IITJEE PHYSICS SAMPLE PAPER - I SOLUTIONS SECTION – I Straight Objective Type Tdt 1. M v 2 Tdt u 2g Tdt l 2 M u 2 v gl Tdt v T1 Mv T1 Mg ..... (i) C Ma T1 M ..... (ii) A a ..... (iii) B Mg v2 l a From (i) and (ii) 3M v 2 M u 2 v u 3 a u2 9l gl 9l g . 9 (b) 2. All the equipotential surfaces of the field between the sphere and the plate are convex down ward. Hence on any straight line parallel to plate the points farther from the sphere will have potential lower than those closer to sphere. (b) 3. Torque about O is zero as well angular momentum hence (d) 4. Mg ( H y) 1 M (vc )2 2 2Mv1 Mv c 2v c v 1 M (v 2 2 1 Ic w 2 v12 ) 3 Solving (i), (ii) and (iii) v 2 4 (d) V V3 V2 1 1 2 = 0. 1 Mv12 2 .....(i) v y v1 ..... (ii) ..... (iii) g (H y) v dy dt 4g (H 3 y) t 3H g 5. 6. Both upper half and lower half will have same effective area of R2 so charge in flux will be same and induced emf will have 2 some value. But since the resistance is different due to which current must be different but ring is as a whole is closed circuit so electric field will be generated to make the current flow in both parts to be same. E R i10r 0 ..... (i) E R ir 0 ..... (ii) 2E R 9ir 0 E R 2b 9ir i E 11r 11r 2 R 2 9r Rb 9 = E Rb 2 R 11r 22 (b) a cos w0t a cos wt cos w0t h ( w w0 ) 2 a cos w0t Q (c) 8. i 10r E – + – + a [cos(w w0 ) cos( w w0 )] 2 Highest possible energy for photon corresponds to frequency w w0 hence. E KEmax 7. 10 r r 2h gh WST r h gh , Wg 2 r 2 gh 2 Heat = Q = WST Wg = and h 2 (a) 2 a 2 h2 vs t1 t1 t2 t2 2 a vp vs (a) vp vs h 2 1 a vp h vs 2S r g Q 2 S2 g E SECTION II Reasoning Type 9. PV = nRT P nR T V V 1 ( slope) v1 = constant P 1 2 v2 = constant T (b) 10. Potential at E and K are different due to which current flows between E and K make current flow between AB and CD also possible. (d) 11. Assertion and Reason correct and correct explanation. (a) 12. Acceleration relative to cart parallel to incline is always zero only the acceleration perpendicular to incline will change in different situation due to which change in tension but angle will remain same and string always remain perpendicular to the incline. (d) SECTION III Linked Comprehension Type Passage-I When the temperature of rods in increased there will be increase in their lengths and thereby the springs are compressed, let x1 , x 2 and x 3 be the compression in the three springs respectively. Then L Kx1 L Kx1 T L T x1 x2 x2 2 2Kx2 and 2 Kx2 3Kx3 x1 2 x 2 3x 3 x x1 3 x1 1 L T 2 3 2 9 x1 L T 11 2Kx2 2Kx2 3Kx3 E1 1 2 Kx1 2 1 9 K L 2 11 2 81 KL2 242 T 2 ( T )2 Similarly E2 E3 13. 14. 15. 16. 81 KL2 484 27 KL2 242 2 T2 2 T2 (b) (a) (a) Passage-II The charge q on the small disc can he calculated by applying Gauss law E · ds 0 q Since one side of the small disc is in contact with plate r2 r2 0 2 . q E · r V xv x 0 0 d d (b) 17. Let v s be the steady velocity of the small disc just after its collision with the bottom plate then the steady state kinetic energy k s of the disc just above the bottom plate is given by 1 ks mv s2 2 for each round trip disc gains electrostatic energy by U = 2qV 1 1 k after . k loss k before k after (1 e 2 ) k before 2 Since k s is the energy after collision at the bottom plate and (ks qV mgd ) is the energy before the collision at the top plates total energy loss can be written as 1 ktot 1 ks (1 2 ) (ks qV mgd ) 2 In its steady state U should be compensated by k total 1 2qV 2 1 ks (1 2 ) qV 2 ) (ks qV 2 ks 4 1 2 1 2 mvs 2 1 2 vs 1 [(1 2 2 (1 ) mgd ] 2 2 qV 2 xV 2 m 1 1 2 mgd 2 (2 gd ) mgd ) vs 18. V2 . (c) The disc will lose kinetic energy eventually cease to move when the disc cannot reach the top plate. In other words, the threshold voltage V c can be determined from the condition that velocity of the disc at the top plate is zero. In order to have velocity zero at top plate kinetic energy at top plate must satisfy the relation Ks ks qVc mgd 0 k s is steady state kinetic energy at the bottom plate e2 qVc 1 e2 Vc 1 e2 1 e2 e2 mgd 1 e2 qVc mgd 0 mgd x (d) SECTION IV Matrix Match Type 1. Concave Convex (A) m < 0 |m| < 1 Object as well image is real in case of concave mirror object and image is virtual incase of convex. (B) m < 0 |m| > 1 Both real in case of concave both virtual incase of convex. (C) |m| < 1 m > 0 Concave object virtual image real. Convex object real image virtual. (D) |m| > 1 m > 0 Concave object real image virtual. Convex object virtual, image real. (A) –2, 4; (B) –1, 3; (C) –2, 3; (D) –2, 3 2. q, v, m B F R/2 y0 O D M A R–y0 x B C R 2B mv ; cos qB R y0 R m q (2 B) = 2 ; t AB T R ; t BC m (3 ) qB t AB t BC ; XB 1 XA R sin XC XB BC 4R sin XD XC CD 5R sin m q (B ) m 3 (cos 1 (1 qB XA AB )) 2R sin When velocity be come parallel t FA t AM m qB m q2B m 3 2qB 3m 2qB (A) –2; (B) –1; (C) –3, 4; (D) –1, 2 SECTION V Subjective or Numerical Problems 2T cos D = W + 2w 2 N cos (90 – ) = W 2 N sin =W Taking torque about B T × AB sin 2 1 = w AB sin 2 T × 4 × sin 2 = w × 2 sin A N OB cos + N × r cot (AB = 4 m ; OB = r cosec ) Solving from above r = 3m 2 C N N O 90º 90º– 1. w B W w 2. y d x ; dt a a log sec dy dx tan w (Constant) d2y dx 2 x ; a 1 x sec2 a a dy dx 1 Radius of curvature = dy dx tan dx dt a d2y dt 2 tan d dt aw · sec2 2 d y dx 2 x a 0; x 1 · dx a a Now resultant acceleration a d 2x dt 2 2 d2y dt 2 0 a 2 w 4 sec 4 aw 2 sec 2 x a 2 3 2 = a sec x a x ; x=a a d 2x dt 2 aw ; 2 2 x a w2 R2 a 2 2 1 a sec2 1 2 4a 2 1 =8 2 4 m/sec 2 4 dy dt dy dx aw 2 sec 2 dx dt x a tan x a aw

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