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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.
| 1. |
A particle is thrown above, then correct `v-t` graph will beA. B. C. D. |
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Answer» Correct Answer - a u=u+at `rArrvpropt` (straight line graph) As body goes upward (positive direction) speed of body decreases at uniform rate and becomes zero at the highest point. After that, the body falls downward (negative direction) and speed increases at uniform rate. |
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| 2. |
The position `x` of a particle with respect to time `t` along the x-axis is given by `x=9t^(2)-t^(3)` where `x` is in meter and `t` in second. What will be the position of this particle when it achieves maximum speed along the positive `x` directionA. 54mB. 81mC. 24mD. 32m |
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Answer» Correct Answer - a `x=9t^(2)-t^(3)` ...........(i) `therefore v=(dx)/(dt)=(d)/(dt)(9t^(2)-t^(3))=18-3t^(2)` For speed u to be maximum, the first derivative shou ld be zero and the second derivative shou ld be negative. `therefore (dv)/(dt)=18-6t` and `(d^(2)v)/(dt^(2))=-6` =18-6t t=3s Thus, the speed will be maximum at t=3s. From eqn. (i). `x=9(3)^(2)-(3)^(3)=54m` |
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| 3. |
Two trains are moving with equal speed in opposite directions along two parallel railway tracks. If the wind is blowing with speed `u` along the track so that the relative velocities of the trains with respect to the wind are in the ratio `1:2,` then the speed of each train must beA. 3uB. 2uC. 5uD. 4u |
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Answer» Correct Answer - a `3u-u=2u` Ratio 1:2 `3u+u=4u` Hence speed is 3u. |
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| 4. |
An acroplane is flying horizontally with a velocity of 600 km/h and a height of 1960m. When it is vectrically above a point A on the ground a bomb is released from it. The bomb strikes the ground at point B. the distance AB is:A. 1200mB. 0.33kmC. 333.3kmD. 3.33km |
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Answer» Correct Answer - d Time of flight, `T=sqrt((2h)/(g))=sqrt((2xx1960)/(9.8))=20`s `therefore` distance AB=uT=`600xx((20)/(3600))=3.33km` |
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| 5. |
A grasshopper finds that he can jump a maximum horizontal distance of 0.8m. With what speed can be travel along the road if he spends a negligible time on the ground?A. 2m/sB. 2.8m/sC. 104m/sD. 1m/s |
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Answer» Correct Answer - a `R=(u^(2))/(g)` `rArr u^(2)=0.8xx10` `rArru=sqrt(8)m//s` |
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| 6. |
A block of ice starts sliding down from the top of the inclined roof of a house (angle of inclination of roof=`30^(@)` with the horizontal) along aline of maximum slope. The highest and lowest points of the roof are at heights of 8.1m and 5.6 m respectively from the ground. At what horizontal distance from the starting point will the block hit the ground? Neglect friction. |
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Answer» When the ice block silides down from A to B, it falls vertically through AC=2.5m If v is the velocity with which it reaches B, then `v^(2)=u^(2)+2as`, where u=0,a=`g sin 30^(@)` and s=AB. `thereforev^(2)=2gxx(AC)/(AB)xxAB(sin 30^(@)=(AC)/(AB))` `=2xx9.8xx2.5=49` `therefore v=7ms^(-1)` along AB, Resolve this velocity v into: (i) a horizontal component `=v cos 30^(@)=7xx(sqrt(3)//2)=3.5xxsqrt(3)ms^(-1)` Vertical motion: Let t be the time taken by the block to reach the ground. Apply `s=ut+(1)/(2)g t^(2)` Given `s=5.6m,u=3.5ms^(-1),g=9.8ms^(-2)` `therefore 5.6=3.5t+(1)/(2)xx9.8t^(2)` or `4.9t^(2)+3.5t-5.6=0` Dividing this equation by 0.7, we get `7t^(2)+5t-8=0` `therefore t=(-5+0sqrt((5)^(2)+(4xx7xx8)))/(2xx7)` `therefore t=(-5+-15.78)/(14)` Rejecting the -ve value, `=(-5+15.78)/(14)=(10.78)/(14)=0.77s` Horizontal Motion: Horizontal distance travelled EF =Horizontal velocity xx time `=3.5xxsqrt(3)xx0.77=4.668m` `therefore` Total horizontal distance =DF=DE+EF `=4.33+4.668=8.998m` `(DE=BC=AC cot 30^(@)=2.5xxsqrt(3)=4.33m)` |
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| 7. |
