1. For the second hand of a clock, T = 60 s, r = 4  10 2 m.   = 2 3.14 60  = 1.047  10 1 rad/s and v = 4  10 2  1.047  10 1 = 4.188  10 3 m/s. 2. n = 600 rpm = 600 60 = 10 rps, r = 0.5 m.   = 2  3.142  10 = 62.84 rad/s v = 0.5  62.84 = 31.42 m/s. 3. r = 20  10 2 m, n = 240 60 = 4 rps.   = 2  3.14  4 = 25.12 rad/s and v = 20  10 2  25.12 = 5.024 m/s. 4. T = 60 min. = 3600 s, t = 20 min. = 1200 s,  = ?  = 2 T  = 2 3600  = 1800  rad/s   = t = 1800   1200 = 2.093 rad. 5. r = 5  10 2 m, s = 7.25  10 2 m s = r    = 2 2 7.25 10 5 10     = 1.45 rad 6. v = r   = 2 10 = 0.2 rad/s. Now,  = t = 0.2  15 = 3 rad. 7. T = 24h = 24  3600s,  = 2 T  = 2 3.14 24 3600   = 7.27  10 5 rad/s. 8. v = r   = 300 1500 = 0.2 rad/s. v = 2 r / 5 t  or t = 2 r 5v   t = 2 3.14 1500 5 300    = 6.28 s 9.  = 5 rad/s, r = 50  10 2 m, v = r = 50  10 –2  5 = 2.5 m/s. 10. n 1 = 500 rpm = 500 60 rps, n 2 = 0, t = 10 s,  = ?  = 2 1 2 (n n) t   = 2 3.14 10   500 0 60        = 5.23 rad/s 2 11. n 1 = 600 rpm = 600 60 = 10 rps and n 2 = 1200 rpm = 1200 60 = 20 rps, t = 20 s.  = 2 1 2 (n n) t   = 2 (20 10) 20    =  rad/s 2 Now,  1 = 2  10 = 20 rad/s and  2 = 2  20 = 40 rad/s  Using, 2 2 2 1 2     we get,  = 2 2 2 1 2     = 2 (2) (400 100) 2    = 2  300 rad  No. of revolutions = 2  300 2 = 300 12. n 1 = 100 rpm = 100 10 60 6  rps,   1 = 2n 1 = 2  10 6 = 20 6  rad/s  2 = 0, t = 15 s,  = 2 rad Using,  = 2 1 t    = 0 20 / 6 15

support material for cbse class 12

Course objective: The course aims to introduce numerical methods used for the solution of engineering problems. The course emphasizes algorithm development and programming and application to realistic engineering problems.