WEEKLY TEST-2 GZRS-1902 (JEE ADVANCED PATTERN)

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1 WEEKLY TEST- GZRS-90 (JEE DVNCED PTTERN) Test Date: 0--07

2 [ ] WT- (dv) GZRS-90_ (B) PHYSICS ngle between a and p is : p a = cos a.p a p = cos 8 (6 9) (64 6) = cos 4 50 Clearly both are not perpendicular, hence accelerated circular motion. (C) a a c a t tan = a a c t a a c t = tan 0 =. (D) V V 4. (B) Work done depends upon frame of reference. 5. a s, 0 s dv v s ds, vdv sds, v0 0 v0 s s = v 0 (B) Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

3 WT- (dv) GZRS-90_0..07 [ ] 6. Time to cross river (t) = B 0.4 v sin 5sin mr BC (v cos v )t mr r (5cos ) 5sin 5sin 5cos 5 sin 5 cos 50 sin cos 5sin4 4 sin = 5 5 (C) 7. (B) 8. () u = 5gR v = u gr = 5gR gr = gr Tangential acceleration at B is a t = g (downwards) Centripetal acceleration at B is u v B a C = R v = g Total acceleration will be a = C t a a = g 0 9. (B) s W = K Force is along negative x-axis and displacement is along + x-axis W = negative Hence K = negative Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

4 [ 4 ] WT- (dv) GZRS-90_ (B) Take the mass m as a point mass. t the instant when the pendulum collides with the nail, m has a velocity v = g. The angular momentum of the mass with respect to the point at which the nail locates is conserved during the collision. Then the velocity of the mass is still v at the instant after the collision and the motion thereafter is such that the mass is constrained to rotate around the nail. Under the critical condition that the mass can just swing completely round in a circle, the gravitational force when the mass is at the top of the circle. Let the velocity of the mass at this instant be v, and we have or mv d = mg, v = ( d)g The energy equation or mv mv = mg( d ), g = ( d)g + 4( d)g then gives the minimum distance as d = 5. (D). (B) For stone thrown from top of the tower y 40 ( v sin ) t gt (i) 40m v 0 For stone thrown from bottom of the tower v 0 gt ( v sin ) t (ii) 0 m x By (i) and (ii) 40 v sin t v sin t (iii) But So, t 0 0 v (iv) cos v cos 40 ( v 0 sin ) v cos 40 0 tan tan ( v 0 sin ) v cos Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

5 WT- (dv) GZRS-90_0..07 [ 5 ] tan tan (v) Given = 0 60 Putting t = 0 0 v cos By equation (iv), tan v sin 0 in equation (ii) gt 0 v cos tan gt 0, v 0 cos0 40 t = s 0 0 m/s 0 v = 0 m/s cos60. Because resultant velocity is always perpendicular to line joining C and boat, so path is circular with center at C. (C) 4. Let any time the velocity of boat with respect to river makes an angle with CP. Since along CP, net velocity is zero u cos u sin cos sin C u resultant u P V resultant = u sin ucos u cos u cos u cos. ngular velocity / 0 d cos t 0 u d u cos u cos, d d dt d t ln u d u cos dt d 5. () 6. () 7. (B) (B) Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

6 [ 6 ] WT- (dv) GZRS-90_ () 9. () w T sin 80 F and sin 0. (B) 5w T sin 80 7 T F 0 sin w 4 5T T ' F sin 90 T ' 5 w 4 5w 5 ()-, (B)- 4, (C)-, (D)- Let maximum speed of motorbike = v v v = 60 m/s So acceleration of motorbike 5 w Maximum separation = shaded area 40 OD Separation at t = (), (B)-, (C)- 4, (D) - If maximum extension is x m, By conservation of energy, m kx m 00 x m = m x m gx m 00 T C 5 0 w F F T T Let extension in spring is x when velocity of m is 0/ m/s. The velocity of m B will also be 0 m/s. v(m/s) v 40m/s B Motorbike C Car t(s) O D 8 7 Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

