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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.
| 1401. |
Three particles A, B and C are thrown from the top of a tower with the same speed. A is thrown up, B is thrown down and C is horizontally. They hit the ground with speeds `v_(A),v_(B) and v_(C)` respectively then,A. `v_(A)=v_(B)=v_(C)`B. `v_(A) gt v_(B) gt v_(C)`C. `v_(A)=v_(B) gt v_(C)`D. `v_(A) gt v_(B)=v_(C)` |
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Answer» Correct Answer - A |
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| 1402. |
For four particle A,B,C,D, the velocities of one with respect to other are given as `vecV_(DC)` is `20(m)/(s)` towards north, `vecV_(BC)` is `20(m)/(s)` towards east and `vecV_(BA)` is `20(m)/(s)` towards south. Then `vecV_(DA)` isA. `20m//s` towards northB. `20m//s` towards southC. `20m//s` towards eastD. `20m//s` towards west |
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Answer» Correct Answer - D |
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| 1403. |
A particle is released from rest from a tower of height 3h. The ratio of time intervals for fall of equal height h i.e. `t_(1):t_(2):t_(3)` is :A. `5:3:1`B. `3:2:1`C. `9:4:1`D. `1:(sqrt(2)-1):(sqrt(3):sqrt(2))` |
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Answer» Correct Answer - D |
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| 1404. |
A body is allowed to fall from a height of `10 m`. If the time taken for the first `50 m` is `t_(1)` and for the remaining `50 s`,is `t_(2)`. Which is correct?A. `t_(1)=t_(2)`B. `t_(1) le t_(2)`C. `t_(1) lt t_(2)`D. `t_(1).t_(2)` |
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Answer» Correct Answer - B `s=(1)/(2) g t_(1)^(2)` or `t_(1)^(2) =(50 xx 2)/(g) =(100)/(g)` or `t_(1) =(10)/(sqrtg)` and `100 =(1)/(2) g t^(2)` or `t =(10 sqrt2)/(sqrtg)` `t_(2) =t-t_(1) =(10)/(sqrtg) (sqrt2-1) =0. 4t_(1) t_(1)g t t_(2)`. |
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| 1405. |
A body is allowed to fall from a height of `10 m`. If the time taken for the first `50 m` is `t_(1)` and for the remaining `50 s`,is `t_(2)`. The ratio `t_(1)` and `t_(2)`. Is nearly .A. `5:2`B. `3:1`C. `3:2`D. `5: 3` |
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Answer» Correct Answer - A `t_(1)`: `t_(2) =(10)/(sqrtg) xx (sqrtg)/(10 (sqrt2-1)) =(1)/(0.4) =(10)/(4) =(5)/(2)` . |
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| 1406. |
A body is dropped from a balloon moving up wigh a velocity of `4 m s^(-2) ` when the balloon is at a height of `12.5 m` from the ground. The height of the body after `5 s` from the groumd is `(g=9.8 ms^(-2))`.A. `8 m`B. `12 m`C. `18 m`D. `24 m` |
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Answer» Correct Answer - C `s=ut+(1)/(2)at^(2)=4xx5-(1)/(2)xx9.8xx5^(2)=20-122.5=-102.5 m` This shows that body is `102.5 m` below the initial position, i.e. height of the body `=120.5 -1002.5 =18 m` . |
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| 1407. |
A body is allowed to fall from a height of `10 m`. If the time taken for the first `50 m` is `t_(1)` and for the remaining `50 s`,is `t_(2)`. The ratio of time to reach the ground and to reach first half of the distance is .A. `sqrt3: 1`B. `sqrt 2:1`C. `5:2`D. `1: sqrt3` |
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Answer» Correct Answer - B `t`: `t_(1) =(10sqrt2)/(sqrtg) xx (sqrt g)/(10) =sqrt 2`. |
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| 1408. |
Velocity of a projectile at height `15 m` from ground is `v=(20hati+10hatj)m//s`. Here `hati` is in horizontal direction and `hatj` is vertically upwards. Then Horizontal range of the ground is …………`m`A. 60B. 50C. 80D. 70 |
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Answer» Correct Answer - C |
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| 1409. |
Two boats, A and B, move away from a buoy anchored at the middle of a river along the mutually perpendicular straight lines: the boat A along the river, and the boat B across the river. Having moved off an equal distance from the buoy the boats returned. Find the ratio of times of motion of boats `tau_A//tau_B` if the velocity of each boat with respect to water is `eta=1.2` times greater than the stream velocity. |
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Answer» Correct Answer - [1.8] |
