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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.
| 301. |
Which of the following represents a unit vector ?A. `(|vecA|)/A`B. `vecA/(|vecA|)`C. `vecA/vecA`D. `(|vecA|)/(|vecA|)` |
| Answer» Correct Answer - B | |
| 302. |
The component of vector `vecA = 2hati + 3 hatj` along the direction of `(hati - hatj)` isA. `(1)/(sqrt(2))`B. `-(1)/(sqrt(2))`C. `(1)/(2)`D. `-(1)/(2)` |
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Answer» Correct Answer - B Given : `vecA = 2hati + 3hatj , vecB = hati - hatj` (say) Component of `vecA` along the direction of `vecB` `= (vecA*vecB)/(B) = ((2hati + 3hatj)*(hati -hatj))/(sqrt(2)) = - (1)/(sqrt(2))` |
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| 303. |
`vecA + vecB` can also be written asA. `vecA - vecB`B. `vecB -vecA`C. `vecB + vecA`D. ` vecB .vecA` |
| Answer» Correct Answer - C | |
| 304. |
The vector quantily among the following isA. massB. TimeC. DistanceD. Displacement |
| Answer» Correct Answer - D | |
| 305. |
If `hatn` is a unit vector in the direction of the vector `vecA`, them :A. `hatn=(vecA)/(|vecA|)`B. `hatn = (|vecA|)/(vecA)`C. `hatn = |vecA|vecA`D. `hatn = vecA` |
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Answer» Correct Answer - A `hatn = (vecA)/(|vecA|)` |
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| 306. |
A boat is moving with a velocity `v_(bw)=5 km//hr` relative to water. At time `t=0`.the boat passes through a piece of cork floating in water while moving down stream.If it turns back at time `t_(1)=30 min`. a) when the boat meet the cork again? b) The distance travelled by the boat during this time. |
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Answer» Let `AB=d`is the distance travelled by boat along down stream in `t_(1)` sec and it returns back and it meets the cork at point `C` after `t_(2)` sec. `:.` Let `AC=x` is the distance travelled by the cork during `(t_(1)+t_(2))` sec. `d=(V_(B)+V_(W))t_(1)`....(1) `d-x=(V_(B)-V_(W))t_(2)`....(2) `x=V_(W)(t_(1)+t_(2))`....(3) Substitute (1) and (3) in (2) we get `t_(1)=t_(2)` `:.` The boat meets the cork again after `T=2t_(1)=60 min` and the distance `(AB+BC)` travelled by the boat before meets the cork is `D=2d-x` `D=2(V_(B)+V_(W))t_(1)-V_(w)2t_(1)` `D=2V_(B)t^(1)+2V_(w)t_(1)-2V_(w)t_(1)` `D=2V_(B)t_(1)=2xx5xx30/60=5km` |
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| 307. |
A particle is projected from ground with some initial velocity making an angle of `45^(@)` with the horizontal. It reaches a height of `7.5 m` above the ground while it travels a horizontal distance of `10 m` from the point of projection. The initial speed of the projection isA. `5 m//s`B. `10 m//s`C. `20 m//s`D. `40 m//s` |
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Answer» Correct Answer - C `y=x tan theta-g/(2u^(2) cos^(2)theta)x^(2)` |
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| 308. |
In the question number 62, the distance from the thrower to the point where the ball returns to the same level isA. 58 mB. 68 mC. 78 mD. 88 m |
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Answer» Correct Answer - C The dsitance from the thrower to the point where the ball returns the same leve is `R=(u^(2)sin2theta)/(g)=((30ms^(-1))^(2)(sin60^(@)))/(10ms^(-2))~~78` |
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| 309. |
The maximum height attained by a ball projected with speed ` 20 ms^(-1)` at an angle `45^(@)` with the horizontal is [take g = ` 10 ms^(-2)`]A. 40 mB. 20 mC. 10 mD. 30 m |
| Answer» Correct Answer - C | |
| 310. |
In the question number 62. the maximum height attained by the ball isA. 11.25 mB. 48.2 mC. 23. 5 mD. 68 m |
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Answer» Correct Answer - A The maximum height is given by `H=(u^(2)sin^(2)theta)/(2g)=((30)^(2)sin^(2)30^(@))/(2xx10)=11.25m` |