7 WT- (dv) GZRS-90_0..07 [ 7 ] m v m B v kx m x 0 0 x 5 00x 00x 4 x 4x 0 (x ) 0 x 0.5 m If acceleration of m and m B is a T kx m a T a B gx T 5a 00 (i) m g T m a 0 0T 0a T 000a (ii) (i) and (ii) 5a 00000a a = 0 When acceleration m B is 4 m/s, the acceleration of m will also be 4 m/s T kx m B a T 00x 0...(i) m g T m a 00T 40 (ii) (i) + (ii) x = x = m 00 5 U kx 00 5 = 4 J () 4, (B), (C),(D) 5 Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

8 [ 8 ] WT- (DV) GZRS-90_ (B) Let dibasic acid is H Silver salt : g g g 0.90 g 0.54 g By POC of g (6 M ) 08 (6 M ) 60 M (C). () 4. (D) Mol. wt H = + 44 = 46 CHEMISTRY Law of multiple proportion is defined for different compounds having same constituent elements PN 0cm of Hg 76.6 or, PN 8cm of Hg atm Now, PV = nrt for N n 60 or, n mole 60 mass of nitrogen 4 8 g % 00.% 0.4 V c = b or, = b or b= litre/mol Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

9 WT- (DV) GZRS-90_0..07 [ 9 ] 4 Now, b 4 r N 000 or, 6N r 5. (D) 6. (C) 7. (C) 8. () 9. (D) 0. (C). (B). (D) 000 or, r 6N / cm t same T, K.E. is equal and P.E. of gas > P.E. of liq. In () and (B) use (z / e) concept for isoelectronic specie. In (C) size of neutral atom is greater than its cation. In (D) Se and s related with 4th period, while Ba + and Cs + related with 6th period. (These are not isoelectronic species) Due to diagonal relationship radius of Li + is close to Mg + ion Difference in vertical height = 5 cm. Vertical difference of height is 5 cm. Now, h. (C) 4. () 5. (C) h 5 h Hg Hg or, h = 0 cm 0 l 0cm sin0º O - ion will resist the addition of another electron due to inter-electronic repulsion. Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

10 [ 0 ] WT- (DV) GZRS-90_ (D) 7. (D) 8. (C) 9. () 40. (C) 4. () 4. () B < S < P < F (Data base) () (R); (B) (S); (C) (P); (D) (Q) () (R); (B) (Q); (C) (P); (D) (S) () (Q); (B) (R); (C) (S); (D) (P) () (R); (B) (S); (C) (Q); (D) (P) Let P is (h, k) then h + k = x y which is a square having centre at (, ). MTHEMTICS x = 4 x x log (x )log 4 (x ) or xlog x or 4. (B) x log Rearranging, we get log4 x. log4 log log4 Let, are the roots of given equation , 5 5 (8 5) H 4 (4 5) Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

11 WT- (DV) GZRS-90_0..07 [ ] 44. (D) 45. (B) gain x x x x sin cos sin cos x x x x sin cos sin cos x tan x x x x x x x x sin cos sin cos sin cos sin cos 46. () x y (0, ) (y ) x (,0) x-axis 47. () 48. (D) 49. (C) y x (0, -) The lines represented by the given equation will be coincident if h ab = 0. Here, a =, b = k, h = 4. Substituting the values, we get (4) k = 0 k = 8. 4 sin cos sin sin Taking logarithm, y y y where y log x log x,, 4 4 x,, 4 Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

