JEE Main 2021 Kinematics and Vector Algebra Past Paper PDF

JEE Main 2021 | Kinematics and Vector Algebra Question 1 (Only one correct answer) 2021 A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another s...

JEE Main 2021 | Kinematics and Vector Algebra Question 1 (Only one correct answer) 2021 A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25 m below the top. Both stones reach the bottom of building simultaneously. The height of the building is: (a) 35 m (b) 25 m (c) 50 m (d) 45 m Question 2 (Only one correct answer) 2021 A bomb is dropped by a fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a : (a) hyperbola (b) straight line vertically down the plane (c) parabola in a direction opposite to the motion of plane (d) parabola in the direction of motion of plane Question 3 (Only one correct answer) 2021 The trajectory of a projectile in a vertical plane is y = αx − βx 2. where α and β are constants and x & y are respectively the horizontal and vertical distance of the projectile from the point of projection. The angle of projection θ and the maximum height attained H are respectively given by: 2 4α (a) tan −1 α, β 2 α (b) tan −1 β, 2β 2 β α (c) tan −1 ( ), α β  2 α (d) tan −1 α, 4β Question 4 (Only one correct answer) 2021 Water drops are falling from a nozzle of a shower onto the floor from a height of 9.8 m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor. (a) 2.94 m (b) 4.18 m (c) 2.45 m (d) 7.35 m Question 5 (Only one correct answer) 2021 A helicopter is flying horizontally with a speed v at an altitude h has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped ? −−−−−−−−− 2 2ghv + 1 (a) √ 2 h −−−−−−−−− (b) √2v2 hg + h2 −−−− 2gh (c) √ + h 2 2 v −−−−−−−−− 2 2v h (d) √ + h 2 g Question 6 (Only one correct answer) 2021 → → → → Statement-I : Two forces ( P + Q) and ( P − Q) where → → −−−−−−−−− (P ⊥ Q) , when act at an angle θ1 each other, the magnitude of their resultant is √3(P 2 2 + Q ) , −−−−−−−−− when they act at an angle θ2 , then magnitude of their resultant becomes √2(P 2 2 + Q ). This is possible only when θ1 < θ2. : In the situation given above. θ1 and θ2 ∘ ∘ Statement-II = 60 = 90 In the light of the above statement, choose the most appropriate answer from the options given below : (a) Statement-I is true but Statement-II is false. (b) Both Statement-I and Statement-II are true.  (c) Statement-I is false but Statement-II is true. (d) Both Statement-I and Statement-II are false. Question 7 (Integer type question) 2021 A particle is moving with constant acceleration ′ a′. Following graph shows v2 versus x (displacement) plot. The acceleration of the particle is........ m/s2. Question 8 (Only one correct answer) 2021 The instantaneous velocity of a particle moving in a straight line is given as v = αt + βt 2 , where α and β are constants. The distance travelled by the particle between 1 s and 2 s is: α β (a) + 2 3 3 7 (b) α + β 2 3 3 7 (c) α + β 2 2 (d) 3α + 7β Question 9 (Only one correct answer) 2021 A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t