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| 311. |
Assertion: For motion in two or three diemensions, velocity and acceleration vecotrs must have any angle between `0^(@) and 90^(@)` between them. Reason: For such motion velocity and acceleration of an object is always in the opposite direction. |
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Answer» Correct Answer - D In one dimension, the velocity and the acceleration of an object are always along the same straight line either in the same direction or in the opposite direction. However, for motion in two or three dimensions velocity and acceleration vectors may have any angle from `0^(@)` to `180^(@)` between tehm. |
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| 312. |
(i) What does `|(dv)/(dt)|` and `(d|V|)/(dt)` represent ? (ii) Can these be equal ? (iii) Can `(d|V)/(dt)= 0` while `|(dV)/(dt)ne0` ? (iv) Can `(d|V|)/(dt)ne 0` while `|(dv)/(dt)|=0` ? |
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Answer» (i) `|(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`. (ii) These two are equal in case of one dimensional motion (without change in direction). (iii) In case of uniform circular motion speed remains constant while velocity changes. Hence, `(d|V|)/(dt)=0` while `|(dv)/(dt)|ne0`. (iv) `(d|V|)/(dt)ne 0` implies that speed of particle is not constant. Velocity cannot remain constant if speed is changing. Hence, `|(dV)/(dt)|` cannot be zero in this case. So, it is not possible to be have `|(dv)/(dt)|=0` while `(d|V|)/(dt)ne 0` |
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| 313. |
Match the following columns. `|{:(,"Column I",,"Column II"),((A),"d v/dt",(p),"Acceleration"),((B),"d|v|/dt",(q),"Magnitude of acceleration"),((C ),(dr)/(dt),(r ),"Velocity"),((D),|(dr)/(dt)|,(s),"Magnitude of velocity"),(,,(t),"Rate of change of speed"):}|` |
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Answer» Correct Answer - A::B::C::D `(d|v|)/(dt)` is rate of change of speed of the particle. `(d|r|)/(dt)` in the rate by which distance of particle from the origin is changing. |
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| 314. |
The pathe of a projectile isA. CircularB. parabolicC. LinearD. Hyperbolic |
| Answer» Correct Answer - B | |
| 315. |
Two cyclists cycle along circular tracks of radii `R_(1) and R_(2)` at uniform rates. If both of them take same time to complete one revolution, then their angualr speeds are in the ratioA. `R_(1) :R_(2)`B. `R_(2) : R_(1)`C. `1:1`D. `R_(1)R_(2):1` |
| Answer» Correct Answer - B | |
| 316. |
The maximum number of rectangualr components in which a vector can be resolved in a plane, isA. InfiniteB. FourC. TwoD. One |
| Answer» Correct Answer - C | |
| 317. |
Which of the following is not a projectile ?A. An aircraft taking offB. A bullet fired from a rifleC. A ball thrown horizontally from a roofD. A football kicked by a player |
| Answer» Correct Answer - A | |
| 318. |
What is the angle between velocity vector and acceleration vector in unitorm circular motion ?A. `0^(@)`B. `180^(@)`C. `90^(@)`D. `45^(@)` |
| Answer» Correct Answer - C | |
| 319. |
if ` vecP a+ vecQ =vec0` , then which of the following is necessarily true ?A. `vecP =vec0`B. `vecP =- vecQ`C. `vecQ = 0`D. `vecP = vecQ` |
| Answer» Correct Answer - A | |
| 320. |
The resultant of ` vecp and vecq` makes angle `alpha "with " vecp and beta " with " vecq` . ThenA. `alpha lt beta`B. `alpha lt beta, p lt q`C. `alpha lt beta, " if " p gt q`D. `alpha lt beta, " if " p = q ` |
| Answer» Correct Answer - 3 | |
| 321. |
Two bodies are projected at angles `30^(@)` and `60^(@)` to the horizontal from the ground such that the maximum heights reached by them are equal then a) Their times of flight are equal b) Their horizontal ranges are equal c) The ratio of their initial speeds of projection is `sqrt3:1` d) Both take same time to reach the maximum height.A. `a,b,c` and `d` are correctB. only `a,b` and `c` are correctC. only `a` and `c` are correctD. only `a,c` and `d` are correct |
