InterviewSolution
Saved Bookmarks
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.
| 451. |
A particle starts moving rectilinearly at time `t=0` such that its velocity `v` changes with time `t` according to the equation `v=t^(2)-t`, where `t` is in seconds and `v` in ` s^(-1)`. Find the time interval for which the particle retards. |
|
Answer» Acceleration of the particle, `vec a=(d vec v)/(dt)=2t-1` The particle retards when acceleration is oppostie to velocity. Hence, acceleration vector and velocity vector should be opposite to each other or the dot product of `vec a` and `vec v` should be negative. `rArr vec a. vec v lt 0` `rArr (2t-1)(t^(2)-t) lt 0` `rArr t (2t-1)(t-1) lt 0` `t` is always positive. `because (2t-1)(t-1) lt 0` `rArr` Either `2t-1 lt 0` or ` t-1 lt 0` `rArr t lt (1)/(2) s` and `t lt 1 s`. This is not possible. or `2t-1 lt 0` and `t-1 lt 0 rArr t g t (1)/(2)s` and `tlt 1 s`. Hencec, the required time interval is `(1)/(2)s g t t g t1 s`. |
|
| 452. |
The position of a particle moving along x-axis is related to time `t` as follow: `x=2 t^(2)-t^(3)`, where `x` is in meters and `t` is in seconds. a. What is the maximum positive displacement of the particle along the `x` axis and at what instant does it attain it? b. Describe the motion of the particle. c. What is the destamce covered in the first three seconds? d. What is its desplacement in the first four seconds ? e. What is the particle`s average speed and average veloctry in the first `3` seconds ? f. What particles instantaneous acceleration at the instant of its maximum positiv `x` displacement? ltbtgt g. What is the average acceleration between the interval `t=2` to `t=4 s`?. |
|
Answer» The derivative of `x` w.r.t. time will give velocity of the particle in relation with time. `v=(dx)/(dt)=4t-3t^(2)` ..(i) The double derivative of `x` w.r.t. time will give velocity of the particle in relation with time. `a=(d^(2)x)/(dt^(2))=4-6t`.........(ii) a. For minimum and maximum displacement, `dx//dt=0`. Thus, `(dx)/(dt)=4t-3t^(2)=0` `t(4-3t)=0 rArr` either `t=0` or `t=(4)/(3) s` Also, for maxima, double derivativ `x` should be negative, i.e.,`(d^(2)x)/(dt^(2)) lt0`. `(d^(2)x)/(dt^(2))=4-6t` At`t=0: (d^(2)x)/(dt^(2))=4-(0)=4`(positive) And at t=(4)/(3) s: (d^(2)x)/(dt^(2))=4-6((4)/(3))=4-8=-4`(negative) (negative) Threfore, the porticle is at maximum displacement at `t=(4)/(3) s` and the corresponding displacement is `x_(max)=2((4)/(3))^(2)-((4)/(3))^(3)=(32)/(27)m`. |
|
| 453. |
A particle moving with uniform acceleration from `A` to `B` along a straight line has velcities `v_(1)` and `v_(2)` at `A` and `B` respectively. If `C` is the mid-point between `A` and `B` then determine the velocity of the particle at `C` . . |
|
Answer» Let `v` be the velocity of the particle at `c`. Assume the acceleration of the particle to be `A` and distance between `A` and `B` be `x`. To find the velocity `c`, consider the motion from `A`and `C`: Applying `v^(2)-u^(2)=2` as, we get `v^(2)-v_(1)^(2)=2a(x)/(2)` `rArr v^(2)-v_(1)^(2)=ax`........ (i) Appying the same equation from `C` to `B`. we get` `V_(2)^(2)=2a(x)/(2) rArr v_(2)^(2)-v^(2)=ax` ......(ii) From (i) and (ii), we get `v=sqrt(v_(1)^(2)+v_(2)^(2))/(2)`. |
|
| 454. |
Two trains `P` and `Q` are moving along parllel tracks same uniform speed of `20` m `s^(-1)`. The driver of train `P` decides to overtake train `Q` and accelerate his frain by `1 ms^(-2)`, After `50 s`, train `P` crosses the engine of train `Q`. Find out what was the distance between the two trains initially. provided the length each is `400 m`. |
|
