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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.
| 251. |
If a body starts from rest and travels 120 cm in the 6 second then what is the accelerationA. `0.20 ms^(-2)`B. `0.027 ms^(-2)`C. `0.218 ms^(-2)`D. `0.03 ms^(-2)` |
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Answer» Correct Answer - C `S_(n)=u+(a)/(2)(2n-1) rArr 1.2=0+(a)/(2)(2xx6-1)` `rArr a=(1.2xx2)/(11)=0.218 ms^(-2)` |
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| 252. |
Two projectiles are projected simultaneously from the top and bottom of a vertical tower of height `h` at angles `45^(@)` and `60^(@)` above horizontal respectively.Body strike at the same point on ground at distance `20m` from the foot of the tower after same time. The ratio of the speed of the projectile projected from the top and the speed of the projectile projected from the bottom of tower isA. `1:sqrt(2)`B. `1:sqrt(3)`C. `sqrt(5):1`D. `sqrt(7) :1` |
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Answer» Correct Answer - A `R_(1)=(U cos theta)t=U/sqrt2sqrtsqrt3` `U/sqrt(2)2sqrtsqrt3=20,U=(10sqrt2)/(root(4)(3)) U/V=1/sqrt2` |
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| 253. |
Two bodies `A and B` are projected from the same place in same vertical plane with velocities `v_(1) and v_(2)`.Form a long inclined plane as shown Find the ratio of their times of flight. A. `(v_(1)sin theta)/v_(2)`B. `(2v_(1)sin theta)/v_(2)`C. `(v_(1)sin theta)/(2v_(2))`D. `(v_(1)cos theta)/v_(2)` |
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Answer» Correct Answer - A `T_(1)/T_(2)=(v_(1)sin theta)/v^(2)` |
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| 254. |
A boat takes `4 hrs` to travel certain distance in a river in down stream and it takes `6 hrs` to travel the same distance in upstream. Then the time taken by the boat to travel the same distance in still water isA. `4.8 hrs`B. `9.8 hrs`C. `24 hrs`D. `10 hrs` |
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Answer» Correct Answer - A `t_(1)=d/(V_(B)+V_(W)),t_(2)=d/(V_(B)-V_(W)),t_(3)=d/V_(B)` |
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| 255. |
A point moves with uniform acceleration and `v_(1), v_(2)`, and `v_(3)` denote the average velocities in the three successive intervals of time `t_(1).t_(2)`, and `t_(3)` Which of the following Relations is correct?.A. `(upsilon_(1)-upsilon_(2)):(upsilon_(2)-upsilon_(3))=(t_(1)-t_(2)):(t_(2)+t_(3))`B. `(upsilon_(1)-upsilon_(2)):(upsilon_(2)-upsilon_(3))=(t_(1)+t_(2)):(t_(2)+t_(3))`C. `(ipsilon_(1)-upsilon_(2)):(upsilon_(2)-upsilon_(3))=(t_(1)-t_(2)):(t_(1)-t_(3))`D. `(upsilon_(1)-upsilon_(2)):(upsilon_(2)-upsilon_(2)-upsilon_(3))=(t_(1)-t_(2)):(t_(2)-t_(3))` |
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Answer» Correct Answer - B Average velocity in uniformly accelerated motion is given by, `upsilon_(av)=(s)/(t)=(ut+(1)/(2)at^(2))/(t)=u+(1)/(2)at` Now, `upsilon_(1)=u+(1)/(2)at_(1), upsilon_(2)=(u+at_(1))+(1)/(2)at_(2)` and `upsilon_(3)=u+a(t_(1)+t_(2))+(1)/(2)at_(3)` `therefore upsilon_(1)-upsilon_(2): upsilon_(2)-upsilon_(3)=(t_(1)+t_(2)):(t_(2)+t_(3))` |
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| 256. |
An object starts from rest, and moves under the acceleration ` veca = 4hati` . Its position after 3 s is given by ` vecr = 7 hati + 4hatj` . What is its initial position ? |
| Answer» Correct Answer - ` -11 hati + 4hatj` | |
| 257. |
A particle moves with constant acceleration along a straight line streaing from rest. The percentage increase in its displacement during the 4th second compared to that in the 3rd second isA. `33%`B. `40%`C. `66%`D. `77%` |