12 [ ] WT- (DV) GZRS-90_ () an r a t r r,r 0,,,...n. a a n r r a (n r)d {a rd}. a (n r)d {a rd} (n r)d a nd The sum is n n (n r)d Sn tr a r0 r 0 nd = n (n) (n )... d a nd = n(n )d n(n ) a a a a n n 5. (B) Image of (, ) in line x + y = is () (), (, ) 5. (C) So line BC passes through (, ) and, 5 5. Equation of line BC is y = / 5 / 5 (x + ) 7x + y + 4 = 0 Vertex B is point of intersection of 7x + y + 4 = 0 and x + y = i.e. B = ( 5/, 9/) 5. (C) 9 sin sin sin = 5 sin sin.sin = 4 cos cos cos cos cos cos = 8 sin 7 8 sin (B) cos cos.cos... cos = sin 0 56 sin sin sin 0 Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

13 WT- (DV) GZRS-90_0..07 [ ] 8 sin 56 8 sin sin sin (C) x 56. (D) = y 4 = x =, y = 9 = = 57. () x 5 = y = x = 9, y = 5 = 6 = p a a (p )d, (P) q b a (q )d, r c a (r )d a (q r) 0 p (Q) R = ak r R s t = a s t k (s t)(r ) S t r = a t r k (s )(t r) T r s = a r s k (t )(r s) R s t S t r T r s =. (R) x y z.y z x z x y = (R m ) (n p)d (R n ) (p m)d (R p ) (m n)d = 58. (B) (S) a(b c)loga = loga abc c b = (r q)d(log (p )logr) = 0 abc (P) Reduce the expression in the form of l cos msin whose maximum is l m Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

14 [ 4 ] WT- (DV) GZRS-90_0..07 (Q) cos cos cos( ) cos( ) max. (cos + cos) = cos(+) (R) sin + sin = sin( + ) cos( ) maximum value = cos( ) (S) 0, tan cot 59. (C) tan cot cos( ) cos( ) 4sin ( ) (P) Let P be the point (, ), then = 0 mid point of OP is,. Locus of, is 4x +4y + 4x + 4y = 0 i.e., x + y + x + y = 0g =, f = g + f = (Q) Centres of the circle are (, ), (5, -6). Equation of C C is y - = i.e., x + y 4 = 0 8 (x ) 4 Equation of radical axis is 8x 6y 56 = 0 i.e., x y 7 = 0 Points of intersection is (, ). (R) Let length of common chord be a, then 9 a 6 a 5 6 a 5 9 a 4 6 a = a a 9 a = 8 00(9 a ) = 4 i.e., 00a = a k a k = Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

15 WT- (DV) GZRS-90_0..07 [ 5 ] 60. (C) (S) Equation of common chord is 6x + 4y + p + q = 0 common chord pass through centre (, 6) of circle x + y + 4x + y + p = 0 (P) p + q = 6 x + y 6 will represent the shaded region as shown in figure. (, 0) (, 0) x (0, ) required area = 4. (Q) = sq. units. D, CD is perpendicular B. k =. y C (, k) B(0, ) D (, 0) x k = 5. slope of BC = 4 4m =. Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

16 [ 6 ] WT- (DV) GZRS-90_0..07 (R) rea of triangle BC = 0 square units D C (4, 6) O B(0, 4) E C can not be at D and E. four positions are possible two above B and two below B. (S) 0 =. = (0,) y H (,) (, ) G B( ) x x- coordinate of B =. C( ) Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna- Helpline No. : /4/6/7

PHASE TEST-1 RB-1804 TO 1806, RBK-1802 & 1803 RBS-1801 & 1802 (JEE ADVANCED PATTERN)

PHASE TEST-1 RB-1804 TO 1806, RBK-1802 & 1803 RBS-1801 & 1802 (JEE ADVANCED PATTERN) PHASE TEST- RB-804 TO 806, RBK-80 & 80 RBS-80 & 80 (JEE ADVANCED PATTERN) Test Date: 0-09-07 [ ] PHASE TEST- (ADV) RB-804-806, RBK-80-80 & RBS-80-80_0.09.07. (B) En = ; Ea = y IP EA IP y EN I.P. y. (A)