JEE-Physics Example#9 Solve with regards to significant figure (i) 908 + 2.76 (ii) 999 – 989 (iii) 4.0 × 10–4 – 2.5 × 10–6 (iv) 4.0 × 10–4 – 2.5 × 10–5 (v) 6.75 × 103 + 4.52 × 102 (vi) 625 ÷ 125 Solution Ans. (i) 911 (ii) 10.0 (iii) 4.0 × 10–4 (iv) 3.8 × 10–4 (v) 7.20 × 103 (vi) 5.00 Example#10 A scale is calibrated to centimeters and the following measurements are estimated by the scale. Find out the significant digits. (i) 200 m (ii) 92.80 m (iii) 80.26 m (iv) 8.23 cm (v) 8.921 mm (vi) 6.001 m Solution Ans. (i) 3 (ii) 4 (iii) 4 (iv) 2 (v) 1 (vi) 4 Example#11 Ans. (4.0 ± 0.3) ms–2 An object covers (16.0 ± 0.4) m distance in (4.0 ± 0.2) s. Find out its speed. Solution Speed v = distance = 16.0 = 4.0 m/s; Error in speed v s t v 0.4 0.2 4.0 0 .3 m /s time 4.0 s t 16.0 4.0 Example#12 Students I , J ,J and I perform an experiment for measuring the acceleration due to gravity (g) using a simple 1 13 2 pendulum. they use different lengths of the pendulum and record time for different number of oscillations. The observations are shown in the table. Least count for length = 0.1 cm, Least count for time = 1s Students Length of the No. of Time period pendulum (cm) oscillations (n) of pendulum (s) I 1 100.0 20 20 400.0 10 40 J1 100.0 10 20 J 400.0 20 40 3 I 2 If P ,P ,P and P are the % error in g for students I ,J ,J and I respectively then- 123 4 113 2 (A) P = P (B) P is maximum (C) P is minimum (D) P = P 13 3 4 24 Solution Ans. (B,C) T 2 g T–2 = g 2T . g g T Therefore P 2 T 1 00 P1 0.1 2(1) × 100 = 0.6%, P2 0.1 2(1) × 100 = 0.42% T 100 400 400 400 P3 0.1 2(1) × 100 = 1.1%, P4 0.1 2(1) × 100 = 0.28% NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Practical Physics\\Eng\\Sheet.p65 100 200 400 800 Example#13 The length of a cylinder is measured with a metre rod having least count 0.1 cm. Its diameter is measured with vernier callipers having least count 0.01 cm. Given the length is 5.0 cm and diameter is 2.00 cm. Find the percentage error in the calculated value of volume. Solution Ans. 3 V = r2h D 2h V 2 D h V 100 2 0.01 0.1 100 = 3% 4 V D h V 2.00 5.0 20 E
EXERCISE–1 JEE-Physics CHECK YOUR GRASP MCQ's (only one correct answer) 1 . Significant figures in 3400 are- (A) 2 (B) 5 (C) 6 (D) 7 2 . The percentage errors in the measurement of mass and speed are 2% and 3% respectively. How much will be the maximum error in the estimate of kinetic energy obtained by measuring mass and speed ? (A) 11% (B) 8% (C) 5% (D) 1% 3 . The density of a cube is measured by measuring its mass and the length of its side. If the maximum errors in the measurement of mass and length are 4% and 3% respectively, the maximum error in the measurement of the density is - (A) 9% (B) 13% (C) 12% (D) 7% ab2 4 . An experiment measures quantities a, b and c, and X is calculated from X = c3 . If the percentage error in a, b and c are ±1%, ±3% and ±2% respectively, the percentage error in X will be – (A) ±13% (B) ±7% (C) ±4% (D) ±1% 5 . If error in measuring diameter of a circle is 4%, the error in the radius of the circle would be (A) 2% (B) 8% (C) 4% (D) 1% 6 . If a, b, c are the percentage errors in the measurement of A, B and C, then percentage error in ABC would be approximately - (A) abc (B) a + b + c (C) ab + bc + ac abc (D) bca 7 . The diameter of a wire is measured with a screw gauze having least count 0.01 mm. Which of the following correctly expresses the diameter – (A) 0.20 cm (B) 0.002 m (C) 2.00 mm (D) 0.2 cm 8 . While measuring acceleration due to gravity by a simple pendulum a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of the time period. His percentage error in the measurement of the value of g will be - (A) 2% (B) 4% (C) 7% (D) 10% 9 . A student measured the diameter of a wire using a screw gauge with least count 0.001 cm and listed the measurements. The correct measurement is – (A) 5.3 cm (B) 5.32 cm (C) 5.320 cm (D) 5.3200 cm 1 0 . The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate. If the maximum error in the measurement of force and length are respectively 4% and 2%, the maximum error in the measurement of pressure is – (A) 1% (B) 2% (C) 6% (D) 8% NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Practical Physics\\Eng\\Sheet.p65 1 1 . When a copper sphere is heated, maximum percentage change will be observed in– (A) radius (B) area (C) volume (D) none of these 1 2 . The period of oscillation of a simple pendulum in the experiment is recorded as 2.63s, 2.56s, 2.42s, 2.71s and 2.80s respectively. The average absolute error is (A) 0.1s (B) 0.11s (C) 0.01s (D) 1.0s 1 3 . The significant digits in 200.40 are (A) 4 (B) 5 (C) 2 (D) 3 14. A scientist performs an experiment in order to measure a certain physical quantity and takes 100 observations. E He repeats the same experiment and takes 400 observations, by doing so (A) The possible error remains same (B) The possible error is doubled (C) The possible error is halved (D) The possible error is reduced to one fourth 21
JEE-Physics 1 5 . A quantity is represented by X = Ma Lb Tc. The percentage error in measurement of M, L and T are %, % and % respectively. The percentage error in X would be (A) ( a + b + c) % (B) (a – b + c) % (C) ( a – b– c) % (D) None of these 1 6 . If error in measuring diameter of a circle is 4 %, the error in circumference of the circle would be :- (A) 2% (B) 8% (C) 4% (D) 1% 1 7 . A wire has a mass (0.3±0.003) g, radius (0.5±0.005) mm and length (6 ± 0.06) cm. The maximum percentage error in the measurement of its density is – (A) 1 (B) 2 (C) 3 (D) 4 1 8 . The length of a cylinder is measured with a metre rod having least count 0.1 cm. Its diameter is measured with vernier callipers having least count 0.01 cm. Given the length is 5.0 cm and diameter is 2.00 cm. The percentage error in the calculated value of volume will be – (A) 2% (B) 1% (C) 3% (D) 4% 1 9 . The volume of a sphere is 1.76 cm3. The volume of 25 such spheres taking into account the significant figure is- (A) 0.44 × 102 cm3 (B) 44.0 cm3 (C) 44 cm3 (D) 44.00 cm3 2 0 . What is the fractional error in g calculated from T 2 ? Given that fractional errors in T and are ± x and g ± y respectively. (B) x – y (C) 2x + y (D) 2x – y) (A) x + y 2 1 . The resistance is R V where V = 100 ± 5 Volts and I = 10 ± 0.2 amperes. What is the total error in R ? I (A) 5% (B) 7% (C) 5.2% (D) 5 % 2 22 . The length, breadth and thickness of a strip are (10.0 ± 0.1)cm, (1.00 ± 0.01) cm and (0.100 ± 0.001) cm respectively. The most probable error in its volume will be (A) ± 0.03 cm3 (B) ± 0.111 cm3 (C) ± 0.012 cm3 (D) none of these 2 3 . The external and internal radius of a hollow cylinder are measured to be (4.23 ± 0.01) cm and (3.89 ± 0.01) cm. The thickness of the wall of the cylinder is :- (A) (0.34 ± 0.02)cm (B) (0.17 ± 0.02)cm (C) (0.17 ± 0.01)cm (D) (0.34 ± 0.01)cm 2 4 . The radius of a disc is 1.2 cm. Its area according to idea of significant figures, will be given by:- (A) 4.5216 cm2 (B) 4.521 cm2 (C) 4.52 cm2 (D) 4.5 cm2 2 5 . The length , breadth b and thickness t of a block of wood were measured with the help of a measuring scale. The results with permissible errors are = 15.12 0.01 cm, b = 10.15 0.01 cm, t = 5.28 0.01 cm. The percentage error in volume upto proper significant figures is – NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Practical Physics\\Eng\\Sheet.p65 (A) 0.28% (B) 0.36% (C) 0.48 % (D) 0.64% 2 6 . The following observations were taken for determining surface tension of water by capillary tube method: Diameter of capillary D = 1.25 × 10–2 m Rise of water in capillary, h = 1.45 × 10–2 m. Taking g = 9.80 m/s2 and using the relation T = (rhg/2) × 103 N/m, what is the possible error in surface tension. T– (A) 0.16% (B) 1.6% (C) 16% (D) 2.4% 2 7 . The least count of a stop watch is 1/5 sec. The time of 20 oscillations of a pendulum is measured to be 25 s. What is the maximum percentage error in this measurement (A) 8% (B) 1% (C) 0.8% (D) 16% 2 8 . The area of a rectangle of size 1.23 × 2.345 cm is (A) 2.88 cm2 (B) 2.884 cm2 (C) 2.9 cm2 (C) 2.88435 cm2 22 E
JEE-Physics 2 9 . What is vernier constant (A) It is the value of the one main scale division by the total number of divisions on the main scale. (B) It is the value of one vernier scale division divided by the total number of division on the vernier scale. (C) It is the difference between value of one main scale division and one vernier scale division (D) It is not the least count of vernier scale. 3 0 . The vernier of a circular scale is divided into 30 divisions which coincide against 29 divisions of main scale. Each main scale division is 0.5°. The least count of the instrument is – (A) 10' (B) 0.1' (C) 1' (D) 30' 3 1 . What is the reading of micrometer screw gauge shown in figure 012 35 mm 30 25 (A) 2.30 mm (B) 2.29 mm (C) 2.36 mm (D) 2.41 mm 3 2 . In a vernier calliper, N divisions of vernier scale coincide with (N – 1) divisions of main scale (in which 1 division represents 1mm). The least count of the instrument in cm. should be (A) N (B) N – 1 1 1 (C) 10N (D) N 1 3 3 . A vernier callipers having 1 main scale division = 0.1 cm is designed to have a least count of 0.02 cm. If n be the number of divisions on vernier scale and m be the length of vernier scale, then (A) n= 10, m = 0.5 cm (B) n=9, m= 0.4 cm (C) n=10, m = 0.8 cm (D) n=10, m= 0.2 cm 3 4 . In a vernier callipers, N divisions of the main scale coincide with N+m divisions of the vernier scale. What is the value of m for which the instrument has minimum least count? (A) 1 (B) N (C) infinity (D) N/2 3 5 . In a vernier callipers the main scale and the vernier scale are made up different materials. When the room temperature increases by T°C, it is found the reading of the instrument remains the same. Earlier it was observed that the front edge of the wooden rod placed for measurement crossed the Nth main scale division and N+2 MSD coincided with the 2nd VSD. Initially, 10 VSD coincided with 9 MSD. If coefficient of linear expansion of the main scale is 1 and that of the vernier scale is 2 then what is the value of 2? (Ignore the expansion of the rod on heating) (A) 1.8 /N (B) 1.8/ (N+3.8) (C) 1.8/ (N–2) (D) 1.8/N+2 3 6 . Consider the MB shown in the diagram, let the resistance X have temperature coefficient 1 and the resistance from the RB have the temperature coefficient 2. Let the reading of the meter scale be 10 cm from the LHS. If the temperature of the two resistance increase by small temperature T then what is the shift in the position of the null point? Neglect all the other changes in the bridge due to temperature rise A X B C90 D G NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Practical Physics\\Eng\\Sheet.p65 (A) 9(1 - 2)T E OF (D) 1/9 (1 - 2)T 10cm (C) 1/9 (1 + 2)T (B) 9 (1 + 2)T 3 7 . For a post office box, the graph of galvanometer deflection versus R (resistance pulled out of RB) for the ratio 100 : 1 is given as shown. A careless student pulls out two non consecutive values R marked in the graph. Find the value of unknown resistance Deflection (in division) (A) 3.2 ohm 5 326 R (D) None 320 E -3 (C) 3.206 ohm (B) 3.24 ohm 23
JEE-Physics 3 8 . Identify which of the following diagrams represent the internal construction of the coils wound in a resistance box or PO box ? P P P P (C) (A) (B) (D) 3 9 . In a meter bridge set up, which of the following should be the properties of the one meter long wire? (A) High resistivity and low temperature coefficient (B) Low resistivity and low temperature coefficient (C) low resistivity and high temperature coefficient (D) High resistivity and high temperature coefficient MCQ's (Multiple correct answer) 4 0 . In the Searle's experiment, after every step of loading, why should we wait for two minutes before taking the readings? (More than one correct (A) So that the wire can have its desired change in length (B) So that the wire an attain room temperature (C) So that vertical oscillations can get subsided (D) So that the wire has no change in its radius Comprehension Internal micrometer is a measuring instrument used to measure internal diameter (ID) of a large cylinder bore with high accuracy. Construction is shown in figure. There is one fixed rod B (to the right in figure) and one moving rod A (to the left in figure). It is based on the principle of advancement of a screw when it is rotated in a nut with internal threads. Main scale reading can be directly seen on the hub which is fixed with respect to rod B. When the cap is rotated, rod A moves in or out depending on direction of rotation. The circular scale reading is seen by checking which division of circular scale coincides with the reference line. Cap Ratchet Main scale A 5 B P Q 0 0 10 45 Hub Circular scale Internal diameter (being measured) This is to be multiplied by LC to get circular scale reading. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Practical Physics\\Eng\\Sheet.p65 pitch Least count = value of 1 circular scale division number of division on circular scale Length of rod A is chosen to match the ID (PQ) to be measured. Zero error is checked by taking reading between standard blocks fixed at nominal value of ID to be measured. Zero error is positive if cap end is on the right side of the main scale and negative it is on the left side. Zero error checking Standard blocks 1 . In an internal micrometer, main scale division is of 0.5 mm and there are 50 divisions in circular scale. The least count of the instrument is - (A) 0.005 mm (B) 0.001 mm (C) 0.05 mm (D) 0.01 mm 2 . In the above instrument, while measuring an internal diameter. ID is set of 321 mm with no zero error. If cap end is after 7th division and 17th division of main scale coincides with the reference line, the ID is- 24 E
