InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 351. |
The distance travelled by a particle starting from rest and moving with an acceleration `(4)/(3) ms^-2`, in the third second is.A. `(10)/(3)m`B. `(19)/(3)/(m)`C. 6 mD. 4 m |
|
Answer» Correct Answer - A `S_(n)=u+(a)/(2)(2n-1)` `rArr S_(3)=0+(4//3)/(2)(2xx3-1)` `rArr S_(3)=(10)/(3)m` |
|
| 352. |
A swimmer is capable of swimming `1.65 ms^(-1)` in still water.If she swims directily across a `180m` wide river whose current is `0.85 ms^(-1)`,how far downstreams (from a point opposite her starting point ) will she reach?A. `92.7m`B. `40m`C. `48m`D. `20m` |
|
Answer» Correct Answer - A `x=V_(W).d/V_(B)` |
|
| 353. |
A body starting from rest has an acceleration of `4ms^(-2)` . Calculate distnce travelled by it in 5th second. |
|
Answer» Given, `u=O, a=4 ms^(-2)` Distance travelled in 5th second `s_(n)=u+(1)/(2)a(2n-1)` `s_(5)=0+(1)/(2)xx4(2xx5-1)=(1)/(2)xx4(9)=(36)/(2)=18 m` |
|
| 354. |
Two cars start off a race with velocity `2 ms^(-1)` and `4ms^(-2)` respectively. What is the length of the path if they reach the final point at the same time ? |
|
Answer» Let both particles reach at same position in same time t then from `s=ut+(1)/(2)at^(2)` For 1st particle, `s=4(t)+(1)/(2)(1)t^(2)=4t+(t^(2))/(2)`, For 2nd particle, `s=2(t)+(1)/(2)t^(2)=2t+t^(2)` Equations, we get `4t+(t^(2))/(2)=2t+t^(2)rArr t=4s` Substituting value of t in above equations, we get `s=4(4)+(1)/(2)(1)(4)^(2)=16+8=24m` |
|
| 355. |
A boatman finds that he can save `6 s` in crossing a river by the quickest path than by the shortest path. If the velocity of the boat and the river be, respectively, `17 ms^-1 and 8 ms^-1`, find the river width.A. `675 m`B. `765 m`C. `567 m`D. `657 m` |
|
Answer» Correct Answer - B Given `t_(1)-t_(2)=6,t_(1)=d/sqrt(V_(B)^(2)-V_(W)^(2)),t_(2)=d/V_(2)` |
|
| 356. |
On an open ground, a motorist follows a track that turns to his left by an angle of `60^(@)` after every `500 m`. Starting from a given turn, The path followed by the motorist is a regular hexagon with side `500 m`, as shown in the given figure specify the displacement of the motorist at the end of eighth turn.A. `3000 m`B. `1500 m`C. `0 m`D. `866 m` |
| Answer» Correct Answer - D | |
| 357. |
Which of the following qauntities is dependent of the choice of orientation of coordinates axes?A. `vecA + vecB`B. `A_(x) + B_(y)`C. `|vecA + vecB|`D. Angle between `vecA and vecB` |
|
Answer» Correct Answer - B A vector, its magnitude and the angle between two vectors do not depend on the choice of the orientation of the coordinate axes. `vecA + vecB, |vecA + vecB|`,angle between `vecA and vecB` are independent of the orientation of the coordinate axes. but the quantity `A_(x) + B_(y)` depends upon the magnitude of the component along x and y-axes, soit will change with change in coordinate axes. |
|
| 358. |
At high altitude , a body explodes at rest into two equal fragments with one fragment receiving horizontal velocity of `10 m//s`. Time taken by the two radius vectors connecting of explosion to fragments to make `90^(@)` isA. `1s`B. `2s`C. `1.5s`D. `1.7s` |
