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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 451. |
Give an example of situation in which distance is equal to displacement. |
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Answer» A man climbing a ladder Explanation: Distance is a scalar quantity (it has magnitude only) while displacement is a vector quantity (it has magnitude and direction both) Hence if we take any example considering a motion in which direction does not change throughout the motion and we will always get Distance = Displacement. |
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| 452. |
What is meant by displacement? |
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Answer» The displacement of an object is the shortest distance travelled between the initial and final position of the object.
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| 453. |
Define Distance. |
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Answer» The distance travelled by an object is the length of actual path travelled by the object during the motion. |
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| 454. |
Define Motion. |
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Answer» Motion means movement. The motion of an object is perceived when its position changes continuously with respect to some stationary object. |
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| 455. |
Two balls are thrown simultaneously from ground with same velocity of `10 ms^(-1)` but different angles of projection with horizontally. Both balls fall at same distance `5sqrt(3)m` from point of projection. What is the time interval between balls striking the ground ?A. `(sqrt(3)-1)s`B. `(sqrt(3)+1)s`C. `sqrt(3)s`D. 1s |
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Answer» Correct Answer - A Given, `5sqrt(3) = ((10)^(2) sin 2 theta)/(g)` or `sin 2 theta = (sqrt(3))/(2)` `:. 2 theta = 60^(@)` or `theta = 30^(@)` Two different angles of projection are therefore `theta` and `90^(@) - theta` or `30^(@)` and `60^(@)`. `:. T_(1) = (2u sin 30^(@))/(g) = 1s` `:. T_(2) = (2u sin 60^(@))/(g) = sqrt(3)s` `:. Deltat = T_(2) - T_(1) = (sqrt(3)-1)s` |
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| 456. |
Which of the following is correct ?A) Distance ≥ Displacement B) \(\cfrac{Distance}{Displacement}\) ≥ 1C) Both A & B D) None |
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Answer» C) Both A & B |
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| 457. |
Second law of motion is ……………A) v = u + at B) s = ut + 1/2 at2C) v2 – u2 = 2as D) None of these |
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Answer» B) s = ut + 1/2 at2 |
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| 458. |
This graph represents the body is moving withA) uniform acceleration B) uniform speed C) uniform deceleration D) constant speed |
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Answer» A) uniform acceleration |
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| 459. |
Which graph represents a state of rest for an object? |
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Answer» With increase in time distance remains constant which means it is not moving. |
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| 460. |
In a two dimensional motion,instantaneous speed `v_(0)` is a positive constant.Then which of the following are necessarily true?A. The acceleration of the particle is zeroB. The acceleration of the particle is boundedC. The acceleration of the particle is necessarily in the plane of motionD. The particle must be undergoing a uniform circular motion. |
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Answer» Correct Answer - C As given motion is two dimensional motion and given that instantaneous speed `v_(0)` is positive constant. Acceleration is rate of change of velocity (instantaneous speed) hence it will also be in the plane of motion. |
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| 461. |
How do we understand motion? |
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Answer» A body is said to be in motion when its position is changing continuously with time relative to an observer. |
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| 462. |
What answer may the passenger give to the driver? |
