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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.
| 151. |
The greatest and least resultant of two forces are `7N` and `3N` respectively. If each of the force is increased by `3 N` and applied at `60^(@)`.The magnitude of the resultant isA. `7 N`B. `3 N`C. `10 N`D. `sqrt129N` |
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Answer» Correct Answer - D `P+Q=7,P-Q=3,P^(1)=P+3,Q^(1)=Q+3` `R=sqrt(P^(1^(2))+Q^(1^(2))+2P^(1)Q^(1)costheta)` |
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| 152. |
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` `t_(1)=(2u sin alpha_(1))/(g)`,`t_(2)=(2u sin alpha_(2))/(g)` |
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| 153. |
Eleven forces each equal to `5N` act on a particle simultaneously.If each force makes an angle `30^(@)` with the next one, the resultant of all forces isA. `15 N`B. `55 N`C. `5 N`D. zero |
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Answer» Correct Answer - C No. of forces`=11, n=360/theta=12` polygon formed with `1 side` absent, resultant is closing side. |
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| 154. |
If the angle between two vectors `vecA and vecB " is " 90^(@)` thenA. `vecA = 2vecB `B. `vecA - vecB = vec0`C. `|vecA + vecB| = |vecA -vecB| `D. `vecA + vecB = vec0` |
| Answer» Correct Answer - C | |
| 155. |
If the frequency of an object in unifrom circular motion is doubled, its acceleration becomesA. Two timesB. Four timesC. HalfD. one fourth |
| Answer» Correct Answer - B | |
| 156. |
A particle moves on the curve ` y = x^(2)/4 ` where x=t/2. x and y are meausured in metre and t in second. At t=4s the velocity of particel isA. `1/sqrt2 m//s`B. `sqrt3 m//s`C. `sqrt2 m//s`D. `2sqrt2 m//s` |
| Answer» Correct Answer - A | |
| 157. |
The ceiling of a hall is 40m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of `56ms^(-1)` without hitting the ceiling of the hall isA. `25^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)` |
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Answer» Correct Answer - B Here, `u = 56 ms^(-1)` Let `theta` be the angle of projection with the horizontal to have maximum range, with maximum height = 40 m Maximum height, `H = (u^(2) sin^(2) theta)/(2g)` ` 40 = ((56)^(2) sin^(2) theta)/(2 xx 9.8)` `sin^(2) theta = (2xx 9.8 xx 40)/((56)^(2)) = (1)/(4) or sin theta = (1)/(2) = sin^(-1) ((1)/(2)) = 30^(@)` |
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| 158. |
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 third turn.A. `500 m`B. `250 m`C. `1000 m`D. `1500 m` |
| Answer» Correct Answer - C | |
| 159. |
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 sixth turn.A. `3000 m`B. `1500 m`C. `0 m`D. `1000 m` |
| Answer» Correct Answer - C | |
| 160. |
Calculate the angular speed of the hour hand of a clock . |
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Answer» The hour completes one round in 12 hours. One round make an angular displacement ` 2pi` ` omega = (Deltatheta)/(Deltat)` `= ( 2 pi rad)/( 12 hr) ` ` ( 2pi)/(12xx 3600) "rad"//s^(-1)` `pi/21600 " rad" s^(-1)` |
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| 161. |
Find the angular speed of the minute hand of a clock. |
| Answer» Correct Answer - 0.0017 ` rad s^(-1)` | |
| 162. |
the length of seconds hand of a clock is `10 cm`. The speed of the tip of the hand is ……… .A. `pi/6000`B. `pi/18000`C. `pi/36000`D. `pi/1200` |
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Answer» Correct Answer - B `v=lomega` |
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| 163. |
Which of the following graphs cannot possibly represent one dimensional motion of a particle. .A. I and IIB. II and IIIC. II and IVD. All four |
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Answer» Correct Answer - D I is not possible because total distance covered by a particle cannot decrease with time. II is not possible because at a particular time, position cannot have two values. III is not possible because at a particle time, valocity cannot have two values. IV is not possible speed can never be negative. |
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| 164. |
A person throws a bottle into a dustbin at the same height as he is `2m` away at an angle of `45^(@)`.The velocity of thrown isA. `g`B. `sqrtg`C. `2g`D. `sqrt(2g)` |
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Answer» Correct Answer - D `R=(u^(2)sin 2theta)/g,2=u^(2)/g,u=sqrt(2g)` |