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Answer» Correct Answer - B We know that `s_(nth)=u+(1)/(2)a(2n-1)` `s_(3rd)=0+(1)/(2)a(2xx3-1)=(5)/(2)a " " ("for " n= 3s)` `s_(4rd)=0+(1)/(2)a(2xx4-1)=(7)/(2)a" " ("for " n=4s)` So, the percentage increase `=(s_(4th)-s_(3rd))/(s_(3rd))xx100` `=((7)/(2)a-(5)/(2)a)/((5)/(2)a)xx100=((2a)/(2))/((5)/(2)a)xx100=2xx20=40%` |
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| 258. |
Two particles are projected simultaneously with the same speed `v` in the same vertical plane with angles of elevation `theta`, and `2theta`, where `theta lt 45^(@)`.At what time will velocities be parallel?A. `t=v/(g) "tan"theta/2 "cosec"(3theta)/2`B. `t=v/(g) "cos"theta/2 "cot"(3theta)/2`C. `t=v/(g)"cos"theta/2 "tan"(3theta)/2`D. `t=v/(g) "cos" theta/2 "cosec"(3theta)/2` |
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Answer» Correct Answer - D Velocity of particle after time `t` is `vecV_(1)=(v cos theta hati+v sin theta hatj)-(ghatj)t` `vecV_(2)=(v cos 2theta hati+v sin 2theta hatj)-(ghatj)t` To be parallel of `V_(1)` and `V_(2)` `rArr (v cos theta)/(v cos 2theta)=(v sin theta-"gt")/(v sin 2theta-"gt")` Solving the above equation, we get, `t=v/gcos(theta/2)cosec(3theta)/2` |
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| 259. |
Velocity and acceleration of a particle at time t = 0 are `u=(2hat(i)+3hat(j))m//s` and `a=(4hat(i)+2hat(j))m//s^(2)` respectively. Find the velocity and displacement of particle at t = 2 s. |
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Answer» Here, acceleration `a=(4hat(i)+2hat(j))m//s^(2)` is constant. So, we can apply `v=u+at` and `s=ut+(1)/(2)at^(2)` Substituting the proper values, we get `v=(2hat(i)+3hat(j))+(2)(4hat(i)+2hat(j))` `=(10hat(i)+7hat(j)) m//s` and `s=(2)(2hat(i)+3hat(j))+(1)/(2)(2)^(2)(4hat(i)+2hat(j))` `=(12hat(i)+10hat(j))m` Therefore, velocity and displacement of particle at t = 2s are `(10hat(i)+7hat(j))m//s` and `(12hat(i)+10hat(j))m` respectively. |
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| 260. |
Two particles are projected from the same point, with the same speed, in the same vertical plane, at different angles with the horizontal.A frame of references is fixed to one particle.The position vector of the other particle as observed from this frame is `vecr`Which of the following statements are corrects?A. direction of `vecr` does not changeB. `vecr` changes in magnitudes and direction with timeC. The magnitude of `vecr` increases linearly with timeD. The direction of `vecr` changes with time, its magnitude may or may not change, depending on the angles of projection |
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Answer» Correct Answer - A::C Let `u`= the speed of projection, `theta_(1)` and `theta_(2)`= the angles of projection. Let `vecr_(1)` and `vecr_(2)` be the position vectors of the two particles in a ground frame. `vecr_(1)` `=(u cos theta_(1)),thati+[(usintheta_(1))t-1/2"gt"^(2)]hatj` `vecr_(2)=(u cos theta_(2))t hati+[(usintheta_(2))t-1/2"gt"^(2)]hatj` The position vector of one particle with respect to another is `vecr=vecr_(1)-vecr_(2)=[u (cos theta_(1)-cos theta_(2))t]hati+[u(sin theta_(1)-sin theta_(2)t]hatj=at hati+bt hatj=[ahati+bhatj]t`, where `a` and `b` are constants. |
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| 261. |
Two inclined planes OA and OB having inclination (with horizontal) `30^(@) and 60^(@)`, respectively, intersect each other at O as shown in figure. A particle is projected from point P with velocity `u = 10sqrt3 ms^(-1)` along a direction perpendicular to plane OA. If the particle strikes plane OB perpendicularly at Q, calculate The vertical height h of P from O,A. The time of flight `2s`B. The velocity with which the particle strikes the plane `OB=10 m//s`C. The height of the point `P` from point `O` is `5m`D. The distance `PQ=20m` |
