Physics-DPP-3-Kinematics

PHYSICS Topic: KINEMATICS Date: 1. An elevator in which a man is standing is moving upwards with a speed of 10 m/sec. If the man drops a coin from a height of 2.45 metre, it reaches the floor of the elevator after a time : (g = 9.8 m/sec2) (b) s (c) 2 s (d) 1 s 2 2. If the velocity of a particle is given by v = (180 – 16x) 1/2 m/s then its acceleration will be : (a) zero (b) 8m/s2 (c) – 8m/s2 (d) – 4m/s2 3. A body is projected with a velocity u. If passes through a certain point above the ground after t1 sec. The time after which the body passes through the same point during the return journey is : u 2 ÷ æç u ö÷ æçu2 ö÷ æçu2 ö÷ (a) èç g – t1 ø÷ (b) 2çè g – t1 ø÷ (c) 3çè g – t1 ø÷ (d) 3ç g 2 – t1 ø÷ 4. An express train is moving with velocity v1 , meanwhile the driver of this train comes to know that an another train is also moving ahead with a velocity v2 on the same track. To void collision the driver applies retardation a. Then what will be the minimum time to void collision : v2 – v2 1 æv – v ö (a) t = v1 – v2 (b) t = 1 2 (c) t = ç 2 1 ÷ (d) none of these a 2 2 çè a ø÷ 5. A particle starts moving in a straight line with a constant acceleration a. At a time t1 seconds after the beginning of motion, the acceleration changes sign, remaining the same in magnitude. Determine the time from the beginning of motion, till it returns to the starting point : (a) t1 (2 + 2)sec (b) t2 (1 + 2)sec (c) t1 2 sec (d) 2 2 t1 sec 6. In the one-dimensional motion of a particle, the relation between position x and time t is given by x2 + 2x = t Choose the correct statement : (a) The retardation of the particle is 1 4(x + 1)3 (b) The uniform acceleration of the particle is 1 (x + 1)3 (c) The uniform velocity of the particle is 1 (x + 1)3 (d) The particle has a variable acceleration of 4t + 6. 7. A body freely falling from rest has a velocity v after it falls through distance h. The distance it has to fall down further for its velocity to become double is : (a) h (b) 2h (c) 3h (d) 4h 8. An object starts from rest at t = 0 & accelerates at a rate given by a = 6t then its velocity & displacement after 2 seconds will be (a) 10 m/sec, 8 m (b) 12 m/sec, 4 m (c) 4 m/sec, 6 m (d) 12 m/sec, 8 m PHYSICS Topic: KINEMATICS Date: Answer Key 1. (b) 2. (c) 3. (b) 4. (a) 5. (a) 6. (a) 7. (a) 8. (d)

Comments

Post a Comment

Popular posts from this blog

PHYSICS-15-10- 11th (PQRS) SOLUTION

XI (PQRS) PHYSICS REVIEW TEST-5 Select the correct alternative. (Only one is correct) [15 × 3 = 45] There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Q.31 Two trains, which are moving along different tracks in opposite directions towards each other, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when the trains are 300 m apart. Graphs given below show their velocities as function of time as the trains slow down. The separation between the trains when both have stopped, is: (A) 120 m (B) 280 m (C) 60 m (D*) 20 m 1 [Sol: x1 = 2 1 × 10 × 40 = 200 m; x2 = – 2 × 8 × 20 = –80 m  xreletive = x1 + x2 = 280  separation = 300 – 280 = 20 m (D) ] Q.32 One end of a thin uniform rod of length L and mass M is riveted to the centre of a uniform circular disc of radius r and mass 2 M so that the rod is normal to the disc. The centre of mass of the combination from the centre of the disc is at dist
Read more

8-Circular Motion

8.1 (D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards. The free body diagram of the bead is For the bead not to slide upwards. m2 (r sin 45°) cos 45° – mg sin 45° < N (1) where N = mg cos 45° + m2 (r sin 45º) sin 45° (2) From 1 and 2 we get.  = rad / s. 8.2 (C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv2 = mg sin  (1) r Applying conservation of energy as the block moves from A to B.. 1 mv2 = mg (r cos  – r sin ) (2) 2 Solving 1 and 2 we get 3 sin  = 2 cos  8.3 (A) As the mass is at the verge of slipping ∴ mg sin37 –  mg cos37 = m2r 6 – 8 = 4.5   = 3 16 8.4 (B) As when they collide vt  1  72v 2  2 2  25R   R = vt  t =   5R 6v Now angle covered by A =   vt R 11 Put t  angle covered by A = 6 8.5 (C) The acceleration vector s
Read more

4. THEORY-Current Electrictricity

CURRENT ELECTRICITY  1 . ELECTRIC CURRENT (a) Time rate of flow of charge through a cross sectional area is called Current. if q charge flows in time interval t then average current is given by q Iav = t and Instantaneous current i =. lim q  dq t0 t dt (b) Direction of current is along the direction of flow of positive charge or opposite to the direction of flow of negative charge. But the current is a scalar quantity. i i q+ velocity q– velocity SI unit of current is ampere and 1 Ampere = 1 coloumb/sec 1 coloumb/sec = 1A 2 . CONDUCTOR In some materials, the outer electrons of each atom or molecule are only weakly bound to it. These electrons are almost free to move throughout the body of the material and are called free electrons. They are also known as conduction electrons. When such a material is placed in an electric field, the free electrons drift in a direction opposite to the field. Such materials are called conductors. 3 . INSULATOR Another class o
Read more