An open merry go round rotates at an angular velocity `omega`.A person stands in it at a distance `r` from the rotational axis.It is raining and the rain drops falls vertically at a velocity `v_(0)`.How should the person hold an umbrella to protect himself from the rain in the best way.Angle made by umbrella with the vertical isA. `cot alpha=(v_(0))/(romega)`B. `tan alpha=(v_(0))/(romega)`C. `cot alpha=(romega)/(v_(0))`D. `tan alpha=(v_(0))/(romega)` |
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Answer» Correct Answer - a Let the velocity of the drops above the preson rel. to the merry go round be at an angle `alpha` to the vertical. This angle can be determined from the velocity triangle `v_(0)=v_("rel")+v_(m.g.r.)`, `v_("rel")=v_(0)-v_(m.g.r)` `v_(m.g.r)=romega, cot alpha=(v_(0))/(romega)` |
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| 8. |
Three balls A,B and C are projected from ground with same speed at same angle with the horizontal. The balls A,B and C collie with the wall during a flight in air and all three collide perpendicu lary and eleastically with the wall as shown in Fig. 5.119 it the time taken by the ball A to fall back on ground is 4 seconds and that by ball B is 2 seconds. Then the time taken by the ball C to reach the ground after projection will be: |
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Answer» Correct Answer - 6 `T=(2u sin theta)/(g)=4s`.....(i) Since in second case going up time is 1s and coming back time 1s. So in third case particle will take 3s to hit the plane and 3s to come back so total time in case 3 is 6s. |
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| 9. |
Particle projected from tower fo heigh 10m as shown in figure. Find the time (in sec) after which particle will hit the ground. |
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Answer» Correct Answer - 2 `u_(y)=v sin 45^(@)=5sqrt(2)xx(1)/sqrt(2)=5ms^(-1)` Using, `y=u_(y)t+(1)/(2)a_(y)t^(2)` `-10=5t+(1)/(2)(-10)t^(2)` `t^(2)-t=2=0` `(r-2)(t+1)=0` `therefore` t=2 (rejecting -ve time) |
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| 10. |
A ball is projected from the ground with velocity v such that its ranege is maximum. |
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Answer» Correct Answer - `ararrp;brarrq;crarrr;drarrs` The range of projectile is maximum. `therefore theta=45^(@)` and `H=(v^(2)sin^(2)45^(@))/(2g)=(v^(2))/(4g)` (a) Using, `v_(y)^(2)-u_(y)^(2)=2a_(y)Y` `v_(y)^(2)-(v^(2))/(2)=2(-g)(H)/(2)` `=-g((v^(2))/(4g))=-(v^(2))/(4)` `therefore v_(y)^(2)=(v^(2))/(4) rArr v_(y)=(v)/(2)` `therefore (a)rarr(p)` (b) At the maximum height, the ball has only horizontal velocity. `therefore v_(x)=(v)/sqrt(2)` `therefore (b)rarr(q)` (c) Horizontal velocity remains the same but the vertical velocity gets reversed. `therefore Deltav_(y)=(v)/sqrt(2)-(-(v)/sqrt(2))=(2v)/sqrt(2)=vsqrt(2)` `therefore (c)rarr(r)` (d) `vecv_(av)=(1)/(2)(vecv_(1)+vec_(f))` `=(1)/(2)[(v)/sqrt(2)(hati+hatj)+(v)/sqrt(2)hati]` `=(1)/(2)[(2v)/sqrt(2)hati+(v)/sqrt(2)hatj]` `therefore |vecv_(av)|=(1)/(2)sqrt(((4v^(2))/(2)+(v^(2))/(2)))=(v)/(2)sqrt((5)/(2))` `therefore (d) rarr(s)` |
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| 11. |
Water is flowing through a horizontal pipe fixed at a height of 2 m above the ground as shown in Fig. 5.90. water strikes ground at a distance of 3m from the pipe. The speed of water as it leaves the pipe is: A. 47m/sB. 4.7m/sC. 9.4m/sD. 4.9m/s |
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Answer» Correct Answer - b `x=uT=usqrt((2h)/(g))` `x=usqrt((2h)/(g))` `3=usqrt((2xx2)/(9.8))` `u=(3)/(2)sqrt(9.8)=4.7m//s` |
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| 12. |
During volcanic eruption chunks of slid rock are blasted out of the volcano . (a) At what initial speed Wou ld a valcanic object have to be ejected at `37^(@)` to the horizontal from the vent A in order to fall at B as shown in Fig. 5.22? (b) what is the time of flight? (g=9.8`m//s^(2)`) |
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Answer» (a) Taking the point A as origin , with upward direction of motion postive and using the equation, `s=ut+(1)/(2)at^(2)` for horizontal motion: `x=u cos thetaxxt` ...(i) and for vertical motion: `y=u sin thetaxxt-(1)/(2)g t^(2)`.....(ii) Substituting r from Eqn. (i) in (ii), `y=xtan theta-(g)/(2)[(x)/(u cos theta)]^(2)` Here `y=3.3xx10^(3)m,x=9.4x10^(3)m` and `theta=37^(@)` So, `-3.3xx10^(3)=9.4xx10^(3) tan 37^(@)-4.9[(9.4xx19^(3))/(u cos 37^(@))]^(2)` or `(9.4xx10^(3))/(u cos 37^(@))=[(10.35xx10^(3))/(4.9)]^(1//2)=46 [tan37^(@)=(3)/(4)]` or `u=(9.4xx10^(3))/(46xx(4//5))=(47000)/(184)=255m//s` (b) Substituting the above value of u in Eqn. (i). `t=(x)/(ucos theta)=(9.4xx10^(3))/(255xx(4//5))` `=(9400)/(204)=46s` |