JEE-Physics (A) 321. 717 mm (B) 321.87 mm (C) 328.17 mm (D) 324.67 mm 3 . During zero setting of the above instrument, the end of the cap is on left side of the zero of main scale (i.e. zero of main scale is not visible) and 41st division of circular scale coincides with the reference line, the zero error is- (A) –0.09 mm (B) +0.41 mm (C) –0.41 mm (D) +0.09 mm Subjective Questions 1 . In a given slide callipers 10 division of its vernier coincides with its 9 main scale divisions. If one main scale division is equal to 0.5 mm then find its least count. 2 . Consider a home made vernier scale as shown in the figure. In this diagram, we are interested in measuring the length of the line PQ. If both the inclines are identical and their angles are equal to then what is the least count of the instrument. P Q 3 . The pitch of a screw gauge is l mm and there are 50 divisions on its cap. When nothing is put in between the studs, 44th division of the circular scale coincides with the reference line zero of the main scale is not visible. When a glass plate is placed between the studs, the main scale reads three divisions and the circular scale reads 26 divisions. Calculate the thickness of the plate. 4 . A short circuit occurs in a telephone cable having a resistance of 0.45 m-1 . The circuit is tested with a Wheatstone bridge. The two resistors in the ratio arms of the Wheatstone bridge network have values of 100 and 1110 respectively. A balance condition is found when the variable resistor has a value of 400. Calculate the distance down the cable, where the short has occurred. 5 . A glass prism of angle A = 60° gives minimum angle of deviation = 30° with the maximum error of 1° when a beam of parallel light passed through the prism during an experiment. Find the permissible error in the measurement of refractive index of the material of the prism. 6 . In a given optical bench, a needle of length 10 cm is used to estimate bench error. The object needle, image needle & lens holder have their reading as shown. x = 1.1 cm; x = 0.8 cm; x = 10.9 cm 01L Estimate the bench errors which are present in image needle holder and object needle holder. Also find the focal length of the convex lens when x = 0.6 cm ; x = 22.5 cm; x = 11.4 cm 0I L 7 . Consider S = x cos () for x = (2.0 ± 0.2) cm, = 53 ± 2°. Find S. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Practical Physics\\Eng\\Sheet.p65 ANSWER KEY Single Choice Questions : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ABBACBCCCDCBBDAC 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 DC DC B A ADB B CA CC AC 33 34 35 36 37 38 39 40 C A B A B D A ABC Comprehension : 1. D 2. D 3. A Subjective Questions : 1. 0.05 mm 2. LC 1 cos 3. R = 3.64 mm 4.40m 5. 5 % cos t 18 E 6. 5.5 ± 0.05 cm 7. S = (1.2 ± 0.18) cm 25
JEE-Physics EXERCISE–2(A) PREVIOUS YEARS QUESTIONS 1 . The 'rad' is the correct unit used to report the measurement of : [AIEEE - 2006] (1) the ability of a beam of gamma ray photons to produce ions in a target (2) the energy delivered by radiation to a target (3) the biological effect of radiation (4) the rate of decay of a radioactive source 2 . An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment distances are meausred by [AIEEE - 2008] (1) a vernier scale provided on the microscope (2) a stanard laboratory scale (3) a meter scale provided on the microscope (4) a screw gauge provided on the microscope 3 . Two full turns of the circular scale of gauge cover a diastance of 1 mm on scale. The total number of divisions on circular scale is 50. Further, it is found that screw gauge has a zero error of -0.03 mm. While measuring the diameter of a thin wire a student notes the main scale reading of 3 mm and the number of circular scale division in line, with the main scale as 35. The diameter of the wire is [AIEEE - 2008] (1) 3.32 mm (2) 3.73 mm (3) 3.67 mm (4) 3.38 mm 4 . In an experiment the angles are required to be measured using an instrument 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a- degree (=0.5°), then the least count of the instrument is :- [AIEEE - 2009] (1) One degree (2) Half degree (3) One minute (4) Half minute 5 . In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45° with the x-axis meets the experimental curve at P. The coordinates of P will be :- [AIEEE - 2009] (1) (ƒ, ƒ) (2) (4ƒ, 4ƒ) (3) (2ƒ, 2ƒ) (4) ƒ , ƒ 2 2 6 . The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10–3 are:- (1) 4, 4, 2 (2) 5, 1, 2 (3) 5, 1, 5 [AIEEE - 2010] (4) 5, 5, 2 7 . A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm. [AIEEE - 2011] Circular scale reading : 52 divisions Given that 1 mm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is :- (1) 0.026 cm (2) 0.005 cm (3) 0.52 cm (4) 0.052 cm 8 . A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading : 58.5 degree [AIEEE - 2012] Vernier scale reading : 09 divisions Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Practical Physics\\Eng\\Sheet.p65 match with 29 divisions of the main scale. The angle of the prism from the above data :] (1) 59 degree (2) 58.59 degree (3) 58.77 degree (4) 58.65 degree 9 . Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is :- [AIEEE - 2012] (1) 3% (2) 6% (3) zero (4) 1% Que. 1 2 3 4 5 6 7 8 9 E Ans. 3 1 4 3 3 2 4 4 2 26
EXERCISE–2(B) JEE-Physics PREVIOUS YEARS QUESTIONS MCQ With One Correct Answer 1 . The edge of a cube is a = 1.2 × 10–2 m. Then its volume will be recorded as [IIT-JEE 2003] (A) 1.7 × 10–6 m3 (B) 1.70 × 10–6 m3 (C) 1.70 × 10–7 m3 (D) 1.78 × 10–6 m3 2 . A wire has a mass (0.3 ± 0.003)g, radius (0.5 ± 0.005) mm and length (6 ± 0.06) cm. The maximum percentage error in the measurement of its density is – [IIT-JEE 2004] (A) 1 (B) 2 (C) 3 (D) 4 3 . For the post office box arrangement to determine the value of unknown resistance, the unknown resistance should be connected between [IIT-JEE 2004] B CD A (A) B and C B1 C1 (D) B and C 11 (B) C and D (C) A and D 4 . In a resonance column method, resonance occurs at two successive level of = 30.7 cm and = 63.2 cm using 12 a tuning fork of f = 512 Hz. What is the maximum error in measuring speed of sound using relations v = f & = 2 ( – ). [IIT-JEE 2005] 21 (A) 256 cm/sec (B) 92 cm/sec (C) 102.4 cm/sec (D) 204.8 cm/sec 5 . Graph of position of image vs position of point object from a convex lens is shown. [IIT-JEE 2006] Then, focal length of the lens is NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Practical Physics\\Eng\\Sheet.p65 (A) 0.50 ± 0.05 cm (B) 0.50 ± 0.10 cm (C) 5.00 ± 0.05 cm (D) 5.00 ± 0.10 cm 6. A student performs an experiment for determination of 4 2 , 1m, and he commits an error of g T2 . For T he takes the time of n oscillations with the stop watch of least count T and he commits a human error of 0.1 s. For which of the following data, the measurement of g will be most accurate ? (A) L = 0.5, T = 0.1, n = 20 (B) L = 0.5, T = 0.1, n = 50 [IIT-JEE 2006] (C) L = 0.5, T = 0.01, n = 20 (D) L = 0.1, T = 0.05, n = 50 E 27
JEE-Physics 7 . The circular scale of a screw gauge has 50 divisions and pitch of 0.5 mm. Find the diameter of sphere. Main scale reading is 2 – [IIT-JEE 2006] (A) 1.2 (B) 1.25 (C) 2.207 (D) 2.25 8 . A student performs an experiment to determine the Young's modulus of a wire, exactly 2m long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. Take g = 9.8 m/s2 (exact). The Young's modulus obtained from the reading is – [IIT-JEE 2007] (A) (2.0 ± 0.3) × 1011 N/m2 (B) (2.0 ± 0.2) × 1011 N/m2 (C) (2.0 ± 0.1) × 1011 N/m2 (D) (2.0 ± 0.05) × 1011 N/m2 9 . In the experiment to determine the speed of sound using a resonance column – [IIT-JEE 2007] (A) prongs of the tuning fork are kept in a vertical plane. (B) prongs of the tuning fork are kept in a horizontal plane. (C) in one of the two resonance observed, the length of the resonating air column is close to the wavelength of sound in air (D) in one of the two resonance observed, the length of the resonating air column is close to half of the wavelength of sound in air 1 0 . In an experiment to determine the focal length (f) of a concave mirror by the u-v method, a student placed the object pin A on the principal axis at a distance x from the pole P. The student looks at the pin and its inverted image from a distance keeping his/her eye in line with PA. When the student shift his/her eye towards left, the image appears to the right of the object pin. Then- [IIT-JEE 2007] (A) x < f (B) f < x < 2f (C) x = 2f (D) x > 2f 1 1 . The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale excatly coincides with one of the main scale divisions. The diameter of the cylinder is :- [JEE Advanced 2013] (A) 5.112 cm (B) 5.124 cm (C) 5.136 cm (D) 5.148 cm 1 2 . Using the expression 2d sin = , one calculates the values of d by measuring the corresponding angles in the range 0 to 90°. The wavelength is exactly known and the error in is constant for all values NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Practical Physics\\Eng\\Sheet.p65 of . As increases from 0° :- [JEE Advanced 2013] (A) the absolute error in d remains constant (B) the absolute error in d increases (C) the fractional error in d remains constant (D) the fractional error in d decreases Multiple Choice Questions 1 . A student performed the experiment of determination of focal length of a concave mirror by u-v method using an optical bench of length 1.5 meter. The focal length of the mirror used is 24 m. The maximum error in the location of the image can 0.2 m. The 5 sets of (u, v) values recorded by the student (in cm) are : (42, 56), (48, 48),(60, 40), (66, 33), (78, 39). The data set(s) that connot come from experiment and is (are) incorrectly recorded is (are ) [IIT-JEE 2009] (A) (42, 56) (B) (48, 48) (C) (66, 33) (D) (78, 39) 28 E
JEE-Physics 2 . A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by resonance and that with the longer air-column is the second resonance. Then, [IIT-JEE 2009] (A) The intensity of the sound heard at the first resonance was more than that at the second resonance (B) the prongs of the tuning fork were kept in a horizontal plane above the resonance tube (C) the amplitude of vibration of the ends of the prongs is typically around 1 cm (D) the length of the air-column at the first resonance was somewaht shorter than 1/4th of the wavelength of the sound in air Subjective Questions 1 . In a vernier callipers, n divisions of its main scale match with (n + 1) divisions on its vernier scale. Each division of the main scale is a units. Using the vernier principle, calculate its least count. [IIT-JEE 2003] 2 . In a Searle's experiment, the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young's modulus of the material of the wire from these data. [IIT-JEE 2004] 3 . Draw the circuit for experimental verification of Ohm's law using a source of variable D.C. voltage, a main resistance of 100 , two galvanometers and two resistance of values 106 and 10–3 respectively. Clearly show the positions of the voltmeter and the ammeter. [IIT-JEE 2004] 4 . The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Find the curved surface area (in cm2) of the wire in appropriate number of significant figures. [IIT-JEE 2004] 5 . The edge of a cube is measured using a vernier calliper. (9 divisions of the main scale is equal to 10 divisions of vernier scale and 1 main scale division is 1 mm). The main scale division reading is 10 and 1 division of vernier scale was found to be coinciding with the main scale. The mass of the cube is 2.736 g. Calculate the density in g/cm3 upto correct significant figures. [IIT-JEE 2005] ANSWER KEY Single Choice Questions : 1. A 2. D 3. C 4. D 5. C 6. D 7. A 8. B 9. A 10.B 11. B 12. D NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Practical Physics\\Eng\\Sheet.p65 2. A, D Multiple Choice Questions : 1. C, D Voltmeter 106 G1 100 Subjective Questions : a 2. 3. Ammeter 1. G3 103 n 1 E 4. 2.6 cm2 5. 2.66 g/cm3 E 29
