s2. Calculate absolute errors and percentage error in g. is x % then x is. 100 14. From Meter Bridge, resistivity of a wire comes from = l D 2s 4L 100  l where  is balance length, D is diameter of wire, S is resistance, L is total length of wire. Find the value of  corresponding to max. permissible error in .(in cm) 15. In Searle’s experiment to find Young’s modulus the diameter of wire is measured as d = 0.05 cm, length of wire is  = 125 cm and when a weight, m = 20.0 kg is put, extension in wire was found to be 0.100 cm. Find maximum mg x % then x is (Use : Y = permissible error in Young’s modulus (Y). is .) (  / 4) d2 x 10 16. In u – v method to find focus distance of a concave mirror, if object distance was found to be 10.0 cm and image 165 % , due to error is u and v measurement. distance was found to be 40.0 cm then the max - permissible error in f is x Then x is 17. A physical quantity x is calculated from the relation x = a2b3 c d . If % error in a, b, c and d are 2%, 1%, 3% and 4% respectively, what is the percentage error in x. 18. If the length of wire is (200.0 cm) and radius of wire as measured from screw gauge is (2.00 mm). Its resistance is 4.25 . If find specific resistance of wire material is X × 10–7  – m then x is. MOCK TEST 1. C 2. A 3. D 4. B 5. B 6. 14. 50 15. 43 16. 100 17. 12 18. 268 A 7. B 8. C 9. D 10. D 11. C 12. B 13. 14