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Answer» Correct Answer - A::B::C::D Consider the motion of particle along the axes shown in figures.We have `u_(x)=u, a_(x)=-g sin 60^(@)` `u_(y)=0, a_(y)=-g cos 60^(@)` (a) As the particle strikes the plane `OB` perpendicularly, `:.v_(x)=0` as `v_(x)=u_(x)+a_(x)t` or `0=u-g sin 60^(@)t rArr t=u/(g sin 60^(@))=(10sqrt3)/(10xxsqrt3/2)=2s` (b) Initial velocity along `y`-axis is zero.The velocity along `y`-axis after `2 s ,v_(y)=u_(y)+a_(y)t` `=0-g cos 60^(@) xx2=-10xx1/2xx2=-10m//s` (c) We have , `v_(x)^(2)=u_(x)^(2)+2a_(x)s` Since `v_(x)=0` and `a_(x)=g sin 60^(@),u=10sqrt3m//s` `:.0=(10sqrt3)^(2)-2xxg sin 60^(@)xx(OQ)` `OQ=(10^(2)xx3)/(2xx10xxsqrt3/2)=10sqrt3m` Distance`PO=0+1/2g sin 30^(@)xx(2)^(2)` `=1/2xx10xx1/2xx4=10m` Therefore height `h` of point `P`, `h=PQ sin 30^(@)=10xx1/2=5m` (d) Distance `PQ=sqrt(PO^(2)+OQ^(2))` `=PQ=sqrt((10)^(2)+(10sqrt3)^(2))=20m` |
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| 262. |
A body is projected at an angle `theta` so that its range is maximum.If `T` is the time of flight then the value of maximum range is (acceleration due to gravity =`g`)A. `(g^(2)T)/2`B. `(gT)/2`C. `(gT^(2))/2`D. `(g^(2)T^(2))/2` |
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Answer» Correct Answer - C for maximum range `theta=45^(@)` `R_(max)=u^(2)/g` when`theta=45^(@),T=(2usintheta)/g` |
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| 263. |
A particle is projected at an angle with horizontal. If `T` is the time of flight and `R` is a horizontal range, then `theta` isA. `cot^(-1)((g T^(2))/(R ))`B. `cot^(-1)((g T^(2))/(2 R ))`C. `tan^(-1)((2R )/(g T^(2)))`D. `tan^(-1)((g T^(2))/(2 R))` |
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Answer» Correct Answer - D `T = (2 u sin theta)/(g), R = (u^(2))/(g).2 sin theta cos theta` `(T^(2))/(R ) = (4 u^(2) sin^(2) theta//g^(2))/(2u^(2) sin theta cos theta//g) = (2 tan theta)/(g)` `tan theta = (g T^(2))/(2 R) rArr theta = tan^(-1)((g T^(2))/(2 R))` |
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| 264. |
A particle is projected from the ground with velocity `u` at angle `theta` with horizontal.The horizontal range,maximum height and time of flight are `R,H` and `T` respectively.Now keeping `u` as fixed, `theta` is varied from `30^(@)` to `60^(@)`.ThenA. `R` will first increase.`H` will increase and `T` will decreaseB. `R` will first increase.than decrease while `H` and `T` both will increaseC. `R` will decrease while `H` and `T` will increaseD. `R` will increase while `H` and `T` will decrease |
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Answer» Correct Answer - B `R prop sin 2 theta,H prop sin^(2) theta` and `T prop sin theta,sin 2theta` will first increase, then decrease.While `sin theta` will only increase. |
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| 265. |
A particle is thrown such that its time of flight is `10 s` and horizontal range is `500 m`. (i) `H_(max) = 125 m` (ii) `u = 50 sqrt(2) m//s` (iii) `theta = 45^(@)` (iv) velocity at highest point `= 50 m//s`A. (i), (ii), (iii)B. (ii), (iii), (iv)C. (i), (ii), (iv)D. all option are correct |
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Answer» Correct Answer - D `T = (2u sin theta)/(g) = 10 rArr u sin theta = 50` `R = u cos theta T rArr 500 = u cos theta xx 10 rArr u cos theta = 50` (i) `H_(max) = (u^(2) sin^(2) theta)/(2g) = ((50)^(2))/(2 xx 10) = 125 m` (ii) `u = sqrt((u sin theta)^(2) + (u cos theta)^(2)) = 50 sqrt(2) m//s` (iii) `tan theta = (u sin theta)/(u sin theta) = 1 rArr theta = 45^(@)` (iv) velocity at highest point `= u cos theta = 50 m//s` |
