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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.
| 551. |
Knowing that block B starts to move downward with a constant velocity of 18cm/sec, the velocity of block A will be: A. 27 cm/sB. 12 cm/sC. 36 cm/sD. 9 cm/s |
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Answer» Correct Answer - B |
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| 552. |
There are two cars on a straight road, marked as x axis. Car A is travelling at a constant speed of `V_(A) = 9 m//s`. Let the position of the Car A, at time `t = 0`, be the origin. Another car B is `L = 40 m` ahead of car A at `t = 0` and starts moving at a constant acceleration of `a = 1 m//s^(2)` (at t = 0). Consider the length of the two cars to be negligible and treat them as point objects. (a) Plot the position–time `(x–t)` graph for the two cars on the same graph. The two graphs intersect at two points. Draw conclusion from this. (b) Determine the maximum lead that car A can have. |
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Answer» Correct Answer - (b) `0.5 m` |
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| 553. |
Two cars travelling towards each other on a straight road at velocity `10 m//s` and `12 m//s` respectively. When they are 150 metre apart, both drivers apply their brakes and each car decelerates at `2 m//s^(2)` until it stops. How far apart will they be when they have both come to a stop? |
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Answer» Correct Answer - [89 m] |
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| 554. |
In the previous question the angle `phi` is equal toA. `theta`B. `tan^(-1) [etan theta]`C. `tan^(-1) [(1)/(e) tan theta]`D. `(1+e) theta` |
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Answer» Correct Answer - C |
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| 555. |
Fig. shows the graph of the x-coordinate of a particle going along the x-axis as a function of time. Find (a) the average velocity during `0` to `10 s,` (b) instantaneous velocity at 2, 5, 8 and 12s. |
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Answer» Correct Answer - A::B (a) Average velocity=`s/t=(x_f-x_i)/t` `=(x_(10 sec)-x_(0 sec))/10=100/10=10m//s` (b) Instantaneous velocity =Slope of x-t graph |
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| 556. |
From the velocity-time plot shown in Fig. find the distance travelled by the particle during the first `40 s.` Also find the average velocity during this period. |
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Answer» Correct Answer - A From 0 to 20s, displacement `s_1`=area under v-t graph `+50 m` (as is negative) From 20 to 40s, displacement `s_2`=area under v-t graph `=-50m` (as v is negative) Total distance travelled =`s_1+|s_2|=100m` Average velocity=`s/t=(s_1+s_2)/t` `=(50-50)/40=0` |
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| 557. |
Two particles are moving along x-axis. Their x-coordinate versus time graph are as shown below. Find velocity of A w.r.t. B |
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Answer» Correct Answer - B `v_A=Slope of A =2/5=0.4 m//s` `v_B=Slope of B=12/5=2.4 m//s` `:. V_(AB)=v_A-v_B=-2 m//s` |
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| 558. |
A body has maximum range `R_1` when projected up the plane. The same body when projected down the inclined plane, it has maximum range `R_2`. Find the maximum horizontal range. Assume equal speed of projection in each case and the body is projected onto the inclined plane in the line of the greatest slope. |
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Answer» Let `theta_0` be the angle of inclined plane with horizontal. Then for upward projection, `R_(max) = (u^2)/(g(1 + sin theta_0)) = R_1` …(i) For downward projection, `R_(max) = (u^2)/(g(1 - sin theta_0)) = R_2` …(ii) For a projection on horizontal surface, we have `R_(max) = (u^2)/(g) = R`(say) ....(iii) To establish a relation between `R,R_1`, and `R_2`, we need to eliminate `theta_0`. From (i) and (ii), we get `(2)/(R) = (1)/(R_1) + (1)/(R_2) rArr R = (2 R_1 R_2)/(R_1 + R_2)`. |