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| 165. |
The launching speed of a certain projectile is five times the speed it has at its maximum height.Its angles of projection isA. `theta=cos^(-1)(0.2)`B. `theta=sin^(-1)(0.2)`C. `theta=tan^(-1)(0.2)`D. `theta=0^(@)` |
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Answer» Correct Answer - A `u=5u sin theta rArr cos theta=1/5` |
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| 166. |
A body lying initially at point (3,7) starts moving with a constant acceleration of `4 hati`. Its position after 3s is given by the coordinatesA. (7,3)B. (7,18)C. (21,7)D. (3,7) |
| Answer» Correct Answer - C | |
| 167. |
A particle starts from the origin of coordinates at time t = 0 and moves in the xy plane with a constant acceleration `alpha` in the y-direction. Its equation of motion is `y= betax^2`. Its velocity component in the x-direction isA. `sqrt((alpha)/(beta))`B. `sqrt((beta)/(alpha))`C. `sqrt((alpha)/(2 beta))`D. `sqrt((beta)/(2 alpha))` |
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Answer» Correct Answer - D `y = alpha x^(2) rArr (d y)/(d t) = alpha.2x (d x)/(d t)` `v_(y) = 2 alpha x v_(x)` `(dv_(y))/(d t) = 2 alpha(x(d v_(x))/(d t) + v_(x).(d x)/(d t))` `a_(y) = 2 alpha(x a_(x) + v_(x)^(2))` `a_(x) = 0, a_(y) = beta` `beta = 2 alpha v_(x)^(2) rArr v_(x) = sqrt((beta)/(2 alpha))` OR This equation is similar to the equation of trajectory of a projectile thrown horizontally from the top of the tower. `y = alpha x^(2), y = (g x^(2))/(2 u^(2)) = (beta x^(2))/(2 u^(2))` `alpha = (beta)/(2 u^(2)) rArr u = sqrt((beta)/(2 alpha))` |
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| 168. |
A particle starts from the origin of coordinates at time t = 0 and moves in the xy plane with a constant acceleration `alpha` in the y-direction. Its equation of motion is `y= betax^2`. Its velocity component in the x-direction isA. VariableB. `sqrt((2alpha)/beta)`C. `alpha/(2beta)`D. `sqrt(alpha/(2beta))` |
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Answer» Correct Answer - D `y=betax^(2)` `rArrV_(y)=2xbetaV_(x)rArr V_(y)=2betaV_(x)rArr alphat=2betat.V_(x)^(2)` `rArrV_(x)^(2)=alpha/(2beta)rArrV_(x)=sqrt(alpha/(2beta))` |
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| 169. |
The driver of a car moving towards a rocket launching pad with a speed of 6 m `s^(-1)` observed that the rocket is moving with speed of 10 m `s^(-1)`. The upward speed of the rocket as seen by the stationery observer is nearlyA. 4 m `s^(-1)`B. 6 m `s^(-1)`C. 8 m `s^(-1)`D. 11 m `s^(-1)` |
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Answer» Correct Answer - D Observed speed, `v = sqrt((10)^(2) + (6)^(2)) = sqrt(136)` ` = 11 ms^(-1)` |
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| 170. |
A mark on the rim of a rotating cicular wheel of 0.50 m radius is moving with a speed of `10 ms^(-1)` . Find its angular speed. |
| Answer» Correct Answer - `20 rad s^(-1)` | |
| 171. |
The component of a vector `r` along X-axis will have maximum value ifA. r is along positive Y-axisB. r is along positive X-axisC. r makes an angle of `45^(@)` with the X-axisD. r is along negative Y-axis |
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Answer» Correct Answer - B Let r makes an angle `theta` with positive x-axis component of r along X-axis `r_(x)=|r| cos theta` `(r_(x))_("maximum")=|r| (cos theta)_("maximum")` `=|r| cos 0^(@)=|r| " "` (`:. Costheta ` is maximum of `theta=0^(@)`) As `" " theta=0^(@)`, r is along positive x-axis. |
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| 172. |
A person throws balls into the air one after the other at an interval ofone second. The next ball is thrown when the velocityof the ball thrown earlier is zero. To what height the ball rise:A. 2 mB. 5 mC. 8 mD. 10 m |
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Answer» Correct Answer - B Time taken to reach maximum height is 1 s. Height = free fall distance in `1 s=(1)/(2)g t^(2)=5 m` |
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| 173. |
Velocity of a body moving a straight line with uniform acceleration (a) reduces by `(3)/(4)` of its initial velocity in time `t_(0)`. The total time of motion of the body till its velocity becomes zero isA. `(4)/(3)t_(0)`B. `(3)/(2)t_(0)`C. `()/(3)t_(0)`D. `(8)/(3)t_(0)` |