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| 266. |
A foot ball is kicked off with an initial speed of `19.6m//s` to have maximum range. Goal keeper standing on the goal line `67.4 m` away in the direction of the kick starts running opposite to the direction of kick to meet the ball at that instant. What must his speed be if he is to catch the ball before it hits the ground? |
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Answer» `R=(u^(2)sin 2theta)/g=((19.6)^(2)sin90)/9.8` or `R=39.2` metre. Man must run `67.4 m-39.2m=28.2m` in the time taken by the ball to come to ground Time taken by the ball. `t=(2u sin theta)/g=(2xx19.6xxsin45^(@))/9.8=4/sqrt2` `t=2sqrt2=2xx1.41=2.82sec`. Velocity of man`=(28.2m)/(2.82sec)=10m//sec`. |
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| 267. |
If a body is thrown with a speed of `19.6m//s` making an angle of `30^(@)` with the horizontal,then the time of flight isA. `1 s`B. `2 s`C. `2sqrt3s`D. `5 s` |
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Answer» Correct Answer - B `T=(2u sin theta)/g` B. 2 s is the answer |
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| 268. |
In the time taken by the projectile to reach from `A` to `B` is `t`. Then the distance `AB` is equal to. .A. `(ut)/sqrt3`B. `sqrt3ut/2`C. `sqrt3ut`D. `2ut` |
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Answer» Correct Answer - A `u_(x)=ucostheta=ucos60^(@)` `x=AC=u_(x).t` , from figure `cos 30^(@)=(AC)/(AB)` |
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| 269. |
The relation between the time of flight of projectile `T_(f)` and the time to reach the maximum height `t_(m)` isA. `T_(f) = 2t_(m)`B. `T_(f) = t_(m)`C. `T_(f) = (t_(m))/(2)`D. `T_(f) = sqrt(2)(t_(m))` |
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Answer» Correct Answer - A Time to reach maximum height ` = t_(m)` Time to reach back to ground ` = t_(m)` Total time of flight, `T_(f) = t_(m) + t_(m) = 2t_(m)` |
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| 270. |
A particle projected at some angle with velocity 50 m/s crosses a 20 m high wall after 4 s from the time of projection. The angle of porjection of the particle isA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `53^(@)` |
| Answer» Correct Answer - A | |
| 271. |
A projectile is thrown at angle `beta` with vertical.It reaches a maximum height `H`.The time taken to reach the hightest point of its path isA. `sqrt(H/g)`B. `sqrt((2H)/g)`C. `sqrt(H/(2g))`D. `sqrt((2H)/(g cos theta))` |
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Answer» Correct Answer - B `t=sqrt((2H)/g)` |
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| 272. |
A particle is projected in `xy` plane with `y`-axis along vertical, the point of projection is origin. The equation of the path is `y=sqrt3x-g/2x^(2)`.where `y` and `x` are in `m`.Then the speed of projection in `ms^(-1)` isA. 2B. `sqrt3`C. 4D. `sqrt3/2` |
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Answer» Correct Answer - A Compare the equation with `y=x tan theta-g/(2u^(2)cos^(2) theta)x^(2)` |
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| 273. |
Galileo, in his book Two new sciences, stated that "for elevations which exceed or fall short of `45^(@)` by equal amounts, the ranges are equal. Prove this statement. |
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Answer» For a projectile launched with velocity `v_(0)` at an angle `theta_(0)`, the range is given by `R=(V_(0)^(2)sin2theta_(0))/g` Now, for angles, `(45^(@)-alpha)`, `2theta_(0)` is `(90^(@) + 2alpha)` and `(90^(@)-2alpha)`, respectively. The value of `sin(90^(@)+2alpha)` and `sin(90^(@)-2alpha)` are the same, equal to that of `cos 2alpha`. Therefore, ranges are equal for elavations which exceed or fall short of `45^(@)` by equal amounts `alpha`. |