|
Answer» Correct Answer - A `t=sqrt(u_(1)u_(2))/g` |
|
| 359. |
It is found that `|A+B|=|A|`,This necessarily implies.A. `vecB=0`B. `vecA,vecB` are antiparallelC. `vecA,vecB` are perpendicularD. `vecA,vecB le0`. |
|
Answer» Correct Answer - A::B Given that `|A+B|=|A|` or `|A+B|^(2)=|A|^(2)` `=|A|^(2)+|B|^(2)+|A||B|cos theta=|A|^(2)` Where `theta` is angle between `A` and `B`. `=|B|=0` or `|B|+2|A|cos theta=0` `=cos theta =-|B|/(2|A|)` If `A` and `B` are antiparallel, then `theta=180^(@)` Hence, from Eq (i) `-1=-|B|/(2|A|)rArr|B|=2|A|` |
|
| 360. |
The distance travelled by a particle is proportional to the squares of time, then the particle travels withA. uniform accelerationB. uniform velocityC. Both of theseD. speed changes |
|
Answer» Correct Answer - A `S prop t^(2)` [Given] `therefore s = kt^(2)` Acceleration, `a=(d^(2)s)/(dt^(2)=2k` [constant] It means the particle travels with uniform acceleration. |
|
| 361. |
The distance travelled by a body is proportional to the square of time. What type of motion this body has ? |
|
Answer» Let x be the distance travelled in time t. Then, `x prop t^(2)` (given) `x = kt^(2)` (Here, k = constant of proportionality) We know that velocity is given `v = (dx)/(dt) = 2 kt` and acceleration is given by `a=(dv)/(dt)=2 k` (constant) Thus, the body has uniform accelerated motion. |
|
| 362. |
The motion of a particle is described by the equation at `u = at`.The distance travelled by the particle in the first 4 secondsA. `4 a`B. `12 a`C. `6 a`D. `8 a` |
|
Answer» Correct Answer - D `s=(1)/(2)at^(2) rArr s=(1)/(2)xxaxx(4)^(2) rArr s=8a` |
|
| 363. |
A body falls from a height `h = 200 m` (at New Delhi). The ratio of distance travelled in each `2 sec` during `t = 0` to `t = 6` seconds of the journey is.A. `1 : 4 : 9`B. `1 : 2 : 4`C. `1 : 3 : 5`D. `1 : 2 :3` |
|
Answer» Correct Answer - C `h_(1)=(1)/(2)xx g xx(2)^(2)=2g` `hArr h_(2)=(1)/(2)xx g xx(4)^(2)-2g=6g` `h_(3)=(1)/(2)xx g xx(6)^(2)-8g=10 g` `therefore h_(1): h_(2): h_(3)=1 : 3 : 5` |
|
| 364. |
Assertion : A body is momentarily at rest at the instant it reverses the direction. Reason : A body cannot have acceleration if its velocity is zero at a given instant of time.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. |
|
Answer» Correct Answer - C When a particle is released from rest, `upsilon = 0` but `a ne 0` |
|
| 365. |
When a motor cyclist takes a `U-turn` in `4s` what is the average angular velocity of the motor cyclist. |
|
Answer» When the motor cyclist takes a U-turn, angular displacement, `theta=pi` rad and `t=4s`. The average angular velocity, `omega=theta/t,alpha_(av)=(Deltaomega)/(Deltat)` `omega=(d theta)/dt,alpha=(d omega)/dt=(d^(2)theta)/dt^(2)=omega(domega)/(d theta)` int d theta=int omega dt, int d omega=int alpha dt,int alpha d theta=int omega d omega` |
|
| 366. |