Answer» Let the initial distance between two trains be `x`. Distance travelled by train `P` in `50 s`: `S_(p)=ut+(1)/(2)at^(2)=20xx50+(1)/(2)xx1xx50^(2)=2250 m` Distance travelled by train `Q` in `50` s: `S_(Q)=ut =20 xx50=100 m` Now `S_(P)-S_(Q) =2250-1000=1250 m`. This distance must be equal to the initial distance between the trains plus the sum of the length of the trains. `x+800 =1250 rArr x=450 m`. |
|
| 455. |
The position of an object moving along x-axis is given by `x = a + bt^2` where `a=8.5 m, b=2.5 m s^-2 and t` is measured in seconds. What is its velocity at `t= 0 s and t = 2.0 s`. What is the average velocity between `t = 2.0 s and t = 4.0 s?` |
|
Answer» Given, `x=a+bt^(2), wher, `a= 8.5 m` and `b=2.5 m//s^(2)` :. Velocity, v9dx)/(dt) =d/9dt) (a+ bt^(2))` At ` |
|
| 456. |
Two particles A and B are moving uniformly in a plane in two concentric circles. The time period of rotation is `T_(A) = 8` minute and `T_(B) = 11` minute respectively for the two particles. At time`t = 0`, the two particles are on a straight line passing through the centre of the circles. The particles are rotating in same sense. Find the minimum time when the two particles will again fall on a straight line passing through the centre. |
|
Answer» Correct Answer - `(88)/(3) min` |
|
| 457. |
A particle moves in xy plane with its position vector changing with time (t) as `vec(r) = (sin t) hati + (cos t) hatj` ( in meter) Find the tangential acceleration of the particle as a function of time. Describe the path of the particle. |
|
Answer» Correct Answer - `a_(t) = 0`; path is circular |
|
| 458. |
Two paper discs are mounted on a rotating vertical shaft. The shaft rotates with a constant angular speed `omega` and the separation between the discs is H. A bullet is fired vertically up so that it pierces through the two discs. It creates holes H1 and H2 in the lower and the upper discs. The angular separation between the two holes (measured with respect to the shaft axis) is `theta`. Find the speed (v) of the bullet. Assume that the speed of the bullet does not change while travelling through distance H and that the discs do not complete even one revolution in the interval the bullet pierces through them. |
|
Answer» Correct Answer - `v = (omega H)/(theta)` |
|
| 459. |
A particle starts from rest and moves with acceleration a which varies with time `t` as `a=kt` where `k` is a costant. The displacement `s` of the particle at time `t` isA. `1/2 kt^(2)`B. `1/2 at^(2)`C. `1/6 at^(2)`D. None |
|
Answer» Correct Answer - C |
|
| 460. |
The linear speed of a particle moving in a circle of radius `R` varies with time as `v = v_0 - kt`, where `k` is a positive constant. At what time the magnitudes of angular velocity and angular acceleration will be equal ? |
|
Answer» We have `v = v_0 - kt` (given). Then, the tangential acceleration of the particle is `a_t = (d v)/(d t) = k` The angular acceleration of the particle is `omega = (v)/( R) = (v_0 - kt)/( R)` If `prop = omega`, we have `(k)/( R) = (v_0 - kt)/( R)` Then, `t = (v_0 - k)/(k)`. |
|
| 461. |
A passenger is standing `d` metres away from a bus. The bus begins to move eith constat acceleration `a. To catch the bus the passenger runs at a constant speed (v) towards the bus, What must so that he may catch the bus. |
|
Answer» Let the passenger catches the bus after time Distance travelled bu the in time (t), ` S_91) =0 + 1/2 at^(2) = 1/2` …(i) Distance travelled by the passenger, (ii) ` S_(2) =v t + 0 v t` The passenger will catch the bus if, ` d+ S_(1) = S_(2)` or ` d + 1/2 at^(2) =v t or 1/2 at^(2) - v t + d =0` or ` t= =([v+- sqrt(v^(2) -2 a d)}])/a` The passenger will catch the bus it 9t) is real, i.e. `v^(2) gt- 2 a d or v gt- sqrt2 ad` Thus minimum speed passenger for catching the bus in ` sqrt 2 ad`. |