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| 274. |
A projectile is projected with the initial velocity `(6i + 8j) m//s`. The horizontal range is `(g = 10 m//s^(2))`A. `96 m`B. `960 m`C. `9.6 m`D. `4.8 m` |
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Answer» Correct Answer - C `vec(u) = 6 hat(i) + 8 hat(j) = u cos theta hat(i) + u sin theta hat(j)` `R = (u^(2))/(g)sin 2 theta = (2)/(g) (u cos theta)(u sin theta)` `= (2)/(10)(6)(8) = 9.6 m` |
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| 275. |
Galileo writes that for angles of projection of a projectile at angles `(45 + theta)` and `(45 - theta)`, the horizontal ranges described by the projectile are in the ratio of (if `theta le 45`)A. `2 : 1`B. `1: 2`C. `1 : 1`D. `2 : 3` |
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Answer» Correct Answer - C `R_(1) = (u^(2))/(g) sin{2(45 + theta)} = (u^(2))/(g) cos 2 theta` `R_(2) = (u^(2))/(g) sin {2(45 - theta)} = (u^(2))/(g) cos 2 theta` `(R_(1))/(R_(2)) = (1)/(1)` |
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| 276. |
Two particles are projected with same speed but at angles of projection `(45^(@)-theta)` and `(45^(@)+theta)`. Then their horizontal ranges are in the ratio ofA. `1:2`B. `2:1`C. `1:1`D. none of the above |
| Answer» Correct Answer - C | |
| 277. |
A stone is thrown horizontally with velocity `g ms^(-1)` from the top of a tower of height `g` metre. The velocity with which it hits the ground is `("in" ms^(-1))`A. `g`B. `2g`C. `sqrt3g`D. `4g` |
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Answer» Correct Answer - C `V=sqrt(u^(2)+2gh)` |
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| 278. |
Four balls `A, B, C` and `D` are projected with equal velocities having angles of projection `15^(@)`, `30^(@)`, `45^(@)` and `60^(@)` with the horizontal, respectively. The ball having the smallest range isA. `A`B. `B`C. `C`D. `D` |
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Answer» Correct Answer - A `R_(A) alpha sin 30^(@), R_(B) alpha sin 60^(@)` `R_( C) alpha sin 90^(@), R_(D) alpha(sin 120^(@) = sin 60^(@))` `R_(A)` is the smallest. |
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| 279. |
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. Then |
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Answer» During the projectile motion, angle at any instant `t` is such that `tan alpha=(u sin theta-g t)/(u cos theta)` For `t=2` seconds, `alpha=30^(@)` `1/sqrt3=(usintheta-2g)/(ucostheta)`...(1) For `t=3` seconds, at the highest point `alpha=0^(@)` `0=(usin theta-3g)/(ucostheta)` `usintheta=3g`....(2) using eq.(1) and eq. (2) `ucos theta=sqrt(3)g`....(3) Eq.(2) `div` eq. (3) give `theta=60^(@)` squaring and adding equation (2) and (3) `u=20sqrt(3) m//s`. |
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| 280. |
A stone is thrown with a speed of `10 ms^(-1)` at an angle of projection `60^(@)`. Find its height above the point of projection when it is at a horizontal distance of 3m from the thrower ? (Take `g = 10 ms^(-2)`) |
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Answer» Considering the equation of trajectory. ` y = ( tan theta_(0)) x - g/(2 (v_(0)^(2) cos^(2) theta_(0))) x^(2)` Here ` theta_(0) = 60^(@)` ` v_(0) = 10 ms ^(-1)` x =3m ` y= ( tan 60^(@))3 - 10/(2 (100 cos^(2) 60^(@)) (3)^(2)` ` = 3 sqrt3 - 9/5 ` ` ( 15 sqrt3 -9)/ 5 ` m = 3.396 m |
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| 281. |