Two thin wood screens `A` and `B` are separated by `200m` a bullet travelling horizontally at speed of `600 m//s` hits the screen `A` penetrates through it and finally emerges out from `B` making holes in `A` and `B` the resistance of air and wood are negligible the difference of heights of the holes in `A ` and `B` is.A. `5 m`B. `49/90m`C. `7/sqrt90m`D. zero |
|
Answer» Correct Answer - B `R=usqrt((2h)/g),h=h_(1)-h_(2)` |
|
| 367. |
A boy aims a gun at a bird from a point, at a horizontal distance of `100m`. If the gun can impart a velocity of `500m//sec` to the bullet, at what height above the bird must he aim his gun in order to hit it? |
|
Answer» `x=vt or 100=500xxt,t=0.2sec`, Now `h=0+1/2xx10xx(0.2)^(2)=0.20m=20cm`. |
|
| 368. |
A particle moves along `x`-axis with constant acceleration and its `x`-position depend on time `t` as shown in the following graph (parabola),then in interval `0` to `4 sec`. A. relation between `x`-coordinate & time is `x=t-t^(2)//4`B. maximum `x`-coordinate is `1m`.C. total distance travelled is `2m`D. average speed is `0.5 m//s` |
|
Answer» Correct Answer - A::B::C::D `(dx)/(dt)=1-t/2 ,1-t/2=0 , t=2s` `x=t-t^(2)/4 , x=1m` total distance`=2m` Average speed `V_(av)=2/24=0.5 m//s`. |
|
| 369. |
A railway compartments is `16 m` long, `2.4 m` wide and `3.2 m` high. It is moving with a velocity `v` A particle moving horizontally with a speed `u` perpendicular to the direction of `v` enters through a hole at an upper corner `A` and strikes the diagonally opposite corner `B`.Assume `g=10 m//s^(2)`. A. `v=20 m//s`B. `u=3 m//s`C. To an observer inside the compartment, the path of the particle is a parabolaD. To a stationary observer outside the compartment, the path of the particle is parabola |
|
Answer» Correct Answer - A::B::C::D (a,b,c,d) consider the vertical motion of the particle after entering the compartment.Let it reach the floor in time `t` `3.2=1/2(10)t^(2) , or t=0.8s` Due to the velocity component `u`,which remain costant, it covers a distance of `2.4 m` in `0.8 s` `u=(2.4m)/(0.8s)=3m//s` Also `v=(16m)/(0.8)=20m//s` |
|
| 370. |
A boat crosses a river with a velocity of `8(km)/(h)`. If the resulting velocity of boat is `10(km)/(h)` then the velocity of river water is |
| Answer» Correct Answer - ` 6 km h^(-1)` | |
| 371. |
An aeroplane is flying with the velocity of `V_(1)=800kmph` relative to the air towards south.A wind with velocity of `V_(2)=15ms^(-1)` is blowing from west to east.What is the velocity of the aeroplane with respect of the earth.A. `221.7ms^(-1)`B. `150ms^(-1)`C. `82ms^(-1)`D. `40ms^(-1)` |
|
Answer» Correct Answer - A `vecv_(AW)=vecv_(1)=800xx5/18hatj` `vecv_(W)=vecv_(2)=15hati,vecv_(W)=vecv_(AW)+vecv_(W)` |
|
| 372. |
A hydrogen is flying eastward with speed ` 12 m s^(-1)` , when wind starts blowing from north to south with speed ` 5 m s ^(-1)` what is the resultant velocity of balloon ? |
| Answer» Correct Answer - `13 m s^(-1) , theta = tan^(-1) (5/12)` South of east. | |
| 373. |
The velocity - time graph for two bodies A and B are shown in figure. Then, the acceleration of A and B are in the ratio A. `sin 25^(@)` to `sin 50^(@)`B. `tan 25^(@)` to `tan 40^(@)`C. `cos 25^(@)` to `cos 50^(@)`D. `tan 25^(@)` to `tan 50^(@)` |
