Tiêu đề Periodic motion & Chemical change-2- Daily-11 (Set-A)-With Solve.pdf ✅

2 e ̈vL ̈v: m‡e©v”P MwZkw3, K1 = 1 2 m 2A 2 = 1 2 mgL 2      = g L Ges A = L Ges K2 = 1 2 mg 3L 2  K2 : K1 = 3 : 1  K2 = 3K1 7. ZvcgvÎv n«vm †c‡j mij †`vj‡Ki †`vjbKvj K‡g ev‡o GKB _v‡K †Kv‡bvwUB bq DËi: K‡g e ̈vL ̈v: Temp  L  T  ZvcgvÎv  Kvh©Kix ˆ`N© ̈  †`vjbKvj  8. †Kv‡bv ̄’v‡b GKwU †m‡KÛ †`vj‡Ki ˆ`N© ̈ 1 m| †h †`vjK H ̄’v‡b cÖwZ wgwb‡U 15 evi †`vj †`q, Zvi ˆ`N© ̈ KZ n‡e? 12.5 m 4 m 6 m 5.5 m DËi: 4 m e ̈vL ̈v: T1 = 2 s; T2 = t N = 60 15 = 4 s  T  L  L2 = T 2 2 T 2 1  L1 =     4 2 2  1 = 16 4  L2 = 4 m 9. mijQw›`Z MwZm¤úbœ KYvi †ÿ‡Î †KvbwU mwVK? miY me©vwaK n‡j, †eM Amxg miY me©wb¤œ n‡j, †eM m‡e©v”P miY me©vwaK n‡j, †eM me©vwaK miY me©wb¤œ n‡j, †eM me©wb¤œ DËi: miY me©wb¤œ n‡j †eM m‡e©v”P e ̈vL ̈v: v =  A 2  x 2 vmax = A n‡e hLb A 2  x 2 m‡e©v”P n‡e Ges GwU ZLbB m¤¢e hLb x = 0| 10. mij Qw›`Z MwZm¤úbœ †Kv‡bv KYvi e ̈eKjbxq mgxKiY : 5 d 2 x dt2 + 80x = 0 n‡j, Gi †KŠwYK K¤úv1⁄4 KZ n‡e? 4 rads1 5 rads–1 2 rads1 25 rads1 DËi: 4 rads1 e ̈vL ̈v: 5 d 2 x dt2 + 80x = 0  d 2 x dt2 + 16x = 0   2 = 16   = 4 rads–1 11. GKwU mij †`vj‡Ki †`vjbKvj 25% e„w× Ki‡Z Gi Kvh©Kix ˆ`N© ̈ KZ ̧Y evov‡Z n‡e? 1.25 ̧Y 0.56 ̧Y 2.25 ̧Y 1.65 ̧Y DËi: 0.56 ̧Y e ̈vL ̈v: T T = L L  T + T  25 100 T = L L      1 + 1 4 T T = L L  5 4 = L L  L L = 25 16  L = 25 16 L  e„w× Ki‡Z n‡e = 25 16 L  L = 9 16 L = 0.56 L = 0.56 ̧Y 12. GKwU mij †`vjbMwZ m¤úbœ e ̄`KYvi fi m Ges K¤úv1⁄4  n‡j, ej aaæeKwU n‡e m  m m 2  m DËi: m 2 e ̈vL ̈v:  = k m   2 = k m  k = m 2 13. 0.01 kg f‡ii GKwU e ̄`KYv GKwU mij‡iLv eivei mij‡`vjb MwZ‡Z `yj‡Q| Gi we ̄Ívi 0.2 m I miY 0.1 m| †`vjKwUi MwZkw3 I wefekw3i AbycvZ KZ? 12 : 1 3 : 1 24 : 1 1 : 12 DËi: 3 : 1 e ̈vL ̈v: MwZkw3 wefekw3 = 1 2 k(A 2  x 2 ) 1 2 kx2 = A 2  x 2 x 2 = (0.2) 2  (0.1) 2 (0.1) 2 = 0.04  0.01 0.01 = 3 1 = 3 : 1