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| 13. |
A shell is projected from a gun with a muzzle velocity `u`.The gun is fitted with a trolley car at an angle `theta` as shown in the fig. If the trolley car is made to move with constant velocity `v` towards right, find the horizontal range of the shell relative to ground. A. `R=(2u sin theta(u cos theta+v))/(g)`B. `R=(2u sin theta(u cos theta-v))/(g)`C. `R=(u sin theta(u cos theta+v))/(2g)`D. `R=(u sin theta(u cos theta+v))/(g)` |
| Answer» Correct Answer - d | |
| 14. |
An artillary piece which consistently shoots its shells with the same muzzle speed has a maximum range R. To hit a target which is `R//2` from the gun and on the same level, the elevation angle of the gun should beA. `15^(@)`B. `45^(@)`C. `30^(@)`D. `60^(@)` |
| Answer» Correct Answer - a | |
| 15. |
An object in projected up the inclined at the angle shown in the figure with an initial velocity of `30ms^(-1)`.The distance `x` up the incline at with the object lands is A. 600mB. 104mC. 60m/sD. 208m |
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Answer» Correct Answer - c From figure: `alpha=30^(@),theta=60^(@)` `R=(2u^(2)costhetasin(theta-alpha))/(g cos ^(2)alpha)` |
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| 16. |
Two inclined planes OA and OB having inclinations `30^@` and `60^@` with the horizontal respectively intersect each other at O, as shown in figure. A particle is projected from point P with velocity `u=10 sqrt(3) m//s` along a direction perpendicular to plane OA. If the particle strikes plane OB perpendicular at Q. Calculate. (a) time of flight, (b) velocity with which the particle strikes the plane OB, (c) height h of point P from point O, (d) distance PQ. (Take `g=10m//s^(2)`) |
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Answer» We have taken x and y directions along OB and OA respectively. Then `u_(x)=u=10sqrt(3)m//s,u_(y)=0` `a_(x)=-g sin 60^(@)=-5sqrt(3)m//s^(2)` and `a_(y)=-g cos 60^(@)=-5m//s^(2)` (a) At point Q, x-component of velocity is zero. Hence substituing in `v_(x)=u_(x)+a_(x)t` `0=10sqrt(3)-5sqrt(3)t` or `t=(10sqrt(3))/(5sqrt(3))=2s` (b) At point Q. `v=v_(y)=u_(y)+a_(y)t` `therefore v=0-(5)xx(2)=-10m//s` Here negative sign impu lies that velocity of particle at Q is along negative y-direction. (c) Distance PO= |displacement of particle along y-direction| Here, PO=`|u_(y)t+(1)/(2)a_(y)t^(2)|=|0-(1)/(2)xx5xx(2)^(2)|` `-|-10m|=10m` Therefore, `h=PO sin 30^(@)=10xx(1)/(2)=5m` (d) Distance OQ=displacement of particle along x-direction or `OQ=10sqrt(3)m` `therefore PQ=sqrt((PO)^(2)+(OQ)^(2))` `=sqrt((10)^(2)+(10sqrt(3))^(2))=20m` |
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| 17. |
The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmospehre) are given by `y=(8t-5t^(2))` metre and x=6t metre, where t is in seconds. The velocity of projection is:A. 8m/sB. 6m/sC. 10m/sD. not obtianed from the data |
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Answer» Correct Answer - c `y=(u_(x))/(3)-(5x^(2))/(36)` Now, `y=x tan theta-(gx^(2))/(2u^(2)cos^(2)theta)` `u=10m//s` |
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| 18. |
A bomber plane moving at a horizontal speed of `20 m//s` releases a bomb at a height of `80 m` above ground as shown. At the same instant a Hunter of negligible height starts running from a point below it, to catch the bomb with speed `10 m//s`. After two seconds he relized that he cannot make it, he stops running and immediately hold his gun and fires in such direction so that just before bomb hits the ground, bullet will hit it. What should be the firing speed of bullet (Take `g = 10 m//s^(2)`) A. 10m/sB. `20sqrt(10)` m/sC. `10sqrt(10)` m/sD. None of these |
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Answer» Correct Answer - b In 2 sec. horizontal distance travelled by bomb=20xx2=40m In 2 sec. vertical distance travelled by bomb `=(1)/(2)xx10xx2^(2)=20m` In 2 sec. horizontal distance travelled by hunter `=10xx2=20m`. Time remaining for bomb to hit ground `=sqrt((2xx80)/(10))-2=2` Let `v_(x)` and `v_(y)` be the velocity components of bu llet along horizontal and vertical direction. thus we use `(2v_(y))/(g)=2 rArr v_(y)=10m//s` and `(20)/(v_(x)-20)=2 rArr v_(x)=30m//s` Thus velocity of firing is `v=sqrt(v_(X)^(2)+v_(y)^(2))=10sqrt(10)m//s` |