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| 559. |
A gun is fired from a moving platform and ranges of the shot are observed to be `R_1 and R_2` when the platform is moving forwards and backwards, respectively, with velocity `v_P`. Find the elevation of the gun `prop` in terms of the given quantities. |
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Answer» Correct Answer - `tan^-1[(g)/(4 v_p ^2) xx (R_1 - R_2)^2/((R_1 + R_2))]`. Horizontal range `= (2 u_x u_y)/(g)` Let the initial horizontal and vertical components of the velocity of the shot be `u and v`, respectively, w.r.t platform. `R_1 = (2 (u + v_P)v)/(g)`, platform moving forward `R_2 = (2(u - v_P)v)/(g)`, platform moving backward `R_1 + R_2 = (4 u v)/(g) and R_1 - R_2 = (4 v_p v)/(g)` Now `(R_1 - R_2)^2 = (16 v_p ^2v^2)/(g^2)` and `((R_1 - R_2)^2)/(R_1 + R_2) = (4 x_p ^2)/(g) xx (v)/(u)` or `(v)/(u) = (g)/(4 v_p^2) ((R_1 - R_2)^2)/((R_1 + R_2))` Elevation of gun `prop = tan^-1((v)/(u)) = tan^-1 [(g)/(4 v_p^2) xx (R_1 - R_2)^2/((R_1 + R_2))]`. |
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| 560. |
A guided missile is fired to strike an object at the same level 38 km away. It may be assumed that it rises vertically 1.5 km and then for the remainder of the flight it follows a parabolic path at an elevation of `45^(@)` . Calculate its velocity at the begining of its parabolic path. |
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Answer» Correct Answer - [2177 km/h] |
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| 561. |
To a person travelling due East with velocity u the wind appears to blow from an angle `alpha` North of East. When he starts travelling due North with velocity 2 u,the wind appears to blow from an angle `beta` North of East. Find the true direction of the wind. |
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Answer» Correct Answer - [`theta WOS` where `tan theta = (1+2 cost alpha)/(2+tan beta)`] |
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| 562. |
Rain is falling with a speed of 4nVs inadirection making an angle of `30^(@)` with vertical towards south. What should be the magnitude & direction of velocity of cyclist to hold his umbrella exactly vertical, so that rain does not wet him:A. 2 m/s towards northB. 4 m/s towards southC. 2 m/s towards southD. 4 m/s towards north |
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Answer» Correct Answer - C |
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| 563. |
The greatest acceleration or deceleration that a train may have is a. The minimum time in which the train may reach form one station to the other seprated by a distance is-A. `sqrt(s/a)`B. `sqrt((2s)/a)`C. `1/2 sqrt(s/a)`D. `2sqrt(s/a)` |
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Answer» Correct Answer - D |
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| 564. |
Two inclined planes `OA` and `OB` of inclinations `alpha` and `beta` equal to `30^(@)` are as shown in the figure. A particle is projected at an angle of `90^(@)` with plane `OA` from point `A` and it strikes the plane `OB` at point `B` normally. Then find the speed of projection `u` in m/s. (Given that `OA=OB=20cm` and `g=10m//s^(2)`) A. `2`B. `5`C. `4`D. `6` |
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Answer» Correct Answer - A |
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| 565. |
A projectile is thrown in the upward direction making an angle of `60^@` with the horizontal direction with a velocity of `150 ms^-1`. Then the time after which its inclination with the horizontal is `45^@` is |
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Answer» Here, ` theta= 60^@, u =150 ms^(-1)` . Let after time (t) its velocity become (v) making an angle ` 45^@` with the horizontal. Since the horizontal component velocity of projectile remains sunchanged during angular projection, henc ` u cos 60^@ = v cos 45^2` or ` v= ( u cos 60^@0/ ( cos 45^@)` Taking vertcal upward motin of projectile from tiem (t), we have, ` u =u sin 60^@, a =- g, t=t` and ` v=v sin 45^@`. ltbRgt As ` v=u + at so v sin 45^2 = u sin 60^@- gt` or ` ( u cos 60^@)/( cos 45^@) xx sin 45^@ = u sin 60^@-gt` ltbRgt or ` t = u/g [ sin 60^@- cos 60^@] = ( 150)/(10)[(sqrt3)/2 - 1/2]` ltbRgt =` 5.49 s`. |