JEE-Physics PRINCIPLES OF COMMUNICATION SYSTEMS 1. INTRODUCTION Communication means transmission of information. Everyone experiences the need to impart or recieve information continuously in the surrounding and for this , we speak, listen,send message by a messenger, use coded signalling methods through smoke or flags or beating of drum etc. and these days we are using telephones, TV, radio, satellite communication etc. The aim of this chapter is to introduce the concepts of communication namely the mode of communication, the need for modulation, production and detection of amplitude modulation. Elements of a Communication System : Every communication system has three essential elements- (i) transmitter (ii) medium/channel (iii) receiver Information Message Transmitter Transmitted Channel Received Message User of source signal signal signal Receiver signal information Noise Transmitter converts the message signal into an electric signal and transmits through channel. The receiver receives the transmitted signal and reconstructs the original message signal to the end user. There are two basic modes of communication: (i) point-to-point and (ii) broadcast. In point-to-point communication mode, communication takes place over a link between a single transmitter and a receiver as in telephony. In the broadcase mode, there are a large number of receivers corresponding to a single transmitter. Radio and television are most common examples of braoadcast mode of communication. However the communication system can be classified as follows : Types of Communication Systems On the basis of nature of On the basis of signal On the basis of On the basis of type Information transmitted (1) Speech Transmission (1) Analog transmission Channel of modulation (2) Digital (Radio) (1) Line communication (1) Continuous wave (2) Picture Transmission (a) Two wire Transmission modulation (TV) (3) Fascimile Transmission line (a) Amplitude (FAX) (b) Coaxial cable (b) Frequency (4) Data Transmission (c) Optical fibre (c) phase (Internet) (2) Space communication (2) Pulse Modulation (a) PAM PPM E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\3. Principles of Communication System.p65 (b) PTM (c) PCM PWM Basic terminology Used in Electronic Communication systems : (i) Transducer. Transducer is the device that converts one form of energy into another. Microphone, photo detectors and piezoelectric sensors are types of transducer. They convert information into electrical signal. (ii) Signal Signal is the information converted in electrical form. Signals can be analog or digital. Sound and picture signals inTV are analog. It is defined as a single–valued function of time which has a unique value at every instant of time. (1) Analog Signal :– A continuously varying signal (Voltage or Current) is called an analog signal. A decimal number with system base 10 is used to deal with analog signal. E 35
JEE-Physics (2) Digital Signal :– A signal that can have only discrete stepwise values is called a digital signal. A binary number system with base 2 is used to deal with digital signals. (iii) Noise : There are unwanted signals that tend to disturb the transmission and processing of message signals. The source of noise can be inside or outside the system. (iv) Transmitter : A transmitter processes the incoming message signal to make it suitable for trans-mission through a channel and subsequent reception. (v) Receiver : A receiver extracts the desired message signals from the received signals at the channel output. (vi) Attenuation : It is the loss of strength of a signals while propagating through a medium. It is like damping of oscillations. (vii) Amplification : It is the process of increasing the amplitude (and therefore the strength) of a signal using an electronic circuit called the amplifier. Amplification is absolutely necessary to compensate for the attenuation of the signal in communication systems. (viii) Range : It is the largest distance between the source and the destination upto which the signal gets received with sufficient strength. (ix) Bandwidth : It is the frequency range over which an equipment operates or the portion of the spectrum occupied by the signal. (x) Modulation : The original low frequency message/information signal connot be transmitted to long distances. So, at the transmitter end, information contained in the low frequency message signal is superimposed on a high frequency wave, which acts as a carrier of the information. This process is known as modulation. (xi) Demodulation : The process of retrieval of original information from the carrier wave at the receiver end is termed as demodulation. This process is the reverse of modulation. (xii) Repeater : A repeater acts as a receiver and a transmitter. A repeater picks up the signal which is comming from the transmitter, amplifies and retransmits it with a change in carrier frequency. Repeaters are necessary to extend the range of a communication system as shown in figure A communication satellite is basically a repeater station in space. Mountain E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\3. Principles of Communication System.p65 BANDWIDTH Use of repeater station to increase the range of communication Bandwidth of signals : Different signals used in a comminication system such as voice, music, picture, computer data etc. all have different ranges of frequency. The difference of maximum and minimum frequency in the range of each signal is called bandwidth of that signal. Bandwidth can be of message signal as well as of transmission medium. (i) Bandwidth for analog signals 36 E
JEE-Physics Bandwdith for some analog sinals are listed below : Signal Frequency range Bandwidth required Speech 300-3100 Hz 3100-300 =2800 Hz Music High frequencies produced 20 kHz by musical instrument Picture Audible range =20 Hz - 20 kHz 4.2 MHz TV 6 MHz - Contains both voice and picture (ii) Bandwidth for digital signal Basically digital signals are rectanglar waves and these can be splitted into a superposition of sinusoidal waves of frequencies 0, 20, 30, 40, n , where n is an integer extending to infinity. This implies that the infinite 0 band width is required to reproduce the rectangular waves. However, for practical purposes, higher harmonics are neglected for limiting the bandwidth Band width of Transmission Medium Different types of transmission media offer different band width in which some of are listed below Frequency Bands S er vice Frequency range Rem arks W ire 750 MHz N ormally operated below 1 (m ost com m on : (Bandwidth) 18 GHz Coaxial Cable) Free space Few hundred kHz to 2 (radio waves) GHz (i) Standa rd AM 540kHz -1600 kHz b roa d ca st (ii) FM 88-108 MHz (iii) Tele vision 54-72 MHz VHF (Very) 76-88 MHz h ig h 174-216 MHz frequencies) TV 420-890 MHz U HF (U ltra h ig h t E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\3. Principles of Communication System.p65 (iv) Ce llular 896-901 MHz frequency) TV m obile 840-935 MHz r a d io Mobile to base 5.925-6.425 GHz S ta tion (v) Satellite 3.7 - 4.2 GHz Base station to Com m unication m ob ile U plinking D ownlinking One single Op tical 1THz-1000 THz optical fibre 3 com m unication (m icrowa ves- offers using fibres ultra violet) bandwidth > 100 GHz E 37
JEE-Physics Propagation of Electromagnetic Waves : In case of radio waves communication, an antenna at the transmitter radiates the electromagnetic waves (em waves). The em waves travel through the space and reach the receiving antenna at the other end. As the em wave travels away from the transmitter, their strength keeps on decreasing. Many factors influence the propagation of em waves including the path they follow. The composition of the earth's atmosphere also plays a vital role in the propagation of em waves, as summarised below. Table 4 Layers of atmosphere and their interaction with the propagating em wave s Atmospheric Height over earth's Exists during Frequencies most stratum (layer) surface (approx) Day and night likely affected 10 km 1. Troposphere VHF (upto several GHz) 2. Ionosphere 65-75 km (i) D (part of Day only Reflects LF, absorbs MF & 100 km HF to some degree stratosphere) Day only Helps surface waves, (ii) E (part of 170-190 km reflects HF Daytime, merges Partially absorbs HF waves stratosphere) 300 km at night, with F2 at night yet allowing themto reach F2 (iii) F1 (Part of 250-400 km Day and night Efficiently reflects HF waves during daytime particularly at night Mesosphere) (iv) F2 (Thermosphere) Ground Wave Propagation : (a) The radio waves which travel through atmosphere following the surface of earth are known as ground waves or surface waves and their propagation is called ground wave propagation or surface wave propagation. (b) The ground wave transmission becomes weaker with increase in frequency because more absorption of ground waves takes place at higher frequency during propagation through atmosphere. (c) The ground wave propagation is suitabel for low and medium frequency i.e. upto 2 or 3 MHz only. (d) The ground wave propagation is generally used for local band broadcasting and is commonly called medium wave. (e) The maximum range of ground or surface wave propagation depends on two factors : (i) The frequency of the radio waves and (ii) Power of the transmitter Sky Wave Propagation : (a) The sky waves are the radio waves of frequency between 2 MHz to 30 MHz. (b) The ionoopheric layer acts as a reflector for a certain range of frequencies (3 to 30 MHz). Electromagnetic E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\3. Principles of Communication System.p65 waves of frequencies higher than 30 MHz penetrate the ionosphere and escape. (c) The highest frequency of radiowaves which when sent straight (i.e. normally) towards the layer of ionosphere gets refelcted from ionosphere and returns to the earth is called critical frequency. . If is given by fc = 9 (Nmax)1/2, where N is the number density of electron/m3. Space wave propagation : (a) The space waves are the radiowaves of very high frequency (i.e. between 30 MHz. to 300 MHz or more). (b) the space waves can travel through atmosphere from transmitter antenna to receiver antenna either derectly or after reflection from ground in the earth's troposphere region. That is why the space wave propagation is also called as tropospherical propagation or line of sight propagation. (c) The range of communication of space wave propagation can be increased by increasing the heights of transmitting and receiving antenna. 38 E
JEE-Physics (d) If the transmitting antenna is at a height h , then you can show that the distance to the horizontal d is given TT as dT 2RhT , where R is the radius of the earth (approximately 6400 km). d is also called the radio T maximum line-of sight distance d between the two antennas having heights h and h above the earth is m TR given by : dM 2RhT 2RhR where h is the height of receiving antenna. R Modulation * It is a process by which any electrical signal called input / baseband or modulating signal, is mounted onto another signal of high frequency which is known as carrier signal. * It is defined as the process by which some characteristic (called parameter) of carrier signal is varied in accordance with the instantaneous value of the baseband signal. * The signal which results from this process is known as modulated signal. Need for Modulation : (i) To av io d i nter ference : If many modulating signals travel directly through the same transmission channel, they will interfere with each other and result in distortion. (ii) To de sign a ntennas of pract icable size : The minimum height of antenna (not of antenna tower) should be /4 where is wavelength of modulating signal. This minimum size becomes impracticable because the frequency of the modulating signal can be upto 5 kHz which corresponds to a wavelength of 3 × 108/5 × 103 = 60 km. This will require an antenna of the minimum height of /4 = 15 km. This size of an antenna is not practical. (iii) Effective Power Radiated by an Antenna : A theoretical study of radiation from a linear antenna (length ) shows that the power radiated is proportional to (frequency)2 i.e. (/)2. For a good transmission, we need high powers and hence this also points out to the need of using high frequency transmission. The above discussion suggests that there is a need for translating the original low frequency baseband message signal into high frequency wave before transmission. In doing so, we take the help of a high frequency signal, which we already know now, is known as the carrier wave, and a process known as modulation which attaches information to it. The carrier wave may be continuous (sinusoidal) or in the form of pulses, as shown in figure Time period T =2T Pulse duration E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\3. Principles of Communication System.p65 Amplitude Pulse Pulse fall amplitude Time Pulse (a) rise (b) Carrier wave : Sinusoidal A sinusoidal carrier wave can be represented as c(t) = A sin (ct + ) C where c(t) is the signal strength (voltage or current), A is the amplitude, c (= 2fc) is the angular frequency C and is the initial phase of the carrier wave. Thus, modulation can be affected by varying, any of three parameters, viz A, c and , of the carrier wave can as per the parameter of the message or information c signal. This results in three types of modulation : (i) Amplitude modulation (AM) (ii) Frequency modulation (FM) and (iii) Phase modulation (PM), as shown in figure. E 39