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| 19. |
A particle moves on a circle of radius r with centripetal accelration as function of time as `a_(c)=K^(2)rt^(2)` where k is a positive constant , find the resu ltant acceleration.A. `kt^(2)`B. `kr`C. `krsqrt(k^(2)t^(4)+1)`D. `krsqrt(k^(2)t^(2)-1)` |
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Answer» Correct Answer - c From given equation `omega=kt` `alpha=(domega)/(dt)=k, alpha_(t)=ralpha, alpha=sqrt(a_(c)^(2)+a_(r)^(2))` |
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| 20. |
If the radius of earth is 6400km, calcu late (a) angu lar velocity (b) linear velocity and (c) radial accelration for a point on its equator considering its spin motion alone. What will be the values of above quantities if the pint is at the pole? |
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Answer» As earth makes one rotation on its axis every 24hrs- (a) `omega_(s)=(2pi//T)=(2pi//24xx60xx60)=7.3xx10^(-5)` rad/s (b) `v=4omega=(6.4xx10^(6))(7.3xx10^(-5))=4.7xx10^(2)m//sapprox0.5` km/s (c) `a_(r)=romega^(2)=(6.4xx10^(6))(7.3xx10^(-5))^(2)=3.4xx10^(-2)m//s^(2)` Now, if the point is at pole, `r rarr0` so `v=(romega)` and `a_(r)(=romega^(2))` will become zero while `omega` (chracteristic of rigid body ) will remain unchanged. |
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| 21. |
A gun is fired from a moving platform and the ranges of theshot are observed to be R and S when platform is moving forward or backward respectively with velocity v prove that the elevation of the gun is `tan ^(-1)[(g(R-S)^(2))/(4V^(2)(R+S))]` |
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Answer» Let u and v be the horizontal and vertical components of velocity when the platform is at rest. When the platform is moving forward, the rsu ltant horizontal velocity is `=u+V` and vertical velocity is v. the range R is given by `R=(2(u+V)v)/(g)`.....(i) When the plat form is moving backward with velocity V, the horizontal and vertical components of the velocity of the shot are u-V and v respectively. The range is then `S=(2(u-V)v)/(g)` .......(ii) From Eqns. (i) and (ii), `R+S=4u v//g` and `R-S=4Vv//g` `therefore ((R-S)^(2))/((R+S))=(16J//^(2)v^(2)//g^(2))/(4uv//g)=(4v^(2)v)/(ug)` If `alpha` be the angle of projection, then `tan alpha=v//u` Thus, `((R-S)^(2))/((R+S))=(4V^(2))/(g) tan alpha` `tan alpha=(g(R-S)^(2))/((R+S))` or `alpha=tan^(-1)[(g(R-S)^(2))/(4V^(2)(R+S))]` |
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| 22. |
Mathch Column -I with Column-B: |
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Answer» Correct Answer - `ararrq,r;brarrp;s;crarrq,r;drarrq,r` (a) As the body is accelrating, tis velocity increases. Therefore, the motion will be non-uniform linear motion. `therefore (a)rarr(q,r)` (b) As velocity `(v=72//6=12ms^(-1))` remains the same, the motion is univorm linear motion. Also x-t graph is a straight line inclined to x-axis. `therefore (b)rarr(p,s)` (c) When body is thrown vertically upward, the velocity decreases (while going upward) and then increases (while coming downward). Therefore, the motion is non-uniform linear motion. The motion is under the action of force of gravity, the `therefore (c)rarr(q,r)` (d) When a bu llet is fired into air, it is an example of non-linear motion. Also,linear momentum does not remain conserved as the motion is under the action of force of gravity. `therefore (d)rarr(q,r)` |
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| 23. |
`V_(x)` and `V_(y)` are the horizontal and vertical compounds of velocity with x and y as the corresponding displacements along horizontal and vertical at any time t in a projectile motion in XY co-ordinate system, where g is the acceleration due to gravity. |
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Answer» Correct Answer - `ararrs;brarrs;crarrq;drarrs` In projectile motion horizontal component of velocity remains constant. Hence tis will be straight line II to time axis. y-t graph is parabolic. |
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| 24. |