|
| 462. |
A passenger is standing `d` metres away from a bus. The bus begins to move eith constat acceleration `a. To catch the bus the passenger runs at a constant speed (v) towards the bus, at what minimum speed he must have ,so that he may catch the bus. |
|
Answer» Let the passenger catch the bus after time t. The distance travelled by the bus, `" "s_(1)=0+1/2 at^(2)` …(i) and the distance travelled by the passenger `" "s_(2)=ut+0` …(ii) Now the passenger will catch the bus if `d+s_(1)=s_(2)` ...(iii) `rArr d+1/2 at^(2)=ut rArr 1/2 at^(2)-ut+d=0rArr t=([u+-sqrt(u^(2)-2ad)])/a` so the passenger will catch the bus if t is real, i.e., `u^(2) ge 2ad rArr u^(2) ge u^(2)sqrt(2ad)` So the minimum speed of passenger for catching the bus is `sqrt(2ad)`. |
|
| 463. |
A particle moves in the plane x y with constant acceleration a directed along the negative y-axis. The equation of motion of the particle has the form `y = k_(1)x-k_(2)x^(2)`, where `k_(1)` and `k_(2)` are positive constants. Find the velocity of the particle at the origin of coordinates. |
|
Answer» Correct Answer - `[sqrt((k_(1)^(2)+1)(a)/(2k_(2)))]` |
|
| 464. |
A bus begins to move with an accelaration of ` 1 ms^(-1)` . A man who is `48m` behind the bus starts running at ` 10 ms^(-1)` to catch the bus, the man will be able to catch the bus after .A. 8 sB. 10 sC. 12 sD. 14 s |
|
Answer» Correct Answer - A |
|
| 465. |
The velocity-time graph of a linear motion is shown in figure. The displacement & distance from the origin after 8 sec is :- A. `5m, 19m`B. `16 m, 22m`C. `8 m, 19 m`D. `6m, 5m` |
|
Answer» Correct Answer - A Displacement`=1/2[4+2]xx4-1/2[4+3]xx2` `=12-7=5m` Distance `=12+7=19 m` |
|
| 466. |
A car (A) is moving at `60 km h^(-1)` on a straight road, is ahead of car (B) moving in the same direction at `10 ms^(-1)`. Find the velocity of (A) relative to (B) and vice versa. |
|
Answer» Here, `v_(A) =60 km h^(-1)` `=(60 xx 1000 m)/(60 xx 60 s) =(50)/3 m//s :` `v_(B) =10 ms^(-1)` Velocity of (A) value of `v_(AB)` indicates that the person sitting in car (B) sees car (A) moving ahead from him at the rate of `6.67 ms^(-1)`. Velocity of (B) respect to `A, v_(BA)` =v_(B) -v_(A)` `=10 -(50)/3 =- (20)/3 =- 6. 67 ms^(-1)` Negative value of `v_(BA)` indicates that the person sittion in (A) sees the car (B) lagging behind at the rate of `6.67 ms^(-1)`. |
|
| 467. |
Initially car `A` is `10.5 m` ahead of car `B`. Both start moving at time `t = 0` in the same direction along a straight line. The velocity-time graph of two cars is shown in figure. The time when the car `B` will catch the car `A`, will be. A. `t-21` secB. `t=2sqrt(5)` secC. `20` secD. None of these |
|
Answer» Correct Answer - A `S_(B)=S_(A)+10.5` `t^(2)/2=10t+10.5` `t^(2)=20t+21` `t^(2)-20t-21=0` `t=21` sec |
|
| 468. |
State with reasons, whether the following algebraic operations with scalars and vectors ar meaningful. (a) Adding any scalar. 9b0 multiplying any tow sclars (e ) Adding any two vectors (f) Adding a component of a vector to the same vector. |
|
Answer» (a) No because only the scalars of same dimensins can be added. 9b) No, because a scalar connot be added to a verctor. (C ) Yes, when acceleration `vec A` is multiplied by mass (m0, we ger a force ` vec F =m vec a`, which is a meaningful operation. (d) Yes, when power (p) is multiplied by time (t0, we word done ` = Pt`, which is a useful operation. (e) No , because the two vectors of same dimensions can be added. (f) Yes, because both are vectors of the same dimensions. |
