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InterviewSolution
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.
| 651. |
Choose the correct statement for a particle going on a straight line.A. If the position and velocity are in opposite direction, the particle is moving towards the originB. If the acceleration and velocity are in opposite direction. The particle is slowing downC. If the velocity is zero for a time interval the acceleration is zero at any moment within that time intervalD. If the velocity is zero at any instant, then the acceleration must be zero at that instant . |
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Answer» Correct Answer - A::B::C Knowledge based question. |
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| 652. |
What is Acceleration ? |
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Answer» The time rate of change of velocity of an object is called acceleration of the object. (1) It is a vector quantity. It’s direction is same as that of change in velocity (not of the velocity) (2) There are three possible ways by which change in velocity may occur. (3) Dimension : [M0 L1 T–2] (4) Unit : metre/second2 (S. I.); cm/second2 (C. G. S.) |
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| 653. |
A particle is moving with a uniform acceleration `4 m s^(-2)` for time `2 second` and then `5 m s^(-20` for time `3 seconds`. What is the average acceleration of particle during motion. |
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Answer» Hence, `a_(1) =4 ms^(-2) , t_(1) =2 s, a_(2) =5 ms^(-2) , t_(2) =3 s` `a_(av) =(a_(1)t_(1) +a_(2) t_(2))/(t_(1) +t_(2)) =(4xx 2+ 5 xx 3)/(2 =3) = 23/5 =4.6 ms^(-2)`. |
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| 654. |
What is the angle between `vec(A)` and `vec(B)`, if `vec(A)` and `vec(B)` are the adjacent sides of a parallelogram drawn from a common point and the area of the parallelogram is `AB//2`? |
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Answer» Area of paralleloogram `= | vec A xx vec B|` `=AB sin theta 1/2 AB (Given )` :. Sin theta =1/2 sin 30^(@)` or ` theta =30^(@)`. |
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| 655. |
A vector `vec A` prints vertically upward and vec B` points towards. What is the direction of ` vec A xx vec b` ? |
| Answer» The direction of ` vec A xx vec B` is along north according to Reght handed screw rule. | |
| 656. |
Difference between instantaneous speed and instantaneous velocity. |
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Answer» (a) Instantaneous velocity is always tangential to the path followed by the particle. (b) A particle may have constant instantaneous speed but variable instantaneous velocity. (c) The magnitude of instantaneous velocity is equal to the instantaneous speed. (d) If a particle is moving with constant velocity then its average velocity and instantaneous velocity are always equal. (e) If displacement is given as a function of time, then time derivative of displacement will give velocity. |
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| 657. |
What are the difference between Uniform velocity and Non-uniform velocity ? |
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Answer» Uniform velocity : A particle is said to have uniform velocity, if magnitudes as well as direction of its velocity remains same and this is possible only when the particles moves in same straight line without reversing its direction. Non-uniform velocity : A particle is said to have non-uniform velocity, if either of magnitude or direction of velocity changes (or both changes). |
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| 658. |
Comparison between distance and displacement. |
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Answer» (i) Distance > Displacement. (ii) For a moving particle distance can never be negative or zero while displacement can be i.e., Distance > 0 but Displacement > = or < 0. (iii) For motion between two points displacement is single valued while distance depends on actual path and so can have many values. (iv) For a moving particle distance can never decrease with time while displacement can. Decrease in displacement with time means body is moving towards the initial position. (v) In general magnitude of displacement is not equal to distance. However, it can be so if the motion is along a straight line without change in direction. |
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| 659. |
What are Velocity and its types? |