seconds, the total distance travelled is 2αβ (a) t 2 (α + β) αβ (b) t 2 2(α + β) αβ (c) t 2 4(α + β)  4αβ (d) t 2 (α + β) Question 10 (Only one correct answer) 2021 → → → The angle between vector ( A ) and ( A − B) is : A (a) tan−1 ( ) 0.7B B ⎛ ⎞ − ⎜ 2 ⎟ (b) tan−1 ⎜ ⎜ ⎟ – ⎟ ⎜ √3 ⎟ A − B ⎝ ⎠ 2 – √3B (c) tan−1 ( ) 2A − B B cos θ (d) tan−1 ( ) A − B sin θ Question 11 (Only one correct answer) 2021 → A force F = (40^ i + 10^ j) N acts on a body of mass 5 kg. If the body starts from rest, its position → vector r at time t = 10 s , will be: (a) (100^i + 100^ j) m (b) (400^i + 400^ j) m (c) (100^i + 400^ j) m (d) (400^i + 100^ j) m Question 12 (Only one correct answer) 2021 A ball is thrown up with a certain velocity so that it reaches a height h. Find the ratio of the two h different times of the ball reaching in both the directions. 3 – √2 − 1 (a) – √2 + 1  1 (b) 3 – – √3 − √2 (c) – – √3 + √2 – √3 − 1 (d) – √3 + 1 Question 13 (Integer type question) 2021 A bullet of mass 0.1 kg is fired on a wooden block to pierce through it, but it stops after moving a distance of 50 cm into it. If the velocity of bullet before hitting the wood is 10 m/s and it slows down with uniform deceleration, then the magnitude of effective retarding force on the bullet is ′ x′ N. The value of ′ x′ to the nearest integer is........... Question 14 (Only one correct answer) 2021 If the velocity-time graph has the shape AM B, what would be the shape of the corresponding acceleration - time graph ? (a) (a) (a) (a) Question 15 (Only one correct answer) 2021 A particle of mass M originally at rest is subjected to a force whose direction is constant but 2 t − T magnitude varies with time according to the relation F = F 0 [1 − ( ) ] T Where F0 and T are constants. The force acts only for the time interval 2T. The velocity v of the particle after time 2T is : 2F0 T (a) M F0 T (b) 2M 4F0 T (c) 3M F0 T (d) 3M Question 16 (Only one correct answer) 2021 Consider two satellites S1 and S2 with periods of revolution 1 hr. and 8 hr. respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S1 to the angular velocity of satellite S2 is : (a) 8 : 1 (b) 2 : 1 (c) 1 : 8 (d) 1 : 4 Question 17 (Integer type question) 2021 A swimmer wants to cross a river from point A to point B. Line AB makes an angle of 30∘ with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle θ with the line AB should be.......... ∘ , so that the swimmer reaches point B. Question 18 (Only one correct answer) 2021 An engine of a train, moving with uniform acceleration, passes the signal-post with velocity u and the last compartment with velocity v. The velocity with which middle point the train passes the signal post is v − u (a) 2 −−−−−−− 2 v + u2 (b) √ 2 −−−−−−− 2 v − u2 (c) √ 2 u + v (d) 2 Question 19 (Integer type question) 2021 Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3 s with the same initial velocity of 35 m/s , then these balls collide at a height of......... m. (take g = 10 m/s ) 2 Question 20 (Only one correct answer) 