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Answer» Correct Answer - A `(u)/(4)=u-at_(0) " " (therefore a=(3u)/(4t_(0)))` or `(u)/(a)=(4)/(3)t_(0)` Now, `0=u-at` or `t=(u)/(a)=(4)/(3)t_(0)` |
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| 174. |
Give examples where a. the velocity of a particle is zero buts its acceleration is not zero. b.the velocity is opposite in direction to the acceleration, c. the velocity is perpendicular to the acceleration. |
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Answer» (i) A particle thrown upwards has its velocity in opposite direction to its acceleration (g, dowanwards). (ii) When the particle is released from rest from a certain height, its velocity is zero, while acceleration is g acting pendulum velocity is zero, while acceleration is not zero. (iii) In uniform circular motion velocity is perpendicular to its radial or centripetal acceleration. |
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| 175. |
The velocity-time graph of particle comes out to be a non-linear curve. The motion isA. uniform velocity motionB. uniformly accelerated motionC. non-uniform accelerated motionD. Nothing can be said about the motion |
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Answer» Correct Answer - C Velocity - time graph gives the instaneous value of velocity at any instant. For non-uniformly accelerated motion, `upsilon-t` graph is non-liner. |
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| 176. |
A person reaches on a point directly opposite on the other bank of a river.The velocity of the water in the river is `4 m//s`and the velocity of the person in still water is `5 m//s`.If the width of the river is `84.6m`, time taken to cross the river in seconds isA. `28.2`B. `9.4`C. 2D. `84.6` |
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Answer» Correct Answer - A Given, velocity of water `upsilon_(w)=4 ms^(-1)` Velocity of person `upsilon_(p)=5 ms^(-1)`, width of the river `= 84.6 m` We know that time taken, `t=(s)/(sqrt(upsilon_(w)^(2)-upsilon_(p)^(2)))` `rArr t=(84.6)/(sqrt(25-16))` `rArr t=(84.6)/(3)~=28.25` |
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| 177. |
A person reaches on a point directly opposite on the other bank of a river.The velocity of the water in the river is `4 m//s`and the velocity of the person in still water is `5 m//s`.If the width of the river is `84.6m`, time taken to cross the river in seconds isA. 9.4B. 2C. 84.6D. 28.2 |
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Answer» Correct Answer - D `t=d/sqrt(v_(b)^(2)-v_(w)^(2))` |
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| 178. |
It is found that `|A+B|=|A|`,This necessarily implies.A. `vecB = 0 `B. `vecA, vecB` are antiparallelC. `vecA, vecB` are perpendicularD. `vecA * vecB le 0 ` |
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Answer» Correct Answer - B If `|vecA + vecB| = |vecA|`, then either `vecB = - or vecB = -2vecA.` Both are satisfied when `vecA and vecB` are anti-parallel. |
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| 179. |
It is found that `|A+B|=|A|`,This necessarily implies.A. B=0B. A, B are antiparallelC. A, B are perpendicularD. `A.B le 0` |
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Answer» Correct Answer - A::B Given that `" " |A+B|=|A| or |A+B|^(2)=|A|^(2)` `rArr " " |A|^(2)+|B|^(2)+2|A||B|cos theta=|A|^(2)` where `theta` is angle between A and B. `rArr " " |B|(|B|+2|A|cos theta)=0` `rArr " " |B|=0 or |B|+2|A|cos theta=0` `rArr " " 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|` Hence, correct answer will be either |B|=0 or B are antiparallel provided |B|=2|A| |
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| 180. |
Three vectors `vecA,vecB` and `vecC` add up to zero.Find which is false.A. `(vecAxxvecB)xxvecC` is not zero unless `vecB,vecC` are parallelB. `(vecAxxvecB).vecC` is not zero unless `vecB,vecC` are parallelC. If `vecA,vecB,vecC` define a plane, `(vecAxxvecB)xxvecC` is in that planeD. `(vecAxxvecB).vecC =|vecA||vecB||vecC|rarrC^(2)=A^(2)+B^(2)` |
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Answer» Correct Answer - B::D 2) `(AxxB). C=(BxxA).C=0` Whatever be the positions of `A,B` and `C`.if `B||C`, then `BxxC=0`, then `(BxxC)xxC=0.3` 4) If `C^(2)=A^(2)+B^(2)`, then angles between `A` and `B` is `90^(@)` |
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| 181. |
