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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.
| 1251. |
A ball is thrown vertically upwards. Which of the following graph/graphs represent velocity time graph of the ball during its flight ( air resistance is neglected).A. (a) B. (b) C. (c ) D. (d) |
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Answer» Correct Answer - A When a body is thrown vertically upwards with velocity (u) then its velocity (v) after time (t) si ` v= u - g t ` It is a straight line for ` v-t` graph, with begative slope. Thus option (a) is true. |
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| 1252. |
Among the four graphs shown in the figure there is only one graph for which average velocity over the time interval (0,T) can vanish for a suitably chosen `T`. Which one is it ?A. .B. .C. .D. . |
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Answer» Correct Answer - B In graph (b) , for one value of displacement, ther are two timeings. As a result of it, for one time the average velocity is positve and for other tiem is equivalnt negative. Due to it, the average velocity for the two timeings (equal to time period) can vanish. |
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| 1253. |
The diresction of crntripetal accleration is ………………………….. . |
| Answer» along the radius derected towards the centre of circular path. | |
| 1254. |
Which of the following graphs may not repersent variation of distance (S) with respect ot time (t) ?A. (a) B. (b) C. (c ) D. (d) |
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Answer» Correct Answer - C Groph (c ) is not possible as the distance shown there is begative, which is not possible. The other graphs are possible. |
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| 1255. |
In abave questions ` 13 and 14`, we have carefully distinguished between average speed and magnitude of average velocity. No such distainction is necessary when we considedr speed and magnitude of velocity. The instantneoud speed if alwary equal to the magnitude of nistantaneous velocity. Why ? |
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Answer» Instantaneoud speed ` (v_(ins)` of the paticle at an instant is the firsta dericative of the distance with respect to time at theat instant of time i.e. ` v_(ins) = (dx)/(dt)` . Since in instantaneous speed we taken taken only a small interval of time (dt) durig which direction of motion of a body is not supposed to change, hence there is no difference between total path length and magnitude of displacement fro small interval of time (dt) Hence, instantaneous speed is alwary equal to magnitued of instantaneous velocity. |
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| 1256. |
A man walks on a straight road from his home to a moardet `3 km` away with a speed of`6 km//h`. Finding the market closed, he instantly turns and walks back with a speed of `9 km//h`. What is the (a) magnitude of average velocity and (b) averge speed of the man, over the interval of time (i) `0 to 30 min`, (ii) `0to 50 min`, (iii) `0 to 40 min`? |
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Answer» Time taken by man ti go fro m his home to market, `t_(1) =(distance )/(speed) =(3 km )/(6 km //h) =1/2 h =30 min` Time taken by man to go from market to home, `t_(2) =(3 km)/(9 km//h) =1/3 h =20 min` Total time taken`=t_(1) +t_(2) =1/2 +1/3 =-5/6 h` `=50 min` (i) `0 to 30 min` (a) Average velocity `=(displacement )/(time)` `=(3.0 km)/((1//2) h) =6 km //h` (b) Average speed =(path lenght)/(time) =(3.0 km )/((1//2) h)` `=6 km //h` (ii) `0 to 50 min` Total path lenth coverad `=3.0 +3.0` `=6.0 km` Total displacement `=(displacement 0 /(time)` Total displacement `=zero (a) Average velocity `=(displacrmrnt)/(time)` `=0/((5//6)h) =0` (b) Average speed `=(total path length)/(time)` `=(6 km )/((5//6)j) =7.2 km //h` (iii) `0 to 40 min` Distance moved in `30 min (from home to market) =3.0 km` Distance moved in `10 min (from market to home) with speed `9 km //h` `=9xx (10)/(60) =1.5 km` . So, displacement `=3.0 -1.5 =1.5 km` Total path length travelled `3.0+ 1.5` `=4.5 km` (a) Average velocity`=(1.5km)/((400//60)h) =2.25 km //k` (b) Average speed `=(4.5 km)/9(40.60)h) =6.75 km //h`. |