2021 − −→ − −→ − −→ −→ − −→ The resultant of these forces OP , OQ , OR, OS and OT is approximately........... N. [Take – – √3 = 1.7, √2 = 1.4. Given ^ i and ^ j unit vectors along x, y -axis] (a) 9.25^i + 5^ j (b) 3^i + 15^ j (c) −1.5^i − 15.5^ j (d) 2.5^i − 14.5^ j Question 21 (Only one correct answer) 2021 A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time t1. If he remains stationary on a moving escalator then the escalator takes him up in time t2. The time taken by him to walk up on the moving escalator will be :  (a) t2 − t1 t1 t2 (b) t2 − t1 t1 t2 (c) t2 + t1 t1 + t2 (d) 2 Question 22 (Only one correct answer) 2021 A scooter accelerates from rest for time t1 at constant rate a1 and then retards at constant rate a2 for t1 time t2 and comes to rest. The correct value of will be: t2 a2 (a) a1 a1 + a2 (b) a1 a1 (c) a2 a1 + a2 (d) a2 Question 23 (Integer type question) 2021 −−−−−−− −− If the velocity of a body related to displacement x is given by v = √5000 + 24x m/s , then the acceleration of the body is........... m/s2 Question 24 (Only one correct answer) 2021 A player kicks a football with an initial speed of 25 at an angle of 45 from the ground. What −1 ∘ ms are the maximum height and the time taken by the football to reach at the highest point during motion? (Take g = 10 ms −2 ) (a) hmax = 15.625 m, T = 1.77 s (b) hmax = 3.54 m, T = 0.125 s (c) hmax = 10 m, T = 2.5 s (d) hmax = 15.625 m, T = 3.54 s Question 25 (Only one correct → answer) → → → 2021 Two vectors X and Y have equal magnitude. The magnitude of ( X − Y ) is n times the → → → → magnitude of ( X + Y ). The angle between X and Y is : 2 n + 1 (a) cos −1 ( ) −n2 − 1 2 n − 1 (b) cos −1 ( ) −n2 − 1 2 −n − 1 (c) cos−1 ( ) 2 n − 1 2 n + 1 (d) cos −1 ( ) 2 n − 1 Question 26 (Integer type question) 2021 A body of mass 2 kg moves under a force of (2^ i + 3^ ^) j + 5k N. It starts from rest and was at the origin initially. After 4 s, its new coordinates are (8, b, 20). The value of b is........ (Round off to the Nearest Integer) Question 27 (Only one correct answer) 2021 The relation between time t and distance x for a moving body is given as t = mx 2 + nx , where m and n are constants. The retardation of the motion is: (Where v stands for velocity) (a) 2mnv3 (b) 2nv3 (c) 2mv3 (d) 2n2 v3 Question 28 (Only one correct answer) 2021 A balloon was moving upwards with a uniform velocity of 10 m/s. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground when object strikes the ground was around. (Takes the value of g as 10 m/s 2 ) (a) 125 m (b) 200 m (c) 300 m (d) 250 m Question 29 (Only one correct answer) 2021 The position, velocity and acceleration of a particle moving with a constant acceleration can be represented by : (a) (a) (a) (a) Question 30 (Only one correct answer) 2021 – A butterfly is flying with a velocity 4√2 m/s in North-East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 seconds is : – (a) 12√2 m (b) 15 m (c) 3 m (d) 20 m Question 31 (Integer type question) 2021 In a spring gun having spring