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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.
| 1201. |
A particle has the following displacements in succession (i) `12 m` towards East (ii) ` 5 m` towards North and (ii) ` 6 m` vertically. Find the magniude of the resultant displacement. |
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Answer» ` x= 12 m , y=m , z = 6 m`. ltbRgt Then, ` r= (x^2 +y^2 +z^2)^(1//2)` ltbRgt ` = ( 12^2 + 5^2 + 6^2 ^(1//2)` `= (12^2 + 5^2 + 6^2)^(1//2)= 14.32 m` . |
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| 1202. |
Which of the following sets of displacements might be capable of returning a car to its starting point?A. 4, 6, 8 and 15 kmB. 10, 30, 50 and 120 kmC. 5, 10, 30 and 50 kmD. 40, 50, 75 and 200 km |
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Answer» Correct Answer - A |
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| 1203. |
A ball is kiched at an angle ` 30^(@)` with the verical. If the horizontal componet of its velocity is `20 ms^(-1)`, find the maximum hight and hrizontal range. Use `= 10 ms^(-2)`. |
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Answer» Here, theta =90^(@)-30^(@) =60^(@)` Horizontal velocity, ` u cos 60^(@) =20 ms^(-1)` or `u =(20)/( cos 60^(@)) =(20)/(1//2) =40 ms^(-1)` Maximum height , `H=(u^(2) sin^(2) theta ) /(2 g) =(40^(2) sin ^260^(@))/(2 xx 10)` `=(1600)/(2 g) =(40^(2) sin ^(2) 60^(@))/(10)` `=(17600)/(10) (sqrt 3//2) =238.6 m` |
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| 1204. |
One body is thrown at an angke ` theta` with the horizontal and another similar body is thrown at an angle `theta` with the vertical direction from the same point with same velocity` 40 ms^(-1)`. The second body reaches ` 50` metrea higher than the first body. Deteramine their individual heights. Take `g= 10 ms^(-2)`. |
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Answer» For the first body, angle of projection is `theta` with the horzontal directin. Velocity of projection `u=40 m//s` Maximum heitght, `H_(1) =(u^(2) sin^(2) theta) /(2 g) =((40)^(2) sin ^(2) theta)/(2xx 10)` .(i) For the second body, angle of projection ` (90^(@) -theta) with the horizontal direction, velocity of projection, `u=40 ms^(-1)`. Maximu height reached, `H_(2) =H_(1) +50 `...(ii) Where `H_(2) =((40)^(2) sin^(2) (90^(@)-theta))/(2xx 10) =((40)^(2) cos^(2) theta)/(2 xx 10)` Form (iii), wr have :. H_(1) + 50 =((40)^(2) cos^(2) theta)/(2 xx 10) =((40)^(2) cos^(2) theta)/(2 xx 10)` Adding (i) and (iii), we have `2 H_(1) + 50 =((40)^(2))/((40))^(2)/(2 xx 10) [sin ^(2) theta + cos^(2) theta]` `=((40)^(2))/(20) =80 or `2 H_(1) =80 -50 =30` or `H_(1) =15 m` Height of the first body, H_(1) =15 m` Height of the second body, `H_(2) =15 + 50 =65 m`. |
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| 1205. |
A partacle is thrown in horizontal direction with speed u from a point P, the top of a tower shown in figure.1.49 at a vertical height `h` above the inclined plane of inclination `theta`. Find the speed with which the particle is thrown so that it strikes the plane normally. Also find the distance from the foot of the tower where the particle will strike. |
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Answer» Correct Answer - `"[" [sqrt((2gh)/(2 + cot^(2) theta)), (2h)/(sin theta (2 + cot^(2) theta))] "]"` |
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| 1206. |
A partacle is thrown in horizontal direction with speed u from a point P, the top of a tower shown in figure.1.49 at a vertical height `h` above the inclined plane of inclination `theta`. Find the speed with which the particle is thrown so that it strikes the plane normally. Also find the distance from the foot of the tower where the particle will strike. |
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Answer» Correct Answer - `[u lt "[" [sqrt((1)/(2) dg (sqrt13 - 1))] " ]"` |
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| 1207. |
A particle is moving on a circular path of radius R with uniform angular speed `omega`. The magnitude of average velocityof particle during time `t=0` to `t=(2pi)/(3 omega)`:A. `sqrt(3)/2 (omega R)/pi`B. `3/2 (omegaR)/pi`C. `(3sqrt(3))/(2) (omegaR)/pi`D. `2/3 (omegaR)/(pi)` |
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Answer» Correct Answer - C |
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| 1208. |
The velocityof a particle that moves in the positive X- direction varies with its position, as shown in figure-1.22. Find its acceleration in `m//s^(2)` where `x = 6m` |