Two bodies are projected from the same point with same speed in the directions making an angle `alpha_(1)` and `alpha_(2)` with the horizontal and strike at same point in the horizontal plane through a point of projection. If `t_(1)` and `t_(2)` are their time of flights.Then `(t_(1)^(2)-t_(2)^(2))/(t_(1)^(2)+t_(2)^(2))`A. `(tan (alpha_(1)-alpha_(2))/(tan (alpha_(1)+alpha_(2)))`B. `(sin (alpha_(1)+alpha_(2)))/(sin(alpha_(1)-alpha_(2)))`C. `(sin(alpha_(1)-alpha_(2)))/(sin(alpha_(1)+alpha_(2)))`D. `(sin^(2)(alpha_(1)-alpha_(2)))/(sin^(2)(alpha_(1)+alpha_(2)))` |
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Answer» Correct Answer - c `alpha_(1)+alpha_(2)=90^(@)` `rArrsin(alpha_(1)+alpha_(2))=1` `r_(a)=(2usin alpha_(1))/(g). T_(2)=(2u sin alpha_(2))/(g)` |
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| 25. |
A body is dropped from a plane moving with constant horizontal velociy. The path of the body as seen by a person on the plane will beA. straight lineB. parabolicC. hyperbolicD. none of these |
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Answer» Correct Answer - a As sen by the person in plane path is straight line as horizontal velocity for plane and person remains same. |
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| 26. |
A motor boat has a speed of `5 m//s`.At time `t=0`,its position vector relative to a origin is `(-11hati+16hatj) m`,having the aim of getting as close as possible to a steamer.At time `t=0`,the steamer is at the point `(4hati+36hatj) m` and is moving with constant velocity `(10hati-5hatj) m//s`.Find the direction in which the motorboat must steerA. `3hati+3haj`B. `3hati+4hatj`C. `4hati+3hatj`D. `4hati+4hatj` |
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Answer» Correct Answer - c In order to approach as close as possible to steamer, the direction of motion of motorboat shou ld be perpendicu lar to the relative motion. Let the optimum velocity of motorboat is `vecv_(B)=ahati+bhatj` Velocity of steamer is `vecv_(S)=10hati-5hatj` `vecv_(BS)=vecv_(S)=(a-10)hati+(b+5)hatj` Now, `vecv_(B)botvecv_(BS)` or `vecv_(B).vecv_(BS)=0` or `a(a-10)+b(b+5)=0` ...........(i) But we know that speed of the motorboat is 5m/s. so `a^(2)+b^(2)=25` ..............(ii) Solving Eqns. (i) and (ii), we get a=0 or 4 When a=0,b=-5 and a=4,b=3. Hence either `vecv_(B)=-5hatj` or `vecv_(B)=4hati+3hatj`. However a diagram shows that when `vecv_(B)=-5hatj`. the motorboat is moving futher aways form the steamer . So, `vecv_(B)=4hati+3hatj`. (ii) Now at time `t vecr_(B)=-(1hati+16hatj)+(4hati+3hatj)t` and `vecr_(S)=(4hati+36hatj)+(10hati-5hatj)t` If the motorboat can reach the steamer, there is a value of t for which `vecr_(B)=vecr_(S)` equating the coefficient of `hati`, we get t=-5/2 negative time implies that the motorboat cannot ever react the steamer. |
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| 27. |
A particle moves in x-y plane according to ru le x`=a sin omegat` and y=`a cos omegat`. The particles follows:A. an elliptical pathB. a circu lar pathC. a parabolic pathD. a straight line path equally inclined to x and y-axes |
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Answer» Correct Answer - b `(x^(2))/(a^(2))=sin2 omegat` and `(y^(2))/(a^(2))=cos^(2)omegat` `therefore (x^(2))/(a^(2))+(y^(2))/(a^(2))=1` `x^(2)+y^(2)=a^(2)` Thus, particle follows circu lar path. |
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| 28. |
A stone is thrown with a velocity V making an angle `theta` with the horizontal. The horizontal distance covered by it, before it falls to the ground, is maximum when `theta` is equal to:A. `0^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)` |
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Answer» Correct Answer - c `R=(u^(2)sin 2theta)/(g)rArr2theta=90^(@)rArrtheta=45^(@)` |
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| 29. |
The gretest height to which a man can throw a stone is h. the greatest distance to which he can throw will be:A. h/2B. hC. 2hD. 4h |
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Answer» Correct Answer - c `h_("max")=(u^(2))/(2g),R_("max")=(u^(2))/(g)=2h` |
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| 30. |
In case of projectle motion if two projectles A and B are projected with same speed at angles `15^(@)` and `75^(@)` respectively to the horizontal then:A. `H_(A)gtH_(B)`B. `H_(A)ltH_(B)`C. `T_(A)gtT_(B)`D. `T_(A)ltT_(B)` |
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Answer» Correct Answer - b,d `H=(u^(2)sin^(2)theta)/(2g),T=(2usin theta)/(g)` |