constant 100 N /m a small ball ′ B′ of mass 100 g is put in its barrel (as shown in figure) by compressing the spring through 0.05 m. There should be a box placed at a distance ′ d ′ on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of 2 m above the ground. The value of d is............ m. (g = 10 m/s 2 ) Question 32 (Only one correct answer) 2021 → → → → Two vectors P and Q have equal magnitudes. If the magnitude of P + Q is n times the → → → → magnitude of P − Q , then angle between P and Q is: n − 1 (a) cos−1 ( ) n + 1 n − 1 (b) sin −1 ( ) n + 1 2 n − 1 (c) cos −1 ( ) 2 n + 1 2 n − 1 (d) sin −1 ( ) 2 n + 1 Question 33 (Only one correct answer) 2021 Water droplets are coming from an open tap at a particular rate. The spacing between droplets observed at 4 second after its fall to the next droplet is 34.3. At what rate the droplets are coming th m from the tap? (Take g = 9.8 m/s 2 ) (a) 1 drops/seconds (b) 1 drops/7 seconds (c) 2 drops/seconds (d) 3 drops/2 seconds Question 34 (Only one correct answer) 2021 The velocity of a particle is v = v0 + gt + F t 2. Its position is x = 0 then its displacement after time (t = 1) is : g (a) v0 + + F 2 (b) v0 + 2g + 3F (c) v0 + g + F g F (d) v0 + + 2 3 Question 35 (Only one correct answer) 2021 A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always  81 rising to of the height through which it falls. Find the average speed of the ball. (take 100 g = 10 ms −2 ) (a) 3.50 ms −1 (b) 2.0 ms −1 (c) 2.50 ms −1 (d) 3.0 ms −1 Question 36 (Only one correct answer) 2021 The velocity - displacement graph of a particle is shown in the figure. The acceleration - displacement graph of the same particle is represented by : (a) (a) (a) (a) Question 37 (Only one correct answer) 2021 − −→ − −→ − −→ The magnitude of vector OA , OB and OC in the given figure are equal. The direction of − −→ − −→ − −→ OA + OB − OC with x-axis will be:  – – (1 + √3 − √2) (a) tan −1 – – (1 − √3 − √2) – – (1 − √3 − √2) (b) tan −1 – – (1 + √3 + √2) – – (√3 − 1 + √2) (c) tan−1 – – (1 + √3 + √2) – – (√3 − 1 + √2) (d) tan−1 – – (1 + √3 − √2) Question 38 (Integer type question) 2021 → → → → → → If , the angle and is θ. The value of ′ θ′ will be.......... ∘. ∘ ∘ P × Q = Q × P P Q (0 < θ < 360 ) Question 39 (Only one correct answer) 2021 Assertion A : If A, B, C, D are four points on a semi-circular arc with center at ′ O′ such that − −→ − −→ − −→ − −→ − −→ − −→ − −→ − −→ − −→ |AB | = |BC | = |CD| , then AB + AC + AD = 4AO + OB + OC − −→ − −→ − −→ − −→ − −→ Reason R : Polygon law of vector addition yields AB + BC + CD = AD = 2AO In the light of the above statements, choose the most appropriate answer from the options given below : (a) A is not correct but R is correct (b) Both A and R are correct but R is not the correct explanation of A (c) A is correct but R is not correct (d) Both A and R are correct and R is the correct explanation of A. Question 40 (Integer type question) 2021 A person is swimming with a speed of 10 at an angle of 120 with the flow and reaches to a ∘ m/s point directly opposite on the other side the river. The speed of the flow is ′ x′ m/s. The value of ′ x′ to the nearest integer is.............. Question 41 (Only one correct answer) 2021 Match List-I with List-II Choose the correct answer from the options given below (a) (a) → (iii), (b) → (ii), (c) → (iv), (d) → (i) (b) (a) → (iv), (b) → (iii), (c) → (i), (d) → (ii) (c) (a) → (i), (b) → (iv), (c) → (ii), (d) → (iii) (d) (a) → (iv), (b) → (i), (c) → (iii), (d) → (ii) Question 42 (Integer type question) 2021 A swimmer can win with velocity of 12 km/h in still water. Water flowing in river has velocity 6 km/h. The direction with respect to the direction of flow of river water he should swim in order to reach the point on the other bank just opposite to his staring point is.........∘. (Round off to the Nearest Integer) (Find the angle in degrees) Question 43 (Only one correct answer) 2021 → → → → → → → → If A and B are two vectors satisfying the relation A. B = |A × B |. Then the value of | A − B| will be : −−−−−−−−−−−−− (a) √A2 2 + B + 2AB −−−−−−−−−−− –−−− (b) √A2 2 + B − √2AB −−−−−−− (c) √A2 + B 2 −−−−−−−−−−− –−−− (d) √A2 2 + B + √2AB Question 44 (Only one correct answer) 2021 → A mosquito is moving with a velocity V = 0.5t ^ 2 i + 3t^ ^ m/s j + 9k and accelerating in uniform conditions. What will be the direction of mosquito after 2 s ? 2 (a) tan−1 ( ) from y -axis 3 5 (b) tan−1 ( ) from y -axis 2  5 (c) tan−1 ( ) from x -axis 2 2 (d) tan−1 ( ) from x -axis 3 Question 45 (Only one correct answer) 2021 In an octagon ABCDEF GH of equal side, what is the sum of − −→ − −→ − −→ − −→ − −→ − −→ −→ − − −→ AB + AC + AD + AE + AF + AG + AH. If AO = 2^ i + 3^ ^ j − 4k. (a) 16^i + 24^ ^ j + 32k (b) 16^i + 24^ ^ j − 32k (c) 16^i − 24^ ^ j + 32k (d) −16^i − 24^ ^ j + 32k Answer 1 Correct answers is D Solution: For 1 particle st 1 2 H = g(t) 2 1 ′ 2 5 = g(t ) 2 ′ t = 5 sec  For 2 particle nd 1 2 H − 25 = g(t − 1) 2 1 1 g 2 2 g(t) − 25 = g(t) + − gt 2 2 2 t = 3 sec 1 H = × 10 × 9 = 45 m 2 Answer 2 Correct answers is B Solution: Horizontal component of velocity of bomb & fighter jet are same So, bomb will remains just below the jet, path is straight line w.r.t. pilot. Answer 3 Correct answers is D Solution: x 2 y = αx − βx , y = αx (1 − ) α/β x Compare with y = x tan θ (1 − ) R −1 tan θ = α ⟹ θ = tan α R 4 = H tan θ 2 R tan θ α H = = 4 4β Answer 4 Correct answers is D Solution:  1 For first drop. h = g(2n) 2 2 1 For second drop. h′ = g(n) 2 2 h 4 = ′ h 1 h 9.8 ′ h = = 4 4 So height of second drop 9.8 ′ H = h − h = 9.8 − 4 3 = × 9.8 = 7.35 m 4 Answer 5 Correct answers is D Solution: The aero plane imparts v velocity to the food packet horizontally when dropping. −− − 2h Horizontal range of food packet = v × √ g −− − 2h Also. Horizontal distance. travelled by aero plane = v × √ g thus, vertical distance between aero plane and person = h −−− −−−−−− 2 2v h distance = √ + h 2 g Answer 6 Correct answers is B Solution: → → → → We can determine the magnitude of ( P + Q ) and ( P − Q ) either