Narayana IIT/NEET Academy INDIA IIT_XI-IC_SPARK 2016_P1 Date: Max.Marks: 186

Size: px

Start display at page:

Download "Narayana IIT/NEET Academy INDIA IIT_XI-IC_SPARK 2016_P1 Date: Max.Marks: 186"

Garey Jesse Maxwell
5 years ago
Views:

1 Narayana IIT/NEET Academy INDIA IIT_XI-IC_SPARK 6_P Date: --7 Max.Marks: 86 KEY SHEET PHYSICS D A C C 5 A 6 A,C 7 A,C 8 A,C 9 B,C B,C ABCD ABCD BC CHEMISTRY 9 C B B D C ABC 5 BCD 6 ACD 7 ABC 8 BCD 9 BCD ABC ABD MATHS 7 C 8 C 9 B C D ABC ABC AB 5 ABD 6 AB 7 AB 8 ABCD 9 AC

2 IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & Sol s.. P g h P v P gh g h P v v gh. v gh SOLUTIONS PHYSICS Velocity of efflux of water Momentum of ejected water h v g gh F av Torque of these forces about central line av R av R aghr.. Viscous force F area So let F ka F k A A (i) and T k A (ii) Dividing Eq.(ii) by Eq. (i) we get, T A FA F A A T A A. Q Q inf low outflow 6 6 Q Q 6 m / s 5. (a) Mass of water = (Volume)(density) Page

3 IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & Sol s m AH Velocity of efflux, m H.. (i) A v gh g m mg A A Thrust force on the container due to draining out of liquid from the bottom is given by, F=(density of liquid) (area of hole)(velocity of efflux) F av mg F A / v A / A mg F 5 Acceleration of the container, a F / m g / 5 a=g/5 (b) Velocity of efflux when 75% liquid has been drained out H m i.e., height of liquid, h A m v gh g A v mg A. 6. A, C 7. A,C 8. This is similar to the case as if a tank is filled with a liquid upto a height of h. Page

IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & Sol s In that case range becomes maximum when hole in punched at the centre and the maximum range is equal to the level of liquid in the

4 IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & Sol s In that case range becomes maximum when hole in punched at the centre and the maximum range is equal to the level of liquid in the tank. 9. F = 6rv F v and F r or F A /. Viscous force = weight 6rv r g Substituting the values we can find. Further, v r It r is halved, terminal velocity will remain th. Page

5 IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & Sol s Alg Axg Al a Page 5

6 O - O - CHEMISTRY O - IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & Sol s O - Si O Si O Si O - 9. O -.. Conceptual 5. O - O - Y NaBO7 NH Cl BN NaCl BO H O X HO B O NH Z A BCl gas BN HCl Y (a) Maximum Cl H atom can lie in a plane (b) Maximum 6-atom can lie in a plane (H t, B, H t, H t, B) (c) Bond strength (B H b, B) > (B - H t ) (d) Ht Bˆ Ht H ˆ b B Hb 6. Phenomenon of inert effect increases downward 7. Conceptual. Conceptual. Conceptual Me Me Si. Me Me O Si O O Si Me Me Page 6

7 IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & Sol s 7. SP a t ; SP a t t t SP t SP a t a ; t MATHEMATICS SP SP a V T iˆ Tˆ n ˆj iˆ 8. j V n T Direction ofv on n y n T dy dt dt y T T ; T dx dt dt Given dx dx dt u ; but x T ; T dt dt dt dt dt When P(, ) then T = u =. ; dt dt dy dy dt dt tt 9. t and t t t Hence point t, t and t, t i.e. (, ) and (, ). t t tt t t Equation of the following P parallel to AQ y at x at t put y = x = at at t at a t a at a a at = twice the focal distance of P. t t ; t tt t t 8 for real root t. (a) Equation of the variable circles (x h)(x ) + (y k)(y ) = x + y ( + h) x (k + ) y + k + h = As x intercept = Page 7

8 IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & Sol s g = C h k h (h + ) =k + 8h (h ) = k Locus is (x ) = y L.R. = (b) N = (, -) (c) Figure is a square Area 8sq. Unit. Equation of AB y = x solving it y = x y = (y + ) y y = but y + y + y = but y + y = y = - putting in y = x; x = Hence coordinates c are (, -) sum of the x and y coordinates of C are (a) is correct Obviously normal at C passes through the lower end of the latus rectum (b) is correct x x x Again centroid of ΔABC Now solving y = x with y = x (x ) = x x 8x + = x + x = ; also x = centroid of the ΔABC, Again equation of the normal at C y + = -(-/)(x ) y + = x x y = hence gradient of chord at C is (d) is incorrect. m m m m and eliminate m. ; Thus, the locus is y 6x, which is a parabola, 5. The equation of normal to y x is y mx m m. As it passes through (9, 6) 69m m m ; m 7m 6 m m m 6 m m m m,, Normal is y x or yx or y x ; a, b, d are the correct option. 6. If y mx c is tangent to y x, then mx c x has equal roots m m m c c ; y mx is tangent to y x ; This is also tangent to y x m ; mx x x m x m x has equal roots. m 8m 6 m 6 m, ; y or y x are the tangents. 7. Shifting the origin to (-,, we can write the equation of parabola as Where X x and Y y Y X Page 8

9 IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & Sol s An equation of normal at point, t t on the parabola is Y tx t9 t Its slope = it Now t t 5 Thus, required are,, and 5,, 8. An equation of normal to the parabola y x is y mx m m It will pass through 5,b if b 5m m m m m b Let f m m m b; m R f m m m m Now f m m, Also f m if m f m if m f m if m Max And Min f m f b f m f b b Equation () will have three real roots if b and b i.e., if b If b, then f m has exactly one positive root and other two roots are imaginary. If f m then f m has exactly one negative root and other two roots are imaginary. 9. Equation of circle is x a y a To obtain points of intersection () and y ax, put y ax in (), so that x a ax a x a a as x x a Thus points of intersection are a,a and a, a, and tangents to the parabola at these points are y x a and y x a T : ty = x + t, tan t A AN PN t t A = t = (t ) / i.e. t [, ] & A max occurs when t = A max = 6 Thus X = 6 X = SS = T (y x)(y x ) = (yy (x + x )) (y x)( + ) = [y (x )] = (y x + ) (y x) = (y x + ) Solving with the line x = we get (y 8) = (y ) or (y 8) = y y + or y + y 7 = Where y + y = and y y = - 7 Now y y y y y y or y y 7 7 Page 9

10 y y 7 6 Thus, A = 6 B = A + B = 6 + = 8 5. Conceptual IIT_XI-IC_SPARK- JEE-Adv_6_P_Key & Sol s Page

NARAYANA I I T / N E E T A C A D E M Y. C o m m o n P r a c t i c e T e s t 1 3 XI-IC SPARK Date: PHYSICS CHEMISTRY MATHEMATICS

. (C). (C) 3. (A) 4. (B) 5. (C) 6. (A) 7. (B) 8. (A) 9. (B) 0. (B). (D). (D) 3. (A) 4. (C) 5. (D) NARAYANA I I T / N E E T A C A D E M Y XIS-IC-IIT-SPARK (8..7) C o m m o n P r a c t i c e T e s t 3 XI-IC