Narayana IIT Academy INDIA Sec: Sr. IIT-IZ GTA-7 Date: Time: 02:00 PM to 05:00 PM 2012_P2 Max.Marks:198

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1 Narayana IIT Academy INDIA Sec: Sr. IIT-IZ GTA-7 Date: --7 Time: : PM to 5: PM _P Max.Marks:98 KEY SHEET PHYSICS D C D 4 A 5 A 6 A 7 A 8 C 9 C A A C D 4 5 AC 6 A 7 C 8 AD 9 ACD D CHEMISTY D A 4 A 5 6 C 7 A 8 A 9 D D D D 4 C 5 CD 6 AD 7 AD 8 AC 9 AD 4 C MATHS 4) C 4) 4) D 44) 45) A 46) 47) C 48) 49) C 5) C 5) C 5) 5) 54) A 55) ACD 56) CD 57) A 58) ACD 59) C 6) C

2 x L. x --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s PHYSICS x mx x L Mass of total rod = L Acceleration F a L Tension at x x, Tx mxa T F x x L x L T x F x x L L U T T Adx A Ay x x U Ay U 9F L 5yA xl x T dx. A min s (half life) x A C 6 min s (4 half lies) A C min s (half lies) N N. for no deiation 4. L q d mv x x 8 q L L mv Sec: Sr. IIT-IZ Page

3 _Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s Object x.5. x x x x 4.7 x 4 4 x 4 7 x 4cm 5. C / C / C / Initial Condition Final by total conseration of charge the final charge on each capacitor is C / 9. So heat loss in each capacitor is u C Z me n max least and least Z ; = eydberg constant, reduced mass of atom/ion So hydrogen atom. 7. At instant when the two waes get superposed completely as they are identical and symmetrical, resultant displacement of medium particle eerywhere on wae shape is zero, but elocity of each elements on wae shape gets added and becomes double of initial alue. Sec: Sr. IIT-IZ Page

4 --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s Elastic P.E. eerywhere on wae shape is zero, but KE will be equal to Sum of initial oscillation energy of indiidual waes. So KE in superimposed wae shape instantaneously will be E. 8. Horizontal force can be gien by the area of pressure Vs depth graph. g g 4 g Pressure Vs depth g g 5 g g 5 4 g g 9. L L L M 5M 9M 4 4. Horizontal component of L is rotating. So dl Lxd &. dl d M M C L dt dt Mg T ma () l T ma () l Mg ma ma dv Mg l m M dt l m M dv dt Mg Sec: Sr. IIT-IZ Page 4

5 l m M Mg ln l Mg t. Maximum pressure charge at h g h h f --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s g T T h h Thg T Q u ar a r Q E 4 r Q 4 E r 4 E a u r a r E ar r 4 a du a ar 4r r dr F d t f V V d t F F V d t df V V dv A dyg sin d A dy g dv dy d dy C g dv y dy Velocity gradient at y h is zero Sec: Sr. IIT-IZ Page 5

6 C g h --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s dv g h y dy V g y h y y g y V h y y 4 V g h y y 4 y Now P g sin x z dy V y g 4 h P g xz hy y dy g x z h h h 8 x z h 8 g P x zh g 8. Consider elocity-time graph for which slope gies acceleration and area under the cure gies displacement. 9. I V I x C tan x L tan If x L xc 45 and current and oltage is in same phase so it is state resonance. Sec: Sr. IIT-IZ Page 6

7 . Mass of ball is gien kg let mass of ball is n kg. Now n kg c.m kg n n n n before collision --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s 5 line of impact Components of elocity along line of collision and to it before collision As ball goes at 6 after collision in G-frame, it elocity along line of impact becomes zero. Now applying L.M.C. on ball & system along line of impact. n line of impact Components of elocity of two ball after collision line of impact n all goes at from original direction of motion in C frame Use are obsering motion of in C-frame. final line of motion of 6 6 C C n n n n n Mass of ball kg. Sec: Sr. IIT-IZ Page 7

8 . D. A. CONCEPTUAL HO (aq) HO( ) O (g) t/ 5hrs --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s CHEMISTY t 5hrs hrs 5hrs H O n HO t n H O decomposed 4 n t 5hrs etween 6 th hours to 5 th hours n W 4. A C 7. A 8. A 9. D O formed 8 ngt 8 5 cal 8 8 Magnitude of work = 5 cal n H O t hrs n H O t 5hrs 4 Sec: Sr. IIT-IZ Page 8