algebraically or graphically. Algebraically: −−−−−−−− −−−−−−−−−−−−−−− − → → → → → → → → → → → → → → |P | √ 2 √ + Q = |P + Q| = ( P + Q ). ( P + Q ) = P. P + Q. Q + 2 P. Q  → → Now, P. Q = P Q cos 90 = 0 → → −−−−−−− So, | P + Q | = √P 2 + Q 2 → → −−−−−−− Similarly, | P − Q | = √P 2 + Q2 Now, let's derive the same graphically − → − → F1 and F2 at θ1 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−− − 2 2 2 2 2 2 F net = √P 1 + Q + P + Q + 2(P + Q ) cos θ1 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−− − 2 2 2 2 2 2 F net 2 = √P + Q + P + Q + 2(P + Q ) cos θ2 −−−−−−−− − IfF net 1 = √3(P 2 2 + Q ) −−−−−−−−−−−−−−−−−−−−−−−− − 2 2 2 2 = √2(P + Q ) + 2(P + Q ) cos θ1 2 2 P + Q ⟹ θ1 = 2 2 2(P + Q ) ∘ ⟹ θ1 = 60 −−−−−−−− − 2 2 F net 2 = √2(P + Q ) −−−−−−−−−−−−−−−−−−−−−−−− − 2 2 2 2 = √2(P + Q ) + 2(P + Q ) cos θ2 ⟹ cos θ2 = 0 ∘ ⟹ θ2 = 90 Answer 7 Solution: 2 v = 2x + 20 dv ⟹ 2v = 2 dx 2 2 ⟹ a = = 1 m/s 2 Answer 8 Correct answers is B Solution: dx = vdt 2 2 2 x = ∫ αtdt + ∫ βt dt 1 1  2 2 2 3 αt βt x = ( ) + ( ) ; 2 3 1 1 3 7 x = α + β 2 3 Answer 9 Correct answers is B Solution: V0 = αt1 = βt2. V0 V0 t = t1 + t2 = + α β αβ V0 = ( )t α + β S1 + S2 = S = Area of V − t curve 1 1 αβ 1 αβ 2 = (t)V0 = (t) ( )t = ( )t 2 2 α + β 2 α + β Answer 10 Correct answers is C Solution: ∘ B sin 60 tan ϕ = ∘ A − B cos 60 ∘ B sin 60 −1 ϕ = tan ( ) ∘ A − B cos 60 – √3B −1 = tan ( ) 2A − B Answer 11 Correct answers is D  Solution: → a = 8^ i + 2^ j → → 1→ 2 s = u t + a t 2 → 1 s = (8^ i + 2^ j ) × 100 2 → s = 400^ i + 100^ j Answer 12 Correct answers is C Solution: −−− V = √2gh h −−−− 1 2 = √2ght − gt 3 2 −−−− 2h 2 gt − 2√2ght + = 0 3 −−−−−−−−−−−− −−− 2h 2√2gh ± √8gh − 4g × 3 t = 2g − −−−− −−− 16gh 2√2gh ± √ 3 t = 2g −− − −−− gh 2√2gh ± 4√ 3 t = 2g −− − −−− gh 2√2gh − 4√ t1 3 = −− − t2 gh −−− 2√2gh + 4√ 3 – 4 2√2 − – – – t1 √3 √3 − √2 = = – – t2 – 4 √3 + √2 2√2 + – √3 Answer 13 Solution:  2 2 v = u + 2as 0 = 100 − 2 × a × 0.5 2 a = 100 m/s Retarding force = ma = 0.1 × 100 = 10 N Answer 14 Correct answers is A Solution: Answer 15 Correct answers is C Solution: F Acceleration of ball a = m 2 F0 t − T a = [1 − ( ) ] m T 2 dv F0 t − T = [1 − ( ) ] dt m T 2 v 2T F0 t − T ∫ dv = ∫ [1 − ( ) ] dt 0 m 0 T 2T F0 1 3 v = [1 − (t − T ) ] 2 m 3T 0 3 F0 1 (−T ) 3 v = [(2T − (2T − T ) ) − (0 − )] 2 2 m 3T 3T 4F o T v = 3M Answer 16 Correct answers is A Solution: 2π ω1 = = 2π T1 2π 2π ω2 = = T2 8 ω1 ⟹ = 8 ω2 Answer 17  Solution: → → → V M = V M,R + V R VM should be along line AB → → As | V M,R | = | V R| → → ∴ VM should be along angle bisector of angle between V M,R and V R ∘ ∴ θ = 30 Answer 18 Correct answers is B Solution: 2 2 v = vc + 2al [length of train = 2l] 2 2 vc = u + 2al 2 2 2 2 v − vc = vc − u −−−−−−− 2 2 v + u vc = √ 2 Answer 19 Solution: S1 = S2 1 1 2 2 35t + (−g)t = 35(t − 3) + (−g)(t − 3) 2 2 1 1 2 2 35t − gt = 35t − 35 × 3 − g(t − 6t + 9) 2 2 35 × 3 + 45 = 30t 150 t = = 5 30 1 height h = 35 × 5 − × 10 × 5 2 2 h = 175 − 125 = 50 m Answer 20 Correct answers is A Solution: → Resultant ( R ) ∘ ∘ ∘ ∘ ∘ = ^ i (10 cos 30 + 20 cos 60 − 15 cos 30 − 15 cos 45 + 20 cos 45 ) ∘ ∘ ∘ ∘ ∘ + ^ j (10 sin 30 + 20 sin 60 + 15 sin 30 − 15 sin 45 − 20 sin 45 ) = 9.25^ i + 5^ j Answer 21 Correct answers is C Solution: Suppose length of escalator = L L Speed of man w.r.t. escalator = t1 L Speed of escalator = t2 L L Speed of man w.r.t. ground when escalator is moving = + t1 t2 L t1 t2 Time taken by the man to walk on the moving escalator = = L L t1 + t2 + t1 t2 Answer 22 Correct answers is A Solution: Vmax Vmax tan α = ; a1 = t1 t1 Vmax Vmax tan β = ; a2 = t2 t2 a1 t2 t1 a2 = ; ⟹ = a2 t1 t2 a1 Answer 23 Solution: vdv a = dx −−−−−−− −− 1 v = √5000 + 24x × −−−−−−− −− × 24 2√5000 + 24x 2 a = 12 m/s Answer 24 Correct answers is A Solution:  ∘ θ = 45 2 2 u sin θ H = 2g 1 2 (25) × ( ) 2 = 2 × 10 125 = m 8 T and time t = 2 1 25 ( –) u sin θ √2 = = g 10 25 5 = – = – s 10√2 2√2 Answer 25 Correct answers is B Solution: → → → → | X − Y | = n| X + Y | → → → → 2 2 |X | + |Y | − 2| X || Y | cos θ = → → → → 2 2 2 n [| X | + |Y | + 2| X || Y | cos θ] → → As | X | = |Y | → → 2 2 2| X | − 2| X | cos θ = → → 2 2 2 2 2n | X | + 2n | X | cos θ 2 2 1 − cos θ = n + n cos θ 2 1 − n cos θ = 2 1 + n 2 n − 1 −1 θ = cos ( ) 2 −n − 1 Answer 26 Solution: F 2^ i + 3^ ^ j + 5k a = = m 2 → → → 1→ 2 rf − ri = u t + a t 2 1 2^ i + 3^ ^ j + 5k 2 x^ i + y^ ^ = j + zk × (4) 2 2 x^ i + y^ ^ = 8^ j + zk i + 12^ ^ j + 20k b = 12 Answer 27 Correct answers is C Solution: 2 t = mx + nx Differentiating w.r.t. t 1 = 2mxv + nv 1 = v(2mx + n) Again differentiating w.r.t. t dv × (2mx + n) + 2mv 2 = 0 ; dt 3 a = −2mv Answer 28 Correct answers is A Solution: For stone 1 2 75 = −10t + gt 2 2 75 = −10t + 5t 2 t − 2t − 15 = 0 t = 5 sec Height of balloon H = vt + 75 H = 10 × 5 + 75 = 125 m. Answer 29 Correct answers is D Solution: a = constant dv = constant = a dt v = u + at  dx = v = u + at dt 1 2 x = ut + at 2 Answer 30 Correct answers is B Solution: → D = VF ,G × T = [4^ i + 4^ j + (−^ j )] × 3s → | D | = 15 m Answer 31 Solution: By energy conservation 1 1 2 2 kx = mv 2 2 −− − k ⟹ v = x√ m −−−− 100 −− v = 0.05 × √ = 0.5√10 m/s 0.1 −−−− −−− − − 2H 2 × 2 2 Time of flight of ball T = √ = √ = −− sec g 10 √10 Range of ball s = ut −− 2 d = 0.5√10 × ( −−) = 1 m √10 Answer 32 Correct answers is C Solution: → → → → |P + Q | = n| P − Q| → → → → 2 2 |P | + |Q| + 2| P || Q | cos θ → → → → 2 2 2 = n (| P | + |Q| − 2| P || Q | cos θ) 2 2 + 2 cos θ = n (2 − 2 cos θ) 2 n − 1 −1 θ = cos ( ) 2 n + 1 Answer 33 Correct answers is A Solution: 1 Let next drop after t distance travelled by 1 drop in 4 , is S1 ( should be st 2 sec sec = at = 78.4 m t 2 less then 4 sec ) distance travelled by succeeding drop in 4 − t sec 1 2 S2 = a(4 − t) 2 S1 − S2 = 34.3 2 78.4 − 4.9(4 − t) = 34.3 (4 − t) 2 = 9 ;4−t = 3 t = 1 sec Answer 34 Correct answers is D Solution: 2 v = v0 + gt + F t dS 2 = v0 + gt + F t dt S 1 2 ∫ dS = ∫ (v0 + gt + F t )dt 0 0 g F S = v0 + + 2 3 Answer 35 Correct answers is C Solution:  2 4 S = h + (e h + e h+........ ) × 2 2 2 1 2e 1 + e 2 = h + 2e h = h [1 + ] = h( ) 2 2 2 1 − e 1 − e 1 − e −− − −− − 2h 2h t = √ + 2 × √ [1 + e+......... ] g g −− − −− − 2h 1 2h 1 + e = √ [1 + 2 × ] = √ ( ) g 1 − e g 1 − e 2 h 1 + e 1 − e < V > = ( ) × ( ) −− − 2 2h 1 − e 1 + e √ g 81 −− − −−− −−− 1 + ( ) 2 gh 1 + e 10 × 5 100 = √ = √ ≈ 2.5 −−−− 2 2 (1 + e)2 2 81 (1 + √ ) 100 Answer 36 Correct answers is C Solution: dv a = v( ) (where dv/dx is negative) dx vis decreasing So, a will increase, hence correct option is (c). Answer 37 Correct answers is B Solution: − −→ ∘ ∘ OA = R[cos 30 ^i + sin 30 ^j] − −→ ∘ ∘ OB = R[cos 60 ^i + (− sin 60 )^ j] − −→ ∘ ∘ OC = R[− cos 45 ^i + sin 45 ^j] − −→ − −→ − −→ OA + OB − OC – 1 √3 1 = R[ + + ^ –] i 2 2 √2 – 1 √3 1 +R [ − − ^ –] j 2 2 √2 angle with the x-axis – 1 √3 1 − − – 2 2 √2 tan α = – 1 √3 1 + + – 2 2 √2  – – 1 − √3 − √2 = – – 1 + √3 + √2 – – 1 − √3 − √2 −1 α = tan [ – –] 1 + √ 3 + √2 Answer 38 Solution: → → → → P × Q = Q × P → → → → P × Q = −( P × Q) → → 2( P × Q) = 0 P Q sin θ = 0 ∘ ∘ ∘ θ = 180 (0 < θ < 360 ) Answer 39 Correct answers is B Solution: − −→ − −→ − −→ AB = AO + OB...(i) − −→ − −→ − −→ AC = AO + OC...(ii) − −→ − −→ AD = 2AO...(iii) Adding (i), (ii) and (iii) we get − −→ − −→ − −→ − −→ − −→ − −→ AB + AC + AD = 4AO + OB + OC − −→ − −→ − −→ − −→ − −→ And as per the polygon law of vector addition AB + BC + CD = AD = 2AO Both A and R are correct but R is not the correct explanation of A Answer 40 Solution: Vr 1 ∘ sin 30 = ; Vr = × 10 Vm,r 2 Vr = 5 m/s Answer 41 Correct answers is B Solution:  → → → (I) A + C = B → → → (II) A + B + C = 0 → → → (III) A − B − C = 0 → → → (IV) A + B − C = 0 Answer 42 Solution: VR = 6 km/hr VM,R = 12 km/hr VM should be along AB ∴ VR − VM,R sin θ = 0 6 − 12 sin θ = 0 6 1 sin θ = = 12 2 ∘ θ = 30 ∘ α = 90 + θ = 120 Answer 43 Correct answers is B Solution: → → → → A. B = |A × B | ⟹ AB cos θ = AB sin θ ∘ ∴ θ = 45 → → −−−−−−−−−−−−−−−−− −∘ ∴ | A − B | = √A2 + B2 − 2AB cos 45 −−−−−−−−−−− –−−− = √A2 + B2 − √2AB Answer 44 Correct answers is D Solution: → = 0.5t ^i + 3t^ ^ 2 V j + 9k → V at t = 2 = 2^ i + 6^ ^ j + 9k −− − 2 √117 angle with x-axis cos−1 = tan −1 11 2 −− 6 √85 angle with y-axis cos−1 = tan −1 11 6 None of the option is matching. Answer 45 Correct answers is B Solution: − −→ AO = 2^ i + 3^ ^ j − 4k − −→ − −→ − −→ − −→ − −→ − −→ −→ − AB + AC + AD + AE + AF + AG + AH − −→ − −→ − −→ AC = AB + BC − −→ − −→ − −→ − −→ − −→ − −→ − −→ ⟹ AB + (AB + BC ) + (AB + BC + CD) + (AB + − −→ − −→ − −→ − −→ − −→ − −→ − −→ − −→ BC + CD + DE ) + (AB + BC + CD + DE + EF )+ − −→ − −→ − −→ − −→ − −→ − −→ − −→ − −→ (AB + BC + CD + DE + EF + F G) + (AB + BC + − −→ − −→ − −→ − −→ −→ − CD + DE + EF + FG + GH ) − −→ − −→ − −→ − −→ −→ − − −→ [EF = −AB , F G = −BC , GH = −CD, −→ − − −→ HA = −DE ] − −→ − −→ − −→ − −→ − −→ − −→ − −→ ⟹ AB + (AB + BC ) + (AB + BC + CD) + (AB + − −→ − −→ − −→ − −→ − −→ − −→ −− → BC + CD + DE ) + (BC + CD + DE ) + (CD+ − −→ − −→ DE ) + (DE ) − −→ − −→ − −→ − −→ − −→ − −→ ⟹ 4 × (AB + BC + CD + DE ) = 4 × AE = 4 × 2AO ⟹ 8(2^ i + 3^ ^ ) = 16^ j − 4k i + 24^ ^ j − 32k