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.
| 201. |
A particle moves in a straight line with a constant acceleration. It passing through a distance 135 m in t second. The value of t (in second) isA. 12B. 9C. 10D. `1.8` |
|
Answer» Correct Answer - B `upsilon^(2)=u^(2)+2as` `a=(upsilon^(2)-u^(2))/(2s)=((20)^(2)-(10)^(2))/(2xx135)=(300)/(270)=(10)/(9)ms^(-2)` From first equation of motion `upsilon=u+at rArr t(upsilon-u)/(a)=(20-10)/(10//9)=(10)/(10//9)=09s` |
|
| 202. |
If the sum of two unit vectors is a unit vector,then find the magnitude of their differences.A. 1 unitB. 2 unitC. `sqrt3` unitD. zero |
|
Answer» Correct Answer - C `R=2P "cos" (theta)/(2) ,S=2P "sin" (theta)/(2)` |
|
| 203. |
The maximum resultant of two concurrent forces is `10N` and their minimum resultant is `4N`. The magnitude of large force isA. `5N`B. `7N`C. `3N`D. `14N` |
|
Answer» Correct Answer - B `F_(1)+F_(2)=10,F_(1)-F_(2)=4` |
|
| 204. |
A boat moves perpendicular to the bank with a velocity of `7.2 km//h`.The current carries it `150 m` downstreamk.find the velocity of the current (The width of the river is `0.5 km`).A. `0.4 ms^(-1)`B. `1.2 ms^(-1)`C. `0.5 ms^(-1)`D. `0.6 ms^(-1)` |
|
Answer» Correct Answer - D `x=V_(W).d/V_(B)` |
|
| 205. |
A vehicle travels half the distance (L) with speed ` V_1` and the other half with speed ` V_2`, then its average speed is .A. `(upsilon_(1)+upsilon_(2))/(2)`B. `(2upsilon_(1)+upsilon_(2))/(upsilon_(1)+upsilon_(2))`C. `(2upsilon_(1)upsilon_(2))/(upsilon_(1)+upsilon_(2))`D. `(L(upsilon_(1)+upsilon_(2)))/(upsilon_(1)upsilon_(2))` |
|
Answer» Correct Answer - C Time taken to travel first half distance `t_(1)=(l//2)/(upsilon_(1))=(l)/(2upsilon_(1))` Time taken to travel second half distance `t_(2)=(l)/(2upsilon_(2))` Total time = `t_(1)+t_(2)=(l)/(2upsilon_(1))+(l)/(2upsilon_(2))=(l)/(2)[(1)/(upsilon_(1))+(1)/(upsilon_(2))]` We know that, `upsilon_(av)` = Average speed `=("total distance")/("total time")=(l)/((1)/(2)[(1)/(v_(1))+(1)/(v_(2))])=(2upsilon_(1)upsilon_(2))/(upsilon_(1)+upsilon_(2))` |
|
| 206. |
A golfer standing on level ground hits a ball with a velocity of `52 m s^-1` at an angle `theta` above the horizontal. If `tan theta = 5//12`, then find the time for which then ball is atleast `15 m` above the ground `(take g = 10 m s^-2)`. |
| Answer» `y=u sin alpha.t-1/2 "gt"^(2)` | |
| 207. |
In the question number 52, the speed with which the stone hits the ground isA. 15 m `s^(-1)`B. 90 m `s^(-1)`C. 99 m `s^(-1)`D. 49 m `s^(-1)` |
|
Answer» Correct Answer - C Motion along horizontal direction, `downarrow +ve` `u_(x) = 15 ms^(-1), a_(x) = 0` `v_(x) = u_(x) + a_(x)t = 15 + 0 xx 10 = 15 ms^(-1)` Motion along vertical direction, `u_(y) = 0, a_(y) = g` `v_(y) = u_(y) + a_(y)t = 0 + 9.8 xx 10 = 98 ms^(-1)` `therefore` Speed of the stone when it hits the ground is `v = sqrt(v_(x)^(2) + v_(y)^(2)) = sqrt((15)^(2) + (98)^(2)) ~~ 99 ms^(-1)` |
|
| 208. |
A particle moves in a of radius R from `A` to `B`, as shown in the figure. Find the distance and displacement covered. A. `(pi R)/(3)`B. `(pi R)/(2)`C. `(pi R)/(4)`D. `pi R` |