Two balls are thrown from an inclined plane at angle of projection `alpha` with the plane, one up the incline and other down the incline as shown in figure (`T` stands for total time of flight): A. `h_(1)=h_(2)=(v_(0)^(2)sin^(2)alpha)/(2g cos theta)`B. `T_(1)=T_(2)=(2v_(0)sin alpha)/(g cos theta)`C. `R_(2)-R_(1)=g(sin theta)T_(1)^(2)`D. `v_(1)=v_(2)` |
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Answer» Correct Answer - A::B::C::D `h_(1)h_(2)=(V_(0)^(2)sin ^(2)alpha)/(2g cos alpha)` `T_(1)T_(2)=(2V_(0)sin alpha)/(g cos theta)` `R_(1)=V_(0)cos alphaT_(1)-1/2g sin theta T_(1)^(2)` `R_(2)=V_(0)cos alphaT_(1)-1/2g sin theta T_(1)^(2)` `rArr R_(2)-R_(1)=g sin thetaT_(1)^(2)` |
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| 282. |
The displacement x of an object is given as a funstion of time, `x=2t+3t^(2)`. The instantaneous velocity of the object at t = 2 s isA. `16 ms^(-1)`B. `14 ms^(-1)`C. `10 ms^(-1)`D. `12 ms^(-1)` |
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Answer» Correct Answer - B `x=2t+3t^(2)` `upsilon=(dx)/(dt)=2+6t` For `t=2s, upsilon=2+6(2)=14m//s` |
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| 283. |
The velocity of a particle is given by `v=(2t^(2)-3t+10)ms^(-1)`. Find the instantaneous acceleration at t = 5 s. |
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Answer» Ghiven, `v=(2t^(2)-3t+10)ms^(-1)` `rArr a_("in")=(dv)/(dt)=(4t-3)ms^(-2)` If `t = 5s, a_("in")=5xx4-3=17 ms^(-2)` |
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| 284. |
The position of a particle is given by `vecr = 3t hati + 2t^(2) hatj + 5hatk`, where t is in seconds and the coefficients have the proper units for `vecr` to be in metres. The direction of velocity of the particle at `t = 1` s isA. `53^(@)` with x-axisB. `37^(@)` with x-axisC. `30^(@)` with y-axisD. `60^(@)` with y-axis |
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Answer» Correct Answer - A Given : `vecr = 3thati +2t^(2)hatj + 5hatk` Velocity, `vecv = (dvecr)/(dt) = (d)/(dt) (3thati +2t^(2)hatj+5hatk) = 3hati + 4t hatj ms^(-1)` Let `theta` be angel which the direction of `vecv` makes with the x-axis.Then `tan theta = (v_(y))/(v_(x)) = (4t)/(3) = (4)/(3) or theta = tan^(-1)((4)/(3)) = 53^(@)` |
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| 285. |
Velocity-time equation of a particle moving in a straight line is, `v=(10+2t+3t^2)` (SI units) Find (a) displacement of particle from the mean position at time `t=1s,` if it is given that displacement is 20m at time `t=0`. (b) acceleration-time equation. |
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Answer» (i) The given euation can be written as, `v=(ds)/(dt)=(10+2t+3t^(2))` or `ds = (10+2t+3t^(2))dt` or `int_(20)^(s)ds= int_(0)^(1)(10+2t+3t^(2))dt` or `s=20 = [10t+t^(2)+t^(3)]_(0)^(1)` or `s=20+12=32 m` (ii) Acceleration - time equation can be obtained by differentiating the given equaiton w.r,t, time, Thus, `a=(dv)/(dt)=(d)/(dt)(10+2t+3t^(2)) or `a=2+6t` |
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| 286. |
The motion of a particle along a straight line is described by the function `x=(2t -3)^2,` where x is in metres and t is in seconds. Find (a) the position, velocity and acceleration at `t=2 s.` (b) the velocity of the particle at origin. |
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Answer» (i) Posiotion of the particle, `x = (2t-3)^(2)` At `t=2s, x=(2xx2-3)^(2)=1.0 m` Velocity, `v=(dx)/(dt)=4(2t-3)ms^(-1)` At `t = 2 s, v= 4(2xx2-3)4ms^(-1)` and acceleration of the particle, `a=(dv)/(dt)=8ms^(-2)` (ii) At origin, x = 0 or `(2t-3)^(2)=0` Velocity of the particle at origin, `v=4xx0=0` |
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| 287. |