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Answer» Correct Answer - `[- 1.5 m//s^(2)]` |
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| 1209. |
Is magnitude of the displacement of an object and total distance covered by it in certain time intrval same ? Explain. |
| Answer» Not necessatily. If an object covers a complete circular track of radius ® then its sisplacement is zero but distance travlled is ` 2 pi r`. | |
| 1210. |
A particle moves in a straight line. Figure-1.21 shows the distance traversed by the particle as a function of time `t`.Using the graph, find (a)the average velocity of the point during the time of motion, (b) the maximum velocity, (c) the time `t = t_(0)` at which the instantaneous velocity is equal to the mean velocity averaged over the first `t_(0)` seconds. |
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Answer» Correct Answer - [(a) 10 cm/s, (b) 25 cm/s, (c) 16 s] |
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| 1211. |
velocity and acceleration of a particle at some instant are `v=(3 hati-4 hatj+2 hatk) m//s and a=(2 hati+hatj-2hatk) m//s^2` (a) What is the value of dot product of v and a at the given instant? (b) What is the angle between v and a, acute, obtuse or `90^@`? (c) At the given instant, whether speed of the particle is increasing, decreasing or constant? |
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Answer» Correct Answer - A::B::C::D (a) `v.a=6-4-4=-2m^2//s^3` (b) Since dot product is negative. So angle between v and a is obtuse. (c) As angle between v and a at this instant is obtuse, speed is decreasing. |
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| 1212. |
A point moves in ` x-y` plane according to the law ` x= 4 sin 6 t ` and ` y= 4 ( 1- cos 6 t)`. Fimd distance traversed by the particle in ` 5 seconds`, when (x0 and (y) are in metres. |
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Answer» ` V-x = (dx)/(dT) = 4 xx 6 sin cos 6 t` or ` t = 24 cos 6 t` ltbRgt ` v_y = (dy)/(dt) =4 xx 6 sin 6 t= 24 cos 6 t` ltBrgt :. Velocity, ` v = sqrt (v_x^2 +v_y^2)` ` =sqrt( ( 24 cos 6 t) ^2 + (24 sin 6 t )^2 ) = 24 ms^(-2)`. ltbRgt It means, velcity of point is constanct. The distance travelled by particle in ` 5 sec` ` = 24 xx 5 = 120 m`. |
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| 1213. |
Find the resultant forece of the following forces which act upon a particle. (a) ` 30 N due east ` (b) ` 20 N`1 due North (c ) ` 50 N` due West (d) ` 40 N` due South. |
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Answer» Efferctive force along West, ` A = 50 - 30 = 20 N` Effective force along South, B= 40 - 20 = 20 N` :. ` R sqrt (A^2 + B^@) = sqrt ( 20^2 + 20^2) = 20 sqrt 2 N` : ` tan beta = B/A = (20)/(20) = 1 ` or beta 45^@` . |
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| 1214. |
A body is dropped from the top of the tower and falls freely. The velocity of the body after `n` seconds is proportional to .A. `n^(2)`B. `n`C. `2n-1`D. `2n^(2) -1` |
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Answer» Correct Answer - B `S=(1)/(2)gn^(2) rArr S prop n^(2)` `S=(a)/(2)(2n-1) rArr S prop(2n-1)`. `V=gn rArr v prop n`. |
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| 1215. |
As soon as a car just starts from rest in a certain dercation, a scooter moveing with a uniform speed overtakes the car. Their velocity-time graph is shown in . Calculate . a. The difference between the distances travlled by the car and the scooter in `15 s`, b. The distance of car and scooter from the starting point at that instant. |
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Answer» a. The distance travekked by car in `15 `s `=Area` of `Delta OAC` `=(1)/(2) xx 15 xx 45 =337. m` Distance travelled by scooter in `15 s` `=Area` of retangle `OCEF` `=15xx30=450 m` Thus, difference between distance travelled by them `=450 m-337.5 m= 112.5 m` b. Let aftre time `t` from start car will catch up the scooter. In time `t`, the distance travelled by them are equal. Distance travelled by car `=(1)/(2)xx 1545+ 45(t-15)` Distance travelled by scooter `=30t` `(1)/(2) xx 15 xx 45 +45 (t-15)=30t` which gives `t=22.5 s` Distance travelled by car or scooter in `22.5 s=30xx22.5` `=675 m` So the catr catches the scoter when both are at `67.5 m` from the starting point. |
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| 1216. |