JEE-Physics 1 0 (a) -1 0 0.5 1 1.5 2 2.5 3 1 m(t) 0 (b) -1 0 0.5 1 1.5 2 2.5 3 2 cm(t)for AM 0 (c) -2 10 0.5 1 1.5 2 2.5 3 cm(t)for FM 0 (d) -1 0 0.5 1 1.5 2 2.5 3 1 cm(t)for PM 0 0.5 (e) -1 2 2.5 3 0 Time Modulation of a carrier wave : (a) a sinusoidal carrier wave (b) a modulating signal : (c) amplitude modulatin : (d) frequency modulation : and (e) phase modulation Carrier Wave Pulses : Similarly, the significant characteristics of a pulse are : Pulse Amplitude, Pulse duration or pulse Width, and pulse Position (denoting the time of rise or fall of the pulse amplitude) Hence, different types of pulse modulation are (a) pulse amplitude modulation (PAM), (b) Pulse duration modulation (PDM) or pulse width modulation (PWM), and (c) Pulse position modulation (PPM). Ex.1 A separate high freq. wave (i.e. carrier wave) is needed in modulation why ? A ns . This is because we cannot change any of the characteristics (amplitude, frequency or phase) of the audio signal as this would change the message to be communicated. So keeping the audio signal same, the amplitude of freq. or phase of the high freq. carrier wave is modified in accordance with the modulating (i.e. audio signal) signal. Amplitude Modulation : In amplitude modulation the amplitude of the carrier is varied in accordance with the information signals. Let c(t) = A sinc t represent carrier wave and m(t) = A sin m t represent the message or the modulating c m signal where m = 2fm is the angular frequency of the message signal. The modulated signal c (t) can be m written as c (t) = (A + A sin mt) sin ct E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\3. Principles of Communication System.p65 m c m Ac Am sin m t c t .....(1) 1 Ac sin Note that the modulated signal now contains the message signal & it can be written as : c (t) = A sinct + µA sinmt sinct ......(2) m c c Here µ = A /A is the modulation index mc In paractice, µ is kept 1 to avoid distortion. Using the trignomatric relation sinA sinB = ½ (cos (A – B) – cos (A + B), we can write c (t) of eq. (15.4)as m c (t) = A sinct + µA c cos(c – m) t – µA c cos(c + m) t ......(3) m c 2 2 40 E
JEE-Physics Here c – m and c + m are respectively called the called the lower side and upper side frequencies. The modulated signal now consists of the carrier wave of frequency c plus two sinusoidal waves each with a frequency slightly different from, know as side bands. The frequency spectrum of the amplitude modulated signal is shown in figure : Ac Amplitude µAc 2 (e-m) e (e+m) in radians As long as the broadcast frequencies (carrier waves) are sufficiently spaced out so that sidebands do not overlap, different stations can operate without interfering with each other. E x . 2 A message signal of frequency 10 kHz and peak voltage of 10 volts is used to modulate a carrier of frequency 1 Mhz and peak voltage of 20 volts. Determine (a) modulation index, (b) the side bands produced. Sol. (a) Modulation index = 10/20 = 0.5 (b) The side bands are at (1000 + 10 kHz) = 1010 kHz and (1000–10 kHz) = 990 kHz. Producation of Amplitude modulated Wave : Ampitude modulation can be produced by a veriety of methods. A conceptually simple method is shown in the block diagram of figure. m(t) x(t) SQUARE y(t) BANDPASS Amsin mt + FILTER AMWave LAW DEVICE CENTRED (Modulating Singna) Bx(t) + Cx(t)2 AT c c (t) Ac sinct (carrier) Here the modulating signal A sin mt is added to the carrier signal A sin c t to produce the signal x (t). m c This signal x (t) = A sin m t + A sinct is passed through a square law device which is a non-linear device m c which produces an output E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\3. Principles of Communication System.p65 y(t) = B x (t) + Cx2 (t) ......(4) where B and C are constants. Thus, y(t) = BA sinm t + BA sin ct m c + C [A 2 sin2 m t + A2 sin2 ct + 2A A s i n mt sinct] ......(5) m c m c = BA s i n m t + BA sin ct m c C A 2 A 2 C A 2 cos 2m t C A 2 cos 2c t m c m c 2 2 2 + CA A cos ( – ) t – CA A cos ( + ) t ......(6) mc cm mc cm where the trigonometric relations sin2A = (1 – cos2A)/2 and the relation for sinA sinB mentioned earlier are used. 41 E
JEE-Physics In equation (6), there is a dc term C/2 ( A 2 A 2 ) and sinusoids of frequencies m, 2m, c – m and c m c + m . The output of the band pass filter therefore is of the same form as equation (3) and is therefore an AM wave. It is to be mentioned that the modulated signal connot be transmitted as such. The modulator is to be followed by a power amplifier which provides the necesary power and then the modulated signal is fed to an antenna of appropriate size for radiation as shown in figure. m(t) AMPLITUDE POWER TRANSMITTING Message signal MODULATOR AMPLIIFIER ANTENNA Carrier Detection of Amplitude Modulated Wave : The transmitted message gets attenuated in propagating through the channel. The receiving antenna is therefore to be followed by an amplifier and a detector. In addition, to facilitate further processing, the carrier frequency is usually changed to a lower frequency by what is called an intermediate frequency (IF) stage preceding the detection. The detected signal may not be strong enough to be made use of and hence in required to be amplified. A block diagram of a typical receiver is shown in figure. RECEIVING ANTENNA AMPLIFIER IF STAGE DETECTOR AMPLIFIER OUTPUT E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\3. Principles of Communication System.p65 Detection is the process of recovering the modulating signal from the modulated carrier wave. We just saw that the modulated carrier wave contains the frequencies c and c ± m . In order to obtain the original message signal m(t) of angular frequency m, a simple method is shown in the from of a block diagram in figure. 42 E
JEE-Physics AM Wave RECTIFIER ENVELOPE m(t) (a) OUTPUT DETECTOR (c) (b) time time time Rectified wave Output (without RF component) AMinput wave The modulated signal of the form given in (a) of above figure is passed through a rectifier to produce the output shown in (b). This envelope of signal (b) is the message signal In order to retrieve m(t), the signal is passed through an envelope detector (which may consist of a simple RC circuit). The internet Students must be quite familiar with internet these days. The information provided by different these days. The information provided by different bodies all over the world is centralised at one place. Which is then used by anyone having a computer and internet facility. It's main uses are : (a) Emial (b) File transfer (c) WWW - World Wide Web (d) E-commerce (e) Chatting Facsimile (FA X) : FAX is abbreviation for facsimile which means exact reproduction. A fax machine sends a printed document or a photograph besides speech, music or coded data from one place to another by data communication system which, as we already know, consists of three elements. Tra nsmit ter Transmission channel Receiver E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\3. Principles of Communication System.p65 It scans the contents of a document (as an image, not test) to create electronic signals. These signals arethen sent to the destination (another FAX machine) in an orderly manner using telephone lines. At the destination, the signals are reconverted into a replica of the original document. Mobile telephony The concept of mobile telephony was developed first in 1970's and it was fully implemented in the following decade. The central concept of this system is to divide the serice area into a suitable number of cells centred on an office called MTSO (Mobile Telephone Switching Office). Each cell contains a low-power transmitter called a base station and castomers. When a mobile receiver crosses the coverage area of one base station, it is necessary for the mobile user to be transferred to another base station. This procedure is called handover or handoff. This process is carried out very rapidly, to the extent that the consumer does not even notice it. Mobile telephones operate typically in the UHF range of frequencies (about 800-950 MHz). E 43
JEE-Physics SOLVED EXAMPLE Ex.1 Suppose a pure Si crystal has 5 × 1028 atoms m–3. It is doped by 1 ppm concentration of pentavalent Sol As. Calculate the number of electrons and hole. (Given that ni = 1·5 × 1016 m–3.) Ex.2 FGH JKIHere ne ND = Sol. 5 1028 = 5 × 1022 m–3 Ex.3 106 Sol. e jnh = Ex.4 n2 = 1 5 1016 2 = 2 25 1032 = 4·5 × 109 m–3 Sol. 5 1022 5 1022 i Ex.5 Sol. ne Ex.6 Pure Si at 300K has equal electron (n ) and hole (n ) concentrations of 1.5 × 1016 m–3. Doping in indium eh increases n to 4.5 × 1022 m–3. Calculate n is in the doped Si. he n2 (1.5 1016 )2 At thermal equilibrium n n = n2 n= i = 4.5 1022 = 5 × 109 m–3 eh i e n h The diagram shows a piece of pure semiconductor S, in series with a variable resistor R, and a source of constant voltage V. Would you increase or decrease the value of R to keep the reading of ammeter (A) constant, when semiconductor S is heated ? Give reason. Value of R should be increased with the increase in temperature of semiconductor as circuit resistance decreases and current tends to increase. A p–n junction diode can withstand current up to 10 mA under forward bias, the diode has a potential drop of 0.5V across it which is assumed to be independent of current what is the maximum voltage of battery used to forward bias the diode when a resistance of 200 is connected in series with it. Potential drop is 0.5 and current is 10 mA. so resistance of diode is 0.5 Total resistance of circuit is = 200 + 50 = 250 = = 50 10 10 3 So maximum current is Vmax. =10 × 10–3 Vmax.= 2.5 volt 250 If current in given circuit is 0.1A then calculate resistance of P–N junction. 5 E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\3. Principles of Communication System.p65 Let resistance of PN junction be R then I = R 30 10 = 0.1 R = 10 What is the value of current I in given circuits (a) (b) (c) 2.7 0.7 20 (c) 18 6 Sol. (a) I = 1 103 = 2mA (b) I = 10 10 = 1 A I = 500 = 24 mA 44 E
JEE-Physics Ex.7 For half wave rectifier if load resistance R is 2k and P–N junction resistance R is 2k determine rectification Sol. L d Ex.8 efficiency. Sol. GF IJ 40.6 RL 2k Ex.9 H K b gHWR Rd RL Sol. = 40.6 × 2 2 k = 20.3 % Calculate the value of V0 and I if the Si diode and the Ge diode start conducting at 0·7V and 0·3 volt respectively, in the given circuit. If the Ge diode connection be reversed, What will be the new values of V0 and I ? V0 = 12 – 0·3 = 11·7 V Current I = V0 117 = 2·34 mA R = 5 103 L On reversing the connections of Ge diode the output voltage V0' = 12 – 0·7=11·3 V The current in load I´ = V '0 113 = 2·26 mA RL = 5 103 A common emitter amplifier has a voltage gain of 50, an input impedance of 200 and an output impedance of 400. Calculate the power gain of the amplifier. F IVoltage gain = () R 0 = 50 = 50 R i = 50 200 = 25 HG KJR i R0 400 FHG IKJ GFH JKIPower gian = (2) R0 = (25)2 400 = 625 × 2 = 1250 Ri 200 E x . 1 0 An n-p-n transistor in a common emitter mode is used as a simple voltage amplifier with a collector connected to load resistance RL and to the base through a resistance RB. The collector-emitter voltage VCE = 4V, the base-emitter voltage VBE = 0.6V, current through collector is 4 mA and the current amplification factor = 100. Calculate the values of RL and RB. Sol. Given, ic = 4 mA applying Kirchhoff's second law in loop 1, E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\3. Principles of Communication System.p65 VCE = 8 – iC RL RL = 8 VCE 84 = 1× 103 = 1 k iC = 4 10 3 IB IC 4mA = 4 × 10–5 A 100 VBE = 8 – IBRB RB = 8 VBE 8 0.6 = 185 k IB 4 105 Ex.11 A transistor has a current amplification factor (current gain) of 50. In a CE amplifier circuit, the collector Sol. resistance is chosen as 5k and the input resistance is 1k. Calculate the output voltage if input voltage E is 0·01V. FGH JKIFor transistor amplifier V0 = F I5 RC Vi GH JK= (50) 1 (0·01) = 2·5 V RB 45