|
Answer» Correct Answer - D The slope of velocity-time graph gives acceleration, i.e., `(a_(A))/(a_(B))=(tan theta_(1))/(tan theta_(2))` `rArr (a_(A))/(a_(B))=(tan 25^(@))/(tan 50^(@))` |
|
| 374. |
The following figure-1.105 shows the velocity-time graph ofa body. According to this, at the point B: A. the is zeroB. there is at force towards motionC. there is a force which opposes motionD. there is only gravitational force |
|
Answer» Correct Answer - C In region B, velocity is decreasing. So, a force will act which opposes the motion of a body. |
|
| 375. |
The velocity - time graph is shown in the figure, for a particle. The acceleration of particle is A. `22.5 ms^(-2)`B. `5 ms^(-2)`C. `-5 ms^(-2)`D. `-3 ms^(-2)` |
|
Answer» Correct Answer - C a = slope of `upsilon - t` graph `=-(15)/(3)=-5 ms^(-2)` |
|
| 376. |
Assertion: Vector addition is commutative. Reason: Two vectors may be added graphically using head- to-tail method or parallelogram method. |
|
Answer» Correct Answer - B Vector addition is commutative i.e., `vecA + vecB = vecB + vecA`, where `vecA and vecB` are two vectors Two vectors `vecA and vecB` may be added graphically using head-to-tail method or parallelogram method. |
|
| 377. |
Assertion: Two vectors are said to be equal if , and only if, they have the same magnitude and the same dirction. Reason: Addition and subtraction of scalars make sense only for quantities with same units. |
|
Answer» Correct Answer - B Addition and subtraction of scalars make sense only for quantities with same unit. But you can multiply and divide scalars of different units. |
|
| 378. |
A ball dropped from the top of a tower covers a distance `7x` in the last second of its journey, where `x` is the distance covered in the first second. How much time does it take to reach to ground?.A. 3 sB. 4 sC. 5 sD. 6 s |
|
Answer» Correct Answer - B In first second distance travelled, `x=(1)/(2)xx g xx t^(2)=5m` (for `g=10 ms^(-2))` in last second `7x=35 m` Now, `S_(t)th=(u+at-(1)/(2)a)` `rArr 35=0+10xx t-(1)/(2)xx10` `therefore t=4s` |
|
| 379. |
In the question number 1, the angle `phi` which the velocity vector of stone makes with horizontal just before hitting the ground is given byA. `tan phi = 2 tan theta`B. `tan phi = 2 cot theta`C. `tan phi = sqrt(2) tan theta`D. `tan phi = sqrt(2) cot theta` |
|
Answer» Correct Answer - C Let `u_(y)andv_(y)` be initial and final vertical components of velocity respectively. `:.u_(y)^(2)=2gHandv_(y)^(2)=4gH:.v_(y)=sqrt2u_(y)` Final velocity makes angle `(phi)`with horizontal. `:.tanphi=(v_(y))/(u_(x))=sqrt(2)(u_(y))/(u_(x))=sqrt(2)tantheta` |
|
| 380. |
In the question number 56, the maximum horizontal distance covered by the ball will beA. `160sqrt(3)` mB. `140sqrt(3)` mC. `120sqrt(3)` mD. `100sqrt(3)` m |
|
Answer» Correct Answer - A Horizontal range, `R = (u^(2)sin 2theta)/(g)` `R = ((56)^(2) sin60^(@))/(9.8) = (56 xx 56 xxsqrt(3))/(9.8 xx 2) = 160sqrt(3)m` |
|
| 381. |