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Answer» Rate of change of position i.e., rate of displacement with time is called velocity. (i) It is a scalar quantity having symbol v. (ii) Dimension : [M0 L1T–1] (iii) Unit : metre/second (S. I.), cm/second (C. G. S.) Types : (a) Uniform velocity (b) Non-uniform velocity (c) Average velocity (d) Instantaneous velocity |
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| 660. |
What are the speed ? Give its types. |
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Answer» Rate of distance covered with time is called speed. (i) It is a scalar quantity having symbol v. (ii) Dimension : [M0L1T–1] (iii) Unit : metre/second (S.I.), cm/second (C. G. S.) Types of speed : (a) Uniform speed (b) Non-uniform (variable) speed (c) Average speed
(d) Instantaneous speed |
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| 661. |
The slope of straight line joining two points on velocity-time graph of an object having bobuniform motion gives…………………………………… fro the given interval of time. |
| Answer» Correct Answer - average accleration | |
| 662. |
Two junglemen are standing at the two opposite banks of a river of width `l` facing each other. One of the m starts beating a drum and sound reaches to the other one after time `t_(1)` the stats. Then second one starts beating the drum and now first one hear the sound after time `t_(2)`. Calculate the velocity of sound relative to air and the velocity of wind, if it is blowing from first bank to the other bank at right angle to the river flow |
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Answer» Correct Answer - `[(1)/(2)((1)/(t_(1)) + (1)/(t_(1))), (l)/(2) ((1)/(t_(1)) - (1)/(t_(2)))]` |
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| 663. |
A train(T) is running uniformly on a straight track. A car is travelling with constant speed along section AB of the road which is parallel to the rails. The driver of the car notices that the train is having a speed of `7 m//s` with respect to him. The car maintains the speed but takes a right turn at B and travels along BC. Now the driver of the car finds that the speed of train relative of him is `13 m//s`. Find the possible speeds of the car. |
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Answer» Correct Answer - `5 m//s, 12 m//s` |
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| 664. |
A physics professor was driving a Maruti car which has its rear wind screen inclined at `theta = 37^(@)` to the horizontal. Suddenly it started raining with rain drops falling vertically. After some time the rain stopped and the professor found that the rear wind shield was absolutely dry. He knew that, during the period it was raining, his car was moving at a constant speed of `V_(c) = 20 km//hr. [tan 37^(@) = 0.75]` (a) The professor calculated the maximum speed of vertically falling raindrops as `V_(max)`. What is value of Vmax that he obtained. (b) Plot the minimum driving speed of the car vs. angle of rear wind screen with horizontal `(theta)` so as to keep rain off the rear glass. Assume that rain drops fall at constant speed `V_(r)` |
| Answer» Correct Answer - (a) `V_(max) = 12 km//hr` | |
| 665. |
A particle is projected from the ground with an initial velocity of ` 20 m//s` at angle fo ` 30^@` with the horizontal. What is the magnitude of chang in velocity in `0.5 scond` ? ( g= 10 m//s^2)`. |
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Answer» The change in velocity is due to vertical motion of projectile which will be given by ` | Deltavec v| = g Delta t = 10 xx 0.5 =5 m//s`. |
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| 666. |
A bullet fired at an angle of ` 60^@` with the vertical hits the grond at a distance of `2.5 km`., Calculate the distance at which the bullet will hit the grund when fired at an angle of ` 45^@` with vertical. Assuning the speed to be the same. |
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Answer» Angel or projection with horizontal, `theta_1 = 90^@- 60^@= 30^@` Horizontal range, ` R_1 = 2.5 km = 2500 m` As` R_1 (u^2 sin 2 hteta)/g , so , `2500 = (u^2 sin 2 xxx 30^@)/g` or 1 u^2/g = ( 2500)/( sin 60^@) = ( 2500)/( sqrt 3//2) = ( 2500 xx2)/( sqrt 3)` When angle of projection is ` 45^@` with vertical, the angle or projection with horizontal, ` theta_2 = 90^2 - 45= 45^@`. Horizontal range, ` R-2 = (u^ sin 2 xx 45^@)/g = u^@/g = ( 2500 xx2)/ (sqrt3)` `= 2886.8 m= 2. 89 km`. |
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| 667. |
A bullet fired at an angle of ` 30^(@)` with the horizontal hits the ground ` 3 sqrt 3 km ` away. Can wr hit a target at a distance of `6 sqrt 2km ` by adjustion its angle of projection? |