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| 1410. |
A man running on the horizontal road at ` 8 km h^(-1)` find the rain appears to be falling vertically. He incresases his speed to `12 km h^(-1)` and find that the drops make angle ` 30^2` with the vertical. Fin dthe speed and direction of the rain with respedt to the road. |
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Answer» Correct Answer - `[4 sqrt7 kph]` |
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| 1411. |
A particle is projected in `x-y` plane with `y-`axis along vertical, the point of projection being origin. The equation of projectile is `y = sqrt(3) x - (gx^(2))/(2)`. The angle of projectile is ……………..and initial velocity si ………………… . |
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Answer» Correct Answer - `60,2 m//sec` `y = sqrt(3) x - (gx^(2))/(2)` comparing with `y = tan theta.x - (1)/(2) (gx^(2))/(u^(2)cos^(2) theta)` `tan theta = sqrt(3) rArr theta = 60^(@)` `u^(2) cos^(2) theta = 1 rArr u^(2) xx (1)/(4) = 1` `u = 2m//s` |
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| 1412. |
Two persons are pulling a heavy block with the help of horizontal inextensible strings. At the instant shown, the velocities of the two persons are `v_(1)` and `v_(2)` directed along the respective strings with the strings making an angle of `60^(@)` between them. (a) Find the speed of the block at the instant shown. (b) For what ratio of `v_(1)` and `v_(2)` the instantaneous velocity of the block will be along the direction of `v_(1)`. |
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Answer» Correct Answer - (a) `(2)/(sqrt(3)) sqrt(v_(1)^(2) +v_(2)^(2) - v_(1)v_(2))` (b) `(v_(1))/(v_(2)) = 2` |
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| 1413. |
A projectile is thrown from ground at a speed `v_(0)` at an angle `alpha` to the horizontal. Consider point of projection as origin, horizontal direction as X axis and vertically upward as Y axis. Let t be the time when the velocity vector of the projectile becomes perpendicular to its position vector. (a) Write a quadratic equation in t. (b) What is the maximum angle a for which the distance of projectile from the point of projection always keeps on increasing? [Hint: Start from the equation you obtained in part (a)] |
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Answer» Correct Answer - (a) `t^(2) - (3v_(0)sin alpha)/(g) t + (2v_(0)^(2))/(g^(2)) = 0` (b) `sin^(-1).(8)/(9)` |
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| 1414. |
Two friends start bikes from one corner of as quarefield of edge L towards the diagonally opposite corner in the same time t: They both start from the same place and take different routes. One travels along the diagonal with constant acceleration a, and the other accelerates momentarily and then travels along the edge of the field with constant speed v. What is the relationship between a and v ? |
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Answer» Correct Answer - `[a = (v^(2))/(sqrt(2L)) ]` |
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| 1415. |
A food package was dropped from an aircraft flying horizontally. 6 s before it hit the ground, it was at a height of `780 m`, and had travelled a distance of 1 km horizontally. Find the speed and the altitude of the aircraft. |
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Answer» Correct Answer - `360 km//hr, 1.28 km` `780=u sin theta xx6+1/2xxgxx36` `780-180=u sin theta xx6` using `theta=600/6=100 m//sec` i.e. food package dropped before 10 secs `1000=uxx10 rArr u=100 m//s` `:. H=(gxx(16)^(2))/2=1280 m`. |
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| 1416. |
A relief food package is dropped from a airplane which is moving horizontal with a velocity of `30 m s^-1` at a height `h = 50 m`. Find the (a) time of flight of the package, (b) location of the point of striking of the food package, (c ) velocity of the package at the time of striking the ground, and (d) displacement of the food package. . |
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Answer» `u_x = 30 m s^-1, u_y = 0 ms^-1, and h = 50 m`. `h = u_y t + (1)/(2) g t^2` `rArr (50 xx 2)/(10) = t^2 rArr t = sqrt(10) s` Distance (horizontally) covered in `t = sqrt(10)) s` `x = u_x t = 30 sqrt(10)) m` `v_x = u_x = 30 m s^-1` `v_y = u_y + "gt" = 10 sqrt(10)) m s^-1` `v = sqrt(900 + 1000) = 10 sqrt(19) m s^-1` Displacement `= sqrt(h^2 + x^2)) = sqrt(2500 + 9000)) = 10 sqrt(115)) m`. |