A particle moves along a straight line. Its position at any instant is given by `x = 32t-(8t^3)/3` where x is in metres and t in seconds. Find the acceleration of the particle at the instant when particle is at rest.A. `-16 ms^-2`B. `-32 ms^-2`C. `32 ms^-2`D. `16 ms^-2` |
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Answer» Correct Answer - B `v=(dx)/(dt)=32-8t^2` `v=0` at `t=2s` `a=(dv)/(dt)=-16t` at `t=2s, a=032m//s^2` |
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| 1217. |
It takes one minute for a passenger standing on an escalator to reach the top. If the escalator does not move it takes him `3` minute to walk up. How long will it take for the passenger to arrive at the top if he walks up the moving escalator ?A. `30 sec`B. `45sec`C. `40 sec`D. `35 sec` |
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Answer» Correct Answer - B `V_(esclator) = (x)/(1) rArr V_(esclator) x, v_(man) = (x)/(3)` `t = (x)/(v_(esclator) +v_(man)) rArr t = (x)/(x+(x)/(3))` `rArr t = (3)/(4)min rArr t = 45sec` |
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| 1218. |
A person moves 30m north and then 20 m towards east and finally `30sqrt(2)`m in south-west direction. The displacement of the person from the origin will beA. 14 m south westB. 20 m southC. 10 m westD. 15 m east |
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Answer» Correct Answer - C |
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| 1219. |
A man can swim at a speed of 3 km/h in still water. He wants t cross a 500 m wide river flowing at 2 km/h. He flow while swimming. A. Find the time he takes to cross the river. b.At what point on the opposite bank will he arrive? |
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Answer» Correct Answer - [`sin^(-1) (3//7)` from the normal direction , 12.65 min] |
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| 1220. |
A man can swim at a speed of `3 km h^-1` in still water. He wants to cross a `500-m` wide river flowing at `2 km h^-1`. He keeps himself always at an angle to `120^@` with the river flow while swimming. The time taken to cross the river is.A. `(3)/(2) h`B. `(1)/(6) h`C. `(1)/(3 sqrt(3)) h`D. none |
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Answer» Correct Answer - C ( c) `t = (d)/( v sin theta) =(0.5 km)/(3 sin 120^@ km h^-1)=(1)/(3 sqrt(3)) h` `x = (u + v cos theta) t = (2 + 3 cos 120^@)(1)/(3 sqrt(3)) = (1)/(6 sqrt(3))`. |
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| 1221. |
A plane flying horizontally at `100 m s^-1` releases an object which reaches the ground in `10 s`. At what angle with horizontal it hits the ground ?A. `55^@`B. `45^@`C. `60^@`D. `75^@` |
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Answer» Correct Answer - B (b) `v_x = u_x = 100 ms6^-1, v_y = u_y + a_y t = 0 + 10 xx 10` `tan theta = (v_y)/(v_x) = (100)/(100) = 1 rArr theta = 45^@`. |
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| 1222. |
Velocity of a particle at some instant is `v=(3hat i + 4hat j + 5hat k) m//s`. Find speed of the particle at this instant. |
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Answer» Correct Answer - B Magnitude of velocity vector at any instant of time is the speed of particle. Hence, `Speed= v or |v| = sqrt(3^2 + 4^2 + 5^2 ) = 5sqrt2 m//s` |
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| 1223. |
Two particle are projected from the same point on ground simultaneously with speed `20m//s` and `20//sqrt(3)m//s` at angle `30^(@)` and `60^(@)` with the horizontal in the same direction. The maximum distance between them till both of them strikes the ground is approximately `(g=10m//s^(2))`A. `23.1m`B. `16.4m`C. `30.2m`D. `10.4m` |
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Answer» Correct Answer - A |
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| 1224. |
A man standing on the edge of a cliff throws a stone straight up with initial speed (u) and then throws another stone straight down with same initial speed and from the same position. Find the relation of the speeds. The stones would have attained when they hit ground at the base of the cliff. |
| Answer» Since a stone throw vertically up with speed (u) returns back the same point with the same speed, hence, both the stones will strike the ground with the same speed. Hence, the ratio of the speeds of the two stones will be ` 1: 1. | |
| 1225. |
A particle moves from position `A` to position `B` in a path as shown in If the poit vectors `vec r_(1)` and `vec r_(2)` making an angle `theta` between them are give, find the magnitude of displacement. . |