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| 566. |
(Figure 5.11) shows two positions `A and B` at the same height `h` above the ground. If the maximum height os the projectile is `H`, then determine the time `t` elapses between the positions `A and B` in terms of `H`. . |
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Answer» Let `T` be the time of flight. We can now wire `T = (2 u sin theta)/(g)` `T^2 = (4 u^2 sin^2 theta)/(g^2) = (8)/(g) ((u^2 sin^2 theta)/(g)) = (8 H)/(g)` Hence, we can write in similar way, `t^2 = (8(H - h))/(g)` Thus, `t = sqrt((8)/(g) (H - h))`. |
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| 567. |
A projectile is thrown in the upward direction making an angle of `60^@` with the horizontal direction with a velocity of `150 ms^-1`. Then the time after which its inclination with the horizontal is `45^@` isA. `15 (sqrt(3) - 1) s`B. `15 (sqrt(3) + 1) s`C. `7.5 (sqrt(3) - 1) s`D. `7.5 (sqrt(3) + 1) s` |
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Answer» Correct Answer - C ( c) At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then `150 xx (1)/(2) = v xx (1)/(sqrt(2))` or `v = (150)/(sqrt(2)) ms^-1` Initially : `u_y = u sin 60^@ = (150 sqrt(3))/(2) ms^-1` Finally : `v_y = v sin 45^@ = (150)/(sqrt(2)) xx (1)/(sqrt(2)) = (150)/(2) ms^-1` But `v_y = u_y + a_y t` or `(150)/(2) = (150 sqrt(3))/(2) - 10 t` `10 t = (150)/(2) (sqrt(3) - 1)` or `t = 7.5 (sqrt(3)- 1)`. |
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| 568. |
A man crosses a river in a boat. If he cross the river in minimum time he takes `10 min` with a drift `120 m.` If he crosses the river taking shortest path, he takes `12.5 min,` find (a) width of the river (b) velocity of the boat with respect to water (c) speed of the current |
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Answer» Correct Answer - `200m, 20m//min, 12m//min` For minimum time `t = (d)/(v_(B)) rArr 10 = (d)/(v_(B)) …(1)` driff `120 = v_(w) xx 10 rArr v_(w) = 12m//min` For shortest path `sqrt(v_(B)^(2)-v_(A)^(2)) = (d)/(12.5) rArr sqrt(v_(B)^(2)-12^(2)) = (2d)/(25) …(2)` from (1) & (2) `sqrt(v_(B)^(2)-144) = (2xx10V_(B))/(25) rArr v_(B)^(2-) 144 = (16V_(B)^(2))/(25)` `(3)/(5) v_(B) = 12 rArr v_(B) = 20m//s rArr d = 10v_(B) = 200m` |
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| 569. |
At a height `0.4 m` from the ground the velocity of a projectile in vector form is `vec v = (6 hat i+ 2 hat j) ms ^-1`. The angle of projection isA. `45^@`B. `60^@`C. `30^@`D. `tan^-1 (3//4)` |
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Answer» Correct Answer - C ( c) `v^2 = u^2 - 2 g h` or `u^2 = v^2 + 2gh` or `u_x^2 + u_y^2 = v_x^2 + v_y^2 + 2gh , u_x = v_x` So, `u_y^2 = v_y^2 + 2gh` or `u_y^2 = (2)^2 + 2 xx 10 xx 0.4 = 12` `u_y = (sqrt(12)) = 2 (sqrt(3))^-1, u_x = v_x = 6 ms^-1` `tan theta = (u_y)/(u_x) = (2sqrt(3))/(6) = (1)/(sqrt(3)) rArr theta = 30^@`. |
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| 570. |
A snapshot of apetrol engine is given in which piston is moving downwards with velocity `40sqrt(3)` m/s. Find the angular velocity of the shaft: A. 400 rad/sB. 300 rad/sC. 200 rad/sD. 500 rad/s |
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Answer» Correct Answer - A |
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| 571. |
A particle moving in the positive x-direction has initial velocity `v_(0)`. The particle undergoes retardation `kv^(2)` , where vis its instantaneous velocity. The velocity of the particle as a function of time is given by:A. `v=v_(0)//(1+kv_(0)t)`B. `v=(2v_(0))/(1+kt)`C. `v=v_(0)/(kt)`D. `v=v_(0)/((1+k^(2)v_(0)^(2)t))` |
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Answer» Correct Answer - A |
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| 572. |
Three particles are projected upwards with initial speeds `10 m//s,20 m//s,30 m//s`. The displacements covered by them in their last second of motion are `x_(1),x_(2),x_(3)` then:A. `x_(1):x_(2):x_(3)=1:2:3`B. `x_(1):x_(2):x_(3)=1:4:9`C. `x_(1):x_(2):x_(3)=1:5:7`D. none of these |