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| 31. |
A man swims with avelocity `v_(mw)` in still water. When the water moves with a velocity `u_(w)(ltv_("new")^(2))` the man crosses the river to and fro in minimum time `T_(1)`. If the man intends to cross the river perpendicu larly, he takes time `T_(2)` for to and fro journey. Now he swims in the donwstream and comes back to his initial position by swimming upstream along the shore. For to and fro journey along the shore, the man takes a time `T_(3)`. find the relation between `T_(1),T_(2)` and `T_(3)` assuming equal distance overed by the man in each case. |
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Answer» As derived earler, the minimum time of crossiing the river is `t_(m)=(d)/(v_(mw))`, where , d=width of the river Henc,e the total time for to an fro journey is `T_(1)=2t_(m)=(2d)/(v_(mw))` For the minimum drift, the time of crossing the river is `t_(1)=(d)/(v_(m))` where, `v_(m)=sqrt(v_(mw)^(2)-v_(2)^(2))` Hence, the total time required for to and fro motion is `T_(2)=2t_(1)=(2d)/(sqrt(v_(mw)^(2)-v_(2)^(2)))` When the man moves in downstream, his velocity is `v_(m_(1))=v_(mw)+v_(w)` and when he moves in upstream his velocity is `v_(m_(2))=v_(mw)-v_(w)`. Hence, the total time for to and fro journey is `T_(3)=r_("down")+t_("up")`. where, `t_("down")=(d)/(v_(m_(1)))=(d)/(v_(mw)+v_(w))` and `t_("up")=(d)/(v_(m_(2)))=(d)/(v_(mw)+v_(w))` Hence, `T_(3)=(2dv_("mw"))/(v_("mw")^(2)-v_(2)^(2))` Comparing the above expressions of `T_(1),T_(2)` and `T_(3)`, we have `T_(2)^(2)=T_(1)T_(3)` |
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| 32. |
A particle is thrown with speed `u` at an angle `alpha` with horizontal from the ground. After how much time, the velocity of particle will make an angle `beta` with horizontal.A. `u cos alpha`B. `u cos alpha "sec" beta`C. `u cos alpha cos beta`D. `u sec alpha cos beta` |
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Answer» Correct Answer - b `u cos alpha=v cos beta` `v =(u cos alpha sec beta)` |
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| 33. |
The x and y displacement of a particle in the x-y plane at any instant are given by `x=alphaT^(@)` and `y=2alphaT` where a is a constant. The velocity of the particle at any instant is given by:A. `4asqrt(T^(2)+4)`B. `2asqrt(T^(2)+1)`C. `4asqrt(T^(2)+1)`D. `(a)/(2)sqrt(T^(2)+4)` |
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Answer» Correct Answer - b `v=sqrt(((dx)/(dt))^(2)+((du)/(dt))^(2))` |
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| 34. |
If co-ordinates of a moving point at time t are given by x=a `(l+sint)` and `y=a(1-cost)`, then:A. the slope of accelration time graph is zeroB. the slope of velocity-time graph is constnatC. the direction of motion makes an angle t/2 with x-axisD. all of the above |
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Answer» Correct Answer - d `(dx)/(dt)=a(1+cos t),(dy)/(dt)=sint` `v=sqrt(((dx)/(dt))^(2)+((dy)/(dt))^(2))` `a=sqrt(((d^(2)x)/(dt^(2)))^(2)+((d^(2)y)/(dt^(2)))^(2))` |
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| 35. |
The maximum height attained by a projectile is increased by 5%. Keeping the angle of projection constant, what is the percentage increases in horizontal range?A. 0.05B. 0.1C. 0.15D. 0.2 |
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Answer» Correct Answer - a `theta=`constant. `H_("max")prop(u^(2))/(g),Rprop(u^(2))/(g)rArr HporpR` |
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| 36. |
A water fountain on the ground sprinkles water all around it . If the speed of water coming out of the fountain is ` v` , the total area around the fountain that gets wet is :A. `(piv^(2))/(g^(2))`B. `(piv^(2))/(g)`C. `(piv^(4))/(g^(2))`D. `(piv^(2))/(2g)` |
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Answer» Correct Answer - c Range, `R=(u^(2))/(g)` `therefore` Maximum area `=piR^(2)=R((v^(4))/(g^(2)))` |
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| 37. |
If a force is applied at an angle to a body moving along a straight line:A. the body continues to move in the direction of forceB. the body continues to move in its initial direction of motionC. the body moves in a fixed direction other than that of force and initial motion.D. the body moves in a direction other than that of force and initial motion wich varies with time. |
