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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.
| 101. |
A body projected horizontally from the top of a tower follows `y=20x^(2)` parabola equation where `x,y` are in `m``(g=10 m//s^(2))`.Then the velocity of the projectile is `(ms^(-1))`A. `0.2`B. `0.3`C. `0.4`D. `0.5` |
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Answer» Correct Answer - D `y=g/(2u^(2))x^(2)` |
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| 102. |
An object is projected horizontally from a top of the tower of height `h`.The line joining the point of projection and point of striking on the ground makes an angle `45^(@)` with ground,Then with what velocity the object strikes the groundA. `sqrt((1lgh)/2)`B. `sqrt((9lgh)/2)`C. `sqrt((7lgh)/2)`D. `sqrt((5lgh)/2)` |
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Answer» Correct Answer - D `Tantheta=h/R,R=usqrt((2h)/g),V=sqrt(u^(2)+2gh)` |
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| 103. |
A stone is thrown vertically upwards with an initial speed `u` from the top of a tower, reaches the ground with a speed `3 u`. The height of the tower is :A. `3u^(2)//g`B. `4u^(2)//g`C. `6u^(2)//g`D. `9u^(2)//g` |
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Answer» Correct Answer - B `upsilon^(2)=u^(2)+2gh` `rArr (3u)^(2)=(-u)^(2)+2gh rArr h=(4u^(2))/(g) darr + ve` |
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| 104. |
The velocit-time graph of a body moving in a straight line is shown in Fig. 2 (d) . 32. Find the displacement and the distance travelled by the body in ` 6 seconds`. .A. 8m, 16mB. 16m, 32mC. 16m, 16mD. 8m, 18m |
| Answer» Correct Answer - A | |
| 105. |
A particle is moving such that its position coordinates `(x, y)` are `(2m, 3m)` at time `t=0, (6m, 7m)` at time `t=2 s`, and `(13 m, 14m)` at time `t=5 s`. Average velocity vector`(vec(V)_(av))` from `t=0` to `t=5 s` isA. `(1)/(5)(13hat(i)+14hat(j))`B. `(7)/(3)(hat(i)+hat(j))`C. `(hat(i)+hat(j))`D. `(11)/(5)(hat(i)+hat(j))` |
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Answer» Correct Answer - D Average velocity, `upsilon_(av)=("Displacement"(Delta r))/("Time taken"(Delta t))` Displacement in t = 0 to 5s is `Delta r=(!3-2)hat(i)+(14-3)hat(j)=11hat(i)+11hat(j)` `v_(av)=(Delta r)/(Delta r)=(11)/(5)(hat(i)+hat(j))` |
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| 106. |
(a) Earth can be thought of as a sphere of radius ` 64 00 km`. Any object (or a person ) is performing circula motion around the axis os earth due to earths rotation (period 1 day ). What is acceleration o object on the surface of th earth 9at equator ) towards its centre ? What is its altitude ` theta` ? How does these accelerations compare with `g=9.8 m//s^2 `? (b) Earth also moves in circular orbit around sum every year withon orbital radius of ` 1.5 xx 10 ^(11) m`. What is the acceleration of earth ( or any object on the surface of the earth ) towards the centre of the sum ? How dies thsi acceleration comparte with ` g=9.8 ms^2 `? |
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Answer» (a) Radius of the earth (R)=`6400km=6.4xx10^(6) m` Time period (T) = 1 day `=24xx60xx60 s=86400 s` Centripetal acceleration `(a_(c))=omega^(2)=R=R((2pi)/(T))^(2)=(4 pi^(2)R)/(T)` `=(4xx(22//7)^(2)xx6.4xx10^(6))/((24xx60xx60)^(2))` `=(4xx484xx64xx10^(6))/(49xx(24xx3600)^(2))` `=0.034m//s^(2)` At equator, `" "` latitude `theta = 0^(@)` `:." " (a_(c))/(g)=(0.034)/(9.8)=(1)/(288)` (b) Orbital radius of the earth around the sun (R) `=1.5xx10^(11)m` Time period =1 Yr =365 day `=365xx 24xx60xx60 s = 3.15 xx10^(7) s` Centripetal acceleration `(a_(c))=R omega^(2)=(4pi^(2)R)/(T^(2))` `=(4xx(22//7)^(2)xx1.5xx10^(11))/((3.15xx10^(7))^(2))` `=5.97xx10^(-3)m//s^(2)` `:. " " (a_(c))/(g)=(5.97xx10^(-3))/(9.8)=(1)/(1641)` |
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| 107. |
The position of an object is given by ` vecr= ( 9 thati + 4t^(3)hatj) ` m. find its velocity at time t= 1s. |
| Answer» Correct Answer - 15 m/s atom angle of `53^(@)` with x-axis . | |
| 108. |