At time `t=0`, the position vector of a particle moving in the `x-y` plane is `5hatj m`.By time `t=0`.`62 sec`,its position vector has become `(5.1hati+0.4 hatj)m`.with the data answer the following questions. The magnitude of the average velocity during the above time interval.A. `.0206m//sec`B. `0.206m//sec`C. `20.6m//sec`D. `2.06m//sec` |
|
Answer» Correct Answer - C `x=at rArr V_(x)=a rArr a_(x)=0` `y=at(1-at) rArr V_(y)=a-a2alphat rArr a_(y)=-2alphaa` Again `y=ax/a(1-x/a.alpha)rArry=x-(x^(2)alpha)/a` If `vecV` and `veca` makes an angle `45^(@)` with each other `vecV` must be making an angle `45^(@)` with `X`-axis So `V_(x)=V_(y) rArr a=a-2aalphat_(0) rArr t_(0)=1/alpha` Displacement `vec(Deltax)=(.1hati+4hatj)m` `vecV_(avg)=` Average velocity `((.1hati+.4hatj)/.02)m//sec=(5hati+20hatj)m//sec` So `|vecV_(avg)|=sqrt425 m//sec =20.6 m` `theta= "tan"^(-1)20/5= tan^(-1)4` |
|
| 382. |
In the question number 38, the speed of the particle at this time isA. 16 m `s^(-1)`B. 26 m `s^(-1)`C. 36 m `s^(-1)`D. 46 m `s^(-1)` |
|
Answer» Correct Answer - B Using eqn. (i) from question number 38, Velocity, `vecv = (dvecr)/(dt) = (d)/(dt) (5t +1.5t^(2)) hati + 1t^(2)hatj = (5+3t)hati +2thatj` At` = 6s, vecv = 23 hati + 12 hatj` The speed of the particle is `|vecv| = sqrt((23)^(2) + (12)^(2)) = 26` m `s^(-1)` |
|
| 383. |
Unit vector does not have anyA. DirectionB. MagnitudeC. UnitD. All to these |
| Answer» Correct Answer - C | |
| 384. |
In the question number 35, the acceleration of the particle at `t = 1` isA. `2hatjm s^(-2)`B. `-2 hatj ms^(-2)`C. `4 hatj ms^(-2)`D. `-4hatj ms^(-2)` |
|
Answer» Correct Answer - C From the previous question, `vecv = 3hati + 4t hatj ms^(-1)` `therefore ` Acceleration, `veca = (dvecv)/(dt) = 4hatj ms^(-2)` Acceleration of the particle remains constant at all times. |
|
| 385. |
The displacement of a particle from a point having position vector ` 2hati + 4hatj` to another point having position vector ` 5hatj + 1hatj` isA. 3 unitsB. `3sqrt2` unitsC. 5 unitsD. ` 5sqrt3`units |
| Answer» Correct Answer - B | |
| 386. |
At time `t=0`, the position vector of a particle moving in the `x-y` plane is `5hatj m`.By time `t=0`.`62 sec`,its position vector has become `(5.1hati+0.4 hatj)m`.with the data answer the following questions. The angle `theta` made by the average velocity with the positive `x` axisA. `tan^(-1)(2)`B. `tan^(-1)(3)`C. `tan^(-1)(1)`D. `tan^(-1)(4)` |
|
Answer» Correct Answer - D `x=at rArr V_(x)=a rArr a_(x)=0` `y=at(1-at) rArr V_(y)=a-a2alphat rArr a_(y)=-2alphaa` Again `y=ax/a(1-x/a.alpha)rArry=x-(x^(2)alpha)/a` If `vecV` and `veca` makes an angle `45^(@)` with each other `vecV` must be making an angle `45^(@)` with `X`-axis So `V_(x)=V_(y) rArr a=a-2aalphat_(0) rArr t_(0)=1/alpha` Displacement `vec(Deltax)=(.1hati+4hatj)m` `vecV_(avg)=` Average velocity `((.1hati+.4hatj)/.02)m//sec=(5hati+20hatj)m//sec` So `|vecV_(avg)|=sqrt425 m//sec =20.6 m` `theta= "tan"^(-1)20/5= tan^(-1)4` |
|
| 387. |