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| 1417. |
A projectile is thrown with speed u making angle `theta` with horizontal at `t=0`. It just crosses the two points at equal height at time `t=1` s and `t=3` sec respectively. Calculate maximum height attained by it. `(g=10 m//s^(2))` |
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Answer» Correct Answer - 20 m `u sin thetaxx1-1/2 g(1)^(2)=u sin thetaxx3-1/2xxgxx(3)^(2)` `2u sin theta=40 rArr u sin theta=20 m//s` Max. height `=(u^(2)sin^(2) theta)/(2g)=(20xx20)/2-=20 m` |
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| 1418. |
A truck is moving with a constant velocity of `54 km h^-1`. In which direction (angle with the direction of motion of truck) should a stone be projected up with a velocity of `20 ms^-1`, from the floor of the truck of the truck, so as to appear at right angles to the truck, for a person standing on earth ? |
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Answer» Let the stone be projected at an angle `prop` to the direction of motion of truck, with a speed of `20 m s^-1`. Since the resultant displacement along horizontal is zero, the velocity along horizontal = 0 `15 + 20 cos prop = 0` `cos prop = -(3)/(4) rArr prop = cos^-1 ((-3)/(4))`. |
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| 1419. |
How long will a plane take to fly around a square with side a with the wind blowing at a velocity u, in the two cases (a) the direction of the wind coincides with one of the sides (b) the direction of the wind coincides with one diagonal of the square. The velocity of the plane in still air is `v gt u`.. |
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Answer» Correct Answer - [(a) `(2a(v+sqrt(v^(2)-u^(2))))/(v^(2)-u^(2))`, (b) `(4asqrt((v^(2)-u^(2)//2)))/(v^(2)-u^(2))`] |
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| 1420. |
Time taken by the particle to reach from `A` to `B` is `t`. Then the distance `AB` is equal to A. `(ut)/(sqrt(3))`B. `(sqrt(3)ut)/2`C. `sqrt(3)ut`D. `2 ut` |
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Answer» Correct Answer - A |
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| 1421. |
A body is projected with velocity `u` at an angle of projection `theta` with the horizontal. The direction of velocity of the body makes angle `30^@` with the horizontal at `t = 2 s` and then after `1 s` it reaches the maximum height. ThenA. `u = 20 sqrt(3) m s^-1`,B. `theta = 60^@`C. `theta = 30^@`D. `u = 10 sqrt(3) m s^-1` |
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Answer» Correct Answer - A::B (a.,b.) Time of ascent `= 2 + 1 = 3 s` `rArr (u sin theta)/(g) = 3 rArr u sin theta = 30` and `tan beta = (u sin theta - "gt")/(u cos theta) rArr tan 30^@ = (30 - 10 xx 2)/(u cos theta)` `rArr u cos theta = 10 sqrt(3))` From here `u = sqrt((10 sqrt(3))^2 + 10^2) =20 sqrt(3) ms^-1` and `tan theta = (30)/(10 sqrt(3)) = sqrt(3)) rArr theta = 60^@`. |
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| 1422. |
A particle is projected at point A from an inclination plane with inclination angle `theta` as shown in figure. The magnitude of projection velocity is `vecu` and its direction is perpendicular to the plane. After some time it passes from point B which is in the same horizontal level of A, with velocity `vecv`. Then the angle between `vecu and vecv` will be |
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Answer» Correct Answer - `[u = [(gR(1+3sin^(2)beta))/(2sin beta)]^(1//2)]` |
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| 1423. |
Two lines AB and CD intersect at O at an inclination `alpha`, as shown in figure. If they move out parallel to themselves with the speed v, find the speed of O. |
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Answer» Correct Answer - `[v cosec (alpha)/(2)]` |
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| 1424. |
Asserion: Magnitude of the resultant of two vectors may be less than the magnitude of either vector. Reason: The resultant of two vectors is obtained by means of law of parallelogram of Vectors.A. (a) Statement-1 is true , Statement-2 is true , Statvement -2 is correct explanation of Statement-1 .B. (b) Statement-1 is true , statement -2 is true , statement -2 is not coerrecrt explanation of Statement-1.C. (C ) Statement-1 is true , Statement-2 is false.D. (d) Statement-1 is false , Statement-2 is true. |