The displacement of a particle moving in a straight line is described by the relation `s=6+12t-2t^(2)`. Here `s` is in metre and `t` in second. The distance covered by the particle in first `5s` isA. 20 mB. 32 mC. 24 mD. 26 m |
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Answer» Correct Answer - D `upsilon=(ds)/(dt)=12-4t` Comparing with `upsilon=u+at, u=12 ms^(-1)` and `a=-4 ms^(-2)` Velocity will become zero at time, `0=12-4t_(0)` or `t_(0)=3 s`. Since, the given time t = 5 s is greater than `t_(0)=3s` `"distance" gt|"displacement"|` Distance `d=|s_(0-t_(0))|+|s_(t-t_(0))|=(u^(2))/(2|a|)+(1)/(2)|a|(t-t_(0))^(2)` `=((12)^(2))/(8)+(1)/(2)xx4xx(2)^(2)=26 m` |
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| 288. |
A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point. .A. AB. BC. CD. D |
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Answer» Correct Answer - C In the given distance-time graph at the point C, slope of the graph is maximum thus at this point C instantaneous velocity of the particle is maximum. |
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| 289. |
The range of a projectile, when launched at an angle of ` 15^(@)` with the horizontal is 1.5 km. what is the range of the projectile, when launched at an angle of ` 45^(@)` to the horizontal with the same speed ?A. 1.5 kmB. 3.0 kmC. 6.0 kmD. 0.75 km |
| Answer» Correct Answer - C | |
| 290. |
From a point `A` on bank of a channel with still water a person must get to a point `B` on the opposite bank.All the distances are shown in figure.The person uses a boat to travel across the channel and then walks along the bank of point `B`.The velocity of the boat is `v_(1)` and the velocity of the walking person is `v_(2)`.Prove that the fastest way for the person to get from `A` to `B` is to select the angles `alpha_(1)` and `alpha_(2)` in such a manner that A. `(sin alpha_(1))/(sinalpha_(2))=v_(2)/v_(1)`B. `(sin alpha_(1))/(sinalpha_(2))=v_(1)/v_(2)`C. `(cos alpha_(1))/(cos alpha_(2))=v_(2)/v_(1)`D. `(cos alpha_(2))/(cos alpha_(1))=v_(1)/v_(2)` |
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Answer» Correct Answer - A `S_(1)=x/(sin alpha_(1)),S_(2)=(d-x)/(sin alpha_(2)),t_(1)=S_(1)/v_(1)=x/(v_(1)sin alpha_(1))` and `t_(2)=S_(2)/v_(2)=(d-x)/(v_(2)sin alpha_(2))` `t=x[1/(v_(1)sin alpha_(1))-1/(v_(2)sin alpha_(2))]+d/(v_(2)sin alpha_(2))` For `t` be minimum `(dt)/(dx)=0` or `1/(v_(1)sin alpha_(1))=1/(v_(2)sin alpha_(2)),v_(1)/v_(2)=sin alpha_(2)/sin alpha_(1)` |
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| 291. |
An insect trapped in a circular groove of radius, 12 cm moves along the groove steadily and completes 7 revolutions in 100s. What is the angular speed and the linear speed of the motion ? |
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Answer» Correct Answer - ` omega = 0.44` rad/s v = 5.3 cm ` s^(-1)` |
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| 292. |
An insect trapped in a circular groove of radius `12 cm` moves along the groove steadily and complete `7` revolutions in `100` seconds.The linear speed of the motion in `cm//s`A. `5.3`B. `4`C. `3`D. `5` |
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Answer» Correct Answer - A `v=s/t=(2pirN)/t` |
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| 293. |
An insect trapped in circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100s. The linear speed of the insect isA. 4.3 cm `s^(-1)`B. 5.3 cm `s^(-1)`C. 6.3 cm `s^(-1)`D. 7.3 cm `s^(-1)` |
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Answer» Correct Answer - B Here, r=12 cm, frequency, `v=(7)/(100)rps` The angular speed of the insect is `omega=2piv=2pixx(7)/(100)=0.44rads^(-1)` The linear speed of the insect is `v=omegar=0.44xx12=5.3cms^(-1)` |