|
Answer» Correct Answer - A Distance = Length `AB=2pi Rxx(60)/(360)=(pi R)/(3)` |
|
| 209. |
When two objects move uniformly towards each other, they get `4` metres closer each second and when they move uniformly in the same direction with original speed, they get `4` metres closer each `5s`. Find their individual speeds. |
|
Answer» Let the speeds be `v_(1)` and `v_(2)` and let `v_(1)gt v_(2)`. In First case: Relative velocity, `v_(1)+v_(2)=4/1=4 m//s`…..(1) In Second case: Relative velocity, `v_(1)-v_(2)=4/5=0.8 m//s`…..(2) solving eqns. (1) and (2), we get `v_(1)=2.4ms^(-1),v_(2)=1.6ms^(-1)` |
|
| 210. |
A partachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at `2 ms^(-2)`. At what height did he bail outA. 293 mB. 111 mC. 91 mD. 182 m |
|
Answer» Correct Answer - A After bailing out from point A partachutist falls freely under gravity. The velcity acquired by it will be `upsilon`. From `upsilon^(2)+2as=0+2xx9.8xx50=980` `["As "u=0, a=9.8 ms^(-2), s=50 m]` At point B, parachute opens and it moves with retardation of `2 ms^(-2)` are reach at ground (Point C with velocity of `3 ms^(-1)`). For the part BC by applying the equastion `upsilon^(2)=u^(2)+2as` `upsilon=3 ms^(-1), u = sqrt(989)ms^(-1), a=-2 ms^(-2), s=h` `rArr (3)^(2)=(sqrt(980))^(2)+2xx(-2)xxh` `rArr 9=980-4h` `rArr h=(980-9)/(4)=(971)/(4)=242.7 ~== 243m` So, the total height by which parachutist bails out `=50+243=293m` |
|
| 211. |
When two bodies approach each other with the different speeds, the distance between them decreases by `120 m` for every `1 min`.The speeds of the bodies areA. `2 m//s` and `0.5 m//s`B. `3m//s & 2m//s`C. `1.75m//s & 0.25 m//s`D. `2.5m//s & 0.5m//s` |
|
Answer» Correct Answer - C `v_(1)+v_(2)=120/60`,verify options |
|
| 212. |
If `O` is at equilibrium then the values of the tension `T_(1)` and `T_(2)` are, (`20 N` is acting vertically downwards at `O`). A. `20N, 30N`B. `20sqrt3N,20N`C. `20sqrt3N,20sqrt3N`D. `10N,30N` |
|
Answer» Correct Answer - B `T_(1)sin 30^(@)=T_(2)cos 30^(@),T_(1)=sqrt3T_(2)....(1)` `T_(1)cos 30^(@)=20+T_(2)` ...(2) , soving (1) and (2) |
|
| 213. |
If `P` is in equilibrium then `T_(1)/T_(2)` is A. `sqrt3`B. 2C. `1/sqrt3`D. `1/2` |
|
Answer» Correct Answer - C `T_(1) cos30^(@)=T_(2) cos60^(@)` |
|
| 214. |
Person aiming to reach the exactly opposite point on the bank of a stream is swimming with a speed of `0.5ms^(-1)` at an angle of `120^(@)` with the direction of flow of water.The speed of water in the stream isA. `1ms^(-1)`B. `0.25ms^(-1)`C. `0.67ms^(-1)`D. `3ms^(-1)` |
|
Answer» Correct Answer - B `sin 30=v_(w)/v_(m)` |
|
| 215. |
A cricket fielder can throw the cricket ball with a speed ` v_0`. If he throws the ball while running with speed (u) at angle ` theta` to the horizontal, find (b) what will be time of flight ? (c ) what is the distance (horizontal range) form the point of projection at which the ball will land ? (d) find ` theta` at which he should throw the ball that would maxmise the horizontal range range as found in (c ). (e) how does ` theta` for maximum range change if ` u gt v_0, u=v_0, ult v_0` ? (f) how does ` theta` in (e) compare with that for ` u=0 (i.e., 45^@) ?A. `T=(2v_(0))/g`B. `T=(2(u+v_(0)sin theta))/g`C. `T=(2v_(0)sin theta)/g`D. `T=(2u)/g` |