| Answer» Correct Answer - D | |
| 322. |
Two particls are projected in air with speed u at angles `theta_(1) and theta_(2)` (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then which one of the following is correct? where `T_(1) and T_(2)` are the time of flight.A. `theta_(1) gt theta_(2)`B. `theta_(1) = theta_(2)`C. `T_(1) lt T_(2)`D. `T_(1) = T_(2)` |
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Answer» Correct Answer - A Height, `H = (u^(2) sin^(2) theta)/(2g)` For the same speed, `H prop sin^(2)theta` `because H_(1) gt H_(2)` (Given) `therefore sin^(2) theta_(1) gt sintheta_(2) rArr T_(1) gt T_(2)` |
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| 323. |
Two particles are projected in air with speed `v_(0)` at angles `theta_(1)` and `theta_(2)` (both acute) to the horizontal,respectively.If the height reached by the first particle greater than that of the second,then thick the right choicesA. angle of projection:`theta_(1)gt theta_(2)`B. time of flight:`T_(2) gt T_(1)`C. horizontal range:`R_(1) gt R_(2)`D. total energy : `U_(1) gt U_(2)` |
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Answer» Correct Answer - A::B::C `H_(1)gtH_(2)` `sin theta_(1)-sin theta_(2)` or `theta_(1)gttheta_(2)` `T_(1)gtT_(2)` `R_(1)/R_(2)=(sin 2theta_(1))/(sin 2theta_(2))gt1` `U_(1)=KE+PE=1/2m_(1)v_(0)^(2)` `U_(2)=KE+PE=1/2m_(2)v_(0)^(2)` If `m_(1)=m_(2)` then `U_(1)=U_(2)` `m_(1)gtm_(2)` then `U_(1)gtU_(2)` `m_(1)ltm_(2)` then `U_(1)ltU_(2)` |
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| 324. |
Two particles are projected simultaneously in the same vertical plane, from the with speed `u_(1)` and `u_(2)` at angle of projection `theta_(1)` and `theta_(2)` respectively with the horizontal. The path followed by one, as seen by other (as long as both are in flight), isA. an inclined straight lineB. a vertical straight line if `u_(1) cos theta_(1) = u_cos theta_(2)`C. both (1), (2)D. none |
| Answer» Correct Answer - C | |
| 325. |
Three forces start acting simultaneously on a particle moving the velocity `vecV`. The forces are represented in magnitude and direction by the three sides of a triangle `ABC` (as shown). The particle will now move with velocity A. less than `vec V`B. greater than `vec V`C. `|vecV|` in the direction of largest forceD. `vecV` remaining unchanged |
| Answer» Correct Answer - D | |
| 326. |
Two particles P and Q simultaneously start moving from point A with velocities `15 m//s` and `20 m//s` respectively. The two particles move with acceleration equal in magnitude but opposite in direction. When P overtakes Q at point B then its velocity is `30 m//s`, the velocity of Q at point B will beA. `30 ms^(-1)`B. `5 ms^(-1)`C. `20 ms^(-1)`D. `15 ms^(-1)` |
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Answer» Correct Answer - B For `P, 30=15+at` or `at=15ms^(-1)` For `Q, upsilon = 20 -at` or `upsilon = 20-15=5 ms^(-1)` |
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| 327. |
The slopes of wind screen of two cars are `alpha_1=30^@ and alpha_2= 15^@` respectively. At what ratio `v_1/ v_2` of the velocities of the cars will their drivers see the hail stones bounced back by the wind screen on their cars in vertical direction? Assume hail stones fall vertically downwards and collisions to be elastic. |
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Answer» Correct Answer - C `v_("absolute")` in vertically downward `v_(Hc)` after collision vertically upwards since collision is elastic so velocity of tailstones w.r.t car before and after collision will make equal angles. `vecv_(Hc//1)=vecv_(H)-v_(c)=vecv-vecv_(1),beta+90-2beta+alpha_(1)=90` `a_(1)=beta.2beta=2alpha_(1)tan 2beta=tan 2alpha_(1)=v_(1)/(v)` |
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| 328. |