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Answer» From triangle OAB: `vec r+ Delta vec r =vec r_(2) rArr vec r_(2)- vec r_(1) = Delta vec r` Hence, `|vec S|=|Delta vec r|=|vec r_(2)- vec r_(1)|=sqrt(r_(1)^(2)+r_(2)^(2)-2r_(2)r_(2) cos theta)` |
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| 1226. |
A ground to ground projectile is at point `A` at `t=T/3` is at point `B` at `t=(5T)/6` and reaches the ground at `t=T`.The difference in heights between points `A` and`B` is `(gT^(2))/(6x)`. Find value of `x` |
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Answer» Correct Answer - 4 |
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| 1227. |
A particle starts from rest with uniform acceleration and is velocity after `n` seconds `v`. The displacement of the body in last two seconds is.A. `(2v(n-1))/n`B. `(v(n-1))/a`C. `(v(n+1))/n`D. `(2v(2n+1))/a` |
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Answer» Correct Answer - A |
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| 1228. |
A balloon is moving upwards with velocity `10m//s`. It releases a stone which comes down to the ground in `11s`. The height of the balloon from the ground at the moment when the stone was dropped is `(g=10m//s^(2))`A. `495m`B. `592m`C. `362m`D. `500m` |
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Answer» Correct Answer - A |
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| 1229. |
Two nearly identical balls are released simultaneously from the top of a tower. One of the balls fall with a constant acceleration of `g_(1) = 9.80 ms^( –2)` while the other falls with a constant acceleration that is `0.1%` greater than `g_(1)`. [This difference may be attributed to variety of reasons. You may point out few of them]. What is the displacement of the first ball by the time the second one has fallen 1.0 mm farther than the first ball? |
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Answer» Correct Answer - 1m |
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| 1230. |
A boy releases a toy plane from the top of a high hill of height H. Hill is so high that gravity varies with height from ground as `g = g_(0) (1 - (2h)/(R))`, where h is the height from ground and R is the radius of earth. The engine of toy plane accelerates it in horizontal direction with acceleration `a_(x) = bt^(2)`. Find the position from the foot of hill where the plane lands and the time after which it lands. |
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Answer» Correct Answer - `[(bR^(2))/(48 g_(0)^(2)) [ ln ((R + sqrt(2RH - h^(2)))/(R - H))]^(2), sqrt((R)/(2g_(0)))ln ((R + sqrt(2RH - H^(2)))/(R - H))]` |
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| 1231. |
Two particles A and B separated by 10 m at time `t = 0` are moving uniformly. A is moving along line AB at a constant velocity of `4 m//s` and B is moving perpendicular to the velocity of A at a constant velocity of `5 m//s`. After what time the two particles will be nearest to each other? |
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Answer» Correct Answer - `(40)/(41)s` |
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| 1232. |
The position vector of a particle P with respect to a stationary point O change with time according to the law `vec(r) = vec(b) sin omega t + vec(c) cos omega t` where `vec(b)` and `vec(c)` are constant vectors with `b pot c` and `omega` is a positive constant. Find the equation of the path of the particle `y = f (x)`, assuming x an dy axes to coincide with the direction of the vector `vec(b)` and `vec(c)` respectively and to have the origin at the point O |
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Answer» Correct Answer - `[(x^(2))/(b^(2)) + (y^(2))/(c^(2)) = 1]` |
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| 1233. |
The motion of a particle restricted to move in a two dimensional plane is given by `x = 2 cos pi t` and `y = 1 - 4 cos 2 pi t` where x and y are in metres and `t` is in seconds. Show that the path of the particle is a part of parabola `y = 5 - 2 x^(2)`. Find the velcoity and the acceleration of particle at `t = 0` and `t = 1.5 s`. |
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Answer» Correct Answer - [0, 158.98 `m//s^(2)`, `6.2 m//s, 157.75 m//s^(2)`] |
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| 1234. |