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| 1257. |
When a person is standing on earth, the treesand houses appera stationary to mim. However, when he is sitting in a running train all these obects appear to move in bakward direction. Why? |
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Answer» For a person standing on earth, relative velocity of trees and houses with respect to person is zero. For a persong sitttong a runnning train, the relative velocity of trees and houses whith repect to person is negative `[:. V_(TP) =v_(T) -u _(P) =0 =- v_(P)]` That is why the trees and houses appear to move in bachward direction to a person sitting in a running train. |
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| 1258. |
A man walks on a straight road from his home to a market ` 2.5 km` away with a speed of `5 km //h`. Finding the market closed, he instantly turns and walks back with a speed of `7.5 km//h`. What is the (a) magnitude of average velocity and (b) average speed of the man, over the interval of time (i) `0` to 30 min `. (ii) 0 to 50 min (iii) 0 to 40 min ? |
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Answer» Time tanen by man to go from his home to market, ` t _1 = (distance )/(speed) = (2.5)/ 5 = 1/2 h` Time taken by man to go from market to his home, ` T_2 = (2.5) /( 7. 5) = 1/3 h` :. Total time taken ` t_1 + t_2 + 1/2 = 5/6 = 5/6 h = 50 min. (i) to 30 min` (a) Acerage velocity = (dispacement )/(time ) = ( 2.5)/ (1//2) = 5 km// h (b) Avereage speed = (distance ) /(time ) = ( 2.5) /(1//2) = 5 km //h` (ii) to ` 50 min` Time distance travelled` = 2.5 + 2. 5 = 5 km ` Total displacment = zero (a) Average velocity = (displacemt )/( time) =0 (b) Average speed = (distace) /(5//6) = 6 km //h` ( iii) ` 0 to 40 min` . Distance moved in ` 10 min` (from ome to market) = 2.5 km`. Distance moved in ` 10 min (from market to home ) with speed ` 7.5 km//h = 7. 5 xx (10)/ (60) = 1. 2 5 km` so displacemnt` = 2.5 - 1. 25 = 1. 25 km` Distanc travelled `= 2. 5 + 1. 25 = 3. 75 km ` ltbRgt (a) Average velocity `= ( 1.250/ (( 40 //60)) = 1.875 km //h` ltbRgt (b) Average speed ` = (3. 75)/ ( ( 40 // 60 )) =5. 625 km //h`. |
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| 1259. |
A stone falls freely under gravity. It covered distances `h_1, h_2` and `h_3` in the first `5` seconds. The next `5` seconds and the next `5` seconds respectively. The relation between `h_1, h_2` and `h_3` is :A. ` h_1 = h_2 =h_23`B. ` h_1 =2 h_2 =3 h_3`C. ` h_1 = h_2/3 = h_3/5`D. ` h_2 = 3 h_1 `amd ` h_3 = 3 h_2` |
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Answer» Correct Answer - A Here, ` h_1 = 1/2 g (5)^2 = (250/2 g ,` ` (h-1 + h_2 ) = 1/2 g ( 5 +5 )^2 = 50 g` `( h_1 + h_2 = h_3 ) =1/2 g ( 5 + 5 =5)^2 = (225)/2 g` Now ` h_2 = (h_1 = h_2) - h_1 = 50 g- (25)/2 g` ` = ( 75)/2 g= 3 h_1` or ` h_1 = _2/3 ` And ` h_3 = ( h_1 +h_2 + h_3 ) - ( h_1 + h_2)` ` = ( 225)/2 g - 50 = (125)/5 g = 5 h_1` or ` h_1 = h_3 /5` Chocie (c ) is correct . |
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| 1260. |
Figure shows some velocity versus time graphs : , , , Only some of these can be realised in practice.These are:A. Figure-(a)B. Figure-(b)C. Figure-(c)D. Figure-(d) |
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Answer» Correct Answer - B, D |
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| 1261. |
The drawing shows velocity (v) versus time (t) graphs for two cyclists moving along the same straight segment of a highway from the same point. The second cyclist starts moving at `t = 3 min`. At what time do the two cysclist meet ? |
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Answer» Acceleration of cyclist ` A, a_1 = ` slope of straight line `OA = (MC)/4` Aceeleration of cyclist ` B, a_2 =` slope of straight ` Line ` DB = ((MC))/((4-3)) = (MC)/1 ` :. ` a_1 /a_1 = ((MC))//4)/((MC)//1) = 1/4` …(i) Let the two cyclist meet after tiem (t), when time is noted from the statrt of cyclist (A). At meeting point, the distatnce coverted by both the cyclist is the same. So, `1/2 a_1 t^2 = 1/2 a_2(t -3)^2` or ` a_1/a_2 = ((tt-3)^2)/t_2 = 1/4` [from (i)]` or ` (t-3)/t = 1/2 or ` 2 t- 6 ` or ` t= 6 min`. |