|
| 469. |
A ball is thrown from the ground into air. AT a height of ` 9.0 m`, the velcoity is obsecred ot be ` vec v = 7.0 hat i+ 6.0 hat j`. Find the masimum hight to which the ball will rise. ` g= 10 ,s^(-2)`. |
|
Answer» Here, ` v_x = 7 ms^(-1)` , v_y =6.0 ms^(-1) , S= 9.0 m, ` u_y = ?` Taking vertical upward motion of the ball up to given height, we have ` v_y^2 = u_y^2 + 2 a S` ` 6^2 = u_y^2 +2 (- 10) xx 9` or ` u_y^@ = 36 + 180 = 216` Maximum height attanied by the ball, ` h_(max). = u_y^2/(2 g) = ( 216)/(2 xx 10) = 10.8 m`. |
|
| 470. |
The velocity of a projectile when it is at the greatest height is `(sqrt (2//5))` times its velocity when it is at half of its greatest height. Determine its angle of projection. |
|
Answer» Max. height, ` H= (u^2 sin^2 theta)/(2 g)` or ` g H = ( u^2 sin^2 theta)/2` ltbRgt Velocity at highest point, ` v_h =u cos theta` Let ` v_x, v_y` be the horizontal and vetical velocity of projectile at height ` H//@`. The ltbRgt ` v_x = u cos theta` and ` v_y^2 = u^2 sin^2 theta-2 g xx H//2` ltbRgt ` = u^2 sin^@ thta-g H` ` = u^2 sin^2 theta - (u^2 sin^2 theta)/2 = (u^@ sin^2 theta)/2` Effective velocity at height H//2` `=(v_x^2 +v_y^2)^(1//2)` ltbRgt As per queston, ` sqrt 2/5 (v_x^2 +v_y^2)^(1//2) = v_h ` or ` 2/5 (v_x^2+v_y^2) =v_h^2` or ` sin^2 theta = 3 cos^2 theta` or ` tan theta =sqrt 3 cos theta` or ` tan theta = sqrt 3 = tan 60^2 ` or ` theta = 60^@`. |
|
| 471. |
The slope of straight line joining two points on velocity-time graph of an object having non uniform motion gives…………………………………… for the given interval of time. |
| Answer» indstantaneous acceleation | |
| 472. |
One second after the projection, a stone moves at an angle of ` [email protected]` with the horzontal. Two seconds from the start, it is travelling horizontally. Find the angle of projection with the horizontal. (g=10 ms^(-2))`. |
|
Answer» After one second, let ` v_x. v_y` be the horizontal and vertical coponent velocites of the projectile whose initial velcoity of projection (u) and angle of projection os ` theta` then ` v_x = u cos theta ` and ` v_y =u sin theta -g xx 1 =u sin theta-g` As the resultant of ` vec _x and vec v_y` makes an angle ` beta (= 45^@) with the horizontal. so ` tan beta =v_y/v_x = ( u sin theta-g)/( u cos theta) =tan 45^@` or ` u sin theta- g = u ` or ` u (sin theta - cos theta ) = g ` ....(i) After two seconds, the vetical component velocity of projectile becomes zero, since the velocity or projectile is horizontal agter tow seconds . So, ` u sin theta - 2 g = 0 ` or u 2 g/sin theta` From (i) , (2 g)/( sin theta) (sin theta-cos theta) = g` or ` 2 ( 1- cot theta ) = ` or ` 1-cot theta = 1/2` or ` cot theta 1- 1/2=1/2 ` or tan theta = 2 ` or ` thata = tan^(-10 (2)`. |
|
| 473. |
One second after the projection, a stone moves at an angle of ` [email protected]` with the horzontal. Two seconds from the start, it is travelling horizontally. Find the angle of projection with the horizontal. (g=10 ms^(-2))`. |
|
Answer» After one second, let ` v_x. v_y` be the horizontal and vertical coponent velocites of the projectile whose initial velcoity of projection (u) and angle of projection os ` theta` then ` v_x = u cos theta ` and ` v_y =u sin theta -g xx 1 =u sin theta-g` As the resultant of ` vec _x and vec v_y` makes an angle ` beta (= 45^@) with the horizontal. so ` tan beta =v_y/v_x = ( u sin theta-g)/( u cos theta) =tan 45^@` or ` u sin theta- g = u ` or ` u (sin theta - cos theta ) = g ` ....(i) After two seconds, the vetical component velocity of projectile becomes zero, since the velocity or projectile is horizontal agter tow seconds . So, ` u sin theta - 2 g = 0 ` or u 2 g/sin theta` From (i) , (2 g)/( sin theta) (sin theta-cos theta) = g` or ` 2 ( 1- cot theta ) = ` or ` 1-cot theta = 1/2` or ` cot theta 1- 1/2=1/2 ` or tan theta = 2 ` or ` thata = tan^(-10 (2)`. |