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Answer» The car is in motion with respect to the observer on the road, but at rest with respect to the passenger. Because motion is a combined property of the observer and the body which is being observed. |
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| 463. |
Are the terms relative or not? |
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Answer» The terms “longer”, “shorter”; “up” and “down”, etc. are relative to each other. |
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| 464. |
The maximum range of a gun on horizontal terrain is 1km. If `g = 10 ms^(-2)`, what must be the muzzle velocity of the shell ?A. `400 ms^(-1)`B. `200 ms^(-1)`C. `100 ms^(-1)`D. `50 ms^(-1)` |
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Answer» Correct Answer - C `R_(max) = (u^(2))/(g)` at `theta = 45^(@)` `:. U = sqrt(gR_(max)) = 100 ms^(-1)` |
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| 465. |
An object may appear to be moving to one person and the same object may appear to be at rest to another person. This statement isA. always correctB. always falseC. sometimes correct and sometimes falseD. cannot say. |
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Answer» Correct Answer - A The statement is always correct. |
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| 466. |
A cricket ball thrown across a field is a heights `h_(1)` and `h_(2)` from the point of projection at time `t_(1)` and `t_(2)` respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey isA. `((h_(1)t_(2)^(2)-h_(2)t_(1)^(2))/(h_(1)t_(2) -h_(2)t_(1)))`B. `((h_(1)t_(2)^(2)-h_(2)t_(1)^(2))/(h_(1)t_(2)-h_(2)t_(1)))`C. `((h_(1)t_(2)^(2)-h_(2)t_(1)^(2))/(h_(1)t_(2)-h_(2)t_(1)))`D. None of these |
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Answer» Correct Answer - C For vertically moment, `h_(1) = u sin theta t_(1) -(1)/(2) g t_(1)^(2)` (for `h_(1))` `rArr t_(1) = (h_(1)+(1)/(2)g t_(1)^(2))/(u sin theta)` ........(i) `rArr h_(2) = u sin theta t_(2) - (1)/(2) g t_(2)^(2)` (for `h_(2))` `rArr t_(2) = (h_(2)+(1)/(2)g t_(2)^(2))/(u sin theta)` ........(ii) On dividing Eq. (i) by Eq. (ii), we get `(t_(1))/(t_(2)) = (h_(1)+(1)/(2)g t_(1)^(2)//u sin theta)/(h_(2)+(1)/(2)g t_(2)^(2)//u sin theta) rArr h_(1)t_(2)-h_(2)t_(1) = (g)/(2) (t_(1)t_(2)^(2)-t_(1)^(2)t_(2))` The time of flight of the ball, `T = (2u sin theta)/(g) = (2)/(g) (u sin theta)` [from Eq.(i)] `= (2)/(g) ((h_(1)+1//2g t_(1)^(2))/(t_(1))) =(2)/(t_(1)) ((h_(1))/(g)+(t_(1)^(2))/(2))` `=(h_(1))/(t_(1)) xx (2)/(g)+t_(1) = (h_(1))/(t_(1)) xx ((t_(1)t_(2)^(2)-t_(1)^(2)t_(2))/(h_(1)t_(2)-h_(2)t_(1))) +t_(1)` `=(h_(1)t_(1)t_(2)^(2)-h_(1)t_(1)^(2)t_(2)+h_(1)t_(1)^(2)t_(2)-h_(2)t_(1)^(3))/(t_(1)(h_(1)t_(1)h_(2)t_(1)))=((h_(1)t_(2)^(2)-h_(2)t_(1)^(2))/(h_(1)t_(2)-h_(2)t_(1)))` |
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| 467. |
A projectile is thrown at an angle `theta` with the horizontal and its range is `R_(1)`. It is then thrown at an angle `theta` with vertical anf the range is `R_(2)`, thenA. `R_(1) = 4R_(2)`B. `R_(1) = 2R_(2)`C. `R_(1) = R_(2)`D. None of these |
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Answer» Correct Answer - C We know that, `R_(theta) = R_(90^(@)-theta)` |
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| 468. |
For an object thrown at `45^(@)` to the horizontal, the maximum height H and horizontal range R are related asA. `R = 16H`B. `R = 8H`C. `R = 4H`D. `R = 2H` |
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Answer» Correct Answer - C As we know that maximum height of a projectile is given by `H_(max) = (u^(2)sin^(2)theta)/(2g)` where, u = initial velocity of projectile g = acceleration due to gravity and `theta=` angle of projection. As from question, `H_(max) = (v^(2)sin^(2)45^(@))/(2g)`.....(i) (as `u =v, theta = 45^(@))` Now, range of a projectile is given by `R = (u^(2)sin 2 theta)/(g) rArr R = (v^(2)sin (2xx 45^(@)))/(g)` `rArr R = (v^(@) sin 90^(@))/(g)` .......(ii) On dividing Eq. (i) by Eq. (ii). we get `(H_(max))/(R) = (v^(2)sin^(2)45^(@) xx g)/(2g xx v^(2)sin 90^(@)) = (1)/(4 xx1)` `rArr R = 4H_(max) = 4H` |
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| 469. |
An object is thrown along a direction inclined at an angle of `45^(@)` with the horizontal direction. The horizontal range of the particle is equal toA. vertical heightB. twice the vertical heightC. thrice the vertice heightD. four times the vertical height |