9 Sol: T P 6K gas --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s In low pressure range Z = ut in high pressure range Z>. T Q 6K K gas 5 In low pressure range Z = at oyles temperature K. : D.. D. D 4. C. 5. CD 6. AD O (A) O HCN OH N CH NH NaNO HCl OH (C) 7. AD H x 4 9 H H 5 x 6 6 x 8 8 (4) () x 4 5 A) H H correct 8x ) H(4) 4 5 Incorrect Sec: Sr. IIT-IZ Page 9

10 8 9 9 C) H(4) H 8 6 x x Incorrect D) Correct 8. AC 9. AD (ef: NCET ) 4. C 4. Let / sin n x In Vn Vn dx sin x --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s MATHS n / I I cos n x dx n In In... I Vn Vn V V n n V V Vn n / 4. centre(-, -) ; adius 8 Sec: Sr. IIT-IZ Page PC 8 A, is diameter 4. For the point to lie aboe the parabola, b f a a ba We can rearrange to isolate a b : b a unning past a few alues of a, recognize quickly that: If a, b could be between and 9 (inclusie) If a, b could be between and 9 (inclusie)

11 If a, b could be between7 and 9 (inclusie) If a 4, no single-digit integer b satisfies the inequality. Thus, the probability is Slope of tangent of hyperbola xy a dy y dx x Sec: Sr. IIT-IZ Page --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s at point, x y is Let the slope of tangent of cures which makes an angle of / 4 be dy dx dy y dx x y dy x dx dy y y dy dx x x dx 45. Middle term of the series are a 5 and a 6 since a a i 5 a5a6 5 or ab 5 () Since FF is an equilateral triangle therefore tan b..() a 4 From () and () a, b 5 Equation of ellipse is x y First, 6 distinct digits can be selected in C 6 ways. Now the position of smallest digit in them is fixed i.e. position 4. Of the remaining 5 digits, two digits can be selected in 5 C ways. These two digits can be placed to the right of 4 th position in one way only. 47. The remaining three digits to the left of 4 th position are in the required order automatically. So 5 n S C6 C n k r r f k a a ae b

12 f a f a f a... Now,,..., a a n f a f n a,..., a a f f n,..., a an are in A.P n n are in A.P are in A.P 48. Let A be the initial point and b, c The position ector of X is b and that of Y is --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s be the position ectors of and C respectiely. c b / c b. 4 49& 5 5 & 5 Equation of the line Y is b c r b t 4 The point Z with position ector kc lies on it b c 4 AZ kc b t t k. 4 CZ f x increases in, and f x attains local maximum at x f x f n n N ne n e / x / x n e n nn Sec: Sr. IIT-IZ Page

13 F --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s E C 4,6 D M, 4 CAM A, EM 45 AC is perpendicular to E In AE, AC and EM are altitudes and C is orthocemntre. Slope of A is, So AC is parallel to y-axis And E is parallel to x-axis 5 & 54 9 AP, 4 Q,7C form a triangle A Q P C 9AD 7C C A 4 Q Applying sine rule, Sec: Sr. IIT-IZ Page

14 9ha 4hb 7hc sin A sin sin C a.sin A b.sin c.sin C a k --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s b k c 7k Now apply cosine rule. T 55. A A T A A T T T T A A 56. adius of director circle a b b b 4 4 e Focus lies on the line x y at a distance ae units from centre x y x, y 6, 6 or 6, 6 Sec: Sr. IIT-IZ Page 4

15 57. Take any four numbers Ex: 5,,, --7_Sr.IIT-IZ_JEE-Ad_(_P)_GTA-7_Key & Sol s They can be arranged only in one way such that each number is not smaller that the preceding i.e.,,, 5. Probability 9 C f x f x 4 f x f x 4 f x f x x 4n lim f x f x n /4 n f x lim f x f n 59. f x x x [ ] f x is discontinuous at x,,, 4 7 g x x x x g x 5x x 9x x 6x 4 x x 7 x x x g x is not differentiable at x,,, 6. To is perpendicular to u As u and angle between u and can change infinitely many choice for such w is perpendicular to u u u u If u is in xy plane u and u u Sec: Sr. IIT-IZ Page 5