A car travelling with a velocity of 80 km/h slowed down to 44 km/h in 15 s. The retardation isA. `0.67 ms^(-2)`B. `1 ms^(-2)`C. `1.25 ms^(-2)`D. `1.5 ms^(-2)` |
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Answer» Correct Answer - A `a=("Chamge in velocity")/("Time taken")=(upsilon_(f)-upsilon_(i))/(t)` ltrgt `=((44xx(5)/(18))-(80xx(5)/(18)))/(15)=-0.67m//s^(2)` Negative sign represents the retardation. |
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| 329. |
A hiker stands on the edge of a cliff `490 m` above the ground and throwns a stone horiozontally with an initial speed of `15ms^(-1)` neglecting air resistance.The time taken by the stone to reach the ground in seconds is `(g=9.8ms^(2))`A. 5B. 10C. 1D. 5 |
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Answer» Correct Answer - B `t=sqrt((2h)/g)` |
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| 330. |
A boat stems in a river with velocity ` 2hati + hatj` with respecitve to the ground. The river water flows with a velocity ` -3hati - 4hatj` with respect to the ground. What is the relative velocity of boat w.r.t river water ? |
| Answer» Correct Answer - `5hati + 5hatj` | |
| 331. |
A hiker stands on the edge of a cliff `490 m` above the ground and throwns a stone horiozontally with an initial speed of `15ms^(-1)` neglecting air resistance.The speed with which it hits the ground in `ms^(-1)` is `(g=9.8ms^(2))`A. 9.8B. 99C. 4.9D. 49 |
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Answer» Correct Answer - B `V=sqrt(V_(x)^(2)+V_(y)^(2)),V_(y)^(2)=2gh` |
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| 332. |
The velociies of A and B are `vecv_(A)= 2hati + 4hatj and vecv_(B) = 3hati - 7hatj` , velocity of B as observed by A isA. `5hati-3hatj`B. `hati - 11 hatj`C. `-hati + 11hatj`D. `2hati - 3hatj` |
| Answer» Correct Answer - B | |
| 333. |
Two particles having position vectors `vecr_(1)=(3veci+5 vecj)m` and `vecr_(2)=(-5veci+3 vecj)m` are moving with velocities `vecV_(1)=(4hati-4hatj)ms^(-1)` and `vecV_(2)=(1hati-3hatj)ms^(-1)`. If they collide after `2 ` seconds, the value of `a` isA. `2`B. `4`C. `6`D. `8` |
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Answer» Correct Answer - C `vecr_(1)+vecv_(1)t=vecr_(2)+vecv_(2)t` |
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| 334. |
Suppose that two objects A and B are moving with velocities `vecv_(A)` and `vecv_(B)` (each with respect to some common frame of refrence). Let `vecv_(AB)` represent the velocity of with respect to B. ThenA. `vecv_(AB) + vecv_(BA) = 0`B. `vecv_(AB) - vecv_(BA) = 0`C. `vecv_(AB) = vecv_(A) + vecv_(B)`D. `|vecv_(AB)| ne |vecv_(BA)|` |
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Answer» Correct Answer - A Velocity of object A relative to that of B is `vecv_(AB) = vecv_(A) - vecv_(B)` Velocity of object B relative to that of A is `vecv_(BA) = vecv_(B) - vecv_(A)` `therefore vecv_(AB) = -vecv_(BA) and |vecv_(AB)| = |vecv_(BA)|` |
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| 335. |
From the top of a 490 m high cliff, a boy throws a stone horizontally with a initial speed of ` 15 ms^(-1)`. What is the time taken by the stone to reach the ground. |
| Answer» Correct Answer - t=10s | |
| 336. |
An object is projected with velocity `vecv_(0) = 15hatj + 20hatj` . Considering x along horizontal axis and y along vertical axis. Find its velocity after 2s. |
| Answer» Correct Answer - `15 m s^(-1) ` along horizontal . | |
| 337. |
Find the ratio of maximum horizontal range and the maximum height attained by the projectile. i.e. for ` theta_(0) = 45^(@)` |
| Answer» Correct Answer - `4:1` | |
| 338. |
A player kicks a football obliquely at a speed of `20 m//s` so that its range is maximum. Another player at a distance of `24m` away in the direction of kick starts running at that instant to catch it, the speed with which the second player has to run is `(g=10 ms^(-2))`A. `4 m//s`B. `4sqrt2 m//s`C. `8sqrt2 m//s`D. `8 m//s` |