Two particles `A` and `B` are projected in same vertical plane as shown in figure.Their initial positions `(t=0)`, initial spped and angle of projections are indicated in the diagram.If initial angle of projection `theta_(B)=37^(@)`, what should be initial speed of projection of particle `B`, so that it hits particle `A`.`U_(A)=60m//s` A. `80 m//s`B. `75 m//s`C. `40 m//s`D. `45 m//s` |
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Answer» Correct Answer - A `V_(rel)` in along line joining the particles `60 sin 53` `= V_(B) sin 37 V_(B) = 80 m//s` |
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| 182. |
At a certain height a shell at rest explodes into two equal fragments one of the fragments receives a horizontal velocity `u`.The time interval after which the velocity vectors will be inclined at `120^(@)` to each other isA. `u/(sqrt3g)`B. `(sqrt3u)/g`C. `2u/(sqrt3u)`D. `u/(2sqrt3g)` |
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Answer» Correct Answer - A `t=sqrt(u_(1)u_(2))/gcot(theta//2)` |
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| 183. |
A body projected horizontally with a velocity`v` from a height `h` has a range `R`.With what velocity a body is to be projected horizontally from a height `h//2` to have the same range ?A. `sqrt2 v`B. `2v`C. `6v`D. `8v` |
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Answer» Correct Answer - A `R=u sqrt((2h)/(g))` , given R is same |
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| 184. |
A body is projected horizontally from a height of `78.4 m` with a velocity `10 ms^(-1)`.Its velocity after `3 seconds` is `(g=10 m//s^(2))` (Take direction of projection as `i` and vertically upward direction as `j`)A. `10hati-30hatj`B. `10hati+30hatj`C. `20hati-30hatj`D. `10hati-10sqrt3hatj` |
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Answer» Correct Answer - A `V_(x)=U , V_(y)=-"gt"` |
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| 185. |
What is the linear velocity of a person at equator of the earth due to its spinning motion? (Radius of the earth `=6400km`) |
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Answer» The earth completes one rotation is `24 hour`. Its angular velocity. `omega=(2piN)/t=(2pixx1)/(24xx60xx60)=pi/(43,200)rad s^(-1)` The linear velocity, `v=Romega=6.4xx10^(6)xxpi/(43,000)=465.5m//s` |
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| 186. |
A projectile has the maximum range of `500m`. If the projectile is now thrown up on an inclined plane of `30^(@)` with the same speed, what is the distance covered by it along the inclined plane? |
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Answer» `R_(max)=u^(2)/g` `:.500=u^(2)/g or u=sqrt(500g)` `v^(2)-u^(2)=2as` `0-500g=2xx(-gsin30^(@))x` `x=500m` |
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| 187. |
A projectile is fired from the base of coneshaped hill. The projectile grazes the vertex and strikes the hill again at the base.If `alpha` be the half-angle of the cone, `h` its height, `u` the initial velocity of projection and `theta` angle of projection, then then `theta tan theta`is |
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Answer» Correct Answer - B Here range `=2h tan alpha=(u^(2)sin 2theta)/g` and `h=(u^(2)sin^(2) theta)/2g` Dividing `2tan alpha=(2sin 2theta)/sin^(2)theta rArr tan alpha=2cot theta` `:.u^(2)(2gh)/(sin^(2)theta)=(2gh)/(((2cotalpha)/sqrt(1+4cot^(2)alpha)))=(2gh(1+4cot^(2)alpha))/(4 cot^(2)alpha)=gh(1/2tan^(2)alpha+2)` `tan theta=2cot alpha` |
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| 188. |
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` collide with the wall during a flight in air and all three collide perpendicularly and elastically with the wall as shown in figure.If the time taken by the ball `A` and 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 |
| Answer» Correct Answer - `S=10xx4=40m` | |
| 189. |
Two balls are projected at an angle `theta` and `(90^(@) - theta)` to the horizontal with the same speed. The ratio of their maximum vertical heights isA. ` 1 : 1`B. `tan theta : 1`C. `1 : tan theta`D. `tan^(2)theta : 1` |
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Answer» Correct Answer - D `H_(1) = (u^(2)sin^(2)theta)/(2g)" "...(i)` `H_(2) = (u^(2)sin^(2) (90^(@) -theta))/(2g) = (u^(2) cos^(2)theta)/(2g) " "...(ii)` Divide (i) by (ii), we get `(H_(1))/(H_(2)) = (tan^(2)theta)/(1)` |
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| 190. |
