SOLUTIONS TO CONCEPTS CHAPTER 2

Transcription

1 SOLUTIONS TO CONCPTS CHAPTR 1. As shown in the figure, The angle between A and B = 11 = 9 A = and B = 4m Resultant R = A B ABcos = 5 m Let be the angle between R and A 4 sin9 = tan 1 = tan 1 (4/) = 5 4cos9 B R A Resultant vector makes angle (5 + ) = 7 with -ais.. Angle between A and B is = 6 = A and B = 1 unit R = cos = 19. be the angle between R and A = tan 1 1 sin 1 1 tan 1 1 cos = tan 1 (.6795) = 15 Resultant makes 15 + = 45 angle with -ais.. component of A = 1 cos 45 = 1 / unit component of B = 1 cos 15 = 1 / component of C = 1 cos 15 = 1 / Resultant component = 1 / 1 / + 1 / = 1 / component of A = 1 sin 45 = 1 / unit component of B = 1 sin 15 = 1 / component of C = 1 sin 15 = 1 / Resultant component = 1 / + 1 / 1 / = 1 / Resultant = 1 Tan = component component = tan 1 (1) = 45 = 1 The resultant is 1 unit at 45 with -ais. 4. a 4 i j, b i 4 j a) a 4 = 5 b) b 9 16 = 5 c) a b 7 i 7 j 7 d) a b ( 4)i ˆ ( 4 ) ˆj ˆi ˆj a b 1 ( 1) B 6 45 A 15

2 Chapter- 5. component of OA = cos = component of BC = 1.5 cos 1 =.75 component of D = 1 cos 7 = component of OA = sin = 1 component of BC = 1.5 sin 1 = 1. component of D = 1 sin 7 = 1 R = component of resultant =.75 =.98 m R = resultant component = = 1. m So, R = Resultant = 1.6 m If it makes and angle with positive -ais Tan = component component = tan a = m b = 4 = 1. A m 1.5m 6 O 9 D B 1m a) If R = 1 unit cos = 1 = 18 b) cos = 5 = 9 c) cos = 7 = Angle between them is. 7. AD î.5ĵ 4Kˆ = 6i ˆ.5ˆj C AD = A D = 6. KM Tan = D / A = 1/1 A = tan 1 (1/1) m B 6m 4m.5 km The displacement of the car is 6. km along the distance tan 1 (1/1) with positive -ais. D.5 km 8. In ABC, tan = / and in DC, tan = ( )/4 tan = (/) = ( )/4 = 4 4 = 4 6 = 4 = / ft a) In ABC, AC = AB BC = 1 ft b) In CD, D = 1 (/) = 4/ ft CD = 4 ft. So, C = a) magnitude of displacement = 74 ft b) the components of the displacement vector are 7 ft, 4 ft and ft.. CD D = 4 1 ft c) In AG, A = AG G = ft. 9. Here the displacement vector r 7î 4 ĵ kˆ C A B G D z Y F BC = ft AF = ft D = r

3 1. a is a vector of magnitude 4.5 unit due north. a) a = 4.5 = 1.5 a is along north having magnitude 1.5 units. b) 4 a = = 6 unit 4 a is a vector of magnitude 6 unit due south. 11. a = m, b = m angle between them = 6 a) a b a b cos 6 = 1/ = m b) a b a b sin 6 = / = m. 1. We know that according to polgon law of vector addition, the resultant of these si vectors is zero. Here A = B = C = D = = F (magnitude) So, R = A cos + A cos / + A cos / + A cos / + A cos 4/4 + A cos 5/5 = [As resultant is zero. X component of resultant R = ] = cos + cos / + cos / + cos / + cos 4/ + cos 5/ = Note : Similarl it can be proved that, sin + sin / + sin / + sin / + sin 4/ + sin 5/ = 1. a i j 4k; b i 4 j 5k a b a b ab cos cos 1 ab cos cos A (A B) (claim) As, A B AB sin nˆ Chapter- AB sin nˆ is a vector which is perpendicular to the plane containing A and B, this implies that it is also perpendicular to A. As dot product of two perpendicular vector is zero. Thus A (A B). 15. A i ˆ j ˆ 4kˆ, B 4i ˆ ˆj kˆ ˆi ˆj kˆ A B 4 4 ˆ i(6 1) ˆ j(4 16) k(6 ˆ 1) 6i ˆ 1 ˆ j 6k ˆ. A 6 A5 A 1 A A 4 A 6 = / 16. Given that A, B and C are mutuall perpendicular A B is a vector which direction is perpendicular to the plane containing A and B. Also C is perpendicular to A and B B Angle between C and A B is or 18 (fig.1) So, C ( A B ) = The converse is not true. For eample, if two of the vector are parallel, (fig.), then also C ( A B ) = So, the need not be mutuall perpendicular. C (A B) C A B A.