JEE-Physics E x . 1 2 In a transistor connected in common emitter mode R0 = 4 k, Ri = 1 k, iC = 1 mA and ib = 20 µA. Find the voltage gain. HGF JKI GFH IJK HGF KIJ FHG JIK HFG IKJSol. R0 ic R0 1 103 4 Voltage gain AV = R i = i b R i = 20 106 1 = 200 E x . 1 3 For given CE biasing circuit, if voltage across collector–emitter is 12V and current gain is 100 and base current is 0.04mA then determine the value of collector resistance RC . Sol. VCE = VCC – IC × RC RC = VCC VCE VCC VCE 100 20 12 3 2 k IC IB 0.04 10 E x . 1 4 Write down the actual logic operation carried by the following circuit. Explain your answer. Sol. XOR operation E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\3. Principles of Communication System.p65 Y= AB AB which is boolean expression for XOR gate E 46
JEE-Physics ELASTICITY A body is said to be rigid if the relative positions of its constituent particles remains unchanged when external deforming forces are applied to it. The nearest approach to a rigid body is diamond or carborundum. Actually no body is perfectly rigid and every body can be deformed more or less by the application of suitable forces. All these deformed bodies however regain their original shape or size, when the deforming forces are removed. The property of matter by virtue of which a body tends to regain its original shape and size after the removal of deforming forces is called elasticity. Some terms related to elasticity : • Deforming Force External force which try to change in the length, volume or shape of the body is called deforming force. • Perfectly Elastic Body The body which perfectly regains its original form on removing the external deforming force, is defined as a perfectly elastic body. Ex. : quartz – Very nearly a perfect elastic body. • Plastic Body (a) The body which does not have the property of opposing the deforming force, is known as a plastic body. (b) The bodies which remain in deformed state even after removed of the deforming force are defined as plastic bodies. • Internal restoring force When a external force acts at any substance then due to the intermolecular force there is a internal resistance produced into the substance called internal restoring force. At equilibrium the numerical value of internal restoring force is equal to the external force. STRESS The internal restoring force acting per unit area of cross–section of the deformed body is called stress. Stress Internal restoring force = Fint ernal = Fexternal Area of cross section A A Stress depends on direction of force as well as direction of area of application so it is tensor. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 SI Unit : N–m–2 Dimensions : M1 L–1 T–2 There are three types of stress :- ( a ) Longitudinal Stress : When the stress is normal to the surface of body, then it is known as longitudinal stress. There are two types of longitudinal stress (i) Tensile Stress : The longitudinal stress, produced due to increase in length of a body, is defined as tensile stress. F wall F tensile stress tensile stress E1
JEE-Physics (ii) Compressive stress : The longitudinal stress, produced due to decrease in length of a body, is defined as compressive stress. F Compressive Stress ( b ) Volume Stress If equal normal forces are applied every one surface of a body, then it undergoes change in volume. The force opposing this change in volume per unit area is defined as volume stress. ( c ) Tangential Stre ss or Shear Stre ss When the stress is tangential or parallel to the surface of a body then it is known as shear stress. Due to this stress, the shape of the body changes or it gets twisted. F AF AB B L L F F DC DC STRAIN The ratio of change of any dimension to its original dimension is called strain. change in size of the body Strain = original size of the body It is a unitless and dimensionless quantity. There are three types of strain : Type of strain depends upon the directions of applied force. (a) Longitudinal strain = change in length of the body L = initial length of the body L L L change in volume of the body V node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 ( b ) Volume strain= original volume of the body = V (c) Shear strain A F When a deforming force is applied to a body parallel to its surface then A' B B' its shape (not size) changes. The strain produced in this way is known as shear strain. The strain produced due to change of shape of the body L displacement of upper face is known as shear strain. tan L or = L = distance between two faces F 2 DC E
JEE-Physics Relation Between angle of twist and Angle of shear fixed B When a cylinder of length '' and radius 'r' is fixed at one end and tangential force is applied at the other end, then the cylinder gets twisted. Figure ' shows the angle of shear ABA' and angle of twist AOA'. Arc AA' = r and Arc AA' = so r = r where = angle O twisted A A' of twist, = angle of shear GOLDEN KEY POINTS • When a material is under tensile stress restoring force are caused by intermolecular attraction while under compressive stress, the restoring force are due to intermolecular repulsion. • If the deforming force is inclined to the surface at an angle such that and 90° then both tangential and normal stress are developed. • Linear strain in the direction of force is called longitudinal strain while in a direction perpendicular to force lateral strain. Stress – Strain Graph Breaking Elastic E Y B strength Region C Elastic P Limit node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 Plastic Region PrLiopmiotrtionStress 0 Strain Proportion Limit : The limit in which Hook's law is valid and stress is directly proportional to strain is called proportion limit. Stress Strain Elastic limit : That maximum stress which on removing the deforming force makes the body to recover completely its original state. Yield Point : The point beyond elastic limit, at which the length of wire starts increasing without increasing stress, is defined as the yield point. Breaking Point : The position when the strain becomes so large that the wire breaks down at last, is called breaking point. At this position the stress acting in that wire is called breaking stress and strain is called breaking strain. Elastic after effect We know that some material bodies take some time to regain their original configuration when the deforming force is removed. The delay in regaining the original configuration by the bodies on the removal of deforming force is called elastic after effect. The elastic after effect is negligibly small for quartz fibre and phosphor bronze. For this reason, the suspensions made from quartz and phosphor-bronze are used in galvanometers and electrometers. For glass fibre elastic after effect is very large. It takes hours for glass fibre to return to its original state on removal of deforming force. E3
JEE-Physics Elastic Fatigue : The loss of strength of the material due to repeated strains on the material is called elastic fatigue. That is why bridges are declared unsafe after a long time of their use. Creep : If a small force is applied for a long time then it causes breaking of metal. For example A fan is hung for 200 years then the shaft will break. Elastic Hysteresis : load or stress The strain persists even when the stress is removed. This lagging behind of strain load ldoeacrd eiansicrnegasing is called elastic hysteresis. This is the reason why the values of strain for same stress are different while increasing the load and while decreasing the load. extension or strain Breaking Stress : The stress required to cause actual facture of a material is called the breaking stress Breaking stress = F/A GOLDEN KEY POINTS • Breaking stress also measures the tensile strength. • Metals with small plastic deformation are called brittle. • Metals with large plastic deformation are called ductile. • Elasticity restoring forces are strictly conservative only when the elastic hysteresis is zero. i.e. the loading and unloading stress – strain curves are identical. • The material which have low elastic hysteresis have also low elastic relaxation time. Example 100N Find out longitudinal stress and tangential stress on a fixed block. 300 1m Solution 5m 2m Longitudinal or normal stress 100 sin 300 1 5 2 = 5 N/m2 Tangential stress 100 cos 300 3N / m2 2 5 2 5 Example node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 The breaking stress of aluminium is 7.5 × 108 dyne cm–2. Find the greatest length of aluminium wire that can hang vertically without breaking. Density of aluminium is 2.7 g cm–3. [Given : g = 980 cm s–2] Solution Let be the greatest length of the wire that can hang vertically without breaking. Mass of wire m = cross–sectional area (A) × length () × density (), Weight of wire = mg = Ag This is equal to the maximum force that the wire can withstand. Breaking stress = Ag g 7.5 × 108 = × 2.7 × 980 A 7.5 108 cm = 2.834 × 105cm= 2.834 km 2.7 980 4E
JEE-Physics Hooke's Law If the deformation is small, the stress in a body is proportional to the corresponding strain, this fact is known stress as Hooke's Law. Within elastic limit : stress strain strain =constant This constant is known as modulus of elasticity or coefficient of elasticity. The modulus of elasticity depends only on the type of material used. It does not depend upon the value of stress and strain. Example Find out the shift in point B, C and D FL MgL 10 10 0.1 S o l . LB = LAB = AY AY = 10 7 2.5 1010 =4 × 10–3 m = 4mm 100 0.2 LC = LB + LBC =4 × 10–3 + 107 4 1010 =4 × 10–3 + 5 × 10–3 =9mm 100 0.15 LD=LC + LCD = 9 × 10–3 + 10 7 1 1010 =9 × 10–3 + 15 × 10–3 =24 mm Young's Mo dulus of Elast icit y 'Y' Within elastic limit the ratio of longitudinal stress and longitudinal strain is called Young's modulus of elasticity. longitudinal stress F / A F L Y = longitudinal strain = / L = A Within elastic limit the force acting upon a unit area of a wire by which the length of a wire becomes double, is equivalent to the Young's modulus of elasticity of material of a wire. If L is the length of wire, r is radius and is the increase in length of the wire by suspending a weight Mg at its one end then Young's modulus Mg / r2 MgL of elasticity of the material of wire Y= / L = r2 Unit of Y : N/m2 Dimensions of Y : M1L–1T–2 Increment of length due to own weight Let a rope of mass M and length L is hanged vertically. As the tension of different point on the rope is different. Stress as well as strain will be different at different point. (i) maximum stress at hanging point (ii) minimum stress at lower point dx node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 Consider a dx element of rope at x distance from lower end then tension T M x g L T M xg So stress = A = L A dy Let increase in length of dx is dy then strain = dx M xg So Young modulus of elasticity Y = stress LA M xg = dy / dx L dx = Y dy strain A Mg Mg L2 = Y = MgL dy For full length of rope LA 0L 0 LA 2 2AY xdx =Y [Since the stress is varying linearly we may apply average method to evaluate strain.] E5
JEE-Physics Example A thin uniform metallic rod of length 0.5 m and radius 0.1 m rotates with an angular velocity 400 rad/s in a horizontal plane about a vertical axis passing through one of its ends. Calculate (a) tension in the rod and (b) the elongation of the rod. The density of material of the rod is 104 kg/m3 and the Young's modulus is 2 × 1011 N/ m2. Solution (a) Consider an element of length dr at a distance r from the axis of rotation as shown in figure. The centripetal force acting on this element will be dT dmr2 Adr r2 . r dr L As this force is provided by tension in the rod (due to elasticity), so the tension in the rod at a distance r from the axis of rotation will be due to the centripetal force due to all elements between x = r to x = L i.e., T = L A2rdr 1 A 2 L2 r2 ...(i) 2 r So here T = 1 × 104 × × 10–2 × (400)2 1 2 r2 = 8 106 1 r2 N 2 2 4 (b) Now if dy is the elongation in the element of length dr at position r then strain dy stress T 1 2 = = = [L2 – r2] dr Y AY 2 Y So the elongation of the whole rod 400 2 0.5 3 2 1011 2 104 × L 2Y L L2 r 2 dr = 1 2 L3 1 = 1 10 3 m 3 Y = 3 0 3 Example Find out the elongation in block. If mass, area of cross-section and young modulus of block are m, A and y respectively. smooth F Solution x node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 x dx m' T T T+dT F dx Acceleration, F m m F Fx a = m then T = ma where m = x; T = x m = Tdx Tdx Fxdx F Elongation in element ‘dx’ = Ay , total elongation, = Ay = Ay = 2Ay o o 6 E