|
Answer» Correct Answer - C `y=u_(y)t+1/2a_(y)t^(2)` `rArr 0=v_(0) sin theta T+1/2(-g)T^(2)` |
|
| 216. |
A particle moves along the positive branch of the curve `y = (x^(2))/(2)` where `x = (t^(2))/(2),x` and y are measured in metres and t in second. At `t = 2s`, the velocity of the particle isA. `2hati - 4hatj`B. `2hati + 4hatj`C. `4hatj+ 2hatj`D. ` 4hati- 2hatj` |
| Answer» Correct Answer - B | |
| 217. |
A police jeep is chasing with, velocity of `45 km//h` a thief in another jeep moving with velocity `153 km//h`. Police fires a bullet with muzzle velocity of `180 m//s`. The velocity it will strike the car of the thief is.A. `150 ms^(-1)`B. `27 ms^(-1)`C. `450 ms^(-1)`D. `250 ms^(-1)` |
|
Answer» Correct Answer - A Let `upsilon_(PG)` = velocity of police w.r.t. ground `upsilon_(TG)` =velocity of thief w.r.t. ground `upsilon_(TP)` = velocity of thief w.r.t. police `=upsilon_(TG)-upsilon_(PG)` `=((153-45)/(18))xx5=30 ms^(-1)` `upsilon_(BC)` = velocity of bullet w.r.t. car `=180-30=150 ms^(-1)` |
|
| 218. |
A point traversed 3/4 th of the circle of radius R in time t. The magnitude of the average velocity of the particle in this time interval isA. `(pi R)/(t)`B. `(3pi R)/(2t)`C. `(R sqrt(2))/(t)`D. `(R )/(sqrt(2)t)` |
|
Answer» Correct Answer - C `upsilon_(av)=("Displacement")/("Time")=(sqrt(2)R)/(t)` |
|
| 219. |
The distance traversed by a particle moving along a straight lne is given by `x = 180t+50t^(2)` metre. The acceleration of the particle isA. `180 ms^(-2)`B. `580 ms^(-2)`C. `100 ms^(-2)`D. `50 ms^(-2)` |
|
Answer» Correct Answer - C `x=180t+50t^(2)` `upsilon=(dx)/(dt)=180+100t` `a=(d upsilon)/(dt)=100m//s^(2)` |
|
| 220. |
The displacement-time graph of a particle is as shown below. It indicates that A. the velocity of the particle is constant throughoutB. the acceleration of the particle is constant throughoutC. the particle starts with a constant velocity and is acceleratedD. the motion is retarded and finally the particle stops |
|
Answer» Correct Answer - D At first, the slope is decreasing therefore, the motion is retarded Finally, the displacement becomes constant and the body stops. |
|
| 221. |
The graph between displacement and time for a particle moving with uniform acceleration is aA. straight line with a positive slopeB. parabolaC. ellipseD. straight line parallel to time axis |
|
Answer» Correct Answer - B The graph between displacement displacement and time for a particle moving uniform acceleration is a parabola. |
|
| 222. |
If a stone is thrown up with a velocity of `9.8 ms^(-1)`, then how much time will it take to come back ?A. 1 sB. 2 sC. 3 sD. 4 s |
|
Answer» Correct Answer - B `t=(2u)/(g) rArr t=(2xx9.8)/(10)=1.96 ~~ 2s` |
|
| 223. |