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Answer» Correct Answer - D Distance travelled by each particles in last second of motion i.e. Downwards is equal to the distance travelled by it in first second of its motion i.e. upwards So, `x_(1)=10-1//2 xx10xx1^(2) =5 m` `x_(2)=20 -1//2xx10xx1^(2)= 15 m` `x_(3)=30 -1//2 xx10xx1^(2) =25 m` So, `x_(1):x_(2):x_(3)=1:3:5` |
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| 573. |
A particle is projected vertically upwards with an initial velocity of `40 m//s.` Find the displacement and distance covered by the particle in `6 s.` Take `g= 10 m//s^2.` |
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Answer» Correct Answer - A In 4s, it reaches upto the highest point and then changes its direction of motion. `s=ut+1/2 at^2=(40)(6)+1/2(-10)(6)^2` `=60 m` `d=|s|_(0-4)+|s|_(4-6)=|u^2/(2g)|+|1/2 g(t-t_0)^2|` `=(40)^2/(2xx10)+1/2xx10xx(6-4)^2=100 m` |
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| 574. |
What do you mean by motion in one Dimension? |
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Answer» Position of any point is completely expressed by two factors : Its distance from the observer and its direction with respect to observer. That is why position is characterised by a vector known as position vector. Let point P is in a xy plane and its coordinates are (x, y). Then position vector (r) of point will be xi + yj and if the point P is in a space and its coordinates are (x, y, z) then position vector can be expressed as r = xi + yj + zk. |
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| 575. |
A rectangular farm house has a `1km` difference between its sides. Two farmers simultaneously leave one of the vertex of the rectangle for a point at the opposite vertex. One farmer crosses the farmhouse along its diagonal and other walks along the edge. The speed of each farmer is `4km//hr`. If one of them arrives half an hour earlier then the other then the size of farmhouse is . |
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Answer» Correct Answer - `[3 km xx 4 km]` |
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| 576. |
Acceleration `(a)` -displacement `(s)` graph of a particle moving in a straight line is as shown in the figure. The initial velocity of the particle is zero. The `v-s` graph of the particle would be A. B. C. D. |
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Answer» Correct Answer - C |
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| 577. |
A horizontal wid is blowing with a velocity `v` towards north-east. A man starts running towards north with acceleration `a`. The after which man will feel the wind blowing towards east isA. `v/a`B. `(sqrt(2)V)/a`C. `v/(sqrt(2)a)`D. `(2v)/a` |
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Answer» Correct Answer - C |
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| 578. |
The distance between two moving particles `P` and `Q` at any time is a.If `v_(r)` be their relative velocity and if `u` and `v` be the components of `v_(r)`, along and perpendicular to `PQ`.The closest distance between `P` and `Q` and time that elapses before they arrive at their nearest distance isA. `(av_(1))/(v^(2))`B. `(av_(2))/(v^(2))`C. `(av)/(v_(1)^(2))`D. `(av)/(v_(2)^(2))` |
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Answer» Correct Answer - A |
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| 579. |
Consider a collection of a large number of particles each with speed v. The direction of velocity is randomly distributed in the collection. Show that the magnitude of the relative velocity between a pair of particles averaged over all the pairs in the collection is greater than v.A. zeroB. greater than`v`C. less than `v`D. `v` |
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Answer» Correct Answer - B |
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| 580. |
A train ` 100 m` along is moving with a velocity of ` 60 h^(-1)`. The time it takes to cross the bridge ` 1km long is. |
| Answer» Correct Answer - (b) | |
| 581. |
A man can swim with a speed of `4kmh^(-1)` in still water. He crosses a river 1km wise that flows steadly at `3kmh^(-1)`. If he makes his strokes normal to the river current, how far down the river does he go when he reaches the other bank? |