JEE-Physics Bulk modulus of elasticity 'K' or 'B' Within elastic limit the ratio of the volume stress and the volume strain is called bulk modulus of elasticity. K or B = volume stress = F / A P volume strain V V VV The minus sign indicates a decrease in volume with an increase in stress. Unit of K : N m–2 or pascal Bulk modulus of an ideal gas is process dependence. dP For isothermal process PV = constant PdV + VdP = 0 P= dV / V So bulk modulus = P dP For adiabatic process PV = constant PV 1dV V dP 0 PdV VdP 0 P dV / V So bulk modulus = P dP For any polytropic process PVn = constant nPVn–1dV|VndP=0 PdV+VdP=0 nP = dV / V So bulk modulus = nP 1 Compressibility : The reciprocal of bulk modulus of elasticity is defined as compressibility. C K Modulus of Rigidity '' Within elastic limit the ratio of shearing stress and shearing strain is called modulus of rigidity of a material. shearing stress Ftan gential Ftan gential shearing strain A = A = Note : Angle of shear '' always take in radian node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 Poisson's Ratio ( In elastic limit, the ratio of lateral strain and longitudinal strain is called Poisson's ratio. D d lateral strain D d D L = longitudinal strain = , where = D = D & = L D F E7
JEE-Physics Work done in stretching a wire (Potential energy of a stretched wire) When a wire is stretched, work is done against the interatomic forces, which is stored in the form of elastic potential energy. For a wire of length L stretched by a distance x, the restoring elastic force is : o x L0 L0 F = stress × area = Y L A o The work has to be done against the elastic restoring forces in stretching dx x F YA dW = F dx = L 0 x dx The total work done in stretching the wire from x = 0 to x = is, then YA YA x2 YA ()2 2L o W xdx Lo 2 0 = 0 Lo 11 W = × Y × (strain)2 × original volume= (stress) (strain) (volume) 22 GOLDEN KEY POINTS stressstrain • The value of K is maximum for solids and minimum for gases. • For any ideal rigid body all three elastic modulus are infinite . node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 • is the characteristic of solid material only as the fluids do not have fixed shape. • Potential energy density = area under the stress–strain curve. • Young's modulus = Slope of the stress–strain curve Example A steel wire of 4.0 m in length is stretched through 2.00 mm. The cross–sectional area of the wire is 2.0 mm2. If Young's modulus of steel is 2.0 × 1011 N/m2 find (i) the energy density of wire (ii) the elastic potential energy stored in the wire. Solution (i) The energy density of stretched wire 1 1 1 2 10 3 2 × 2.0 × 1011× = 2 × stress × strain= 2 × Y× (strain)2= = 2.5 × 104 J/m3 2 4 (ii) Elastic potential energy = energy density × volume = 2.5 × 104 × (2.0 × 10–6) × 4.0 J = 20 × 10–2 = 0.20J 8E
JEE-Physics Example Find the depth of lake at which density of water is 1% greater than at the surface. Given compressibility K = 50 × 10–6 /atm. Solution P V P m B = V / V V =– B Patm V We know P = P + hg and m = V = constant P=hg h atm dV dV 0 d dV P 1 hg m = – i.e. = = = Patm+ P V B 100 B V [assuming = constant]; hg B = 1 hg 1 1 105 100K 100 50 106 100 105 100 103 h = 5000 106 1000 10 = 50 m = 2km Example A rubber cube of side 5 cm has one side fixed while a tangential force equal to 1800 N is applied to opposite face. Find the shearing strain and the lateral displacement of the strained face. Modulus of rigidity for rubber is 2.4 × 106 N/m2. Solution Fx x x Stress = = L AL Strain = F = 1800 106 180 = 3 = 0.3 rad A 25 10 4 2.4 = 10 25 24 x as L = 0.3 x = 0.3 × 5 × 10–2 = 1.5 × 10–2 m = 1.5 mm Analogy of Rod as a spring A,Y k Y = stress F or F = AY strain Y = A F F node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 AY = constant, depends on type of material and geometry of rod. F = k where k = AY = equivalent spring constant. For the system of rods shown in figure (a), the replaced spring system is shown in figure (b) [two spring in series]. Figure (c) represents equivalent spring system. A1,Y1 k1=A1Y1 1 (c) keq= k1k2 1 A2,Y2 k1+k2 (b) (a) k2=A2Y2 2 2 E9
JEE-Physics Figure (d) represents another combination of rods and their replaced spring system. A1,Y1,1 A3,Y3,3 k1=A1Y1 1 k3=A3Y3 3 F (d) (e) A2Y2 2 A2,Y2,2 k2= F Example A mass ‘m’ is attached with rods as shown in figure. This mass is slightly stretched and released whether the motion of mass is S.H.M., if yes then find out the time period. 1 A1,Y1 k1 2 A2,Y2 k2 m m Solution k1k2 m m(k1 k2 ) where k = A 1 Y1 and A 2 Y2 Here k = and T = 2 k eq = 2 1 k= eq k1 k2 k1k2 2 1 2 Example A1, y1 Hanger is mass-less. A ball of mass ‘m’ drops from a height ‘h’, which sticks to hanger after striking. Neglect over turning, find out the maximum extension 1 m h in rod. Assuming rod is massless. Let maximum extension be x . A1, y1 max 1 h Sol. Applying energy conservation mg (h + x ) = 1 k1k2 x2 Hanger Enode6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 max 2 k1 k2 k2 where k = A1y1 ; k = A2y2 & K eq A1 A 2 y1 y2 therefore 1 2 k1 1 2 A1y12 A 2 y 1 1 k x2 – 2mgx – 2mgh = 0 eq max max 2mg 4m2g2 8mghkeq mg m2g2 2mgh x = 2 k eq = k 2 k eq max k eq eq 10
Otherway by S.H.M. JEE-Physics keq and v = a2 y2 keq m keq 2mgh m2g2 m a2 y2 2gh = k eq k 2 = a mg 2gh eq y= keq equilibrium mg m2g2 2mgh a position Maximum extension = a + y = keq + k eq k eq Thermal Stress : If temperature of rod is increased by T, then change in length = T; strain = T A,Y, T But due to rigid support, there is no strain. Supports provide force on stresses to keep the length of rod same stress F FF Y = F If T = (+) tive strain FF FF Thermal stress = y strain = Y T If T = (–) tive F Y T F = AY T A node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 Example When composite rod is free, then composite length increases to 2.002 m for temperature 20°C to 120°C. When composite rod is fixed between the support, there is no change in component length find y and of steel, if y = 1.5 × 1013 N/m2 = 1.6 × 10–5/°C. cc steel copper 0.5m 2m Solution = s T + cT 0.002 = [1.5 s + 0.5 × 1.6 ×10–5] × 100 = 1.2 10 5 = 8 × 10–6/°C s c s 1.5 there is no change in component length 11 E
JEE-Physics Final position of junction Initial position of junction For steel //////////// //////////// Steel Copper //////////// For copper x sas T Steel F Fs //////////// Ays cacT F Copper Fc Ayc x = s T – Fs 0 F s T ....(i) s A Ys AYs ...(ii) x = Fc – c T = 0 F = cT A Yc c Ayc Y = Y c 1.5 1013 16 105 (ii)/(i) s c = s 8 106 Y = 3 × 1013 N/m2 s Applications of Elasticity Some of the important applications of the elasticity of the materials are discussed as follows : The material used in bridges lose its elastic strength with time bridges are declared unsafe after long use. To e st imate the maximum height of a mountain : node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 The pressure at the base of the mountain = hg = stress. The elastic limit of a typical rock is 3 × 108 N m–2 The stress must be less than the elastic limits, otherwise the rock begins to flow. h < 3 108 < 3 108 < 104 m ( = 3 × 103 kg m–3 ; g = 10 ms–2) or h = 10 km g 3 103 10 It may be noted that the height of Mount Everest is nearly 9 km. E 12
JEE-Physics Torsion consta nt of a w ire C = r 4 Where is modulus of rigidity r and is radius and length of wire respectively. 2 (a) Toque required for twisting by angle = C 1 (b) Work done in twisting by angle , W = C2. 2 Effect of Temperature on elast icit y When temperature is increased then due to weakness of inter molecular force the elastic properties in general decreases i.e. elastic constant decreases. Plasticity increases with temperature. For example, at ordinary room temperature, carbon is elastic but at high temperature, carbon becomes plastic. Lead is not much elastic at room temperature but when cooled in liquid nitrogen exhibit highly elastic behaviour. For a special kind of steel, elastic constants do not vary appreciably temperature. This steel is called 'INVAR steel'. Effect of Impurity on elasticity Y is slightly increase by impurity. The inter molecular attraction force inside wire effectively increase by impurity due to this external force can be easily opposed. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 E 13
JEE-Physics node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 SURFACE TENSION Surface tension is basically a property of liquid. The liquid surface behaves like a stretched elastic membrane which has a natural tendency to contract and tends to have a minimum possible area. This property of liquid is called surface tension. INTERMOLECULAR FORCES l The force which acts between the atoms or the molecules of different substances is called intermolecular force. This force is of two types. (a) Cohesive force – The force acting between the molecules of one type of molecules of same substance is called cohesive force. (b) Adhesive force – The force acting between different types of molecules or molecules of different substance is called adhesive force. l Intermolecular forces are different from the gravitational forces not obey the inverse–square law l The distance upto which these forces effective, is called molecular range. This distance is nearly 10–9 m. Within this limit this increases very rapidly as the distance decreases. l Molecular range depends on the nature of the substance Examples: • Water Glass : Water wets glass surface but mercury does not. Because when water comes in contact with glass the adhesive force acts between water and glass molecules. As adhesive force is greater than the cohesive force of water molecules, the water molecules, cling with glass surface and surface become wet. In case of mercury adhesive force is less than that of cohesive force and Hg–molecules do not cling with glass surface and surface does not wet with Hg–molecules. • Oil–water : Since Cohesive force of water >Adhesive force oil–water> Cohesive force of oil. (i) If water drop is poured on the surface of oil, it contracts in the form of globule. (ii) If oil drop is poured on the surface of water it spreads to a larger area in the form of thin film. • Ink–paper : Since adhesive force between ink–paper > cohesive force on ink, the ink sticks to the paper. EXPL ANATION OF SURFACE TENSION (Molecular theor y of sur face Tension) Laplace firstly explained the phenomenon of surface tension on the basis of intermolecular forces. According to him surface tension is a molecular phenomenon and its root causes are electromagnetic forces. He explained the cause of surface tension as described under. If the distance between two molecules is less than the molecular range C ( 10–9 m) then they attract each other, but if the distance is more than this then the attraction becomes negligible. If a sphere of radius C with a molecule at centre is drawn, then only those molecules which are enclosed within this sphere can attract or be attracted by the molecule at the centre of the sphere. This sphere is called sphere of molecular activity or sphere of influence. In order to understand the tension acting at the free surface of liquid, let us consider four liquid molecules like A, B, C and D along with their spheres of molecular activity. (a) According to figure D sphere is completely inside liquid. So molecule is attracted equally in all directions and hence resultant force is equal to zero. (b) According to figure sphere of molecule C is just below the liquid surface. So resultant force is equal to zero. (c) The molecule B which is a little below the liquid surface is more attracted downwards due to excess of molecules downwards. Hence the resultant force is acting downwards. 14 E