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Answer» Here, `theta =30^(@)`, `R=3 sqrt 3 km =300 sqrt3 m` As, `R=(u^(2) sin 2 theta)/g` :. 3000 sqrt 3 =(u^(2) sin 2 xx 30^(@))/g` or `u^(2)/g =(3-00 sqrt 3)/(sin 50^(@)) =(3000sqrt 3)/(sqer3//2)` `=6000 m =6 km ` The maximum horizontal rage `=u^(2)/g =6 km`, which is less than `6 sqrt2 km`. Thus the bullet will never hit the target at a distance `6 sqrt 2 km ` by adjusting anlge of projection. |
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| 668. |
An insect crect crawling up a wall crawls ` 6cm` upwards in the first minute but the slides `4 cm `downwards in the next minute. If again crawls up `6 cm` upwards in the third minute but again slides `4 cm` downwards in the forth minute. How long will the insect take to reach a crevice in the wall at a height of `22 cm` from a starting point ?` |
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Answer» Efferctive distance crawled upwards in ` 2 mimute time =6-4 =2 cm` The insect eill craw1 2xx 10 =20 cm` upwards in `2 xx 10 =20 minutes ` In `21 minute ` insect covers `6 cm` upwards and reaches the crevece located at `22 cm`. :. Total time taken `=20 +1 =21 min`. |
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| 669. |
(a) With what velocity `(v_0)` should a ball be projected horizontally from the top of a tower so that the horizontal distance on the ground is `eta H`, where `H` is the height of the tower ? (b) Also determine the speed of the ball when it reaches the ground. . |
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Answer» (a) Given , horizontal distance = `eta H`, where `H` is the height of the tower. To find out the sufficient velocity as asked, we have to use the formula : distance = velocity xx time. Here the time involved will be the time of flight as we are considering projectile motion, which is given by `T = sqrt( 2 H) // g` So, `v_0 T = eta H rArr v_0 sqrt((2 H))/(g)) = eta H rArr v_0 = eta sqrt((g H)/(2))` (b) During the projectile motion, the horizontal component of velocity remains same and its vertical component keeps on changing under the effect of gravity. So horizontal speed, `v_x = v_0` and vertical speed, `v_y = sqrt(2 g H))`. Total speed `= sqrt(v_x^2 + v_y^2))` =`sqrt(v_0^2 + 2 g H))` =`sqrt(eta^2 (g H)/(2) + 2 g H) = sqrt(((eta^2 + 4)/(2)) g H)`. |
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| 670. |
A wet ball is projected horizontally at a speed of `u = 10 m//s` from the top of a tower `h = 31.25 m` high. Water drops detach from the ball at regular intervals of `Deltat = 1.0 s` after the throw. (a) How many drops will detach from the ball before it hits the ground. (b) How far away the drops strike the ground from the point where the ball hits the ground? |
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Answer» Correct Answer - (a) 2 (b) zero |
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| 671. |
A body covers `12 m` in ` 2nd` second and `20 m` in `4 th second` . Find what distansce the body will cover in ` 4 second` agter the ` 5 th second`. |
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Answer» Distance coverd in ` 4 ceconds after the fifth second ` = S_9 - S_5`. The distance covered in (t) seconds is given by `, ` S=ut + 1/2 at^2 ` and Distance coverd in nth second is given by, ` D_n = u + a/ 2 ( 2 n- 1)` . |
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| 672. |
The acceleration of a particle which moves along the positive x-axis varies with its position as shown in figure. If the velocity of the particle is `0.8 ms^(-1)` at x = 0, then velocity of the particle at `x = 1.4 m` is `(in ms^(-1))`. A. `1.6`B. `1.2`C. `1.4`D. None of these |
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Answer» Correct Answer - B |
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| 673. |
A particle is moving along `x`-axis with constant acceleration. At `t=0`, the particle is at `x=3m` and `(dx)/(dt)=+4m//s`. The maximum value of `x` co-ordiante of the particle is observed 2 seconds later. Starting from `t=0` sec, after what time, particle reaches its initial position again?A. `4s`B. `6s`C. `8s`D. `12s` |
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Answer» Correct Answer - A |
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| 674. |
A helicopter is rising vertically up with a velocity of `5 ms^( –1)`. A ball is projected vertically up from the helicopter with a velocity V (relative to the ground). The ball crosses the helicopter 3 second after its projection. Find V. |
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Answer» Correct Answer - `V = 20 ms^(-1)` |