4 17. The particle moves on the straight line PP at speed v. From the figure, OP v (OP)v sin nˆ = v(op) sin ˆn = v(oq) ˆn It can be seen from the figure, OQ = OP sin = OP sin So, whatever ma be the position of the particle, the magnitude and direction of OP v remain constant. OP v is independent of the position P. 18. Give F q q(v B) (v B) So, the direction of v B should be opposite to the direction of. Hence, v should be in the positive z-plane. Again, = vb sin v = B sin For v to be minimum, = 9 and so v min = F/B Chapter- So, the particle must be projected at a minimum speed of /B along +ve z-ais ( = 9 ) as shown in the figure, so that the force is zero. 19. For eample, as shown in the figure, A B B along west B C A along south C along north A B = A B B C B C = But B C. The graph = should be drawn b the student on a graph paper for eact results. To find slope at an point, draw a tangent at the point and etend the line to meet -ais. Then find tan as shown in the figure. It can be checked that, d d Slope = tan = ( ) = 4 d d Where = the -coordinate of the point where the slope is to be measured. 1. = sin So, + = sin ( + ) = sin ( + ) sin = sin = Given that, i = i e t / RC Rate of change of current = When a) t =, di i dt RC b) when t = RC, c) when t = 1 RC, di d i / RC d t / RC ie i e = dt dt dt di i dt RCe di dt i RCe 1 i e RC t / RC Q B O V B A P V B P = = sin C.4

5 . quation i = i e t / RC i = A, R = 6 1 5, C = F = F a) i = e e amp. e b) di i e dt RC t / RC c) At t =.1 sec, i = e 4. = when t =. sec (. /.) 5.8 Amp. e di (. /.) e Amp / sec dt. e Area bounded b the curve, ais with coordinates with = 5 and = 1 is given b, Area = d = 5. Area = ( 1 6 7)d = d = sin d [cos ] = = sin = 115 sq.units. Chapter- = The given function is = e When =, = e = 1 increases, value deceases and onl at =, =. So, the required area can be found out b integrating the function from to. So, Area = mass 7. a b length e d [e ] a) S.I. unit of a = kg/m and SI unit of b = kg/m (from principle of homogeneit of dimensions) b) Let us consider a small element of length d at a distance from the origin as shown in the figure. dm = mass of the element = d = (a + b) d So, mass of the rod = m = dp dt = (1 N) + ( N/S)t momentum is zero at t = momentum at t = 1 sec will be dp = [(1 N) + Ns t]dt p 1 1 dp 1dt (tdt) = 1t L dm (a b) d = 1 1 t = kg m/s. L b bl a al O =1.5

6 9. The change in a function of and the independent variable are related as d = d Taking integration of both sides, d d = c as a function of is represented b =. The number significant digits a) 11 No.of significant digits = 4 b) 1.1 No.of significant digits = 4 c) 1.1 No.of significant digits = 5 d).11 No.of significant digits = 4 c. 1. The metre scale is graduated at ever millimeter. 1 m = 1 mm d. d Chapter- The minimum no.of significant digit ma be 1 (e.g. for measurements like 5 mm, 7 mm etc) and the maimum no.of significant digits ma be 4 (e.g.1 mm) So, the no.of significant digits ma be 1,, or 4.. a) In the value 47, after the digit 4, 7 is present. Its value is greater than 5. So, the net two digits are neglected and the value of 4 is increased b 1. value becomes 5 b) value = 84 c).6 d) value is 8.. Given that, for the clinder Length = l = 4.54 cm, radius = r = 1.75 cm Volume = r l = (4.54) (1.75) Since, the minimum no.of significant digits on a particular term is, the result should have significant digits and others rounded off. So, volume V = r l = (.14) (1.75) (1.75) (4.54) = cm Since, it is to be rounded off to significant digits, V = 4.7 cm. 4. We know that, Average thickness = =.17 mm Rounding off to significant digits, average thickness =.17 mm. 5. As shown in the figure, Actual effective length = (9. +.1) cm But, in the measurement 9. cm, the no. of significant digits is onl. So, the addition must be done b considering onl significant digits of each measurement. So, effective length = = 9.1 cm. r l 9cm.1cm * * * *.6