For a particle performing uniform circular motion, choose the incorrect statement form the following.A. Magnitude of particle velocity (speed) remains constantB. Particle velocity remains directed perpendicular to radius vector.C. Direction of acceleration keeps changing as particles movesD. Angular momentum is constant in magnitude but direction keeps changing. |
|
Answer» Correct Answer - A::B::C In uniform circular motion, magnitude of velocity and acceleration is constant but direction changes continuously. |
|
| 224. |
Following are four different relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incorrect one (s)A. `v_(av)=1/2[v(t_(1))-v(t_(2))]`B. `v_(av)=(r(t_(2))-r(t_(1)))/(t_(2)-t_(1))`C. `r=1/2(v(t_(2))-v(t_(1)t)t_(2)-t_(1))`D. `a_(av)=(v(t_(2))-v(t_(1)))/(t_(2)-t_(1))` |
|
Answer» Correct Answer - C `v=(Deltar)/(Deltat)=(r_(2)-r_(1))/(t_(2)-t_(1))` `a_(av)=(Deltav)/(Deltat)=(v_(2)-v_(1))/(t_(2)-t_(1))` `v_(av)ne(v_(1)+v_(2))/2` `Deltar=r_(2)-r_(1)=(v_(2)-v_(1))(t_(2)-t_(1))`. |
|
| 225. |
A particle is fired from `A` in the diagonal plane of a building of dimension `20m`(length) `xx` `15m` (breadth) `xx` `12.5 m`(height), just clears the roof diagonally & falls on the other side of the building at `B`.It is observed that the particle is travelling at an angle `45^(@)` with the horizontal when it clears the edges `P` and `Q` of the diagonal. Take `g=10m//s^(2)`. The speed of the particle at point `P` will be: A. `5sqrt10m//s`B. `10sqrt5m//s`C. `5sqrt15m//s`D. `5sqrt5m//s` |
|
Answer» Correct Answer - A `PQ` length , `(U^(2) sin 2 theta)/(g) = 25 , U = 5 sqrt(10) m//s` |
|
| 226. |
If x and y components of a vector `vecP` have numberical values 5 and 6 respectively and that of ` vecP + vecQ` have magnitudes 10 and 9, find the magnitude of ` vecQ` |
|
Answer» According to the question ` varP = 5hat + 6hatj` ` and vecP + vecQ = 10 hati + 9hatj` ` Rightarrow vecQ = ( 10hati + 9hatj) - (vecP)` `= (10 hati + 9hatj) + ( -vecP)` `= 10hati + 9 hatj - 5hatj -6hatj` ` (10-5) hatj + ( 9-6) hatj` ` Rightarrow vecQ = 5hatj + 3hatj` `or |vecQ| = sqrt(5^(2) +3^(2))` ` = sqrt(25+9)` ` |vecQ| = sqrt34` Angle that ` vecQ` makes with positive x-axis , ` theta= tan^(-1) (3/5)` ` 30.96^(@)` |
|
| 227. |
a projectile is fired from the surface of the earth with a velocity of `5ms^(-1)` and angle `theta` with the horizontal. Another projectile fired from another planet with a velocity of `3ms^(-1)` at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth.The value of the acceleration due to gravity on the planet is in `ms^(-2)` is given `(g=9.8 ms^(-2))`A. `5.9`B. `16.3`C. `110.8`D. `3.5` |
|
Answer» Correct Answer - D `R = (u_(1)^(2))/(g_(1)) sin 2theta = (u_(2)^(2))/(g_(2))sin 2theta` `g_(2) = ((u_(2))/(u_(1)))^(2) g_(1) = ((3)/(5))^(2) xx 9.8 = 3.5 m//s^(2)` |
|
| 228. |
Two particles move in a uniform gravitational field with an acceleration `g`.At the initial moment the particles were located at same point and moved with velocities `u_(1)=9 ms^(-1)` and `u_(2)=4 ms^(-1)` horizontally in opposite directions.The time between the particles at the moment when their velocity vectors are mutually perpendicular in `s` in (take `g=10 ms^(-2)`)A. `0.36`B. `3.6`C. `0.6`D. 6 |