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| 1262. |
Two trains, each of length ` 100 m`, are running on parallel tracks. One overtakes the other in `20 second ` and one crosses the other in `10 second`. Calculate the velocities of two trains. |
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Answer» Let (u) and (v) be the velocity of two trains (A) and (B). While overtaking, the relatve velcity of train ` A w.r.t. B = -u- v.` Whil crossing, the relative velocity of train ` A w.r.t. B=- + v` . Tpta, dostance tp ne travelled by train ()A while crossing ` = 100 + 100 = 200 m` :. ` 20 = (200)/ ( u-v) ` or ` u- v = 10` and ` 10 = ( 200) /( u+ v) ` or ` u + v= 20` ltbRgt On solving we ger, ltbRgt ` u = 15 ms^(-1)` and ` v= 5 ms^(-1)` . |
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| 1263. |
From an inclined plane a particleis thrown in a direction normal to the surface. Find the ratio of successive ranges of the particle on inclined plane. Consider all collisions as elastic collisions (particle rebounds with the same speed with which it strikes the plane) |
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Answer» Correct Answer - `[1 : 3 : 5]` |
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| 1264. |
Find the constrained relation among the acceleration of blocks A, B and C for the situation shown in figure-1.97. Ratio of radii of step pulley is given as `1:2`. |
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Answer» Correct Answer - `[2a_(A)+a_(C)=4a_(B) (a_(A) uarr; a_(C) larr; a_(B) darr)]` |
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| 1265. |
A particle is projected form a horizontal plane `(x-z` plane) such that its velocity vector at time `t` is gives by `vecV = ahati +(b - ct)hatj`. Its range on the horizontal plane is given byA. `(ba)/c`B. `(2ba)/c`C. `(3ba)/c`D. None |
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Answer» Correct Answer - B `vec(v)=ahat(i)+(b-ct)hat(j)` Time to reach maximum height (when `hat(j)` comp. of velocity becomes zero) `:. B-ct=0rArr t=b/c :.` Time of fight `=(2b)/c` range= horizontal velocity `xx` Time of flight `=axx(2b)/c` |
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| 1266. |
A particle is projected form a horizontal plane `(x-z` plane) such that its velocity vector at time `t` is gives by `vecV = ahati +(b - ct)hatj`. Its range on the horizontal plane is given byA. `(ba)/(c )`B. `(2ba)/(c )`C. `(3ba)/(c )`D. None |
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Answer» Correct Answer - B `{:(vecv=ai+(b-ci)jrArr,,v_(x)=a,v_(y)=b-ct,,):}` `a_(y) = (dv_(y))/(dt) = - c` `R = (2u_(x)u_(y))/(a_(y)) rArr R = (2a.b)/c` |
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| 1267. |
Under what condition, the three vectors (i) cannot giv zero resultant (ii) can give zero resultant ? |
| Answer» (i) When three vectors are not lying in one plance, they can bot produce zero resultant. (ii) When three vectors are lying in a place and are represented in magnitude and direction by the three sides of a triangle taken in the same order, they can prokuce zero resultant. | |
| 1268. |
Can three vectors not in one phane giv e a zero resultant ? Can four vectors do ? |
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Answer» Three vectors which are not in one plane can not givee zero resultant. This is because resultant of two vectors ( in a plane ) line in their plane. If can not balance the third vector which is in a differet plane. The resultant of four coplanar vectors can be zero. if they are represented in magbitude and direction by four sides on a polygon taken in the same order. The resultant of four non-coplanr vectors may be zero.. |
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| 1269. |
What is the angle between (vec A+ vec B) and (vec A xx vec B) ?` |
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Answer» ` (vec A+ vec B) = vec C` (say ) . It lies in the plane of ` vec A and vec B`. (`vec A xx vec B`) = `vec D ` (say). It lies perpendicular to the plane of ` vec A and vec B`. Therefore angle between ` vec Cand vec D` is 90^(@)`. |