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| 675. |
A chain of length L supported at the upper end is hanging vertically. It is released. Determine the interval of time it takes the chain to pass a point 2L below the point of support, if all of the chain is a freely falling body. |
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Answer» Correct Answer - `Deltat = sqrt((2L)/(g)) [sqrt(2)-1]` |
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| 676. |
Choose the wrong statement from the following.A. Zero velocity of a particle does not necessarily mean that its acceleration is zero.B. Zero acceleration of a particle does not necessarily mean that its velocity is zero.C. If speed of a particle is constant, its acceleration must be zero.D. None of these |
| Answer» If speed, that is the magnitude of velocity, is constant, whereas the direction of velocity changes, we cannot say that velocity is constant. Therefore, the particle has non-zero acceleration. | |
| 677. |
A particle is forced to meove on a straight line path. It returns to the starting point after ` 10 seconds. The total distance covered by the particle during this time is ` 20 m`. Which of the foolowing statements are true regarting the motion of the particles .A. the average velocity of the particle is zeroB. the displacement of the particle is zeroC. the average speed of the particle is `2.0 ms^(-1)`D. the displacement of the particle is ` 20 m`. |
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Answer» Correct Answer - A::B::C When a particle whil movign returns to its initial position after a crtain time . Then its displacemnt in tha time is zero . Its average velocity (= diosplacement / time 0 is also zero. But its average speed = (total distance travelled)/(time taken ) = ( 20) /(10) ` ` 2.0 ms^(-1)` |
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| 678. |
A particle moves in `x-y` plane, starting from `A` along straight line path `AB` and `BC` as shown in the graph. When it is at point `P` angle between diections of its average velocity and instantaneous velocity is `[tan37^(@)=3//4]` A. `90^(@)`B. `82^(@)`C. `98^(@)`D. `74^(@)` |
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Answer» Correct Answer - B |
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| 679. |
The figure shows the velocity time graph of the particle which moves along a straight line starting with velocity at `5 m//sec` and coming to rest at `t=30 s`. Then :- A. Distance travelled by the particle is `212.5 m`B. distance covered by the particle when moves with constant velocity is `100 m`C. Velocity of the particle at `t=25s` is `5 m//sec`D. Velocity of the particle at `t=9s` is `8 m//sec` |
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Answer» Correct Answer - A::C::D Area of the curve gives distance. |
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| 680. |
A partial along a straight line whose velocity-displacement graph is as shown in the figure. What is the magnitude of acceleration when displacement is 3 m ? A. `4sqrt3 ms^-2`B. `3sqrt3 ms^-2`C. `sqrt3 ms^-2`D. `4/(sqrt3) ms^-2` |
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Answer» Correct Answer - A `a=v.(dx)/(dt)=(4)(-tan60^@)` `=-4sqrt3 m//s^2` |
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| 681. |
The velocity - displacement graph of a particle moving along a straight line is shown The most suitable acceleration - displacement graph will be |
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Answer» Correct Answer - B from graph `v = v_(0) - (v_(0))/(x_(0)) x …(1)` `a = v(dv)/(dx) ..(2)` `a = (-(v_(0))/(x_(0)))v = (v_(0)-(v_(0))/(x_(0))x)(-(v_(0))/(x_(0)))` `rArr a = ((v_(0))/(x_(0)))^(2) x - (v_(0)^(2))/(x_(0))` |
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| 682. |
If position time graph of particle is sincurve as shown, what will be its velocity-time graph. A. B. C. D. |
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Answer» Correct Answer - C `x =x_(0) sin omega rArr v = (dx)/(dt) = -x_(0) omega cos omega t` |
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| 683. |
Choose the corredct statement :A. the magnitudes of speed and velocity are same when a body travels in a stright line path.B. Average speed ofa moving body can be equal to zero, but its average velocity cannot be equal to zero.C. To describe the velocity , direction is necessary.D. Both (A) and © |
| Answer» Correct Answer - D | |
| 684. |
Two hour is equal to ______ seconds. |
| Answer» Correct Answer - 7200 | |
| 685. |