|
Answer» Correct Answer - C `t=sqrt(u_(1)u_(2))/g` |
|
| 229. |
Speed of an object moving in cirular path of radius 10 m with angular speed 2 rad/s isA. 10 m/sB. 5 m/sC. 20 m/sD. 30 m/s |
| Answer» Correct Answer - C | |
| 230. |
An object moving in a circular path at constant speed has constantA. EnergyB. VelocityC. AccelerationD. Displacement |
| Answer» Correct Answer - A | |
| 231. |
Angular speed of a uniformly circulating body with time period T isA. ` 2piT`B. `(2pi)/T`C. `piT`D. `pi/T` |
| Answer» Correct Answer - B | |
| 232. |
Assertion : Acceleration of a moving particle can change its direction without any change in direction of velocity. Reason : If the direction of change in velocity vector changes, the direction of acceleration vector also changes.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 - D `a=(v_(f)-v_(i))/(Delta t) = (dv)/(dt)` i.e., direction of acceleration is same as that of change in velocity vector, or in the direction of `Delta v`. |
|
| 233. |
A car of mass `m` moves in a horizontal circular path of radius `r` metre. At an instant its speed is `V m//s` and is increasing at a rate of `a m//sec^(2)`.then the acceleration of the car isA. `V^(2)/r`B. `a`C. `sqrt(a^(2)+(V^(2)/r)^(2))`D. `sqrt(a+(V^(2)/r)` |
| Answer» Correct Answer - C | |
| 234. |
Assertion : An object may have varying speed without having varying velocity. Reason : If the velocity is zero at an instant, the acceleration may not be zero at that instant.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 - D If speed varies, then velocity will definitely vary. When a particle is thrown upwards, at highest point `a ne 0` but `upsilon = 0`. |
|
| 235. |
A bullet emerges from a barrel of length `1.2 m` with a speed of `640 ms^(1)`. Assuming constant acceleration, after the gun is fired isA. 4mB. 40mC. 400usD. 1s |
|
Answer» Correct Answer - B `s=12m, upsilon=640 ms^(-1)` `a=?, u=0, t = ? 2as = upsilon^(2)-u^(2)` `rArr 2axx12=640xx640rArr a=(8xx64xx100)/(3)` `upsilon=u+at rArr t=(upsilon)/(a)=(640xx3)/(8xx64xx100)=37.5xx10^(-3)s~~40 ms` |
|
| 236. |
A particle moves on a circle of radius `r` with centripetal acceleration as function of time as `a_c = k^2 r t^2`, where `k` is a positive constant. Find the following quantities as function of time at an instant : (a) The speed of the particle (b) The tangential acceleration of the particle ( c) The resultant acceleration, and (d) Angle made by the resultant with tangential direction.A. `kt^(2)`B. `kr`C. `krsqrt(k^(2)t^(4)+1)`D. `krsqrt(k^(2)t^(2)-1)` |
|
Answer» Correct Answer - C From given eqution `omega=kt`, `alpha=(domega)/(dt)=k,a_(t)=ra,a=sqrt(a_(c)^(2)+a_(t)^(2))` |
|
| 237. |
A particle moves in a circular path such that its speed `v` varies with distance `s` as `v = prop sqrt(s)` , where `prop` is a positive constant. Find the acceleration of the particle after traversing a distance `s`.A. `alpha^(2)sqrt(1/4-S^(2)/R^(2)`B. `alpha^(2)sqrt(1/4+S^(2)/R^(2)`C. `alphasqrt(1/4+S^(2)/R^(2)`D. `alpha^(2)sqrt(1/4+S^(2)/R^(2)` |
|