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| 294. |
A stone is thrown with a velocity `v` at an angle `theta` with the horizontal.Its speed when it makes an angle `beta` with the horizontal isA. `v cos theta`B. `v/(cos beta)`C. `v cos theta cos beta`D. `(v cos theta)/(cos beta)` |
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Answer» Correct Answer - D `V cos theta=V^(1) cos beta , :.V^(1)=(V cos theta)/(cos beta)` |
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| 295. |
If the angle between the vectors `vecA and vecB` is `theta,` the value of the product `(vecB xx vecA) * vecA` is equal toA. `BA^(2) cos theta`B. `BA^(2) sin theta`C. `BA^(2) sin theta cos theta`D. zero |
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Answer» Correct Answer - D We know that scalar triple product is cyclic, `(vec(B)xxvec(A)).vec(A)=(vec(A)xxvec(B)).vec(A)=(vec(A)xxvec(A)).vec(B)=vec(0).vec(B)=0` |
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| 296. |
A particle is thrown with the speed u at an angle `alpha` with the horizontal. When the particle makes an angle ` beta ` the horizontal , its speed will beA. ` u cos alpha `B. `u cos alpha.sec beta`C. ` u cos alpha.cos alpha`D. ` u sec alpha.cos beta` |
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Answer» Correct Answer - Bhe key here is that, for a projectile (defined to be moving only under the influence of gravitational acceleration), there is no acceleration in the horizontal direction. Hence, its horizontal speed is constant during its entire flight (from the moment it leaves the launcher until it hits something, and is no longer a projectile). Horizontal velocity remains same = u cos(alpha) Let particle velocity at time when it makes angle beta be v m/s. Then, Vertical velocity = v cos(beta) horizontal velocity = vertical velocity u cos(alpha) = v cos(beta) v = u cos(alpha)/cos(beta) => v=ucos(alpha)sec(beta) |
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| 297. |
Vectors `vecA` and `vecB` include an angle `theta` between them. If `(vecA + vecB)` and `(vecA- vecB)` respectively subtend angles `alpha and beta` with `vecA`, then `(tan alpha + tan beta)` isA. `((A sin theta))/((A^(2) +B^(2)cos^(2)theta)) `B. `((2AB sin theta))/((A^(2) -B^(2)cos^(2)theta)) `C. `((A^(2) sin^(2) theta))/((A^(2) +B^(2)cos^(2)theta)) `D. `((B^(2) sin^(2) theta))/((A^(2) -B^(2)cos^(2)theta)) ` |
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Answer» Correct Answer - B `tan alpha = (B sin theta)/(A + B cos theta)" "…(i)` where `alpha ` is the angle made by the vector `(vecA + vecB)` with `vecA`. Similarly, `tan beta = (B sin theta)/(A - B cos theta) " "…(ii)` where `beta` is the angle made by the vector `(vecA - vecB)` with `vecA`. Note that the angle between `vecA and (-vecB)` is `(180^(@) - theta)`. Adding (i) and (ii), we get `tan alpha + tan beta = (B sin theta)/(A + b cos theta) + (B sin theta)/(A -Bcostheta)` ` = (AB sin theta - B^(2) sin thetacos theta + AB sin theta + B^(2) sintheta cos theta)/((A + B cos theta)(A-Bcos theta))` ` = (2AB sin theta)/((A^(2) - B^(2) cos ^(2) theta))` |
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| 298. |
What is the resultant of unit vector s ` hatj and hatk` ? [ Hint : ` |hatj| = |hatk| =1`] |
| Answer» Correct Answer - `sqrt2 "at" 45^(@)` with y-axis in z-y plane. | |
| 299. |
One of the rectangular components of a force of 40 N is 20 N. Find the angle it makes with the component. |
| Answer» Correct Answer - `60^(@)` | |
| 300. |
A vector of magnitude 13 makes an angle ` 65.37^(@)` with the x-axis . What is its component along positive y-axis ? |
| Answer» Correct Answer - 11.82 | |