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| 1270. |
Can two vectors of different magitudes be combind to give zero resultant ? |
| Answer» No, because the vector sum of two vectors of different magnitudes can vever be zero. | |
| 1271. |
If `vec A xx vec B= vec C xx vec B`, show that `vec C` need not be equal to vec A`. |
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Answer» Given, ` vec A xx vec B= vec C xx vec B` or ` vec A xx vec B- vec C xx vec B =0` it means if ` vec A != vec C`, then either ` vec B=0 or` ` ( vec A - vec C)` is parallel ot ` vec B`. |
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| 1272. |
Two particles are moving with velocities `v_(1)=hati+thatj+hatk` and `v_(2)=thati+thatj+2hatk m//s` respectively. Time at which they are moving perpendicular to each other is.____________(second) |
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Answer» Correct Answer - 2 |
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| 1273. |
The acceleration-time graph of a particle moving along a straight line is as shown in. At what time the particle acquires its initial velocity? . |
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Answer» Correct Answer - 8 |
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| 1274. |
A lift performs the first part of its ascent with uniform acceleration a and the remaining with uniform retardation `2a.` If t is the time of ascent, find the depth of the shaft. |
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Answer» Correct Answer - 3 |
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| 1275. |
The diagram shows the variatioin of `1//v` (where `v` is velocity of the particle) with respect to time. At time `t=3s` using the details given in the graph, find the instantaneous acceleration (in `m//s^(2)`) |
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Answer» Correct Answer - 3 |
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| 1276. |
A small electric car has a maximum constant acceleration of `1m//s^(2)`, a maximum constant deceleration of `2m//s^(2)` and a maximum speed of `20m//s`. The amount of minimum time it would take to drive this car `1km` starting from rest is `(13n)` second. Find value of `n` |
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Answer» Correct Answer - 5 |
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| 1277. |
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. `4sqrt(d/a)`B. `sqrt((2d)/a)`C. `1/2 sqrt(d/a)`D. `2sqrt(d/a)` |
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Answer» Correct Answer - D |
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| 1278. |
The speed of a car as a function of time is shown in figure (3-E1). Find the distance travelled by the car in 8 seconds and its acceleration. |
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Answer» In the interval 8 sec the velocity changes from 0 to 20 m/s. |
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| 1279. |
On a ` 100 km ` track, a train mvoes the first ` 50 km` with a uniform speed of ` 50 km h^(-10` How fast must the train travel the next ` 50 km ` so as to have average speed ` 60 km h^(-1) for the entire trip ? |
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Answer» Here, ` S_1 = 50 km , v_1 = 50 km h^(-10` ` t_1 = S-1 /v_1 = (50)/(50) = 1 h` Let (v_2 km h^(-1)` be the uniform speed of train while voing from ` 50 km` to ` 100 km `, then time taken is ` t_2 (50) /v_2 h` Given, average speed, ` v_(av) = 60 km h^(-1)` , distance travelled `= 100 km . Therefore, total time taken is ` T= (distance travelled )/( average soeed ) = ( 100) /( 60) = 5/3 h` Here, ` t_1 + t-2 = T` ` :. 1 + 9 50) /v_2 = 5/3 ` or ` (50)/v_2 = 5/3 - 1 = 2/` or ` v_2 = ( 50 xx 3) /2` `= 75 km //h` |
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| 1280. |
The ratio of the numerical values of the average velocity and average speed of a body is always.A. Always less than `1`B. `Always ewual to `1`C. Always more than `1`D. Equal to or than `1` |