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Answer» Correct Answer - D `(R)/(H) = ((u^(2)sin 2theta)/(g))/((u^(2)sin^(2)theta)/(2g)) = (2sin theta cos theta)/(g) xx (2g)/(sin^(2)theta) = 4 cot theta` `rArr R = 4H cot theta` if `theta = 45^(@)` then `R = 4H cot (45^(@)) = 4H` |
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| 470. |
The adjacent distance – time graph indicates A) A particle travels constantly along X-axis.B) Particle is at rest. C) The velocity of the particle increases up to a time t0 and then remains constant. D) The particle travels up to time t(( with constant velocity and then stops. |
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Answer» C) The velocity of the particle increases up to a time t0 and then remains constant. |
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| 471. |
Mehak was moving through the city roads towards her school by a car. She recorded the odometer reading of the car after every five minutes and plotted a graph for distance vs time. She then inferred about the type of motion and found average speed from the graph. (a) Which qualities of Mehak are worth mentioning? (b) What type of motion would she have inferred? (c) How is average speed calculated from the graph? |
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Answer» (a) Mehak is curious, practical, skilled, alert and enthusiastic. (b) Data given is insufficient to decide what kind of motion she infers.Motion can be uniform or non-uniform. (c) Average speed = \(\frac{total\,distance\,travelled}{total\,time\,taken}\) And total distance travelled is given by the area under the velocity v/s time graph Hence, Average speed = \(\frac{area\,under\,the\,curve}{total\,time\,taken}\) |
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| 472. |
A man used his car. The initial and final odometer readings are 4849 and 5549 respectively. The journey time is 70h. What is average speed of the journey? |
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Answer» Distance covered = 5549 – 4849 = 700 km. Time = 70 h. Average speed =\(\frac{Total\,distance}{Time\,taken}\)=\(\frac{700\,km}{70\,h}\)=10 km/h. |
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| 473. |
A man used his car. The initial and final odometer readings are 4849 and 5549 respectively. The journey time is 25h. What is his average speed during the journey? |
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Answer» Distance covered = 5549 – 4849 = 700 km. Time = 25h. Average speed = \(\frac{Total\,distance}{Time \,taken}= \frac{700\,km}{25\,h}=28\,km/h\) |
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| 474. |
The odometer of a car reads ` 2000 km ` at the start of a trip and `2400 km` at the end of the trip . If the trip took `8 h` , calculate the average speed of the car in `km//h and m//s`. |
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Answer» Correct Answer - `50 km//h` ; `13.9 m//s` Distance travelled , `s =` final reading - initial reading `= 2400 - 2000 = 400 km` time taken , `t = 8 h` Average speed `= (s)/(t) = (400)/( 8) = 50 km//h = ( 50 xx 1000 m)/(60 xx 60 s) = 13.9 m//s` |
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| 475. |
Usha swims in a `90 m` long pool. She covers `180 m` in one minute by swimming from one end to the other and back along the same length path. Find the average speed and average speed and average velocity of Usha. |
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Answer» Correct Answer - `3 ms^(-1)` ; zero Here , total distance travelled , ` s = 180 m` , time taken , `t = 1 min. = 60s` Average speed `= (s)/(t) = ( 180)/( 60) = 3 m//s` In swimming from one end to the other and back along the same straight path , displacement `= 0`. Average velocity `= ("displacement")/("time") = zero` |
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| 476. |
A sprinter is running along the circumference of a big sports stadium with constant speed. Which of the following do you think is changing in this case? a) magnitude of acceleration being produced b) distance covered by the sprinter per second c) direction in which the sprinter is running d) centripetal force acting on the sprinter |
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Answer» The correct answer is c) direction in which the sprinter is running |
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| 477. |