|
| 474. |
A boy can throw a ball up to a speed of `u = 30 m//s` . He throws the ball many a times, ensuring that maximum height attained by the ball in each throw is `h = 20 m`. Calculate the maximum horizontal distance at which a ball might have landed from the point of projection. Neglect the height of the boy. `[g = 10 m//s^(2)]` |
|
Answer» Correct Answer - `40 sqrt(5)m` |
|
| 475. |
(a) A boy throws several balls out of the window of his house at different angles to the horizontal. All balls are thrown at speed `u = 10 m//s` and it was found that all of them hit the ground making an angle of `45°` or larger than that with the horizontal. Find the height of the window above the ground [take `g = 10 m//s^(2)]` (b) A gun is mounted on an elevated platform AB. The distance of the gun at A from the edge B is `AB = 960 m`. Height of platform is `OB = 960 m`. The gun can fire shells with a velocity of `u = 100 m//s` at any angle. What is the minimum distance (OP) from the foot of the platform where the shell of gun can reach ? |
|
Answer» Correct Answer - (a) `5m` (b) `480 m` |
|
| 476. |
A particle `P` is projected with velocity `u_1` at an angle of `30^@` with the horizontal. Another particle `Q` is thrown vertically upwards with velocity `u_2` from a point vertically below the highest point of path of `P`. Determine the necessary condition for the two particles to collide at the highest point. . |
|
Answer» Both particles collide at the highest point. It means the vertical distance travelled by both the particle will be equal, i.e., the vertical component of velocity of both particle will be equal `u_1 sin 30^@ = u_2 rArr (u_1)/(2) = u_2 rArr u_1 = 2 u_2`. |
|
| 477. |
A boy throws a ball with speed `u` in a well of depth `14 m` as shown in (Fig. 5.191). On bounce with the bottom of the well, the speed of the ball gets halved. What should be the minimum value of `u( "in" ms^-1)` such that the ball may be able to reach his hands again ? It is given that his hands are at `1 m` height from top of the well while throwing and catching. . |
|
Answer» Correct Answer - `30 m s^-2` Let `v` is the same with which ball collides withb bottom, then `v^2 = u^2 + 2g xx 15` …(i) Now ball rebounds with speed `v//2`. So `(v//2)^2/(2g) = 15 rarr ` to reach the hands of boy `rArr v^2 = 8g xx 15` …(ii) From Eqs. (i) and (ii), `u = 30 ms^-1`. |
|
| 478. |
A rock is launched upward at `45^@`. A bee moves along the trajectory of the rock. What is the magnitude of acceleration `("in" m s^-2)` of the bee at the top point of the trajectory ? For the rock, neglect the air resistance. |
|
Answer» Correct Answer - `20 ms^-2` Velocity of rock at top point `= u cos theta` So, acceleration of rock =`((u cos theta)^2)/( r) = g` `rArr (u^2)/( r) = (g)/(cos^2 theta) = (g)/(cos^2 45^@) = 20 ms^-2` where `r` is the radius of curvature Now acceleration of bee at top point : `a_b = (u^2)/( r) = 20 ms^-2` Hence, `r` is same for both. |
|
| 479. |
A ball is fired from point `P`, with an initial speed of `50 m s^-1` at an angle of `53^@`, with the horizontal. At the same time, a long wall `AB at 200 m` from point `P` starts moving toward `P` with a constant speed of `10 m s^-1`. Find (a) the time when the ball collides with wall `AB`. (b) the coordinate of point `C`, where the ball xollides, taking point `P` as origin. . |