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Answer» Correct Answer - D We know that average velocity=`(displacement)/(time) and average speed `=(destance)/(time )` Sindnce displacement can be less than of equal to distance, average velocity can be less than or equal to average speed. |
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| 1281. |
`B_(1)`,`B_(2)`, and `B_(3)`, are three balloos ascending with velocities `v`, `2v`, and `3v`, respectively, If a bomb is dropped from each when they are at the same height, then.A. Bomb from `B_(1)` reaches ground firstB. Bomb from `B_(2)` reaches ground firstC. Bomb from `B_(3)` reaches ground firstD. They reach the ground simultaneously |
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Answer» Correct Answer - A Bomb `B_(1)` will have less velpcoty upwards on dropping, so it will reach ground first. |
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| 1282. |
A balloon is ascending vertically with an acceelration of `0.2 ms^(-2)` .Two stones are dropped from it at an interval fo `2 s`, the distance between then when the second stone dropped is (tanke g=9.i8 ma^(-2). |
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Answer» When the first stone is dropped from the balloon, let the initial velocity of the balloon be zero. Theis stone will fall under the efferct of gravity alone. The balloon is ascending with asceond stone placed in balloon after ` 2` second is: ` v=u+ at =0 + 0.2 xx 2 =0 0.4 ms^(-10` Upward distanc ecoverd is ` S= ut + 1/2 at^2 =0 + 1/2 xx 0.2 xx 2^2 =0.4 m` The second stone is dropped from ballon after `2 second`. The initial velocy of this stone is ` 0.4`ms^(-1)` up wards. Taking vertcal downward motion of second stone fro time ` 1.5 second`, we have `u =- 0.4 ms^(-1)` , a= 9.8 ms^(-2)` , S=S_1 , t =1.5 s` `S_1 =- 0.4 xx 1.5 + 1/2 xx 9.8 xx (1.5)^@ = 10.425 m` Taking vertical downward motion of first storne for time ``3.5 seconds` [= (2 + 1.5)` second] ,` we have ` u=0, a = 9.8 ms^(-2), S= S_2 , t =3.5 s` S_2 =0 + 1/2 xx 9.8 xx (3.5 )^2 = 60 .025 m` Separation between the two stones after time ` 1.5 cecond` when second stone is dropped `= S_2- S_1 + S= 60.025 -10. 425 + 0.4 = 50 .0 m |
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| 1283. |
What are the two angles of projection of a projectile projected with velocity of `30 m//s`, so that the horizontal range is ` 45 m`. Take, ` g= 10 m//s^(2)`. |
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Answer» Horizontal range `=(u ^(2) sin 2 theta)/g `45=(30^^(2) sin 2 theta)/(10)` `sin 2 theta =(45 xx 10)/(30 xx 30) =1/2 =sin 30^(@) ` or sin 15^(@)` ` 2 theta =30^(@) or 15^(@) or theta =15^(@) or 75%(@)`. |
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| 1284. |
A car is moving horizontally along a straight line with a unifrom velocity of `25 m s^-1`. A projectile is to be fired from this car in such a way that it will return to it after it has moved `100 m`. The speed of the projection must be.A. `10 m s^-1`B. `20 m s^-1`C. `15 m s^-1`D. `25 m s^-1` |
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Answer» Correct Answer - B (b) `T = (100)/(25) = 4 s rArr (2 u sin theta)/(g) = 4 rArr u sin theta = 20 ms^-1`. |
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| 1285. |
A river is flowing from west to east at a speed of ` 5 metres per minute` . A man on the south bank of the river , capable of swimming at `10 metres per minute`, in still water , wants to swim across the river in the shortest time . He should swim in a directionA. due northB. `30^(@)` east of northC. `30^(@)` west of northD. `60^(@)` east of north |
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Answer» Correct Answer - A To cross the river in shortest time one has to swim perpendicular to the river current. |
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| 1286. |
A particle starts from the origin with a velocity of `10 m s^(-1)` and moves with a constant acceleration till the velocity increases to `50 ms^(-1)`. At that instant, the acceleration is suddenly reversed. What will be the velocity of the particle, when it returne to the starticng point?A. ZeroB. `10ms^(-1)`C. `50 ms^(-1)`D. `70 ms^(-1)` |