A 100 m sprinter increases her speed from rest uniformly at the rate of 1 m/s2 upto 40 m and covers the remaining distance with a uniform speed. The sprinter covers the first half of the run in t1s and second half in t2s, then A. t1 > t2 B. t1 < t2 C. t1 = t2 D. information given is incomplete |
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Answer» For the first 40 m: Using the third equation of motion: v2 – u2 = 2] as where: s = distance covered = 40 m v = final velocity = ? m/s u = initial velocity = 0 m/s a = acceleration = 1 m/s2 v2 – u2 = 2 as v2 – 0 = 2 × 1 × 40 v2 = 80 v = 8.94 m/s Using the first equation of motion: v = u + at where: v = final velocity = 8.94 m/s u = initial velocity = 0 m/s a = acceleration = 1 m/s2 t1 = time = ?s v = u + at1 8.94 = 0 + 1 × t1 t1 = 8.94s Here its given the speed is uniform not velocity, therefore acceleration will be 1 m/s2 and speed will be 8.94 m/s everywhere. Using the third equation of motion: v2 – u2 = 2 as where: s = distance covered = 60 m v = final velocity = ?? m/s u = initial velocity = 8.94 m/s a = acceleration = 1 m/s2 v2 – u2 = 2 as v2 – (8.94)2 = 2 × 1 × 60 v2 - 80 = 120 v2 = 120 + 80 = 200 v = 14.14 m/s Using the first equation of motion: v = u + at where: v = final velocity = 14.14 m/s u = initial velocity = 8.94 m/s a = acceleration = 1 m/s2 t2 = time = ?s v = u + at2 14.14 = 8.94 + 1 × t2 t2 = 14.14 – 8.94 t2 = 5.2 secs |
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| 478. |
A sprinter has to cover a total run of 100 m. She increases her speed from rest under a uniform acceleration of 1.0 m/s2 up to three quarters of the total run and covers the last quarter him uniform speed. The time she takes to cover the first half, and to cover the second half of the run will be A. 3.25 s B. 4.25 s C. 5.25 s D. 6.25 s |
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Answer» Using the second equation of motion: s = ut + \(\frac12\)at2 where: s = distance covered = 75 m u = initial velocity = 0 m/s a = acceleration = 1 m/s2 t = time = ts s = ut + \(\frac12\)at2 75 = 0 + \(\frac12\)× 1 × t2 75 × 2 = t2 150 = t2 t = 12.24 s Using the third equation of motion: v2 – u2 = 2 as where: s = distance covered = 75 m v = final velocity = ?? m/s u = initial velocity = 0 m/s a = acceleration = 1 m/s2 v2 – u2 = 2as v2 = 2 × 1 × 75 = 150 v = 12.24 m/s For second part distance is 25 m. Using the second equation of motion: s = ut + \(\frac12\)at2 where: s = distance covered = 25 m u = initial velocity = 12.24 m/s a = acceleration = 1 m/s2 t = time = ts s = ut + \(\frac12\)at2 25 = 12.24t + \(\frac12\)1 x t2 t2 + 24.48t – 50 = 0 t = 1.99 s |
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| 479. |
If a sprinter runs a distance of 100 meters in 9.83 seconds, calculate his average speed in km/h. |
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Answer» Total distance travelled = 100m Total time taken = 9.83 s Average speed = total distance travelled/total time taken = 10.172 m/s Average speed in km/h = 36.62 km/h |
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| 480. |
……………….. is speed in Specified direction. A) Distance B) Velocity C) Acceleration D) Displacement |
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Answer» Correct option is B) Velocity |
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| 481. |
What name is given to the speed in a specified direction ? |
| Answer» Speed of a body in a specified direction is called velocity. | |
| 482. |
A freely falling body crosses the points A, B and C with velocities v, 2v, 3v. Then AB : AC = A) 1 : 2 B) 1 : 3 C) 1 : 1 D) 3 : 8 |
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Answer» Correct option is D) 3 : 8 |
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| 483. |
A body is projected vertically up from ground with a speed \(\sqrt{gh}\). The average speed of total motion isA) \(\sqrt{\cfrac{gh}{2}}\)B) \({\cfrac{\sqrt{gh}}{2}}\)C) \(\sqrt{2gh}\)D) none |
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Answer» B) \({\cfrac{\sqrt{gh}}{2}}\) |
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| 484. |
A body is projected horizontally with a velocity \(\sqrt{29}\) m/s from a height 10 m, the velocity on reaching the ground ……………… m/s.A) \(\sqrt{29}\)B) 10 C) 15 D) 20 |
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Answer» Correct option is C) 15 |