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Answer» Correct Answer - B `R_(max)=u^(2)/g` , distance `=R_(max)-x` `T=(2usintheta)/g` , Distance `=VxxT` |
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| 339. |
A player kicks a ball at an angle of `37^(@)` to the horizontal with an initial speed of ` 15 ms^(-1)` . Find its time of fight. |
| Answer» Correct Answer - 1.84 s | |
| 340. |
If the helicopter is flying at constant velocity then where will it be when the packet strikes the ground. |
| Answer» Correct Answer - Vertically above the packet | |
| 341. |
If `|vecA + vecB| = |vecA - vecB|`, then the angle between `vecA and vecB` will beA. `30^(@)`B. `45^(@)`C. `(60)^(@)`D. `90^(@)` |
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Answer» Correct Answer - D Let `theta` be angle between the vectors `vecA and vecB`, Then `|vecA + vecB| = sqrt(A^(2) + B^(2) + 2AB cos theta)` `|vecA - vecB| = sqrt(A^(2) + B^(2) - 2AB cos theta)` According to given problem `|veA + vecB| = |vecA - vecB|` `therefore sqrt(A^(2) + B^(2) + 2AB cos theta) = sqrt(A^(2) + B^(2) - 2AB cos theta)` Squaring both sides, we get `A^(2) + B^(2) + 2AB cos theta = A^(2) + B^(2) - 2AB cos theta` `therefore 4AB cos theta = 0` As `A ne 0 , B ne 0 therefore cos theta = 0 or theta = 90^(@)` |
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| 342. |
Assertion : If velocity - time equation of a particle moving in a straight line is quadratic in time, then displacement - time equation cannot be linear. Reason : If displacement - time is quadratic in time, then velocity - time is linear.A. If both Assertion and Reason are correct and Reason is the correct explanation of assertion.B. If both Assertion and Reason are correct but Reason in not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - B `upsilon-t overset("Intergration")rarr s-t , s-t overset("Differentitation")rarr upsilon-t` |
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| 343. |
Which one of the following cannot be represented by the three sides of a triangle?A. `5,9,11`B. `5,7,11`C. `7,10,13`D. `3,8,9` |
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Answer» Correct Answer - B To get a closed `Delta^(le)`, the sum of any two vectors in magnitude must be either equal or large in magnitude must be either equal or large In magnitude of the third. |
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| 344. |
The speed of a projectile at its maximum height is `sqrt3//2` times its initial speed. If the range of the projectile is n times the maximum height attained by it, n is equal to :A. `(4)/(3)`B. `2sqrt(3)`C. `4sqrt(3)`D. `(3)/(4)` |
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Answer» Correct Answer - C Let u be the initial speed and `theta` is the angle of the projection. Speed at the maximum height, `v_(H)=ucostheta=(sqrt(3))/(2)u` `:.costheta=(sqrt(3))/(2)ortheta=cos^(-1)((sqrt(3))/(2))=30^(@)` Range, `R=(u^(2)sin2theta)/(g)` and maximum height, `H=(u^(2)sin^(2)theta)/(2g)` As R=PH(Given) `:.(u^(2)sin2theta)/(g)=P(u^(2)sin^(2)theta)/(2g)` `2sinthetacostheta=(P)/(2)sin^(2)thetaortantheta=(4)/(P):.P=(4)/(tan30^(@))=4sqrt3` |
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| 345. |
Two trains are each 50 m long moving parallel towards each other at speeds `10 ms^(-1)` and `15 ms^(-1)` respectively, at what time will they pass each other ?A. 8 sB. 4 sC. 2 sD. 6 s |
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Answer» Correct Answer - B `upsilon_(r )=10+15=25 ms^(-1)` where, `upsilon_(r )` is relative velocity `therefore t=(50+50)/(25)=4s` As both trains are 50m long so distance to be travel is 50+50 = 100m As there velocities are in opposite directions 10m/s and 15m/s Relative velocity is 15+10 = 25m/s time=distance/velocity = 100/25 = 4sec B.