|
Answer» Correct Answer - (a) 5 s; (b) 150 m ; 75 m. (a) Relative horizontal velocity of ball and wall =`50 cos 53^@ + 10 = 40 ms^-1` So time taken: `t = (200)/(40) = 5 s` (b) x coordinate of point C : `x = (50 cos 53^@) 5 = 150 m` To find y coordinates : `y = 50 sin 53^@ xx 5 - (1)/(2) 10 (50)^2 = 75 m`. |
|
| 480. |
Ball `I` is thrown towards a tower at an angle of `60^@` with the horizontal with unknown speed `(u)`. At the same moment, ball `II` is released from the top of tower as shown in (Fig. 5.189). Balls collide after `2 s` and at the moment of collision, the velocity of ball `I` is horizontal. Find the (a) speed `u` (b) distance of point of projection of ball `I` from base of tower `(x)` ( c) height of tower . |
|
Answer» Correct Answer - (a) `(40)/(sqrt(3)) m s^-1` ; (b) `(40)/(sqrt(3)) m ; c. 40 m`. (a) Time of ascent should be `2 s`. So `(u sin 60^@)/(g) = 2 rArr u = (40)/(sqrt(3)) ms^-1` (b) `x = (u cos 60^@) t = (40)/(sqrt(3)) xx (1)/(2) xx 2 = (40)/(sqrt(3)) m` ( c) The height of tower can be found using concept of relative velocity. If the balls have to collide, the initial velocity of ball `I` should be towards ball `II`. For this `(h)/(x) = tan 60^@ rArr h = x tan 60^@ = (40)/(sqrt(3)) sqrt(3) = 40 m` OR : height ascended by ball `I "in" 2 s` : `h_1 = u sin 60^@ xx 2 - (1)/(2) 10 (2)^2 = 20 m` height decended by ball `II in 2 s` , `h_2 = (1)/(2) "gt"^2 = (1)/(2) 10 (2)^2 = 20 m` Now `h = h_1 + h_2 = 20 + 20 = 40 m`. |
|
| 481. |
Under what conditions the directions of sum and differnce of two vectros will be the same. |
| Answer» The direction of sum and differnce of two vectors will be the same, when the two vectors are unequal in vagnitude and acting in the same derction. | |
| 482. |
What is the differnce between the follwing data ? (i) 3 (5 km h^(-1) , west ) (ii) 3 hour (5 km h^(-1), west). |
| Answer» (i) It is the product of a pure muber and a velocity vector, hence the unit of product is same as that of velocity vector i.e. the product is a velocity of magnitude `15 km h^(-1)` towards west. (ii) It is the product of a scalar (time ) and velocity vector. The unit of this product will be hour ` xx km h^(-1) =km `. Thus the product is a dis-placement of magnitude ` 15 km 1 towards west. | |
| 483. |
Sundy the four graphs given below. Answer the follwing questions on the basis of these graphs. In which of the graghs, the particle has more magnitude of velocity at `t_(2)`,A. (i), (ii) and (iv)B. (i) and (iii)C. (ii) and (iii)D. None of the above |
|
Answer» Correct Answer - b In graphs (i) and (iii) magntude of slope is greater at `t_(1)` than that at `t_(2)`, |
|
| 484. |
The velocity-time graph of a particle moving along a straight line is shown is Fig. The rate of acceleration and deceleration is constant and it is equal to `5 m s^(-2)`. If the a average velocity during the motion is `20 m s^(-1)`, Then . The maximum velocity of the particle is .A. `20 m s^(-1)`B. `25 m s^(-1)`C. `30 m s^(-1)`D. `40 m s^(-1)` |
|
Answer» Correct Answer - b Maximum velocity `=5 t=5 xx 5 =25 m s^(-1)`. |
|
| 485. |
The velocity-time graph of a particle moving along a straight line is shown is Fing. The rate of acceleration and deceleration is constant and it is equal to `5 m s^(-2)`. If the a average velocity during the motion is `20 m s^(-1)`, Then . The distace travelled with uniform velcoty is .A. `375 mB. `125 m`C. `300 m`D. `450 m` |
|