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Answer» Correct Answer - d The body will move in a direction other than that of orce which will vary with time. |
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| 38. |
Prove that: The path of a projectile is a straight line. |
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Answer» The co-ordinates of a point on one trajectory relative to other will be `x=x_(2)-x_(1)=(u_(2)cos theta_(@)-u_(1) cos theta_(1))t` and `y=(u_(2)sin theta_(2)t-(1)/(2)g t^(2))-(u_(1)sin tehta_(!)t-(1)/(2)g t^(2))` `(u_(2)sin theta_(2)-u_(1)sin theta_(t))t` So, `(y)/(x)=((u_(2) sin theta_(2)-u_(1)sin theta_(1))/(u_(2)cos theta_(2)-u_(1)cos theta_(1)))=` constt, =m or y=mx which is a straight line |
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| 39. |
A particle is projected from a point of an angle with the horizontal. At any instant t, if p is the linear momentum and E the kinetic energy, then which of the following graph is/are correct?A. B. C. D. |
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Answer» Correct Answer - a (i) `E_(k)=(P^(2))/(2m)rArrE_(k)propP^(2)` Thus, the graph between `E_(k)` and `P^(2)` will be straight line passing through the origin . Hence, option (d) is correct. (ii) `E_(k)=(1)/(2)mv^(2)=(1)/(2)m(v_(X)^(2)+v_(y)^(2))` `=(1)/(2)m(u cos theta)^(2)+(u sin theta-gt)^(2)` `=(1)/(2)m(u^(2)+g^(2)t^(2)-2ugt sin theta)` ........(i) Thus, the graph between `E_(k)` and t will be parabolic with positiv intercept, Hence, option (b) is correct. (iii) `E_(k)=(1)/(2)mv^(2)=(1)/(2)m(u^(2)-2gy)` `E_(k)=-mgy+(1)/(2)mu^(2)` The graph is a straight line with negaive slope and positive intercept. Hence, option (a) is wrong. (iv) From eqn. (i), `E_(k)=(1)/(2)m[u^(2)+g^(2)t^(2)-2ugt sin theta]` `E_(k)=(1)/(2)m[u^(2)+g((x)/(v_(x)))^(2)-2ug((x)/(v_(x)))sintheta]` Thus, the graph between `E_(k)` and x is a parabolic with positive intercept Hence option (c) is correct. |
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| 40. |
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. `4hati-2hatj`C. `4hati+2hatj`D. `2hati+4hatj` |
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Answer» Correct Answer - d `v_(x)=(dx)/(dt)+v_(y)=(dy)/(dt)` `vecv=v_(x)hati+v_(y)hatj` |
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| 41. |
A body has maximum range `R_(1)` when projected up the inclined plane. The same body when projected down, the inclined plane. It has maximum range `R_(2)`. Find its maximum horizontal range. Assume the equal speed of projection in each case and the body is projected onto the gretest slope. A. `R=(2R_(1)R_(2))/(R_(1)-R_(2))`B. `R=(2R_(1)R_(2))/(R_(1)+R_(2)`C. `R=(R_(1)R_(2))/(R_(1)-R_(2))`D. `R=(4R_(1)R_(2))/(R_(1)+R_(2))` |
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Answer» Correct Answer - c For upward projecton, `R_("max")=(V_(0)^(2))/(g(1-sinbeta))=R_(1)` .............(i) For downward projection `R_("min")=(v_(0)^(2))/(g(1+sin beta))=R_(2)` ..........(ii) |
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| 42. |
A particle is moving with velocity ` vecv = k( y hat(i) + x hat(j)) `, where `k` is a constant . The genergal equation for its path isA. `y^(2)=x^(2)+"constant"`B. `y=x^(2)+` constantC. `y^(2)=x+`constantD. xy=constant |
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Answer» Correct Answer - a `vecv=k(yhati+xhatj)` `therefore v_(x)=(dx)/(dt)=ky` `v_(y)=(dy)/(dt)=kx` Dividing, `(dy)/(dx)=(dy//dt)/(dx//dt(x)/(y)` `y dy=x dx` Integrating. `inty dy=int x dx+` constant `(y^(2))/(2)=(x^(2))/(2)+` constant `y^(2)=x^(2)+` constant |
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| 43. |
It was calculated that a shell when fired from a gun with a certain velocity and at an angle of elevation `5pi//36` radius should strike a given target. In actual practice it was found that a hill just intervened in the trajectory. At what angle of elevation should the gun be fired to hit the target ?A. `(5pi)/(36)` radianB. `(7pi)/(36)` radianC. `(11pi)/(36)` radianD. `(13pi)/(36)` radian |
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Answer» Correct Answer - d To hit the same target the projectile can be fixed at complimentary angle i.e., `(pi)/(2)-(5pi)/(36)=(13pi)/(36)` |
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| 44. |
The x and y co-ordinates of a partilce at any time t are given by: `x=7t+4t^(2)` and `y=5t` The acceleration of the particle at 5s is:A. zeroB. `8m//s^(2)`C. `20m//s^(2)`D. `40m//s^(2)` |