At what angle with the horizontal should a ball be thrown so that the range `R` is related to the time of flight as `R = 5 T^2` ? `(Take g = 10 ms6-2)`.A. `30^@`B. `45^@`C. `60^@`D. `90^@` |
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Answer» Correct Answer - B (b) `(R)/(T^2) = g ((sin 2 theta)/(4 sin^2 theta)) = (g)/(2) cot theta = 5 cot theta` Given `(R)/(T_2) = 5 , Hence, 5 = 5 cot theta` or `theta = 45^@`. |
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| 1235. |
(a) What does `|(dv)/(dt)|` and `(d|v|)/(dt)` represent? (b) Can these be equal? |
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Answer» Correct Answer - A::B::C::D (a) `|(dv)/(dt)|` is the magnitude of total acceleration. While `(d|v|)/(dt)` represents the time rate of change of speed (called the tangential acceleration, a component of total acceleration) as `|v|=v.` (b) These two are equal in case of one dimensional motion. |
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| 1236. |
An aeroplane has to go from a point A to another point B, `500 km` away due `30^@` east of north. Wind is blowing due north at a speed of `20 m//s.` The steering-speed of the plane is `150 m//s.` (a) Find the direction in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go fram A to B. |
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Answer» Correct Answer - [(a) `sin^(-1) (1//15)` east of direction Ab, (b) 50 min] |
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| 1237. |
The displacement (in metre) of a particle moving along x-axis is given by `x=18t +5t^(2).` Calculate (i) the instantaneous velocity `t=2 s` (ii) average velocity between `t=2 s` to `t=3 s` (iii) instantaneous acceleration. |
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Answer» Give , `x=18t +5t^(20` (i) Velcity, `v=(dx)/(dt) =d/(dt) (18 t+ 5t^(2) ) =18 + 10 t` At `t=2 s`, the instantaneous velcoty is `v_(i) =18 + 10 xx 2 = 38 ms^(-1)` (ii0 Displacement at `t_(1)=2 s` is. `x_(1) =18 xx 2 xx 5 xx 2^(2) =56 m` Displacement at `t_(2) =3 s` is, `x_(2) =18 xx 3+5xx 3^(2) =99 m` Average velocity, `v_(av) =(x_(2)-x_(1))/(t_(2)-t_(1)) =(99- 56)/(3-2) =43 ms^(-1)` (iii) Acceleration, `a=9dv)/(dt) =d/(dt) (18 +10t) =10 ms^(-2)`. |
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| 1238. |
The odometer of a bus reads 6700 klm when it starts from the station at 9.m. and when it comes back to the sation at 10 p.m. the odometer reading is found to be 6960 km, then the average speed of the bus in the would journey is _____` km h^(-1)`A. 10B. 20C. 30D. 40 |
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Answer» Correct Answer - B Average speed = `("total distance")/("total time") ` `= (6960 - 6700)/13 = 20 km h^(-1)` |
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| 1239. |
When a ball is thrown up vertically with velocity `v_0`, it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocityA. `sqrt3 v_0`B. `3 v_0`C. `9 v_0`D. `3/2 v_0` |
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Answer» Correct Answer - A `h=u^2/(2g) or upropsqrth` |
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| 1240. |
A cricketer can throw a ball to a maximum horizontal distance of 100m. With the same speed how much high above the ground can the cricketer throw the same ball? |
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Answer» Let (u) be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle or projection with whorizontal, ` theta = 45^@. Then, ` R_(max) = u^2 //g` …(i0 In order to study to the motion of the ball along vertcal direction, consider a point on the surface of earth as the origin and vertical upward direction as the positive direction of (Y) axis . Taking motion of the ball along vertical upward direction we have ` u_y = u, a_y =- g. v_y =0, t= ?, y+0 =0 y=?` Also , ` y=y_0 = u_y t = 1/2 a_y t^2` :. ` y=0 + u ( u /g) + 1/2 (- g ) u^2 /g - 1/2 u^2 /g = 1/2 = ( 1000/2 = 50 m` [ from (i) ]` . |
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| 1241. |
A football player kicks a ball at ball at an angle of `30 ^(0)` with the horizontal with an initial speed of `20 m//s`. Assuming that the ball travels in a vertical plance, calculate (a) the time at which the ball reaches the highest point (b) maximum height reached (c ) the horizontal range of the ball (d) the time for which the ball is in air. `g =10 m//s^(2)`. |