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Answer» Correct Answer - C The magnitude of the resultant vector of two given vectors can be less thean the magnitude of individual vectors if the angle between two vectors is in between ` 90^@` to ` 270^@`. |
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| 1425. |
A car travles a distance `A` to `B` at a speed of `40km//h` and returns to `A` at a speed of `30km//h`. (i) What is the average speed of the whole journey? (ii) What is the averege velocity? |
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Answer» (i) Let `AB = s, `time takemn to go form `A` to `B`, `t = (s)/(40)h` and time taken to go form `B` to `A,t_(2) = (s)/(30)h` `:.` total time taken `=` `t_(1) +t_(2) = (S)/(40) +(s)/(30) = ((3+4)s)/(120) = (7s)/(120)h` Total distance travelled `=s +s = 2s` `:.` "Average speed" `= ("total distance travelled")/("total time taken")` `= (2s)/(7s//120) = (120xx2)/(7) =34.3km//h` (ii) Total displacement `=` zero, since the car returns to the original position. Therefore, average velocity `= ("total displacement")/("time taken") = (0)/(2t) = 0` |
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| 1426. |
The distance travelled by a particles `S = 10t^(2) (m)`. Find the value of instantaneous speed at `t = 2` sec. |
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Answer» `V = (dx)/(dt) = (d)/(dt) (10t^(2)) = 10(2t) = 20t` put `t = 2sec. V = 20xx 2 = 40m//s` |
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| 1427. |
In a village Shyam bats for hitting two points A and B on a staircase with his goli from the position P shown in figure. Find the velocities required for P to hit A and B. |
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Answer» Correct Answer - [4.65 m/s, 10.6 m/s] |
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| 1428. |
Two motor cars start from A simultaneously & reach B after 2 hour. The first car travelled half the distance at a speed of `v_(1)=30 km hr^(-1)` & the other half at a speed of `v_(2)=60 km hr^(-1)`. The second car covered the entire with a constant acceleration. At what instant of time, were the speeds of both the vehicles same? Will one of them overtake the other enroute? |
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Answer» Correct Answer - [0.75 hr, 1.5 hr, no overtaking] |
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| 1429. |
Every 10 mintues, two cars start simultaneously from two point `A and B` which are 60 km apart. The cars travel towards each other at 60 kmph. Find graphiclly how many cars will pass by a passenger in one of the cars between `A and B`. |
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Answer» Correct Answer - 11 |
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| 1430. |
A car is moving at a constant speed of 40 km/h along a straight road which heads towards a large vertical wall and makes a sharp `90^0` turn by the side of the wall. A fly flyingat constant speed of 100 km/h, starts from the wall towrds the car t an instant when the car is 20 km away,flies until it reaches the glasspane of the car and returns to teh wll at teh same speed. It continues to fly between the car and teh wall time the car makes the `90^0` turn. a. What is the total distance the fly has travelled during the period?b. How many trips has it made between the car and the wall? |
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Answer» Correct Answer - `[50 km, oo]` |
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| 1431. |
A wall OP is inclined to the horizontal ground at an angle `alpha`. Two particles are projected from points A and B on the ground with same speed (u) in directions making an angle `theta` to the horizontal (see figure). Distance between points A and B is `x_(0) = 24 m`. Both particles hit the wall elastically and fall back on the ground. Time of flight (time required to hit the wall and then fall back on to the ground) for particles projected from A and B are 4 s and 2 s respectively. Both the particles strike the wall perpendicularly and at the same location. [In elastic collision, the velocity component of the particle that is perpendicular to the wall gets reversed without change in magnitude] (a) Calculate maximum height attained by the particle projected from A. (b) Calculate the inclination of the wall to the horizontal `(alpha) [g =10 m//s^(2)]` |
