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PHYSICS Assignment (J-Batch) WA.pdf

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OBJECTIVE SOLVED 1. Pick up the correct statements: (a) area under a – t graph gives velocity (b) area under a – t graph gives change is velocity (c) path of projectile as seen by another projectile is parabola (d) a body, whatever be its motion, is always at rest in a frame of reference fixed to body itself. Solution: dv a dt v2 dv a dt v1 v = Area under a – t graph where v = magnitude of change in velocity. The path of a projectile as seen by another projectile is a straight line becomes, the relative velocity between the particles remains constant. (b) and (d) 2. The angular acceleration of a particle moving along a circular path with uniform speed is (a) uniform but non-zero (b) zero (c) variable (d) such as cannot be predicted from the given information. Solution: As angular speed of the particle is constant and hence angular acceleration is zero. (b) 3. A particle is projected horizontally from the top of a cliff of height H with a speed of curvature of the trajectory at the instant of projection will (a) H/2 (b) H (c) 2H (d) . Solution: Since, → → . The radius Radial acceleration a r g We know a v2 / r v2 r g where r is the radius of curvature. 2gH g r r 2H ( v 2gH ) Hence (C) 4. A body when projected vertically up, covers a total distance D. During the time of its flight t. If there were no gravity, the distance covered by it during the same time is equal to (a) 0 (b) D (c) 2D (d) 4D. Solution: The displacement of the body during the time t as it attains the point of projection S 0 t 2v 0 g v 0 t 1 gt 2 0 2 During the same time t, the body moves in absence of gravity through a distance D v0t , because in absence of gravity g = 0 2v 0 2v 2 D v 0 0 ...(i) g g In presence of gravity the total distance covered is v 2 v 2 D 2H 2 0 0 ...(ii) 2g g (i) (ii) D 2D Hence (c) 5. A particle is projected from point P with velocity 5 2ms1 perpendicular to the surface of a hollow right angle cone whose axis is vertical. It collides at Q normally. The time of the flight of the particle is (a) 1 sec. (b) sec. (c) Solution: t 2 2 sec u 5 2 2 1sec. (d) 2 sec. g sin 10 Hence (a) x 6. The relative velocity of a car ‘A’ with respect to car B is 30 2 m/s due North-East. The velocity of car ‘B’ is 20 m/s due south. The relative velocity of car ‘C’ with respect to car ‘A’ is 10 2 m/s due North-West. The speed of car C and the direction (in terms of angle it makes with the east). (a) 20 m/s, 45º (b) 20 2m / s, 135º (c) 10 Solution: m / s, 45º (d) 10 2 m / s, 135º . Given → 30 2ms1 | vAB | | vB | 20 ms → 1 | vCA | 10 2 ms Now → → → ˆ ˆ E(x) vAB vA vB 30 2(cos 45º i sin 45º j) or, → → vA vB (30iˆ 30ˆj)ms1 ...(i) vB and, → (20ˆj)ms1 ...(ii) and → → → ˆ ˆ vCA vC vA 10 2 (cos 45º i sin 45º