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| 485. |
Manoj was travelling by Metro train while Rajnish was travelling by his car towards the same destination. Manoj was surprised to reach much before Rajnish and said, since his motion was uniform, he could reach on time. However, Rajnish said that Manoj’s motion was also not uniform. (i) Who gave a correct explanation of types of motion? (ii) How did Manoj reach early? (iii) Give the characteristic values of Manoj. |
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Answer» (i) Rajnish gave the correct explanation as for a uniform motion the velocity should be uniform. But the metro has stopped at various stations which means that it had come to rest and its velocity at that point is zero. (ii) Manoj had used metro which is free from the traffic jam and thus manages to reach early. (iii) Manoj is a responsible citizen as he had used a public transport instead of the personal car which helps in conservation of fuel. |
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| 486. |
Ratio of the distance travelled in 1 second, 2 seconds, 3 seconds A) 12 : 22 : 32 B) 1 : 3 : 5 C) 1 : 2 : 3 D) 1: 5 : 9 |
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Answer» A) 12 : 22 : 32 |
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| 487. |
The distance travelled by a particle in time ’t’ is given by s = (2.5 m/s2) t2. Find the average speed of the particle during the time ‘O’ to ‘5’ sec. |
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Answer» The distance travelled by the particle from 0 to 5 sec is s = 2.5 m/s2 × 52 = 62.5 m Average speed during this time v=\(\frac{s}t\)=\(\frac{62.5}5\)=12.5 m/sec. |
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| 488. |
In the above question , calculate (i) distance travelled from `O` to `A`. (ii) distance travelled from `B` to `C`. (iii) total distance travelled by the body in `16 sec`. |
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Answer» Correct Answer - (i) `12 m` ; (ii) `18 m` ; (iii) `66 m` In Fig., (i) Distance travelled from `O` to `A = "area of" Delta OAE = (1)/(2) (OE) xx EA = (1)/(2) xx 4 xx 6 = 12 m` (ii) Distance travelled from `B` to `C = "area of" Delta BCD` `= (1)/(2) DC xx DB = (1)/(2) ( 16 - 10) 6 = 18 m` (iii) We have already calculated in `Q.3`, the distance travelled by the body from `A` to `B = 36 m`. Thus , total distance travelled by the body in ` 16` second `= 12 + 18 + 36 = 66 m` |
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| 489. |
A ship is moving at a speed of `56 km//h`. One second later , it is moving at `58 km//h`. What is its acceleration? |
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Answer» Correct Answer - `0.56 m//s^(2)` Here, ` u = 56 (km)/(h) = ( 56 xx 1000)/( 60 xx 60) = 15.55 m//s , t = 1 s , v = 58 (km)/(h) = ( 58 xx 1000 m)/( 60 xx60 s) = 16.11 m//s` Acceleration , `a = ( v - u)/(t) = ( 16.11 - 15.55)/(1) = 0.56 m//s^(2)` |
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| 490. |
Instantaneous speed can be represented by …………… of the curve at a given instant of time. A) distanceB) midpoint C) slope D) none of these |
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Answer» Correct option is C) slope |
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| 491. |
Identify the types of motion :(a) Movement of the earth around the sun : …………… .(b) Movement of ceiling fan : …………… .(c) A rocket launched from the ground : ……….. .(d) A fish swimming in water : ……… .(e) The plucked string of a sitar :………….. . |
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Answer» (a) periodic, drcular (b) circular (c) linear (d) random (e) oscillatory motion |
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| 492. |
If the initial velocity of a projectile be doubled, keeping the angle of projection same, the maximum height reached by it willA. remain the sameB. be doubledC. become four timesD. be halved |
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Answer» Correct Answer - C `H = (u^(2) sin^(2) theta)/(2g) rArr H prop u^(2)` If intial velocity be doubled, then maximum height reached by the projectile will become four times |
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| 493. |