A particle starts from origin at `t=0` with a constant velocity `5hatj m//s` and moves in `x-y` plane under action of a force which produce a constant acceleration of `(3hati+2hatj)m//s^(2)` the `y`-coordinate of the particle at the instant its `x` co-ordinate is `24 m` in `m` isA. 12 mB. 24 mC. 36 mD. 48 m |
|
Answer» Correct Answer - C The position of the particle is given by `vecr = vecr_(0) = vecv_(0)t + (1)/(2) vecat^(2)`1 where, `vecr_(0)` is the position vector at t = 0 and `vecv_(0)` is the velocity at t = 0 Here, `vecr_(0) = 0, vecv_(0) = 5hati ms^(-1), veca = (3hati + 2hatj) ms^(-2)` `therefore vecr = 5thati +(1)/(2)(3hati +2hatj) t^(2) = (5t + 1.5t^(2))hati + 1t^(2)hatj " "..(i)` Compare it with `vecr = x hati + y hatj,` we get `x = 5t + 1.5t^(2), y = 1t^(2)` `because x = 84 m` `therefore 84 = 5t + 1.5t^(2)` On solving,we get t = 6 s At t = 6 s, y = (1) `(6)^(2) = 36` m |
|
| 388. |
A paarticle starts from origin at ` t=0 ` with a velocity `5.0 hat i m//s` and moves in x-y plane under action of a force which produces a constant acceleration of `( 3.0 hat i + 2.0 j) m//s^(2)`. (a) What is the y-cordinate of the particle at the instant its x-coordinate is `84 m ? (b) What is the speed of the particle at this time? |
|
Answer» The position of the particle is given by `vecr(t)=vecV_(0)t+1/2vecat^(2)` `=5hatit+(1//2)(3hati+2hatj)t^(2)` `=(t+1.5t^(2))hati+t^(2)hatj` Therefore, `x(t)=5t+1.5t^(2),y(t)=t^(2)` `5t+1.5t^(2)=84rArr t=6s` At `t=6s,y=(6)^(2)=36 m` `vecv=(dvecr)/dt=(5+3 t)hati+2t hatj` At `t=6s,vecv=23hati+12hatj` speed `=|vecv|=sqrt(23^(2)+12^(12))=26ms^(-1)` |
|
| 389. |
A particle starts from origin at `t=0` with a constant velocity `5hati m//s` and moves in `x-y` plane under action of a force which produce a constant acceleration of `(3hati+2hatj)m//s^(2)` the `y`-coordinate of the particle at the instant its `x` co-ordinate is `84 m` in `m` isA. `6`B. `36`C. `18`D. `9` |
|
Answer» Correct Answer - B `vecr=ut+1/2at^(2)` equate `x` coordinate to `84` to find time `t` |
|
| 390. |
A body starts with a velocity `(2hati+3hatj+11hatk)m//s` and moves with an acceleration `(5hati+5hatj-5hatk)m//s^(2)`. What is its velocity after `0.2 sec`?A. `7hati+8hatj+6hatk`B. `2hati-3hatj+11hatk`C. `3hati-4hatj-10hatk`D. `3hati+4hatj+10hatk` |
|
Answer» Correct Answer - D `hat(v=)hatu+hata.t` |
|
| 391. |
A car moving at a constant speed of `36 kmph` moves north wards for `20 minutes` then due to west with the same speed for `8(1)/(3)` minutes. What is the average velocity of he car during this run in kmphA. 27.5B. 40.5C. 20.8D. 32.7 |
|
Answer» Correct Answer - A `V_(avg)=(V_(1)t_(1)+V_(2)t_(2))/(t_(1)+t_2)` |
|
| 392. |
A body is prodected with initial velocity` ( 5hati + 12hati)` m//s from origin . Gravity acts in negative y direction. The horizontal rang is (Take g = 10 `m//s^(2)`)A. 12mB. 18 mC. 17 mD. 7 m |
| Answer» Correct Answer - A | |
| 393. |
An object is moving velocity `10 ms^(-1)` . A constant force acts for 4 s object and given it a speed of `2 ms^(-1)` in opposite direction. The acceleration produced isA. `3 ms^(-2)`B. `-3 ms^(-2)`C. `6 ms^(-2)`D. `-6ms^(-2)` |
|