A body is projected up such that its position vector varies with time as `vecr={3thati+(4t-5t^(2))hatj} m`. Here `t` is in second.The time when its `y`-coordinate is zero isA. `3 s`B. `1 s`C. `0.8 s`D. `1.25 s` |
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Answer» Correct Answer - C `4t-5t^(2)=0` |
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| 109. |
The positione of a particle is expressed as ` vecr = ( 4t^(2) hati + 2thatj)` m, where t is time in second. Find the velocity o the particle at t = 3 s |
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Answer» `vecr = ( 4t^(2)hati + 2thatt) m` Velocity ` vecc = (dvecr)/(dt) = hati d/(dt) (4t^(2)) + hatj d/(dt) (2t)` ` vecv = ( 8t) hati + 2hatj` At t = 3 s , velocity is given by ` vecv_(t=3) = (8 xx 3) hati + 2hatj` ` vecv_(t =3) = ( 24hati + 2hatj) m s^(-1)` `= sqrt580` ` = 24.08 m s^(-1)` Direction ` theta = tan ^(-1) 2/24` ` = tan^(-1) 1/12 = 4.76^(@)` Thus the particle has velocity `24.08 m s^(-1)` at an angle ` 4.76^(@)` with x -axis. |
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| 110. |
The resultant of two forces `2P` and `sqrt2P` is `sqrt10P`.The angle between the forces isA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - B `R^(2)=P^(2)+Q^(2)+2PQ cos theta` |
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| 111. |
Assertion : In the s-t diagram as shown in figure, the body starts moving in positive direction but not form s = 0. Reason : At `t = t_(0)`, velocity of body changes its direction of motion.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. |
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Answer» Correct Answer - C Slope of s-t graph = velocity = positive At `t = 0, s ne 0` . Further at `t = t_(0) : s=0, upsilon ne 0`. |
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| 112. |
The displacement-time graph of moving particle is shown below The instantaneous velocity of the particle in negative at the pointA. EB. FC. CD. D |
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Answer» Correct Answer - A `because` Slope of displacement-time graph at the point E is negative. Thus, at the point E the instantaneous velocity of the particle is negative. |
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| 113. |
In a two dimensional motion, instantaneous speed `v_(0)` is a positive constant. Then which of the following are neccessarily true?A. The acceleration of the particle is zero.B. The acceleration of the particle is bounded.C. The acceleration of the particle is necessarily in the plane of motion.D. The particle must be undergoing a uniform circular motion. |
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Answer» Correct Answer - C In two dimensional motion, if instantaneous speed is a postitive constant, then the acceleration of the particle is necessarily in the plane of motion. |
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| 114. |
In a two dimensional motion, instantaneous speed `v_(0)` is a positive constant. Then which of the following are neccessarily true?A. The acceleration of th particle is zeroB. The acceleration of th particle is bounded.C. The acceleration of th particle is necessarly in the plane of motion.D. The particle must be undergoing a uniform circular motion |
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Answer» Correct Answer - C As given motion is two dimensional motion and given that insantaneous speed `v_(0)` is positive constant.Acceleration is rate of change of velocity (instanteous speed hence it will be in the plane of motion. |
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| 115. |
A cylclist is riding with a speed of `27 km h^-1`. As he approaches a circular turn on the road of radius `80 m`, he applies brakes and reduces his speed at the constant rate of `0.5 ms^-2`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?A. `0.5m//s^(2)`B. `0.87m//s^(2)`C. `0.56m//s^(2)`D. `1m//s^(2)` |
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Answer» Correct Answer - C `a=sqrt(a_(c)^(2)+a_(r)^(2))` |
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| 116. |