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| 346. |
The `(x, y, z)` coordinates of two points A and B are given respectively as `(0, 4, -2) and (-2, 8, -4)`. The displacement vector form A to B isA. `-2hati + 4hatj - 2hatk`B. `2hati - 4hatj + 2hatk`C. `2hati+4hatj-2hatk`D. `-2hati-4hatj - 2hatk` |
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Answer» Correct Answer - A Here, `vecr_(A) = 0hati + 4hatj -2hatk, vecr_(B) = -2hati + 8hatj -4hatk` Displacement vector from A to B is given by `vecr = vecr_(B) - vecr_(A) = (-2hati + 8hatj -4hatk)-(0hati + 4hatj -2hatk) = -2hati +4hatj -2hatk` |
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| 347. |
An aircraft executes a horizontal loop of radius 1 km with a steady speed of `900 km h^(-1)`. Compare its centripetal acceleration with the acceleration due to gravity. |
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Answer» Here `r=1km = 10^(3)m, v=900kmh^(-1)= 900 xx 5/18= 250ms^(-1)` Centripetal acceleration = `a_(c) = v^(2)/r=(250)^(2)/(10^(3)) = 62.5 ms^(-2)` Now, `a_(c)/g=62.5/9.8 = 6.38` |
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| 348. |
A body travelling with uniform acceleration crosses two point `A` and `B` with velocities `20 m s^-1` and `30 m s^-1` respectively. The speed of the body at the mid-point of `A` and `B` is.A. `25 ms^(-1)`B. `25.5 ms^(-1)`C. `24 ms^(-1)`D. `10sqrt(6)ms^(-1)` |
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Answer» Correct Answer - B `(30)^(2)=(20)^(2)+2a(2s)` or `2as = 250` Now, `upsilon^(2)=(u)^(2)+2as=upsilon^(2)=(20)^(2)+250` `rArr upsilon^(2)=650` `therefore upsilon=25.5 ms^(-1)` |
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| 349. |
A car starts from rest and accelerates uniformly to a speed of `180 kmh^(-1)` in 10 s. The distance covered by the car in the time interval isA. 200 mB. 300 mC. 500 mD. 250 m |
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Answer» Correct Answer - D `u=0,upsilon=180 kmh^(-1)=50 ms^(-1)` Time taken, `t=10s, a=(upsilon-u)/(t)=(50)/(10)=5 ms^(-2)` Distance covered by the car `s=ut+(1)/(2)at^(2)=0+(1)/(2)xx5xx(10)^(2)` `=(500)/(2)=250m` |
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| 350. |
A particle starts from rest and traverses a distance l with uniform acceleration, then moves uniformly over a further distance 2l and finally comes to rest after moving a further distance 3l under uniform retardation. Assuming entire motion to be rectilinear motion the ratio of average speed over the journey to the maximum speed on its ways isA. `1//5`B. `2//5`C. `3//5`D. `4//5` |
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Answer» Correct Answer - C Let `upsilon_(m)` be the maximum speed, `upsilon_(m)=a_(1)t_(1)` and `upsilon_(m) = sqret(2a_(1)l) " " (because u = 0)` `t_(2)=(2l)/(upsilon_(m))` and `upsilon_(m)=a_(2)t_(3)=sqrt(2a_(2)(3l))` Now, average speed, `upsilon_(av)=(l+2l+3l)/(t_(1)+t_(2)+t_(3))` `upsilon_(av)=(6l)/((upsilon_(m)//a_(1))+(2l//upsilon_(m))+(upsilon_(m)//a_(2)))` `=(6l)/(((upsilon_(m))/(upsilon_(m)^(2)//2l))+((2l)/(upsilon_(m)))+((upsilon_(m))/(upsilon_(m)^(2)//6l)))` `=(6l)/((10l//upsilon_(m)))=(3upsilon_(m))/(5)` `rArr (upsilon_(av))/(upsilon_(m))=(3)/(5)` |
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