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Answer» Correct Answer - D `2ax=(50)^(2)-(10)^(2)` and `2(-a)(-x) =v^(2)-(50)^(2)` This gives `v^(2)-(50)^(2) =(10)^(2)` i.e. `v=70 ms^(-1)`. |
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| 1287. |
A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to `v(x) = beta x^(-2 n)` where `beta` and `n` are constant and `x` is the position of the particle. The acceleration of the particle as a function of `x` is given by.A. (a) ` ` -2 n beta^2 x^(-4n-1)`B. (b) ` `-2 beta^(2)x^(-2n+1)`C. (c ) ` - 2 n beta^2 x^(-4) n+ 1)`D. (d) ` -2 n beta^2 x^(-2n-1)` |
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Answer» Correct Answer - A Here `v = beta x^(-2n) ` ` :. ` (dv)/(dx) =- 2n -1 `, As ` a = (dv)/(dt) = (dv)/(dx) xx (dx)/(dt)` ` or ` a = ( v dv) /(dx) = beta x^(-2n) xx -2 n beta x ^(-2n-1)` ` a=- 2 n beta^2 x^(-4n-1)` . |
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| 1288. |
A particle is rotating in a circle. When it is at point A its speed is V. The speed increases to 2 V by the time the particle moves to B. Find the magnitude of change in velocity of the particle as it travels from A to B. Also, find `vec(V_(A)) vec(DeltaV)`, where `vec(V_(A))` is its velocity at point A and `vec(DeltaV)` is change in velocity as it moves from A to B. |
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Answer» Correct Answer - `sqrt(3)v`, zero |
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| 1289. |
A particle starts from rest moves on a circle with its speed increasing at a constant rate of . Find the angle through which it `0.8 ms^(–2)` would have turned by the time its acceleration becomes `1 ms^(2)`. |
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Answer» Correct Answer - `(3)/(8)` rad |
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| 1290. |
In the arrangement shown in the fig, end A of the string is being pulled with a constant horizontal velocity of `6 m//s`. The block is free to slide on the horizontal surface and all string segments are horizontal. Find the velocity of point P on the thread. |
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Answer» Correct Answer - `2m//s` |
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| 1291. |
In the arrangement shown in the fig, block A is pulled so that it moves horizontally along the line AX with constant velocity u. Block B moves along the incline. Find the time taken by B to reach the pulley P if `u = 1m//s`. The string is inextensible. |
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Answer» Correct Answer - `15.49s` |
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| 1292. |
Two friends A and B are running on a circular track of perimeter equal to 40 m. At time `t = 0` they are at same location running in the same direction. A is running slowly at a uniform speed of `4.5 km//hr` whereas B is running swiftly at a speed of `18 km//hr`. (a) At what time `t_(0)` the two friends will meet again? (b) What is average velocity of A and B for the interval `t = 0` to `t = t_(0)`? |
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Answer» Correct Answer - (a) `t_(0) = (32)/(3)s`; (b) `lt V_(A) gt = lt V_(B) gt = (15sqrt(3))/(8pi) m//s` |
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| 1293. |
Graph of positions (x) vs inverse of velocity `((1)/(v))` for a particle moving on a straight line is as shown. Find the time taken by the particle to move from `x = 3` to `x = 15m`. |
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Answer» Correct Answer - `60s` |
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| 1294. |
A particle is moving along x axis. Its position as a function of time is given by `x = x(t)`. Say whether following statements are true or false. (a) The particle is definitely slowing down if `(d^(2)x)/(dt^(2)) gt 0` and `(dx)/(dt) lt 0` (b) The particle is definitely moving towards the origin if `(d(x^(2)))/(dt) lt 0` |
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Answer» Correct Answer - Both are true |
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| 1295. |