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Answer» Correct Answer - b `(d^(2)x)/(dt^(2))=8, (d^(2)y)/(dt^(@))=0` Hence `a=8m//s^(2)` |
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| 45. |
Two towers AB and CD are situated a distance d apart as shown in Fig. 5.45. AB is 20m high and CD is 30m highfrom the ground . An object of mass m is thrown from the top of AB horizontally with a velocity 10m/s towards CD. simu ltaneously another object of mass 2 m is thrown from the top of CD at an angle of `60^(@)` to the horizotnal towards AB with the same magnitude of initial velocity as that of the first object. the two objects move in the same vertical plane, clollide in mid air and stick to each other, (a) Calcu late the distance d between the towers and (b) fin the position where the objects hit the ground. |
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Answer» Let the two particles collide after time r, then from 2nd equation of motion, i.e., `s=ut+(1)/(2)at^(2)` for vertical motion of A, `h_(1)=(1)/(2)g t^(2)` ........(i) ltbr. While for B, `h_(2)=(u sin 60^(@))t+(1)/(2)g t^(@)` ..........(ii) But as the collision will take place at same level, `30-h_(2)=20-h_(1),,i.e., h_(2)-h_(1)=10` Which, in the light of Eqns. (i) and (ii), gives `10xx((sqrt(3))/(2))t=10, i.e.,t=(2)/sqrt(3)s` (a) Now as horizontal motion is independent of vertical motion, so from 2nd Eqn. of motion. i.e. s=ut+(1/2)`at^(2)` distance travelled by m towards 2m horizotnall. `d_(t)=10xx(2)/sqrt(3)=(20)/sqrt(3)m` and distance travelled by 2m towards m horizontally, `d_(2)=u cos thetaxxt` `=10xx(1)/(2)xx(2)/sqrt(3)=(10)/sqrt(3)m`. So, distance between the towers AB and CD, `d=d_(1)+d_(@)=(20)/sqrt(3)=(30)/sqrt(3)=10sqrt(3)=17 m` (b) Applying conservation of linear momentum along horizontal direction `mu-2mu cos 60^(@)=3mV`, i.e., V=0 i.e., the velocity of combined mass fter collision along horizontal is zero, so after collision it will move vertically down from the point of collision and will hit the ground at a distance `d_(1)=20//sqrt(3)` from AB or at a distance `d_(2)=10//sqrt(3)` from CD between BD. |
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| 46. |
The length of minute hand in a pendu lum clock is 10cm, the speed of lip of the hand is (in m/s):A. `(pi)/(6000)`B. `(pi)/(18000)`C. `(pi)/(3600)`D. `(pi)/(1200)` |
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Answer» Correct Answer - b `V=l omega, omega=(2pi)/(T)` [where =`T=60xx60s=3600`s] |
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| 47. |
A circular disc is rotating about its own axis at the rate of `200` revolutions per minute. Two particles `P,Q` of disc are at distances `5cm,10cm` from axis of rotation. The ratio of angular velocities of `P` and `Q` isA. `1:2`B. `1:1`C. `2:1`D. `4:1` |
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Answer» Correct Answer - b Angu lar velocity `omega`=constant |
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| 48. |
A particle is moving at uniform speed `2 ms^(-1)` along a circle of radius `0.5m`.The centripetal acceleration of particle isA. `1ms^(-2)`B. `2ms^(-2)`C. `4 ms^(-2)`D. `8ms^(-2)` |
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Answer» Correct Answer - d Centripetal accelration , a=`(V^(2))/(r)` |
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| 49. |
Prove that for a projectile fired from level ground at an angle `theta` above the horizontal, the ratio of the maximum height H to the range R is given by `(H)/(R)=(1)/(4)tan theta` |
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Answer» `H=(u^(2)sin ^(2)theta)/(2g)` and `R=(u^(2)sin2 theta)/(g)` `(H)/(R)=(sin ^(2)theta)/(2sin 2theta)=(sin^(2)theta)/(4sin theta cos theta)=(1)/(4) tan theta` |
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| 50. |
A shell is fired vertically from a cannon which is travelling at constant speed u=30km/hr. the projectile leaves the connon with a velocity `v_(r)=20` m/s relative to the cannon . Show that the shell will land on the vehicle at the gun location and calcu late the distance by the vehicle during the flight of shell. |
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Answer» Correct Answer - s=34m `T=(2u)/(g)=(2xx20)/(10)=4` sec Distance travelled by the cannon. `s=30xx4xx(5)/(18)m` `s=120xx(5)/(18)` `s=(100)/(3)m` |
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