Answer» Correct Answer - B `upsilon=u+at` or `-2=10+4a` or `a=-3ms^(-2)` |
|
| 394. |
Velocity of a particle at time `t=0` is `2ms^(-1)`. A constant acceleration of `2ms^(-2)` acts on the particle for `1 second` at an angle of `60^(@)` with its initial velocity. Find the magnitude of velocity at the end of `1 second`.A. `sqrt3 m//s`B. `2sqrt3 m//s`C. `4 m//s`D. `8 m//s` |
|
Answer» Correct Answer - B `vecv=v_(x)hati+v_(y)hatj:v_(x)=u_(x)+a_(x)t,v_(y)=u_(y)+a_(y)t` `a_(x)=a cos theta,a_(y)=a cos theta` |
|
| 395. |
An aeroplane moving in a circular path with a speed `250 km//h`. The change in velocity in half of the revolution is.A. `500km//h`B. `250km//h`C. `120km//h`D. zero |
|
Answer» Correct Answer - A `DeltaV=2V "sin" (theta)/(2)` |
|
| 396. |
The coordinates of a body moving in a plane at any instant of time `t` are `x=alphat^(2)` and `y=betat^(2)`. The speed of the body is. |
|
Answer» `x=alpha t^(2)rArr v_(x)=dx/dt=2alphat` `y=beta t^(2)rArr v_(y)=dy/dt=2betat` `:.` speed `v=sqrt(v_(x)^(2)+v_(y)^(2))=2tsqrt(alpha^(2)+beta^(2))` |
|
| 397. |
A car moving with a velocity of 10m/s can be stopped by the application of a constant force F In a distance of 20m. If the velocity of the car is 30m/s. It can be stopped by this force inA. `(20)/(3)m`B. 20 mC. 60 mD. 180 m |
|
Answer» Correct Answer - D `S prop u^(2)`. If u becomes 3 time, then S will become 9 times i.e., `9xx20=180 m` |
|
| 398. |
The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity `v_0`. The distance travelled by the particle in time t will beA. `v_(0)t+(1)/(6)bt^(3)`B. `v_(0)t+(1)/(3)bt^(3)`C. `v_(0)t+(1)/(3)bt^(2)`D. `v_(0)t+(1)/(2)bt^(2)` |
|
Answer» Correct Answer - A `a=bt` `therefore int_(upsilon_(0))^(upsilon)d upsilon=int adt = int_(0)^(1) bt.dt` `therefore upsilon=upsilon_(0)+(bt^(2))/(2)` Further integrating we get, `s=upsilon_(0)t+(bt^(3))/(6)` |
|
| 399. |
If the velocity v of particle moving along a straight line decreases linearly with its displacement s from `20 ms^(-1)` to a value approaching zero at s = 30 m, then acceleration of the particle at v=10ms^(-1)` is A. `(2)/(3)ms^(-2)`B. `-(2)/(3)ms^(-2)`C. `(20)/(3)ms^(-2)`D. `-(20)/(3)ms^(-2)` |
|
Answer» Correct Answer - D `a=upsilon((d upsilon)/(ds))=(10)(-(20)/(30))=-(20)/(3)ms^(-2)` |
|
| 400. |
An aeroplne `A` is flying horizontally due east at a speed of `400 km//hr`.Passengers in `A`, observe another aeroplane `B` moving perpendicular to direction of motion of `A`.Aeroplane `B` is actually moving in a direction `30^(@)` north of east in the same horizontal plane as shown in the figure.Determine the velocity of `B` A. `400sqrt3hati+400sqrt3hatj`B. `400hati+400/(sqrt(3))hatj`C. `400hati+400hatj`D. `400/sqrt3hati+400hatj` |
|
Answer» Correct Answer - B `vecV_(A)=400hati, vecV_(B)=v_(B) cos 30^(@) hati+v_(B) sin 30^(@) hatj` `vecV_(B)=(v_(B)sqrt3)/2hati+v_(B)/2hatj,` `vecV_(B//A)=vecV_(B)-vecV_(A)` `vecV_(B)=(v_(B)sqrt3/2hati+v_(b)/2hatj)-400hati` `vecV_(B//A)=(v_(B)sqrt3/2hati-400)hati+v_(B)/2hatj` `sqrt3v_(B)-800=0,v_(B)=/800/sqrt3,vecV_(B)=400hati+400/sqrt3hatj` |
|