Two cars `A` and `B` cross a point `P` with velocities `10m//s` and `15m//s`. After that they move with different uniform accelerations and the car `A` overtakes `B` with a speed of `25ms^(-1)`. What is velocity of `B` at that instant?A. `20 ms^(-1)`B. ` 25 ms^(-1)`C. ` 30 ms^(-1)`D. ` 40 ms^(-1)` |
| Answer» Correct Answer - A | |
| 191. |
A man desires to swim across the river in shortest time. The velcoity of river water is ` 3 km h^(-1)` . He can swim in still water at ` 6 km h^(-1)` . At what angle with the velocity of flow of the river should he swim ?A. `30^(@)`B. `60^(@)`C. `90^(@)`D. 120 |
| Answer» Correct Answer - C | |
| 192. |
A particle is projected from the bottom of an inclined plane of inclination `30^@`. At what angle `alpha `(from the horizontal) should the particle be projected to get the maximum range on the inclined plane.A. ` 30^(@)`B. ` 45^(@)`C. ` 15^(@)`D. `75^(@)` |
| Answer» Correct Answer - A | |
| 193. |
A body is projected up from the bottom as inclined plane with velocity `3sqrt3m//sec` which makes an angle `60^(@)` if the horizontal. The angle of projection is `30^(@)` with the plane then the time of flight when it strikes the same plane is `0.1x.`Then the value of `x` is |
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Answer» `T=(2V_(0)sin(theta-alpha))/(g cos alpha)=(2xx3sqrt(3) sin (60-30))/(10 cos 30)` `=(2xx3sqrt3xx1/2)/(10xxsqrt3/2)=0.6=0.1x , x=6` |
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| 194. |
A cricket ball is thrown at a speed of 30 m `s^(-1)` in a direction `30^(@)` above the horizontal. The time taken by the ball to return to the same level isA. 2 sB. 3 sC. 4 sD. 5 s |
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Answer» Correct Answer - B Here, `u=30ms^(-1),theta=30^(@),g=10ms^(-2)` The time taken by the ball to return to the same level is `T=(2usintheta)/(g)=(2xx30ms^(-1)xxsin30^(@))/(10ms^(-2))=3s` |
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| 195. |
A cricket ball is thrown at a speed of ` 28 ms^(-1)` in a direction `30 ^(@)` above the horizontal. Calculate (a)the maximum height (b) the time taken by ball to return ti the same level, and (c )the distance from the thrower to the point wher the ball restance from the throwrt to the point where the ball returns to the same level. |
| Answer» Correct Answer - 10 m | |
| 196. |
Find the velocity of a projectile at the highest point, if it is projected with a speed ` 15 m s^(-1)` in the direction `45^(@)` above horizontla. (take g = ` 10 m s^(-2))` |
| Answer» Correct Answer - `15/sqrt2 m s^(-1)` along horizontal . | |
| 197. |
In case of a projectile motion, what is the angle between the velocity and acceleration at the highest point?A. `0^(@)`B. `45^(@)`C. `90^(@)`D. `180^(@)` |
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Answer» Correct Answer - C At the highest point, velocity is acting horizontally and acceleration ( = acceleration due to gravity) is acting vertically downwards. Therefore, at the highest point the angle between velocity and acceleration is `90^(@)` . |
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| 198. |
If `R` is the horizontal range for `theta` inclination and `h` is the maximum height reached by the projectile, Then maximum range isA. `R^(2)/(h) +2h`B. `R^(2)/(8h)+2h`C. `R^(2)/(8h)+8h`D. `R^(2)/(h)+h` |
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Answer» Correct Answer - B We know that horizontal range, `R=(u^(2)sin2theta)/g` and maximum height `h=(u^(2)sin^(2)theta)/(2g)` `:. (R^(2))/(8h)+2h([(u^(2)sin 2 theta)/(g)]^(2))/(8[(u^(2) sin^(2) theta)/(2g)])+2[(u^(2) sin^(2) theta)/(2g)]` `=(u^(4)(2sin thetacos theta)^(2))/(g^(2)xx8(u^(2)sin^(2)theta)/(2g))+(u^(2)sin^(2)theta)/g` `=u^(2)/(g)(cos^(2)theta+sin^(2)theta)=u^(2)/g=R_(max)` |
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| 199. |
A particles `P` moves with speed `v` along `AB` and `BC`, sides of a square `ABCD`. Another particle `Q` also starts at `A` and moves with the same speed but along `AD` and `DC` of the same square `ABCD`. Then their respective changes in velocities areA. equal in magnitude but different in directionsB. different in magnitude but same in directionsC. different both in magnitude and directionD. same both in magnitude and direction |
| Answer» Correct Answer - A | |
| 200. |
A bus moves over a straight level road with a constant acceleration a. A boy in the bus drops a ball out side. The acceleration of the ball w.r.t the bus and the earth are respectivelyA. `sqrt(a^(2)+g^(2)),g`B. `g,sqrt(a^(2)+g^(2))`C. `a,g`D. `g,a` |
| Answer» Correct Answer - A | |