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Answer» Time to cross the river, ` t= (width of river )/( speed of man ) = ( 1 km ) /( 4 km//h) = 1/4 = 15 min` Didtance moved along the river in time ` t= v_r xx t =3 km // h xx 1/2 = 3/4 km = 750 m`. |
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| 582. |
A motor car moving at a speed of ` 72 km //h` can not come to a stop in less than ` 3.0 s` while for a truch this time interval is ` 5.0s` On a highway the car is behind the truck both moving at `72 km //h` The truck geives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does bot bump onto (collide with) the truck. Human responde time id ` 0.5 s`. |
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Answer» Here, ` u = 72 km//h = 20 m//s , Given, the car is behind the truck. `Retardation of truck ` = (200/5 = 4 m//s^2` Retardation of car ` =- (20)/3 m//s^2` Let car be at a distanc e(x) from truck and (t) be the time taken to cover this distance. As car decelerates only after ` 0.5 s`, so using relation, ` v=u = at , we have ` V_(car) = 20 - ( (20) /3) (t- 0.5)` and `V_(truck) = 20 -4 t` The car will just touch to the truck if or ` 20 - 4 t = 20 - (20)/3 ( t- 0.5) or 4 t = (20)/3 (t- 0.5)` or ` t = 5/3 (t- 0.5 ) or ` 3 t= 5 t- 2.4 or t =2.5 //2 = (5//4) s` Distance travelled by truck in this time (t), ` S_(truck) =50 (5/4) - 1/2 (4) ( 5/4) ^(2) = 21. 875 m` Distanc etravelled by car in time (t). ` S_(car) = (20 xx 0.5 ) + 20 (5/4 - 0 .5 ) - 1/2 ( 200 /3) (5/4 - 0.5 )^2 = 23.125 m` :. ` S_(car) - S_(trick) = 23. 125 - 21. 875 = 1.250 m. |
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| 583. |
The velocity time graph of a uniform motion of a partcile along a st. line id shown in Fig. 2 (a).20. What is the dispacement of the particle in time interval `8 s` to 12 s`? . |
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Answer» Displacement of a particle having uniform motion is equal to the area which velocity-time graph encloses with time axis. Displacement of partcile in time interval `8 s to 12 s (i.e. 4 s) is =velocity xx time interval =5 xx 4 =20 m. |
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| 584. |
Statement I: An object can possess acceleration even at a time when it has uniform speed statement II: It is possible when the direction of momtion keeps changing.A. Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I.B. Statemnt I is true, Statement II is true, Statement II is true, Statement II is false.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
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Answer» Correct Answer - A If the direction of velocity changes (magnitde may or may not change ), we say that velocity changes. If velocity changes, then definitely ther will be acceleration. |
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| 585. |
The dispacement of a body is given by `4s=M +2Nt^(4)`, where `M` and `N` are constants. The velocity of the body at any instant is .A. `(M+2Nt^(4))/(4)`B. `2N`C. ` (M+2N)/(4)`D. `2Nt^(3)` |
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Answer» Correct Answer - D `s=(M+2Nt^(4))/(4) rArr v=(ds)/(dt) =2Nt^(3)` Putting `1 s`, we get `v=2 N` . |
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| 586. |
An aeroplane is flying with velocity ` vec v_p = 100 ms^(-1)` towards East with respect to ground through motionless air and ` vec v_w` is the wind velocity with respect of ground. The total velocity of aeroplane is ` vec v=vec v_P+ vec v_w` The magnitude of the velocity is often called speed . the heading of the plane is the direction in which the nose of the plane points. In fact, it is the direction in which the engine propels the plane . Answer the following questions : If ` theta` is the answer to question ` 15 6`, the total speed of the plane in ` ms^(-1)` is.A. ` 100 sin theta`B. ` 100 cos theta`C. ` 100 cosec theta `D. `100 sec theta` |