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Answer» Correct Answer - (a) `11.25 m` (b) `tan^(-) ((8)/(5))` |
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| 1432. |
From an inclined palne two particles `P,Q` are projected with same speed at same angle `theta`,one up and other down the plane as shown in figure.Which of the following statement(s) is/are correct ? A. The particles will collide the plane with same speedB. The times of flight of each particle are sameC. Both particles strike the plane perpendicularlyD. The particles will collide in mid air if projected simultaneously and time of flight of each particle is less than the time of collision |
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Answer» Correct Answer - B::D |
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| 1433. |
Two paper screens A and B are separated by 150m. A bullet pierces A and B. The hole in B in 15 cm below the hole in A. If the bullet is travelling horizontally at the time of hitting A, then the velocity of the bullet at A is: `(g=10ms^(-2))` |
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Answer» Equation of motion in x- direction, `100 = v xx t` `rArr t = (100)/(v)`…(i) In y - direction, `0.1 = 1//2 xx 10 xx t^2` …(ii) `0.1 = 1//2 xx 10 xx (100//v)^2` From Eqs (i) and (ii), we get `u = 700 m s^-1`. |
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| 1434. |
Two bodies 1 and 2 of different shapes are released on the surface of a deep pond. The mass of the two bodies are `m_(1) = 1 kg` and `m_(2) = 1.2 kg` respectively. While moving through water, the bodies experience resistive force given as `R = bv`, where v is speed of the body and b is a positive constant dependent on shape of the body. For bodies 1 and 2 value of b is `2.5 kg//s` and `3.0 kg//s` respectively. Neglect all other forces apart from gravity and the resistive force, while answering following questions : [Hint : acceleration `= "force"//"mass"]` (i) With what speed `v_(10)` and `v_(20)` will the two bodies hit the bed of the pond. [Take `g = 10 m//s^(2)]` (ii) Which body will acquire speed equal to half the terminal speed in less time. |
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Answer» Correct Answer - (i) `v_(10) = v_(20) = 4m//s` (ii) Both will take same time |
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| 1435. |
An aeroplane takes off at an angle or `30^(@)` to the horizontal. If the component of its velocity along the horizontal is `240 km h^(-1)`. What is the actual velocity. ? Also find the vertical component of its velocity ? |
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Answer» Let (v) be the actual velocity of aroplane while taking off. As per question ` u cos 30^(@) =240` or 1 u= (240)/( cos 30^(@) =9240)/(aqrt 3//2) =(480)/(sqrt 3) = (480 sqrt)/3` `= 160 sqrt 3 km h^(-1)` Vertical component velocity of aeroplace ` =u sin 30^(@) = 160 sqrt 3 xx 1//2` `=80 sqrt 3 km h^(-1)`. |
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| 1436. |
The position verus time graph time graph for a certain particle moving along the x-axis is shown in . Find the average velocity in the time intervals (a) `0` to `2s,` (b) `2 s` to `4 s`, and (c) `4,s` to `7 s`, . |
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Answer» We are given the values of time, but the values of the position are to obtained from the graph correspondin to the given time interval under consideration. We know that the sope of the displacement -time graph represents velocitt. Formula to be used: `v_(av)=(x_(2)-x_(1))/(t_(2)-t_(t))`, where `x_(1)` and `x_(2)` are the initial and final values of time, respitions, respectively, and `t_(1)` and `t_(2)` are the initial and final values of time, frspectively. a. Here `x_(1)=10 m, x_(2)=5 m,t_(1)=2,.t_(2)=4 s`. Therfore, the requited average velocity is giben by `v_(ab)(10-0)/(2-0)= m s^(-1)` b. Here `x_(1)=10 m,x_(2)=5 m,t_(1)=2 s, t_(2)=4 s`. So the repuired value of average velocity is given by `v_(av)=(5-10)/(4-2)=-2.5 m s^(-1)` c. Here `x_(1)=5 m, x_(2)=-5 m, t_(1)=4 s, t_(2)=7 s`. So the repuired value of average velocity is given by `v_(av)=(-5-5)/(7-4)=(-10)/(3)=-3.3 ms^(-1)`. |