JEE-Physics (d) Molecule A is situated at surface so that its sphere of molecular activity is half outside the liquid and half inside. Only down portion has liquid molecules. Hence it experiences a maximum downward force. Thus all the molecules situated between the surface and a plane XY, distant C below the surface, experience a resultant downward cohesive force. When the surface area of liquid is increased molecules from the interior of the liquid rise to the surface. A B liquid surface X C net downward force CY D zero resultant force P As these molecules reach near the surface, work is done against the downward cohesive force. This work is stored in the molecules in the form of potential energy. Thus the potential energy of the molecules lying in the surface is greater than that of the molecules in the interior of the liquid. A system is in stable equilibrium when its potential energy is minimum. Hence in order to have minimum potential energy the liquid surface tends to have minimum number of molecules in it. In other words the surfaces tends to contract to a minimum possible area. This tendency is exhibited as surface tension. GOLDEN KEY POINTS • Surface tension is a scalar quantity. • It acts tangential to liquid surface. • Surface tension is always produced due to cohesive force. • More is the cohesive force, more is the surface tension. • When surface area of liquid is increased, molecules from the interior of the liquid rise to the surface. For this, work is done against the downward cohesive force. As a result, its potential energy increases and internal energy decreases. So on increase in surface area cooling occurs. If liquid temperature remains same, then extra energy may be given by external agency. So the molecules in the surface have some additional energy due to their position. This additional energy per unit area of the surface is called Surface Energy, free surface energy or surface energy density. DEPENDENCY OF SURFACE TENSIONnode6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 l On Cohesive Force Those factors which increase the cohesive force between molecules increase the surface tension and those which decrease the cohesive force between molecules decrease the surface tension. l On Impurities If the impurity is completely soluble then on mixing it in the liquid, its surface tension increases. e.g., on dissolving ionic salts in small quantities in a liquid, its surface tension increases. If the impurity is partially soluble in a liquid then its surface tension decreases because adhesive force between insoluble impurity molecules and liquid molecules decreases cohesive force effectively, e.g. (a) On mixing detergent in water its surface tension decreases. (b) Surface tension of water is more than (alcohol + water) mixture. l On Temperature On increasing temperature surface tension decreases. At critical temperature and boiling point it becomes zero. Note : Surface tension of water is maximum at 4°C E 15
JEE-Physics l On Contamination The dust particles or lubricating materials on the liquid surface decreases its surface tension. l On Electrification The surface tension of a liquid decreases due to electrification because a force starts acting due to it in the outward direction normal to the free surface of liquid. DEFINITION OF SURFACE TENSION Surface tension can be defined in the form of an imaginary line on the surface or by F B relating it to the work done. The force acting per unit length of an imaginary line drawn AF on the free liquid surface at right angles to the line and in the plane of liquid surface, is defined as surface tension. Let an imaginary line AB be drawn in any direction on a liquid surface. The surface on either side of this line exerts a pulling force, which is F perpendicular to line AB. If force is F and length of AB is L then T = L SI UNITS : N/m and J/m2 CGS UNITS : dyne/cm and erg/cm2 Dimensions : M1L0 T–2 Illustrations: Ts in Ts in • When any needle floats on the liquid surface then 2T sin = mg T T Ex. A mosquito sitting on a liquid surface. Tcos Tcos Mg • If the needle is lifted from the liquid surface then required excess F l TT force will be Fexcess = 2T Minimum force required F = mg + 2T min • Required excess force for a circular thick ring (or hollow disc) having internal and external radii r and r is dipped in and taken out from r2 12 r1 liquid F = F + F = T(2 r ) + T(2 r ) = 2T(r1 + r) TT excess 1 2 1 2 2 • Required excess force for a circular ring (r = r = r) node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 12 F = 2 T( r + r) = 4 r T excess TT W • Required excess force for a circular disc (r = 0, r = r) r 12 E F = 2 rT excess T 16
JEE-Physics Surface Energy D C According to molecular theory of surface tension the molecule in the surface have F=2T B some additional energy due to their position. This additional energy per unit area of ST the surface is called ' Surface energy\". Let a liquid film be formed on a wire frame Dx and a straight wire of length can slide on this wire frame as shown in figure. The film A has two surfaces and both the surfaces are in contact with the sliding wire and hence, exert forces of surface tension on it. If T be the surface tension of the solution, each surface will pull the wire parallel to itself with a force T. Thus, net force on the wire due to both the surfaces is 2T. Apply an external force F equal and opposite to it to keep the wire is equilibrium. Thus, F=2T Now, suppose the wire is moved through a small distance dx, the work done by the force is, dW = F dx = (2T) dx But (2) (dx) is the total increase in area of both the surfaces of the film. Let it be dA. Then, dW = T dA T = dW/dA Thus, the surface tension T can also be defined as the work done in increasing the surface area by unity. Further, since there is no change in kinetic energy, the work done by the external force is stored as the potential energy of the new surface. dU T = [as dW = dU] dA Special Cases • Work done (surface energy) in formation of a drop of radius r = Work done against surface tension W = Surface tension T × change in area A = T × 4r2 = 4r2T • Work done (surface energy) in formation of a soap bubble of radius r : W = T × A or W = T × 2 × 4r2 = 8r2T [ soap bubble has two surfaces] Example The length of a needle floating on water is 2.5 cm. Calculate the added force required to pull the needle out of water. [T = 7.2 × 10–2 N/m] Solution The force of surface tension F = T× 2 ( Two free surfaces are there) F = 7.2 × 10–2 × 2 × 2.5 × 10–2 = 3.6 × 10–3 N Example A paper disc of radius R from which a hole of radius r is cut out, is floating in a liquid of surface tension, T. What will be force on the disc due to surface tension? Solution node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 FF T = L = 2(R r) F = 2 (R + r)T Example Calculate the work done against surface tension in blowing a soap bubble from a radius 10 cm to 20 cm if the surface tension of soap solution is 25 × 10–3 N/m. Then compare it with liquid drop of same radius. Solution (i) For soap bubble : Extension in area = 2 × 4r 2 –2 × 4r 2 =8 [(0.2)2– (0.1)2] = 0.24 m2 21 Work done W = surface tension × extension in area = 25 × 10–3 × 0.24 = 6 × 10–3 J 1 (ii) For Liquid Drop : in case of liquid drop only one free surface, so extension in area will be half of soap bubble W2 = W1 = 3 × 10–3 J 2 E 17
JEE-Physics SPLITTING OF BIGGER DROP INTO SMALLER DROPLETS If bigger drop is spitted into smaller droplets then in this process volume of liquid always remain conserved. Let bigger drop has radius R. It is splitted into n smaller drops of radius r then by conservation of volume (i) 4 R 3 n 4 r 3 n R 3 r= R 3 3 r n1/ 3 (ii) Initial surface area = 4R2 and final surface area = n(4r2) Therefore initial surface energy E = 4R2T and final surface energy E = n(4r2T) i f Change in area A = n4r2 – 4R2 = 4 (nr2 – R2) Therefore the amount of surface energy absorbed i.e. E = E – E = 4T (nr2 – R2) f i Magnitude of work done against surface tension i.e. W = 4 (nr2 – R2)T R 1 W = 4R3T 1 1 W = 4T (nr2 – R2) = 4R2 T (n1/3 – 1) = 4R2T r r R In this process temperature of system decreases as energy gets absorbed in increasing surface area. W = J m s = 4R3T 1 1 4R3 T 1 1 = 3T 1 1 r R 4 R3Js r R Js r R 3 Where = liquid density, s = liquid's specific heat Thus in this process area increasing, surface energy increasing, internal energy decreasing, temperature decreasing, and energy is absorbed. Example A big drop is formed by coalescing 1000 small droplets of water .What will be the change in surface energy. What will be the ratio between the total surface energy of the droplets and the surface energy of the big drop ? Solution 4 4R By conservation of volume R3 = 1000 × r3 r = 3 3 10 R 2 Surface energy of 1000 droplets = 1000 × T × 4 10 = 10 (T × 4R2) Surface energy of the big drop = T × 4R2 Surface energy will decrease in the process of formation of bigger drop, hence energy is released and temperature node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 increases . total surface energy of 1000 droplets = 10(T 4R 2 ) 10 surface energy of big drop T 4 R 2 1 Example A water drop of radius 1mm is broken into 106 identical drops. Surface tension of water is 72 dynes/cm. Find the energy spent in this process. Solution As volume of water remains constant, so 4 R 3 = n 4 r3 r = R 3 3 n1 3 Increase in surface area A = n (4r2) – 4R2 = 4 (n1/3 – 1) R2 = 4 (100 – 1)10–6 Energy spent = TA = 4 × 99 × 10–6 × 72 × 10–3 = 89.5 × 10–6 J 18 E
JEE-Physics EXCESS PRESSURE INSIDE A CURVED LIQUID SURFACE The pressure on the concave side of curved liquid surface is greater than that on the convex side. Therefore pressure difference exists across two sides of a curved surface. This pressure difference is called excess pressure. Excess pressure inside the drop Let a drop of radius r having internal and external pressure P and P respectively, so that excess pressure i0 P =(P – P ). ex i 0 If the radius of the drop is changed from r to (r+dr) then dr Work done = F.dr = (PA) dr = P 4r2dr r Pi P0 Change in surface area = r(r + dr)2–4r2= 8rdr So by definition of surface energy T = W 4 r2Pdr = Pi P0 2T A = 8 rdr P = r ex Excess pressure inside soap bubble : P0 Since the soap bubble has two surfaces. The excess pressure will get doubled as compared to a drop r Pi 2T 2T 4T P’ P –P' = r , P' – P = r excess pressure = P –P = r i 0 i0 GOLDEN KEY POINTS • For liquid surface, pressure on concave side is always higher than convex side low high . high low • If a bubble is formed inside a liquid, the pressure inside the bubble is more than the pressure outside the bubble. • A soap film is formed in a wire frame. A loop of thread is lying on the film. If the film inside the loop is broken then the tension in the thread will be 2Tr. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 • In the following arrangement, air will flow from bubble A to B if T and T are opened, because pressure in A 23 is greater than in B. T2 T1 T3 AA B • 2AT The force required to separate two plates of Area A is given by F = d E 19
JEE-Physics Example Prove that If two bubbles of radii r and r coalesce isothermally in vacuum then the radius of new bubble will be 12 r r12 r22 Solution When two bubbles coalesce then number of molecules of air will remain constant and temperature also constant 4T 4 r13 4T 4 r23 4T 4 r 3 r12 r22 3 r2 3 r 3 so n + n = n PV + PV = PV r1 = r= 1 2 11 22 Example Prove that If two bubbles of radii r and r (r < r ) come in contact with each other then the radius of curvature 1 21 2 r1r2 of the common surface r= r2 r1 . Solution P > P Small part of bubbles is in equilibrium r < r 1 2 12 r1 r2 P1 P2 P A – P A = 4T A 4T – 4T = 4T r = r1r2 1 2 r r1 r2 r r2 r1 Example Calculate the excess pressure within a bubble of air of radius 0.1 mm in water. If the bubble had been formed 10 cm below the water surface on a day when the atmospheric pressure was 1.013 × 105 Pa, then what would have been the total pressure inside the bubble? Surface tension of water = 73 × 10–3 N/m Solution Excess pressure P = 2T 2 73 103 =1460 Pa node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 = excess r 0.1 103 The pressure at a depth d, in liquid is P = hdg. Therefore, the total pressure inside the air bubble is 2T P = P + hdg + = (1.013 × 105) + (10 × 10–2 × 103 × 9.8) + 1460 in atm r = 101300 + 980 + 1460 = 103740 = 1.037 × 105 Pa Example The limbs of a manometer consist of uniform capillary tubes of radii are 10–3 & 7.2 × 10–4 m. Find out the correct pressure difference if the level of the liquid in narrower tube stands 0.2 m above that in the broader tube.(Density of liquid =103 kg/m3, Surface tension = 72 × 10–3 N/m) 20 E