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Answer» Correct Answer - D Goven, ` tan theta = (25)/(100) = 1/4` , ` cos theta 4/(sqrt( 4^2 +1^2) = 4/(sqrt17)` Resultant speed of the plance, ` v= sqrt (v_P^2 +v_w^2) = sqrt ( 100^2 + 25^@)` `= sqrt (25^2 +4^2 25^2) = 25 sqrt 17 ms^(-1)` :.` v = (25 sqrt 17)/(cos theta) xx cos theta = (25sqrt17)/( cos theta) xx 4/(sqrt 170 = ( 100)/(cos theta)`. |
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| 587. |
The body will speed up if .A. Velocity and acceleration are in the same direction.B. Velocity and acceleration are in opposite directions.C. Velocity and acceleration are in perpendicular direction.D. Velocity and acceleration are acting at acute angle w.r.t. each other. |
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Answer» Correct Answer - A::D The body will speed up if the anle between velocity and acceleration is acute. |
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| 588. |
Average acceleration is in the direction of .A. Initial velocityB. Findl velocityC. Change in velocityD. Final velocity if initial velcotu is zero. |
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Answer» Correct Answer - C::D Since average acceleration=Change in velocity//time, so averate acceleration is in the diretion of change in velocity . Also if the initial velacity is zero . Then the final velocity and change in velocity will be in the same direction. |
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| 589. |
In the figure, the blocks A, B and C of mass m each have accelerations a1, a2 and a3 respectively. F1 and F2 are external forces of magnitudes 2mg and mg respectively.(a) a1 = a2 = a3 (b) a1 > a3 > a2 (c) a1 = a2, a2 > a3 (d) a1 > a2, a2 = a3 |
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Answer» Correct answer is: (b) a1 > a3 > a2 a1 = g, a2 = g/3, a3 = g/2. |
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| 590. |
An athlete swims the length of `50 m` pool in `20 s` and makes the return trip to the starting position in `22 s`, Determine his averge velocity in a. The first half of the swim b. The second half of the swim c. The round trip. |
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Answer» a. `v_(av_(1)) =(50)/(20) =2.5 m s^(-1)`, b. `v_(av_(2)) =(50)/(22) =2.27 m s^(-1)` c. `v_(av)` is zero for round trip because displacement is zero. |
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| 591. |
A ball thrown up from the ground reaches a maximum height of `20 m` Find: a. Its initial velocity. b. The time taken to reach the highest point. c. Its velocity just before hitting the ground. d. Its desplacement berween `0.5 m` above the ground. |
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Answer» a. `H=20 =(u^(2))/(2g) rArr u=sqrt(2gxx20) =20 m s^(-1)` b. `t_(a) =(u)/(g) =(20)/(10) =2 s` c. On hitting the ground, the velocity will be equal to the initial velocity but in the opposite direction. Hence, answer is `20 m s^(-1)` . d. `S_(1)=20xx 0.5-(1)/(2) 10(0.5)^(2), S_(2) =20xx2.5-(1)/(2) 10 (2.5)^(2)` Required displacement`=S_(2)-S_(1)=10 m` e. `15 =20xx20t-(1)/(2) 10t^(2) rArr t^(2)-4t + 3=0 rArr t=1 s,3 s`. |
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| 592. |
Figure shows the velocity time graph for a particle travelling along a straight line. The magnitude of average velocity (in m/s) of particle during the time interval from `t=0` to `t=6s` is `10 alpha`. Find the value of `alpha`. |
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Answer» Correct Answer - 1 |
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| 593. |
Two bodies `A` and `B` are moving along `y`-axis and `x` -axis as shown. Find the minimum distance between `A` and `B` is subsequent motion (in `m`) |
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Answer» Correct Answer - 8 |
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| 594. |