Harshit and Akanksha both can run at speed v and walk at speed `u(u lt v)`. They together start on a journey to a place that is at a distance equal to L. Akanksha walks half of the distance and runs the second half. Harshit walks for half of his travel time and runs in the other half.A. Who winsB. Draw a graph showing the positions of both Harshit and Akanskha versus time.C. Find Akanksha’s average speed for covering distance L.D. How long does it take Harshit to cover the distance? |
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Answer» Correct Answer - A::B::C::D |
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| 1296. |
A particle has co-ordinates (x, y). Its position vector makes on angle `theta` with positive x direction. In an infinitesimally small interval of time the particle moves such that length of its position vector does not change but angle `theta` increases by `d theta`. Express the change in position vector of the particle in terms of x, y, `d theta` and unit vectors `hati` and `hatj`. |
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Answer» Correct Answer - `vec(Delta)r = (-y hati +x hatj) d theta` |
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| 1297. |
Point A moves uniformly with velocity `nu` so that the vector `v` is continually "aimed" at point B which in its turn moves rectilinearly and uniformly with velocity `upsilon lt nu`. At the initial moment of time `v_|_u` and the points are separated by a distance `l`. How soon will the points converge? |
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Answer» Correct Answer - `[(vl)/(v^(2)-u^(2))]` |
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| 1298. |
A particle starts from the origin at ` t=0` with a velocity of ` 10.0 hat j m//s` and moves in the X-y` plane with a constant accleration of ` ( 8.0 hat I + 2. 0 hat j ) ms^(-2) . (a) At wht time is the s-coordinate of the particle ` 16 m` ? What is the y-coordinate of the particle at that time ? ( b) What is the speed of the particle at that time ? |
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Answer» Here, ` vec u= 10.0 hat j ms^(-1) at t=0`. ` vec a = (d vec v)/(dt) = ( 8. 0 hat I + 2.0 hat j) ms^(-2)` So ` d vec v =( 8. 0 hat I + 2.0 hat j) t ` or ` vec v = vec u + 8.0 t hat I + 2.0 t hat j` As` vec v= (d vec r)/(dt) ` or ` d vec r = vec v dt` so, ` d vec r = (vec u + 8.0 t hat + 2.0 t hat j ) dt` Integrating it within the conditions of motion i.e. as time changes from ` 0` to `t`, displacement changes from ` 0` to ` r`, we have : ` vec r = vec u t + 1/2 xx 8.0 t ^2 hat i + 1/2 xx 2.0 t^2 hat j ` or ` z hat i + y hat j = t + 4. 0 t ^2 hat i + t^2 hat j = 4.0 t^2 hat i + (10 t + t^2) hat j` Here, qwe have, ` = 4.0 t^2` and ` y=10 t+ t^2` :. ` t= ( x//4) ^(1//2)` (b) Velocity of the particle at time (t) is ` vec v= 10 hat j 8.0 t +2.0 t hat j` When ` t= 2 s`, then, ` vec v = 10 hat j + 8.0 xx 2 hat i + 2.0 xx 2 hat j = 16 hat i + 14 hat j` Speed` = | vec v| = sqrt ( 16 ^@ + 14^@ ) = 21. 26 ms^(-10`. |
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| 1299. |
A body covers `10 m` in 4 th ` second and ` 15m` in `16 th` second of its uniformly accelerated motion. How far will it travel I the next `3 seconds . |
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Answer» Here, `D_(4) =10 m , D_(6) =15 ,` `S_(9) -S _(6)=? ` Let (u) and (a) be the initial velocity and unitorm acceleration of the body. As `D(n) =u +a/2 (2 n-1) :. D_(4) +a/2 (2 xx 4-1) =10 ` …(i) and `D_(6) =u +a/2 (2 xx 6 -1) =15` .. (i) Subtravting (i) from (iii), we get `2a=5 or a=(5//2) ms^-2)` Form (i)., `u=10 -(5//2)/2 (8-1) =10 -(35)/4 =5/4 ms^(-1)` Distance travelled in `t seconds is `S=ut + 1/2 at^(20` When `t=9s`, then S_(9) =5/4 xx 1/2 xx 5/2 xx 9^(2) =112. 5 m` When `t=6 s`, `S_(6)=5/4 xx 6 +1/2 xx 5/2 xx 6^(2) =52.5 m` :. Distance travelled in the next three seconds `S_(9 ) S_(6) =112.5 -52 =60.0 m`. |
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| 1300. |
The nature of graph drawn between displacement in tth secong and tiem (t) of a uniformly accelerated motion is. |
| Answer» Correct Answer - (b) | |