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Answer» Here, `theta =30 ^(@), u=20 m//s` (a) Time taken by ball to reach the highest point, `t T/2 =(u sin theta)/g =(20)/(10) xx isn 30^(0) =xx (1//2) =1 s` (b) The maximum height `=(u^(2) sin ^(2) xx theta)/(2g) =((20)^(2) xx sing ^(2) xx sin ^(2) 30^(0))/(2xx10) =5m` (c ) The borixontal rane `=(u^(2) sin ^(2) xx theta)/(g) =((20)^(2) xx sing ^(2) xx sin ^(2) 30^(0))/(10) =34.64 m. (d) The time of flotht `=(2 usin) theta)/g =(2 xx 20xx sin 30^(0))/(10) =2 s. |
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| 1242. |
There is an inclined surface of inclination `theta`. A smooth groove is cut into it forming an angle `alpha` with AB as shown in figure. A steel ball is free to slide a long the groove. If the ball is released from the point O. Find the speed when it comes to A. |
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Answer» Correct Answer - `[sqrt(2gl sin theta sin alpha)]` |
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| 1243. |
A ball is thrown vertically upward. At the highest point of its pathe, what will be its (i) instantaneous velcity and (ii) instan-tanceous accekeratuib ? Conmment. |
| Answer» At the highest point of the path of a veticallyupward gion body. The body is just at rest , so the instaneous velocity is zero but it is having downward acceleration ` g (= 9.8 ms^(-2)` due to gravity pull. | |
| 1244. |
Two cars (A) and (B) are at positions ` 100 m` and ` 200 m` from the origin at time ` t=0`. They start simultaneously with velocities ` 10 ms^(-1) respectively. The car (A) will overtake the car (B) at a distance of . |
| Answer» Correct Answer - (c ) | |
| 1245. |
Three points are located at the vertices of an equilateral triangle each of whose sides measure a. They all start simultaneously with speed v, each aiming at the next in order. How soon will the points converge ? |
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Answer» Correct Answer - 2a/3v |
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| 1246. |
The engine of a motoecycle can produce a maximum acceleration of `5 m//s^(2)`. Its brakes can produce a maximum retardation of `10 m//s^(2)`. What is the minimum time in which the motorcycle can cover a distance of `1.5 km`?A. `5 s`B. `10 s`C. `15 s`D. `30 s` |
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Answer» Correct Answer - D |
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| 1247. |
The retardation experienced by a movign motr bat after its engine is cut-off , at the instant (t) is given by`, ` =-k v^4` , where (k) is a constant. If ` v_0` is the magnitude of velocity at the cut-off , the magnitude fo velocity at time (t) after the cut-off is .A. (a) v_0 /(sqrt( 3 kt v_0^3)`1B. (b) v_0/(sqrt (3 kt v_0^3 + 1 ^(1//3))`C. (c ) ` sqrt ( 3 kt v_0^3)D. (d) (3 kt v_0^3 + 1 )^(1//3)` |
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Answer» Correct Answer - B ` a= (dv)/(dt) =- k v^4 ` or ` - ( dv)/ d v^4) = dt` Integrating both the sides, we have ` - 1/2 int_v_0^v v^(-4) dv - int _0^t dt` or ` -1/k [ v^(-3)/(-3) ]__0^v = t `or 1/(3k)[1/v^3 - 1/v_0^3 ]=t` or ` - 1/v^3 = 3 kt + 1/ v_0^3 = ( 3 kt v_0^3 + 1)/v_0^3` or ` v= v_0/((3 kt v_0^3 +1)^(1//3)`. |
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| 1248. |
The distance traveled by an object along the axes are even by ` x= 2 t^2 , y=t^2-4 t, z=3 t -5`. The initial velocity of the particle is .A. (a) ` 10 unit`B. (b) ` 12 unit`C. (c ) ` 5 unit `D. (d) ` 2 unit` |
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Answer» Correct Answer - C Given, ` s= 2 t^2` so (dx)/(dt) =4 t` y= t^2 -4 t so ` (dy)/(dt) = 2 t-4` ` z=3 t- 5 `so ` (dz)/(dt) =3` velocity `, vec v = (dx)/(dt) hat + (dy)/(dt) hat j + (dz)/(dt) hat k` ` = 4 t hat I + ( 2 t- 4) hat j + 3 hat k` When ` t=0, vec v=- 4 hat j + 3 hat k` so ` v= sqrt (9-4)^2 + (3)^2) = 5 unit` . |
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| 1249. |
A motor cyclist is riding North in still air at `20 km H^(-1)` . If the wind starts blowing westard with a velocity of ` 45.8 km h^(-1)` find the apparent velocity with which the motor cyclist moves, tiem it takes to cover ` 100 km` and its direction of motin. |
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Answer» Here, ` A = 20 km h^(-1) , B = 45. 8 km h^(-1)`, ` theta= 90^@ , r= ? , beta = ?` Time, ` t= distance //resultant velocity`. |
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| 1250. |
A ball is thrown vertically upwards. Which of the following plots represent the speed graph of the ball during its flight if the air resistence is not ignored?A. B. C. D. |
| Answer» Correct Answer - C | |