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| 1437. |
One Kilometer is equal to ______ metre. |
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Answer» Correct Answer - 1000 One kilometer is equal to 1000 meter. |
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| 1438. |
The distance- time graph of a body is as shown in the figure. The part of the graph that represent the unifrom speed of the body is A. OAB. ABC. BCD. Both OA and BC |
| Answer» Correct Answer - D | |
| 1439. |
One microsecond is________ of a second. |
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Answer» Correct Answer - ` (1/10^(6))` One microsecond is one millionth of a second. |
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| 1440. |
A body moves with a unifrom speed of `10 km h^(-1)` for 2h. The average speed of the body is ______ km `h^(-1)`A. 10B. 20C. 5D. 25 |
| Answer» Correct Answer - A | |
| 1441. |
A body is said to be moving with unifrom speed if it covers equal distance in ________ |
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Answer» Correct Answer - equal intervals of time A body is said to move with uniform speed if it covers equal distances in equal intervals of time. |
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| 1442. |
The distance - time graph of a moving vehicle is as shown in the figure. A. the speed of the vehicle is increasing with time.B. the speed of the vehicle is decreasing with time.C. the final speed of the vehicle os zero.D. Graph is not possible. |
| Answer» Correct Answer - D | |
| 1443. |
__________ measures the distance moved by a vehicle. |
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Answer» Correct Answer - Odometer Odomter measures the distance travelled by a vehicle. |
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| 1444. |
Distance versus time graph of an object is as shown in the figure. The average speed of the object is _____ `ms^(-1)` A. 0.08B. `0.5`C. 1D. 2 |
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Answer» Correct Answer - A Average velocity = ` (" total displacement") / ( " total time") = 20/ (4 xx 60)` ` = 0.08 m s^(-1)` |
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| 1445. |
The distance-time graph of the motion of an object moving with a contant speed is a straight line. |
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Answer» Correct Answer - True The distance- time graph of the motion of an object moving with a constant speed is a straight line. |
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| 1446. |
A body moves with a contant speed of `10 m s^(-1)` for 1 hour. The distance travelled by the body is _____A. 36,000 kmB. 36 kmC. 24 kmD. ` 24 xx 10^(3) m` |
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Answer» Correct Answer - B Speed = `( " distance")/(" time")` ,distance= speed `xx` time ` 10 xx 60 xx 60 = 36000 = 36 xx 10^(3)` m 36 km |
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| 1447. |
A body projected upwards from the top of a builing is an example of ______A. uniform motionB. non-uniform motionC. periodic motionD. oscillatory motion |
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Answer» Correct Answer - B A body projected upwards from the top of a building is an example of a non - uniform motion. |
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| 1448. |
The time taken by the bob of a simple pendulum of time period (T) to move from one exrtreme positon to other extreme position is equal to _____ |
| Answer» Correct Answer - `T/2` | |
| 1449. |
It takes 1 s for the bob to moves from one extreme position to other extreme postion. Then the time period of the pendulum is ______ s.A. 1B. 2hC. 3D. 4 |
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Answer» Correct Answer - B Time period = time taken for one oscillation = 2s |
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| 1450. |
An object is moving along the `x`axis with postion as a function of time given by `x =x(t)`. Point `O` is at `x = 0`. Object is definitely moving towardr `O` whenA. `dx//dt lt0`B. `dx//dt gt 0`C. `dx^(2)//dt^(2) lt 0`D. `dx^(2)//dt^(2) gt 0` |
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Answer» Correct Answer - C For particle to move towards orgin `x` and `v` must be of opposite sign. `(d(X^(2)))/(dt) = 2x (dx)/(dt) lt 0` |
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