Assertion A particle is projected with speed u at an angle `theta` with the horizontal. At any time during motion, speed of particle is v at angle `alpha` with the vertical, then `v sin alpha` is always constant throughout the motion. Reason In case of projectile motion, magnitude of radical acceleration at topmost point is maximum.A. If both Asseration and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are correct but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is falseD. If Assertion is false but Reason is true |
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Answer» Correct Answer - B Assertion `v sin alpha =` horizontal component of velocity = constant Reason `a_(r) = sqrt(g^(2)-a_(t)^(2))` At highest point `a_(t) =0`. Therefore, `a_(r)`is maximum. |
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| 494. |
A particle is projected from ground with speed u and at an angle `theta` with horizontal. If at maximum height from ground, the speed of particle is `1//2` times of its initial velocity of projection, then find its maximum height attained.A. `(u^(2))/(g)`B. `(2u^(2))/(g)`C. `(u^(2))/(2g)`D. `(3u^(2))/(8g)` |
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Answer» Correct Answer - D Given at maximum height `u cos theta =(1)/(2) u rArr cos theta = (1)/(2) :. Theta = 60^(@)` `:. H = (u^(2) sin^(2) theta)/(2g) = (u^(2) sin^(2) 60^(@))/(2g) = (3u^(2))/(8g)` |
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| 495. |
A ball of mass m is projected from the ground with an initial velocity u making an angle of `theta` with the vertical. What is the change in velocity between the point of projection and the highest point ?A. `u cos theta` downwardB. `u cos theta` upwardC. `u sin theta` upwardD. `u sin theta` downward |
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Answer» Correct Answer - A `Deltav = aDeltat = a ((T)/(2)) = (-g hatj) ((u cos theta)/(g)) = (-u cos theta) hatj` Therefore, change in velocity is `u cos theta` in downward direction. |
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| 496. |
If a body is projected horizontally from the top of a tower the time taken by it to reach the ground depends upon A) velocity of projection B) height of the tower C) both A & B D) neither of these two |
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Answer» B) height of the tower |
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| 497. |
A particle velocity changes from `(2 hati - 3hatj) ms^(-1)` to `(3hati - 2hatj) ms^(-1)` in 2s. If its mass is 1kg, the acceleraton `(ms^(-2))` isA. `-(hati +hatj)`B. `(hati + hatj)//2`C. zeroD. `(hati -hatj)//2` |
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Answer» Correct Answer - B `a = (v_(t)-v_(i))/(t) = ((3hati -2hatj)-(+2hatj-3hatj))/(2) = (hati +hatj)/(2)` |
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| 498. |
A ball is projected with a velocity `20 sqrt(3) ms^(-1)` at angle `60^(@)` to the horizontal. The time interval after which the velocity vector will make an angle `30^(@)` to the horizontal is (Take, `g = 10 ms^(-2))`A. 5sB. 2sC. 1sD. 3s |
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Answer» Correct Answer - B `tan 30^(@) = (v_(y))/(v_(x)) = (u_(y) - g t)/(u_(x))` `= ((20sqrt(3)sin 60^(@))-10t)/((20sqrt(3)cos 60^(@)))` or `10 = 30 - 10 t` `:. T = 2s` |
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| 499. |
Two particles projected form the same point with same speed u at angles of projection `alpha and beta` strike the horizontal ground at the same point. If `h_1 and h_2` are the maximum heights attained by the projectile, R is the range for both and `t_1 and t_2` are their times of flights, respectively, thenA. `alpha + beta = (pi)/(2)`B. `R = 4 sqrt(h_(1)h_(2))`C. `tan alpha = (t_(1))/(t_(2)) = sqrt(h_(1)h_(2))`D. None of the above |
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Answer» Correct Answer - D (a) Range becomes equal at complimentary angle. Hence `beta = 90^(@) - alpha` (b) `h_(1) = (u^(2) sin^(2)alpha)/(2g)` `rArr h_(2) = (u^(2)cos^(2)alpha)/(2g)` (as `beta = 90 - alpha)` `:. 4sqrt(h_(1)h_(2)) = (2u^(2)sin alpha cos alpha)/(g) = (u^(2)sin 2 alpha)/(g) = R` (c) `(t_(1))/(t_(2)) = ((2u sin alpha//g))/((2u cos alpha//g)) = tan alpha` (d) `sqrt((h_(1))/(h_(2))) = tan alpha` |
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| 500. |
A particle is projected at an angle of `45^(@)` with a velocity of `9.8 ms^(-1)`. The horizontal range will be (Take, `g = 9.8 ms^(-2))`A. `9.8m`B. `4.9m`C. `(9.8)/(sqrt(2))`D. `9.8 sqrt(2)` |
| Answer» Correct Answer - A | |