JEE-Physics Solution If P and P are the pressures in the broader and narrower tubes of radii r and r respectively, the pressure just 12 12 2T 2T below the meniscus in the respective tubes will be P – and P – 1 r1 2 r2 2T 2T 1 1 P1 – P2 So that r1 r2 = hg or P –P = hg –2T r2 r1 12 P – P = 0.2 × 103 × 9.8 –2 × 72 × 10–3 1 1 = 1960– 97 = 1863 Pa 12 7 .2 10 4 14 104 ANGLE OF CONTACT (C) The angle enclosed between the tangent plane at the liquid surface and the tangent plane at the solid surface at the point of contact inside the liquid is defined as the angle of contact. The angle of contact depends the nature of the solid and liquid in contact. l Effect of Temperature on angle of contact On increasing temperature surface tension decreases, thus cosc increases cos c 1 T and c decrease. So on increasing temperature, c decreases. l Effect of Impurities on angle of contact (a) Solute impurities increase surface tension, so cosc decreases and angle of contact c increases. (b) Partially solute impurities decrease surface tension, so angle of contact c decreases. l Effect of Water Proofing Agent Angle of contact increases due to water proofing agent. It gets converted acute to obtuse angle. Table of a ngle of contact of various solid–liquid pair s node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 Solid - Liquid Pair ° C Glass -Normal water 8° Glass -Distilled water 0° Glass - Alcohol 0° Glass - Mercury Paraffin wax - Water 135° Silver - Water 108° 90° Shape of Liquid Surface When a liquid is brought in contact with a solid surface, the surface of the liquid becomes curved near the place of contact. The shape of the surface (concave or convex) depends upon the relative magnitudes of the cohesive force between the liquid molecules and the adhesive force between the molecules of the liquid and the solid. The free surface of a liquid which is near the walls of a vessel and which is curved because of surface tension is known as meniscus. The cohesive force acts at an angle 45° from liquid surface whereas the adhesive force acts at right angles to the solid surface. The relation between the shape of liquid surface, cohesive/adhesive forces, angle of contact, etc are summarised in the table below : E 21
JEE-Physics • Relation between cohesive and adhesive force FA concave surface FA horizontal surface FA convex surface FR FC FR FC FC water water FR mercury silver glass glass FA FC FA FC FA FC 2 2 2 • Shape of meniscus Concave Plane Convex • Angle of contact < 9 0 ° = 9 0 ° > 9 0 ° • Shape of liquid drop C C C • Level of liquid • Wetting property (Acute angle) (Right angle) (Obtuse angle) • Example C C C Liquid rises up nor falls Liquid neither rises Liquid falls Liquid wets the Liquid does not wet Liquid does not wet solid surface the solid surface the solid surface Glass – Water Silver – Water Glass – Mercury CAPILLARY TUBE AND CAPILLARITY A glass tube with fine bore and open at both ends is known as capillary tube. The property by virtue of which a liquid rise or depress in a capillary tube is known as capillarity. Rise or fall of liquid in tubes of narrow bore (capillary tube) is called capillary action. Calculation of Capillary Rise P Enlarged view r C T T P h PC h AB r= radius of the capillary tube R r node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 PC R= Radius of the R = r meniscus cos When a capillary tube is first dipped in a liquid as shown in the figure, the liquid climbs up the walls curving the surface. The liquid continues to rise in the capillary tube until the weight of the liquid column becomes equal to force due to surface tension. Let the radius of the meniscus is R and the radius of the capillary tube is r. The angle of contact is surface tension is T, density of liquid is and the liquid rises to a height h. Now let us consider two points A and B at the same horizontal level as shown. By Pascal's law P = P P = P + gh P – P = gh ( P = P + gh) A B A C A C B C 22 E
JEE-Physics Now, the point C is on the curved meniscus which has P and P as the two pressures on its concave and convex AC sides respectively. PA PC 2T 2T 2T 2T cos gh 2T cos r gh h R r / cos r / cos rg Zurin's Law : The height of rise of liquid in a capillary tube is inversely proportional to the radius of the capillary tube, if T, 1 and g are constant h r or rh = constant. It implies that liquid will rise more in capillary tube of less radius and vice versa. GOLDEN KEY POINTS • For water and clean glass capillary c • If angle of contact is acute then cos is positive, so h is positive and liquid rises. If is obtuse then c c c cosc is negative, so h is negative, therefore liquid depresses. • Rise of liquid in a capillary tube does not obey the law of conservation of energy. • For two capillaries of radii r and r , the capillary rise h and h are such that h2 r1 . The relation h1 r2 12 12 between h and r is graphically represented as h r • Inside a satellite, water will rise upto top level but will not come outside. Radius of curvature(R') increases hR in such a way that final height h' is reduced and given by h ' . (It is in accordance with Zurin's law). R' • If a capillary tube is dipped into a liquid and tilted at an angle from vertical then the vertical height of liquid column remains same whereas the length of liquid column in the capillary tube increases. h h cos node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 h = cos • The height 'h' is measured from the bottom of the meniscus. However, there exist some liquid above this line rg h 1 r 3 also. If correction of this is applied then the formula will be T 2 cos • If a hollow sphere of radius r which has a fine hole, drowned in a vessel upto h depth, then liquid will not enter upto critical height h, given by hg= 2T cos [normally therefore cos r E 23
JEE-Physics Example Calculate the height to which water will rise in a capillary tube of diameter 1 × 10–3 m. [Given : surface tension of water is 0.072 N m–1, angle of contact is 0°, g = 9.8 m s–2 and density of water = 1000 kg m–3] Solution Height of capillary rise h 2T cos 5 2 0.072 cos 00 m 2.94 10 2 m rg 10 4 1000 9.8 Example Water rises to a height of 20 mm in a capillary. If the radius of the capillary is made one third of its previous value then what is the new value of the capillary rise? Solution Since h 2Tcos and for the same liquid and capillaries of difference radii h r = hr rdg 1 1 22 h2 r1 1 3 hence h = 3h = 3 × 20 mm = 60 mm h1 r2 21 1 / 3 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 24 E
JEE-Physics FLUID STATICS Matter exists in three states solid, liquid and gas. Liquids and gases are referred to as fluids. Any state of matter that can flow is a fluid. Study of a fluid at rest is called fluid statics or hydrostatics and the study of fluid in motion is called fluid dynamics of hydrodynamics. Both combined are called fluid mechanics. The intermolecular force in liquids are comparatively weaker than in solids. Therefore, their shape can be changed easily. Thus liquids assume the shape of the container. But their volume (or density) cannot be changed so easily. Liquids are incompressible and have free surface of their own. The intermolecular forces are weakest in gases, so their shape and size can be changed easily. Gases are compress- ible and occupy all the space of the container. DENSITY () Mass per unit volume is defined as density. So density at a point of a fluid is represented as m dm Density is a positive scalar quantity. Lim V0 V dV SI UNIT : kg/m3 CGS UNIT : g/cc Dimensions : [ML–3] RELATIVE DENSITY It is defined as the ratio of the density of the given fluid to the density of pure water at 4°C. density of given liquid Relative density (R.D.) = density of pure water at 4°C Relative density or specific gravity is a unitless and dimensionless positive scalar physical quantity. Being a dimensionless/unitless quantity R.D. of a substance is same in SI and CGS system. SPECIFIC GR AVITY It is defined as the ratio of the specific weight of the given fluid to the specific weight of pure water at 4°C. Specific gravity = specific weight of given liquid m3 ) g = R.D. of the liquid specific weight of pure water at 4°C (9.81 kN w g w node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 Thus specific gravity of a liquid is numerically equal to the relative density of that liquid and for calculation purposes they are used interchangeably. Example A hollow metallic sphere has inner and outer radii, respectively, as 5 cm and 10 cm. If the mass of the sphere is 2.5 kg, find (a) density of the material, (b) relative density of the material of the sphere. Solution The volume of the material of the sphere is 4 4 10 3 5 3 4 3 3 00 3 V r23 0.001 0 .0 00 12 5 r13 3.14 100 3.14 1 4 = × 3.14 × 0.000875 m3 = 0.00367 m3 3 E 25
JEE-Physics M 2.5 (a) Therefore, density of the material of the sphere is = = kg/m3 = 681.2 kg/m3 V 0.00367 681.2 (b) Relative density of the material of the sphere r 1000 = 0.6812 Density of a Mixture of substance in the proportion of mass Let a number of substances of masses M , M , M etc., and densities 1 , 2 , 3 etc. respectively are mixed 1 2 3 together. The total mass of the mixture = M + M + M +.... 123 The total volume = M1 M2 + M3 + .... therefore, the density of the mixture is M1 M2 M 3.... 1 + 2 3 M1 M2 M3 .... 1 2 3 For two substances the density of the mixture 12 M1 M2 2 M 1 1M2 Example Two immiscible of densities 2.5 g/cm3 and 0.8 g/cm3 are taken in the ratio of their masses as 2:3 respectively. Find the average density of the liquid combination. Solution Let masses be 2m g & 3m g, then V = V + V = 2m 3m cm3 12 2.5 0.8 Total mass = 2m + 3m = 5m g. Therefore, the average density = 5m 5m 5 10 av = 2m 3m = = gm/cm3 = 1.09 gm/cm3 V 2 3 9.1 2.5 0.8 2.5 0.8 Density of a mixture of substance in the proportion of volume Suppose that a number of substances of volume V , V , V etc. and densities 1 , 2 , 3 etc. respectively are 1 2 3 mixed. The total mass of the mixture is = 1V1 2V2 3V3 .... The total volume of the mixture is = V1 + V2 + V3+......... Therefore, the density of the mixture is 1V1 2 V2 3 V3 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 V1 V2 V3 .... Therefore, for two substances we can write 1V1 2V2 V1 V2 Example Two miscible liquids of densities 1.2 gm/cc and 1.4 gm/cc are mixed with a proportion ratio of their volumes equal to 3:5 : What is the density of resulting liquid? Solution 1 1V1 2V2 1 V1 2 1.2 3 / 5 1.4 3 .6 7 V1 V2 V2 3 / 5 1 8 = = = = 1.325 V1 / V2 1 26 E
JEE-Physics PRESSURE The pressure P is defined as the magnitude of the normal force acting on a unit surface area. F A P = A here F= normal force on a surface area A F F The pressure is a scalar quantity. This is because hydrostatic pressure is transmitted equally in all directions when force is applied, which shows that a definite direction is not associated with pressure. CONSEQUENCES OF PRESSURE (i) Railway tracks are laid on large sized wooden or iron sleepers. This is because the weight (force) of the train is spread over a large area of the sleeper. This reduces the pressure acting on the ground and hence prevents the yielding of ground under the weight of the train. (ii) A sharp knife is more effective in cutting the objects than a blunt knife. The pressure exerted = Force / area. The sharp knife transmits force over a small area as compared to the blunt knife. Hence the pressure exerted in case of sharp knife is more than in case of blunt knife. (iii) A camel walks easily on sand but a man cannot inspite of the fact that a camel is much heavier than man. This is because the area of camel's feet is large as compared to man's feet. So the pressure exerted by camel on the sand is very small as compared to the pressure exerted by man. Due to large pressure, sand under the feet of man yields and hence he cannot walk easily on sand. TYPES OF PRESSURE There are three types of pressure (i) Atmospheric pressure (P ) (iii) Gauge pressure (P ) (ii) Absolute pressure (P ) 0 gauge abs. • Atmospheric pressure : Force exerted by air column on unit cross–section area of sea level called up to top of atmospheric pressure (Po) atmosphere F air area=1m2 P = =101.3 kN/m2 P = 1.013 × 105 N/m2 column oA o sea level Barometer is used to measure atmospheric pressure. Which was discovered by Torricelli. Atmospheric pressure varies from place to place and at a particular place from time to time. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 • Gauge Pressure : Excess Pressure ( P– P ) measured with the help of pressure measuring instrument called Gauge pressure. atm Patm manometer gas P = hg or P h Pabsolute gauge gauge h Gauge pressure is always measured with help of \"manometer\" E 27
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