Answer» Correct Answer - a `25 (20-t)=25 xx 15 =375 m`. |
|
| 486. |
Explain that a can have zero average velcoity but not zero average speed. |
| Answer» Average velocity=displacement //(total time taken) and average speed =total distance travelled//(total taken ) .If an object completes a circular path of radius (r ) in time (t), then its displacement is zero but distanc etravelled by body is ` 2 pi r`. Therefore, the average velocity of body =zero but average speed of body ` =2 pir//t`. | |
| 487. |
A stone is thrown from the top of a tower of height 50 m with a velocity of 30 m per second at an angle of `30^(@)` above the horizontal. Find (a) the time during which the stone will be in air, (b) the distance from the tower base to where the stone will hit the ground, (c) the speed with which the stone will hit the ground, (d) the angle formed by the trajectory of the stone with the horizontal at the point of hit. |
|
Answer» Correct Answer - [(a) 5.0 s, (b) `75 sqrt3 m` (c) 43.58 m/s, (d) `tan^(-1) ((7)/(3 sqrt3))]` |
|
| 488. |
Two balls are projected simultaneously from the top of a tall building. The first ball is projected horizontally at speed `u_(1) = 10 m//s` and the other one is projected at an angle `theta = tan^(-1) ((4)/(3))` to the horizontal with a velocity `u_(2). [g = 10 m//s^(2)]` (a) Find minimum value of `u_(2) (= u_(0))` so that the velocity vector of the two balls can get perpendicular to each other at some point of time during their course of flight. ltbr? (b) Find the time after which velocities of the two balls become perpendicular if the second one was projected with speed `u_(0)`. |
|
Answer» Correct Answer - (a) `u_(0) = 37.5 m//s` (b) `t = 1.5 m//s` |
|
| 489. |
Referrring to the `v^(2)-s` diagram of a particle, find the displacement of the particle durticle during the last two seconds. . |
|
Answer» Slope of `v^(2)-s` graph` `m=-((2000-1500))/(100-50)=-10` Relation between `v_(2)` and s: `v^(2)=-10 s+C` At `s=50, v^(2)=2000` `rArr 2000=-10xx50+C rArr C=2500` `v=-10 s+2500` `rArr (2v dt)/(dt)=-10(ds)/(dt)` `rArr 2va=-10 v rArr a=-5 ms^(-2) rarr` constant To find initial velocity: Put `s=0, v=u` `rArr u^(2)=2500 rArr u=50 ms^(-1)` Apply `v=u+at` `rArr 0=50-5 T` `rArrT=10 s rarr` time of motion `Displacement during last two seconds : `S=S_(t)=10s-S_(t)=8s` `=[50 xx 10 -(1)/(2)5 (10)^(5)]-[50-(1)/(2)5(8)^(2)]=10m`. |
|
| 490. |
The numuerical value of the ratio of instantaneous velocity to instantaneous spedd is.A. Always less than `1`B. Always equal to `1`C. Always more than `1`D. Equal to or less than `1` |
|
Answer» Correct Answer - B At and instant. The magnitudes of velocity and speed will be equal. |
|
| 491. |
A body moves with unifrom spedd of u `ms ^(-1)` towards east , then the body is said to possess _____ velocity. |
| Answer» Correct Answer - unifrom | |
| 492. |
The ratio of the numerical values of the average velocity and average speed of a body is always. |
| Answer» Correct Answer - ` (d)` | |
| 493. |
A car accelerates from rest at a constant rate `alpha` for some time, after which it decelerates at a constant rate `beta,` to come to rest. If the total time elapsed is t seconds. Then evalute (a) the maximum velocity reached and (b) the total distance travelled.A. `(alpha beta t^(2))/(4(alpha +beta))`B. ` (alpha beta t^(2))/(2(alpha +beta))`C. `(alpha beta t^(2))/(alpha +beta)`D. `(4alpha beta t^(2))/(alpha +beta)` |
|