Answer» Correct Answer - B `a=sqrt(a_(t)^(2)+a_(t)^(2))=sqrt(((dv)/(dt))^(2)+((v^(2))/(R))^(2)` |
|
| 238. |
A vector of magnitude 10 has its rectangular components as 8 and 6 along x and y axes. Find the angles it make with these axes. |
| Answer» Correct Answer - `36.86^(@)` with x-axis , ` 53.14^(@)` with 10 N force. | |
| 239. |
Two swimmers start a race. One who reaches the point `C` first on the other bank wins the race.`A` makes his strokes in a direction of `37^(0)` to the river flow with velocity `5km//hr` relative to water. `B` makes his strokes in a direction `127^(0)` to the river flow with same relative velocity.River is flowing with speed of `2km//hr` and is `100m` wide.speeds of `A` and `B` on the ground are `8km//hr` and `6km//hr` respectively. A. `A` will win the raceB. `B` will win the raceC. the time taken by `A` to reach the point `C` is `165 sec`D. the time taken by `B` to reach the point `C` is `150 sec` |
|
Answer» Correct Answer - A::D time to cross river `=100/(5xxsin 37xx5/18)=120 sec` Drift covered `=(2+5xxcos 37)xx5/18xx120=200` Distance covered by walk `=100` meters and time taken `=20` seconds And similarly for swimmer `B` |
|
| 240. |
The x and y components of a vector are ` 4sqrt3 m` and 4 m respectively. What angle does the vector make with positive x-axis ? |
| Answer» Correct Answer - `30^(@)` | |
| 241. |
A magnitude of vector `vecA,vecB` and `vecC` are respectively `12, 5` and `13` units and `vecA+vecB=vecC` then the angle between `vecA` and `vecB` isA. 5 unit due eastB. 25 unit due westC. 5 units due westD. 25 units due east |
|
Answer» Correct Answer - B `vecA = 5` units due east `therefore - 5hatA = -5`(5 units due east) ` = - 25` units due east = 25 units due west |
|
| 242. |
In the question number 20, a unit vector perpendicular to the direction of `vecA and vecB ` isA. `(-2hati-hatj-hatk)/(sqrt(6))`B. `(2hati+hatj+hatk)/(sqrt(6))`C. `(2hati-hatj-hatk)/(sqrt(6))`D. `(2hati-hatj+hatk)/(sqrt(6))` |
|
Answer» Correct Answer - A Unit vector perpendicular to the direction of `vecA and vecB` is `hatn = ((vecA xx vecB))/(|vecAxx vecB|)` `vecA xx vecB = |{:(hati,,hatj,,hatk),(-2,,3,,1),(1,,2,,-4):}|` ` = hati (-12-2) + hatj(1-8) + hatk(-4-3) = -14hati - 7hatj - 7hatk` `|vecAxxvecB| = sqrt((1-14)^(2) + (-7)^(2) +(-7)^(2)) = 7sqrt(6)` `therefore hatn = (-14hati - 7hatj -7hatk)/(7sqrt(6)) = (-2hati-hatj -hatk)/(sqrt(6))` |
|
| 243. |
A vector has both magnitude and direction. Does that mean anything that has magnitude and direction is necessarily a vector ? The rotation of a body can specified by the direction of the axis of rotation and the angle of rotation about the axis. Does the make any rotation a vector ? |
| Answer» No. Finite rotation of a body about an axis is not a vector because finite rotations do not obey the laws of vector addition. | |
| 244. |
`vecA,vecB,vecC,vecD,vecE` and `vecF`are coplanar vectors having the same magnitude each of `10 units` and angle between successive vectors is `60^(@)` The magnitude of resultant isA. `0` unitsB. `1` unitsC. `2` unitsD. `3` units |