a. Mark the follllowing statements as true offalse. i. A ball thrown vertically up takes moretime to go up than to come down. ii. If a ball starts fallig from the position of rest, then it travels a distance of `25 m` during the third secons of tis fall. iii. A packet dropped from a rising balloon ferst moves upwards and then moves sownward as observed by a stationary observer on the ground. iv. In the absence of air resistance, all bodies fall on the surface of earth at the same rate. b. Fill in the blanks. i. When a body is thrown vertically upwards, at the highest point.........................(both belocity and accelenation are zero//only velocity is zero//ony acceleration is zero). ii. If air drag is not neglected, then which is greater: time of ascent or time of descent? iiii. A body is projected upward. Up to the maximum height time taken will be greater to travel...................... (first half//second half). |
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Answer» a. i. False, time of ascent=time of dwacent. ii. True, `D_(n)+(a)/(2) (2n-1) =0 + (10)/(2) (2xx3-1) =25 m` iii. True, because at the time of dropping the velocity of the packet is upward, and same as that of balloon. iv. True, acceleration due to gravity is same for all bodies. b. i. Only velocity is zero ii. Time of descent iii. Second half`. |
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| 595. |
The `1//v` versus positions graph of a particle is shown in the figure, where `v` is the velocity of the particle. The particle is moving in a straight line aloing positive `x`- axis.Find the time taken by the particle to reach from the point `A` to `B` in second. |
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Answer» Correct Answer - 6 |
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| 596. |
Two cars `A` and `B` are moving on the straigh parallel paths with speeds `36 km h^(-1)` respectively starting from the same point in the same direction. After `20 min`, how much behind is car `A` and from car `B`? |
| Answer» Required separation`(v_(2)-v_(2))t =(72-36) (20)/(60) =12 km`. | |
| 597. |
Two trains `110 m` and `90 m` log respectively, are trunning in opposite directions with velocities `36 km h^(-1)` and `54 km h^(-1)` Find the time taken by the trains to completely cross each other. |
| Answer» `t=("Total length")/("Relative velocity") =(110+90)/((36+54)xx(5//18)) =8 s` ` | |
| 598. |
A railroad flatcar is traveling to the right at a speed of `13.0 ms^(-1)` relative to an observer standing on the groun . Someone is riding a scooter on the flatcar. Corresponding to the relative velocities `18 m s^(-1)` to the right, `3 m s^(-1)` to the left and `0 ms^(-1)` of scooter w.r.t. ground, find thefrlative velocities (magnitude and direction) of scootre w.r.t. the flatcat. . |
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Answer» a. `v_(as) =18 ms^(-1) v_(c) =13 ms^(-1)` `v_(s//c) =v_(s)-v_(c) =18-13 =5 ms^(-1)` b. `v_(s) =-3 ms^(-1) v_(s//c) =v_(s)-v_(c) =-3 -13 =-16 ms^(-1)` `v_(s) =0 v_((s)/(c)) =v_(s)-v_(c) =0-13 =-134 ms^(-1)`. Negative sign indicates left direction. |
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| 599. |
A moving sidewalk in an airport terminal building moves at a speed of `1.0 m s^(-1)` and is `35.0 m` relative to the moving sidewalk, then find the time that she requires to reach the opposite end `a` when she walks in the same direction the sidewalk is moving and `b` when she walks in the opposite derection. |
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Answer» a. When she walks in the same direction, relative velocity of woman w.r.t. fround: `v_(1)=1+1.5 =2.5 ms^(-1)`, Time taken`=(s)/(v_(1)) =(35)/(2.5) =14 s` b. When be she walks in the opposite derection, relative velcity of woman w.r.t. groun: `v_(2)=1.5-12 =0.5 ms^(-1)`. Time taken `=(s)/(v_(2))=(35)/(0.5) =70 s`. |
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| 600. |
Two particles are separated at a horizontal distance `x` as shown in (Fig. 5.57). They are projected at the same time as shown in the figure with different initial speed. Find the time after which the horizontal distance between the particles becomes zero. . |
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Answer» Let `x_1 and x_2` are the horizontal distances travelled by particle `A and B`, respectively, in time `t`. `x_1 = (u)/(sqrt(3)). Cos 30^2 xx t` ….(i) and `x_2 = u cos 60^@ xx t` …(ii) `x_1 + x_2 = (u)/(sqrt(3)). Cos 30^@ xx t + u cos 60^@ xx t = ut` `rArr x = ut rArr t = x//u`. |
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