A body moving with velocity of `10ms^(-1)` increases its velocity to `20ms^(-1)` 2 s. then the arate of change in velocity is ___- |
| Answer» Correct Answer - `5 ms ^(-2)` | |
| 686. |
The time period of a simple pendulum depends on its______ |
| Answer» Correct Answer - length | |
| 687. |
` 20 m s^(-1) ______ km h^(-1)`A. 12.5B. 72C. 50/9D. 32 |
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Answer» Correct Answer - B 1 km h 5/8 ` ms^(-1)` ` 1 ms ^(-1) = 18//5 km h` ` 20 ms^(-1) = 20 18/5 ` km h = 72 km h |
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| 688. |
By using distance-time graph we can find out the average speed of an object. |
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Answer» Correct Answer - True By using distance-time graph we can find out the average speed of an object. |
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| 689. |
The time period of a simple pendulum of length 1 m is _____ second. ( take `pi^(2) = 10.g =10 ms^(-2)`) |
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Answer» Correct Answer - 2 s The time period of a simple pendulum of length 1 m is 2 second. |
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| 690. |
The speed of the tip of a second hand of length 5 cm of a clock is ____ `ms^(-1)`A. 1B. 60C. `5.3 xx 10^(-3)`D. `3.4 xx 10^(-5)` |
| Answer» Correct Answer - C | |
| 691. |
When an object is moving with a unifrom speed, its average speed is equal to its velocity. |
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Answer» Correct Answer - False When an object is moving with constant speed the average speed is not equal to veloctiy. |
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| 692. |
the speed of the bob of a simple pendulum of length 1 m is _____ second. |
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Answer» Correct Answer - mean position The speed of the bob is maximum at its mean position. |
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| 693. |
The time period of a pendulum is independent of ______A. length of the pendulumB. mass of the bobC. shape of the bobD. both (b) and © |
| Answer» Correct Answer - D | |
| 694. |
the distance between two stations is 20 km. if a train moves with a constant speed of `60 km h^(-1)` , then the time taken by the train moves with a constant speed of `60 km h^(-1)` , then the time taken by the rain to reach the next station is _____A. 2 hourB. 20 minC. 20 secD. 40 min |
| Answer» Correct Answer - B | |
| 695. |
The distance -time graph of an object is a straight line parallel to the time axis , then the object is _____.A. at restB. in uniform motionC. moving with a unifrom speedD. moving with a non-uniform speed |
| Answer» Correct Answer - A | |
| 696. |
The distance coverd by an object in a unit time is known as the _______ of the object. |
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Answer» Correct Answer - speed The distance covered by an object in a unit time is known as the speed of the object. |
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| 697. |
A car moves with a speed of `60 km h^(-1) ` for 20 min and then at a speed of `30 km h^(-1)` for the next 20 min. the total distance covered by the car is ____ km.A. 10B. 20C. 30D. `40` |
| Answer» Correct Answer - C | |
| 698. |
The time taken by the pendulum to complete one oscillation is called______ |
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Answer» Correct Answer - time period The time taken by the pendulum to complete one oscillation is called time period . |
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| 699. |
A train is targeted to run from Delhi to Pune at an average speed of 80 kph but due to repairs of track looses 2hr sin the first part of the journey. If then accelerates at a rate of `20 kph^(2)` till the speed reaches 100 kph. Its speed is now maintained till the end ofthe journey. If the train now reaches station in time, find the distance from when it started accelerating ? |
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Answer» Correct Answer - [840 km] |
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| 700. |
A train takes 2 minutes to acquire its full speed 60 kph from rest and 1 minute to come to rest from the full speed. If some where in between two stations 1 km of the track be under repair and the limited speed on this part be fixed to 20 kph, find the late running of the train on account of this repair work, assuming otherwise normal at running of the train between the stations. |
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Answer» Correct Answer - [2min 40 sec] |
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