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- OBJECTIVE SOLVED 1. Pick up the correct statements: (a) area under a – t graph gives velocity (b) area under a – t graph gives change is velocity (c) path of projectile as seen by another projectile is parabola (d) a body, whatever be its motion, is always at rest in a frame of reference fixed to body itself. Solution: a dt dv dt a dv 2 1 v v v = Area under a – t graph where v = magnitude of change in velocity. . The path of a projectile as seen by another projectile is a straight line becomes, the relative velocity between the particles remains constant. (b) and (d) 2. The angular acceleration of a particle moving along a circular path with uniform speed is (a) uniform but non-zero (b) zero (c) variable (d) such as cannot be predicted from the given information. Solution: As angular speed of the particle is constant and hence angular acceleration is zero. (b) 3. A particle is projected horizontally from the top of a cliff of height H with a speed gH 2 . The radius of curvature of the trajectory at the instant of projection will (a) H/2 (b) H (c) 2H (d) . Solution: Since, v g Radial acceleration g ar We know 2 r a v / r 2 v g r where r is the radius of curvature. g r gH 2 ) gH 2 v ( H 2 r Hence (C) 4. A body when projected vertically up, covers a total distance D. During the time of its flight t. If there were no gravity, the distance covered by it during the same time is equal to
- x y Q P 90º (a) 0 (b) D (c) 2D (d) 4D. Solution: The displacement of the body during the time t as it attains the point of projection 0 S 0 gt 2 1 t v 2 0 g v 2 t 0 During the same time t, the body moves in absence of gravity through a distance 0 D v t , because in absence of gravity g = 0 g v 2 g v 2 v D 2 0 0 0 ...(i) In presence of gravity the total distance covered is g v g 2 v 2 H 2 D 2 0 2 0 ...(ii) (i) (ii) D 2 D Hence (c) 5. A particle is projected from point P with velocity 1 ms 2 5 perpendicular to the surface of a hollow right angle cone whose axis is vertical. It collides at Q normally. The time of the flight of the particle is (a) 1 sec. (b) 2 sec. (c) sec 2 2 (d) 2 sec. Solution: . sec 1 10 2 2 5 sin g u t Hence (a) 6. The relative velocity of a car ‘A’ with respect to car B is 2 30 m/s due North-East. The velocity of car ‘B’ is 20 m/s due south. The relative velocity of car ‘C’ with respect to car ‘A’ is 2 10 m/s due North-West. The speed of car C and the direction (in terms of angle it makes with the east). (a) 2 20 m/s, 45º (b) º 135 , s / m 2 20 (c) º 45 , s / m 2 10 (d) º 135 , s / m 2 10 . Solution: Given 1 AB | v | 30 2ms 1 B | v | 20ms 1 CA | v | 10 2 ms Now AB A B ˆ ˆ v v v 30 2(cos45ºi sin 45º j) or, 1 A B ˆ ˆ v v (30i 30j)ms ...(i) and, 1 B ˆ v ( 20j)ms ...(ii) vCA N(Y) vAB E(x) vB 45º 45º
- and CA C A ˆ ˆ v v v 10 2( cos45ºi sin 45º j) or 1 C A ˆ ˆ v v ( 10i 10j)ms ...(iii) solving equation (i) (ii) and (iii) we’ll get C ˆ ˆ v 20i 20j (a) 7. Two particles 1 and 2 are allowed to descend on the two frictionless chord OA and OB of a vertical circle, at the same instant from point O. The ratio of the velocities of the particles 1 and 2 respectively, when they reach on the circumference will be (OB is the diameter) (a) sin (b) tan (c) cos (d) none of these. Solution: cos g a , cos d OA OA Along OA cos d . cos g 2 v2 A Along OB gd 2 v2 B cos v v B A Hence, C is correct. 8. A ball is dropped vertically from a height d above the ground. If it hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air resistance, its velocity v varies with the height h above the ground as (a) d v h (b) d v h (c) d h v (d) d h v . Solution: As the ball falls, at height h the velocity of the ball is zero and at any height ) h d ( g 2 u v , h 2 2 , with decreasing h, v increases. When h = 0, v = velocity is maximum. After the ball collides the floor, its velocity changes in magnitude as wall as direction, as the body goes to a smaller height in bouncing up .The change in velocity takes place within zero height and with no change in time. Hence, (a) is correct choice. A B 1 2 O A B 1 2 O g cos d
- 9. A particle moves along a straight line according to the law c bt 2 at S 2 2 . The acceleration of the particle varies as (a) 3 S (b) 3 / 2 S (c) 2 S (d) 2 / 5 S . Solution: 2 / 1 2 ) c bt 2 at ( S Differentiating, c bt 2 at b at ) b 2 at 2 ( ) c bt 2 at ( 2 1 dt dS 2 2 / 1 2 ) c bt 2 at ( c bt 2 at ) b at )( b at ( a c bt 2 at dt S d 2 2 2 2 2 c bt 2 at c bt 2 at ) b at ( ) c bt 2 at ( a 2 2 2 2 2 2 (ac b ) S S 3 2 2 S 1 dt S d acceleration 3 S (a) 10. A car accelerates from rest at a constant rate for some time, after which it decelerates at a constant rate to come to rest. If the total time elapsed is t, the maximum velocity acquired by car is (a) t ) ( V (b) t ) ( V (c) t ) ( 2 V (d) t ) ( 2 V . Solution: From motion from A to B 1 t V or V t1 From motion from B to C 2 t V 0 or V t2 (t ) 1 v A B C (t ) 2 ) ( V V V t t t 2 1 or t ) ( V (a) 11. A stone A is dropped from rest from a height h above the ground. A second stone B it simultaneous y thrown vertically up from a point on the ground with velocity v. The line of motion of both the stones is same. The values of v which would enable the stone B to meet the stone A midway between their initial positions is (a) 2 gh (b) gh 2 (c) gh (d) gh 2 . Solution: Time of travel of each stone =
- Distance travelled by each stone 2 h For stone 2 h 1 A, gt 2 2 i.e., g h t For stone B, g h g 2 1 g h u gt 2 1 ut 2 h 2 h/2 h A B A B 2 h g h u 2 h or, h g h u gh h g h u The correct option is (c) 12. A stone is dropped from rest from the top of a cliff. A second stone is thrown vertically down with a velocity of 30 m/s two seconds later. At what distance from the top of a cliff do they meet? (a) 60 m (b) 120 m (c) 80 m (d) 44 m. Solution: The two stones meet at distance S from top of cliff t seconds after first stone is dropped. For 1st stone ; gt 2 1 S 2 For 2nd stone 2 ) 2 t ( g 2 1 ) 2 t ( u S . i.e., g 2 gt 2 gt 2 1 u 2 ut gt 2 1 2 2 s 4 20 30 ) 10 30 ( 2 g 2 u ) g u ( 2 t ); g u ( 2 t ) g 2 u ( 0 Distance S at which they meet 16 10 2 1 gt 2 1 2 = 80 m from top of cliff (c) 13. A particle is projected from a point O with velocity u in a direction making an angle upward with the horizontal. At P, it is moving at right angles to its initial direction of projection. Its velocity at P is (a) tan u (b) cot u (c) ec cos u (d) sec u . Solution: cot u v ; cos u sin v ) 90 cos( v (b) 14. A man running at 6 km/hr on a horizontal road observes that the rain hits him at an angle of 30º from the vertical. The actual velocity of rain has magnitude (a) 6 km/hr (b) 3 6 km/hr (c) 3 2 km/hr (d) 2 km/hr u P O 90º (90º- ) v
- Solution: R V velcoity of rain RM V velocity of rain relative to man M V velocity of man Given, M V 6 kmh-1 , RM V at angle 30º from vertical RM V M V R V M V here R RM M V V V from figure, M R V tan30º V R 1 6 V 3 R V 6 3 km/hr (b) 15. The angular acceleration of a particle moving along a circular path with uniform speed is (a) uniform but non-zero (b) zero (c) variable (d) such as cannot be predicted from the given information. Solution: As angular speed of the particle is constant and hence angular acceleration is zero. (b)

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