Answer» Correct Answer - B From `A` to `B` applying `sut +(1)/(2) at^(2)`, we get `S_(1) =0 xx t_(1) +(1)/(2) alph t_(1)^(2) =(1)/(2) (alpha t_(1)) t_(1) =(1)/(2) v_(0)t_(1)` (1) Sinmilarly from `B` to `C`: `S_(2) =v_(0) t_(2) +(1)/(2) (-beta) t_(2)^(2) =v_(0)t_(2) -(1)/(2) (betat_(2)) t_(2) =(1)/(2) v_(0) t_(2) (2) Now total distance travelled is `S_(1) +S_(2) =(1)/(2) v_(0)t_(1) +(1)/(2) v_(0)t_(2) =(1)/(2) v_(0) (t_(1) _t_(2))` `=(1)/(2) (alph beta t)/(2 alpha+beta) t =(alpha beta t^(2))/(2 (alpha+beta) t =(alpha beta t^(2))/(2(alpha +beta)` Also, we can find (i) and (ii) by using the formula, `s=(u+v)/(2) t`. |
|
| 494. |
A car accelerates from rest at a constant rate `alpha` for some time, after which it decelerates at a constant rate `beta,` to come to rest. If the total time elapsed is t seconds. Then evalute (a) the maximum velocity reached and (b) the total distance travelled. |
|
Answer» Correct Answer - A::B::C::D (a) Let the car accelerates for time `t_1` and decelerates for time `t_2.` Then `t=t_1+t_2….(i)` and corresponding velocity-time graph will be as shown in fig. `(DCP_V01_CH6_S01_024_S01.png" width="80%"> From the graph, `alpha`=slope of line `OA=v_(max)/t_1` or `t_1=v_(max)/alpha....(ii)` and `beta` = -slope of line `AB=v_(max)/t_2` or `t_2=v_(max)/beta...(iii)` From Eqs. (i),(ii) and (iii), we get `v_(max)/alpha+v_(max)/beta=t` or `v_(max)((alpha+beta)/(alpha beta))=t` or `v_(max)=(alpha beta t)/(alpha+beta)` (b) Total distance =Total displacement=area under v-t graph `=1/2xxtxxv_(max)` `=1/2xxtxx(alpha beta t)/(alpha+beta)` or Distance = ` 1/2((alpha beta t)/(alpha+beta))` |
|
| 495. |
On a `80 km` track, a train travels `40 km` with a uniform speed of `30 km h^(-1)`. How fast must the train travel the next `40 km h^(-1)` as to have average speed `40 km h^(-1)` for the entire trip? |
|
Answer» Here, `s_(1) =40 km, v_(1) =30 km h^(-1) s_(2) =40 km, v_(av) =40 km h^(-1), v_(2)=?` `v_(av)=(S_(1)+S_(2))/(S_(1)/v_(1) +S_(2)/v_(2)) =((40+40)/((40)/(30) +(40)/v_(2)) =(2xx 30 v_(2))/(30 +v_(2))` or `40 =(60 v_(2) )/(30 + v_(2)) or v_(2) =60 km h^(-1)`. |
|
| 496. |
Two boys simultaneously aim their guns at a bird sitting on a tower. The first boy releases his shot with a speed of 100m/s at an angle of projection of `30^(@)` . The second boy is ahead of the first by a distance of 50 m and releases his shot with a speed of 80m/s. How must he aim his gun so that both the shots hit the bird simultaneously ? What is the distance of the foot of the tower from the two boys and the height of the tower ? With what velocities and when do the two shots hit the bird? |
|
Answer» Correct Answer - `[theta-sin^(-1)((5)/(8))]` |
|
| 497. |
What are positive and negative acceleration in straight line motion? |
|
Answer» If speed of an object increases with time, its acceleration is positive. (Acceleration is in the direction of motion) and if speed of an object decreases with time its acceleration is negative (Acceleration is opposite to the direction of motion). |
|
| 498. |
What are positive and negative acceleration in straight line motion ? |
|
Answer» If speed of an object increases with time, its acceleration is positive. (Acceleration is in the direction of motion) and if speed of an object decreases with time its acceleration is negative (Acceleration is opposite to the direction of motion). |
|
| 499. |
What is the angle between vector (A + B) and vector (A - B) ? |
|
Answer» The angle 90° |
|
| 500. |
`ABC` is a triangle in vertical plane. Its two base angles `/_BAC` and `/_BCA` are `45^(@)` and `"tan"^(-1)1/3` respectively. A particle is projected from point `A` such that is passes through vertices `B` and `C`. Angle of projectio is `theta`. Find the value of `3 tan theta`. |
|
Answer» Correct Answer - 4 |
|