| Answer» Correct Answer - A | |
| 245. |
The distance covered by an object (in meter) is given by `s=8 t^(3)-7t^(2)+5t` Find its speed at t = 2 s. |
|
Answer» As speed `=("Distance")/("Time")` `therefore` Speed `=(ds)/(dt)` `=(d)/(dt)(8t^(3)-7t^(2)+5t)=24t^(2)-14+5` `therefore` Speed object at any instant of time is `(24t^(2)-14t+5)ms^(-1)` At t = 2 Speed `=24(2)^(2)-14(2)+5=73 ms^(-1)` |
|
| 246. |
The x and y coordinates of a particle at any time t are given by `x=7t+4t^2` and `y=5t`, where x and t is seconds. The acceleration of particle at `t=5`s isA. zeroB. `8 ms^(-2)`C. `20 ms^(-2)`D. `40 ms^(-2)` |
|
Answer» Correct Answer - A `a=sqrt(a_(x)^(2)+a_(y)^(2))=[((d^(2)x)/(dt^(2)))^(2)+((d^(2)y)/(dt^(2)))^(2)]^(1//2)` Here `(d^(2)y)/(dt^(2))=0` Hence, `a=(d^(2)x)/(dt^(2))=8 ms^(-2)` |
|
| 247. |
In the s-t equation `(s=10+20 t-5t^(2))` match the following columns. `|{:(,"Column I",,"Column II"),((A),"Distancec travelled in 3s",(p),-20" units"),((B),"Displacement 1 s",(q),15" units"),((C ),"Initial acceleration",(r ),25" units"),((D),"Velocity at 4 s",(s),-10" units"):}|` |
|
Answer» Correct Answer - A::B::C::D At `t = 3s, s=10+20(3)-5(3)^(2)=25` unit `upsilon =(ds)/(dt)=20-10t` At `t=4 s, upsilon = 20-10(4)=-20` unit `a=(d upsilon)/(dt)=-10` units |
|
| 248. |
Statement -1 : A man standing on the ground has to hold his umbrella at an angle ` theta` with the vertical to protect himself from rain. It is possible that the man now starts running on the ground with certain speed, but he still has to hold his umbrella at angle ` theta` with the vertical to protect himself from rain. Statement -2 : Vertical component of velocity of rain w.r.t man does not change. if he starts running on a horizontal ground.A. Statement -1 is Ture, statement -2 is Ture, Statement -2 is a correct explanation for statement -1 .B. Statement -1 is True, statement -2 is True, statement - 2 is NOT a correct explanation for statement -3C. Statement -1 is True, Statement -2 si FalseD. Statement -1 is False ,Statement -2 is True. |
| Answer» Correct Answer - B | |
| 249. |
A particle is dropped under gravity from rest from a height `h(g = 9.8 m//sec^2)` and it travels a distance `9h//25` in the last second, the height `h` is.A. 100 mB. `12.5 m`C. 145 mD. `167.5 m` |
|
Answer» Correct Answer - B Let h be distance covered in t second `rArr h=(1)/(2)g t^(2)` Distance covered in t th second `=(1)/(2)g(2t-1)` `rArr (9h)/(25)=(g)/(2)(2t-1)` From above two equations, `h=122.5 m` |
|
| 250. |
Two projectiles `A` and `B` are fired simultaneously as shown in figure. They collide in air at point at time `t`.Then A. `t(u_(1)cos theta_(1)-u_(2)cos theta_(2))=20`B. `t(u_(1)sin theta_(1)-u_(2)sin theta_(2))=10`C. Both (a) and (b) are correctD. Both (a) and (b) are wrong |
|
Answer» Correct Answer - B `X_(A)=X_(B)` `:. 10+(u_(1)cos theta_(1))t=30-(u_(2)cos theta_(2))t` or `t(u_(1)cos theta_(1)+u_(2)cos theta_(2))=20` `y_(A)=y_(B)` `:. 10+(u_(1)sin theta_(1))t-1/2"gt"^(2)` `=20+(u_(2)sin theta_(2))t-1/2"gt"^(2)` `:. (u_(1)sin theta_(1)-u_(2)sin theta_(2))t=10` |
|