The time of fight of an object projected with speed ` 20 ms^(-1)` at an angle ` 30^(@)` with the horizontal , isA. 1 sB. 4 sC. 2 sD. 6s |
| Answer» Correct Answer - A | |
| 117. |
A rock is launched upward at `45^(@)`.A bee moves along the trajectory of the rock at a constant speed equal to the initial speed of the rock.The magnitude of acceleration of the bee at the top point of the trajectory is `xg`?For the rock, neglect the air resistance.Find the value of `x`. |
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Answer» Correct Answer - B Acceleration of bee `=u^(2)/Rrarr(1)` At highest position, `R=(u/sqrt2)^(2)/g=u^(2)/(2g)rarr(2)` From`(1)&(2)` `a_(bee)=2g :.x=2` (`u=`projection speed of rock) |
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| 118. |
A boy is running over a circular track with uniform speed of `10 ms^(-1)`. What is the average velocity for movement of boy from A to (in `ms^(-1)`) ? A. `(10)/(pi)`B. `(40)/(pi)`C. 10D. None of these |
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Answer» Correct Answer - D `upsilon_(av)=("Displacement")/("Time")=(2R)/((piR//upsilon))=(2upsilon)/(pi)=(20)/(pi)ms^(-1)` |
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| 119. |
Free fall of an object (in vacuum) is a case of motion withA. uniform velocityB. uniform accelerationC. variable accelerationD. constant momentum |
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Answer» Correct Answer - B Free fall of an object (in vacuum) is a case motion with uniform acceleration. |
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| 120. |
The coordinates of a moving particle at any time `t` are given by `x = alpha t^(3)` and `y = beta t^(3)`. The speed of the particle at time `t` is given byA. `sqrt(alpha^(2) + beta^(2))`B. `3t sqrt(alpha^+ beta^(2))`C. `3t^(2) sqrt(alpha^(2) + beta^(2))`D. `t^(2) sqrt(alpha^(2) + beta^(2))` |
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Answer» Correct Answer - C `x = alpha t^(3) rArr v_(x) = (d x)/(d t) = 3 beta t^(2)` `y = beta t^(3) rArr v_(y) = (d y)/(d t) = 3 beta t^(2)` `v = sqrt(v_(x)^(2) + v_(y)^(2)) = 3 t^(2) sqrt(alpha^(2) + beta^(2))` |
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| 121. |
A block of mass `m` in floating in a river flowing with a velocity of `2m//s`.A boat is moving behind the block with a velocity of `5m//s` with respect to the block as shown.From the boat a stone is thrown with a velocity `vecv=v_(1)hati-v_()hatj+v_(3)hatk` with respect to the river such that it hits the block.If `v_(1):v_(2):v_(3)=2sqrt3:2:3` then the velocity of the stone with respect to the ground is `(g=10m//s^(2))` A. `10hati-10/sqrt3hatj+5sqrt3hatk`B. `12hati-10/sqrt3hatj+5sqrt3hatk`C. `10hati-10hatj+5sqrt3hatk`D. `10sqrt3hati-10/sqrt3hatj+5sqrt3hatk` |
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Answer» Correct Answer - B Let `v_(1)=2sqrt3k v_(2)=2k v_(3)=3k, (2k)t=10` `(3k)t-1/2xx10xxt^(2)=0,t=(3k)/5,k=5/sqrt3` `vecv_(b)=vecv_(b//r)+vecv_(r)` |
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| 122. |
A motor boat going down stream comes over a floating body at a point `A`. 60 minutes later it turned back and after some time passed the floating body at a distance of `12 km` from the point `A`.Find the velocity of the stream assuming constant velocity for the motor boat in still water.A. `2 km//hr`B. `3 km//hr`C. `4 km//hr`D. `6 km//hr` |
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Answer» Correct Answer - D `d=(v_(B)+v_(W))t_(1)` --(1) `d-12=(v_(B)-v_(W))t_(2)`--(2) `12=v_(w)(t_(1)+t_(2))`- - - (3) solve above equations |
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| 123. |
Assertion : On a curved path, average speed of a particle can never be equal to average velocity. Reason : Average speed is total distance travelled divided by total time. Whereas average velocity is, final velocity plus initial velocity divided by two.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. |
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Answer» Correct Answer - C On a curved path, distance `gt` displacement `therefore` average speedf `gt` average velocity Further, average velocity `=("Total displacement")/("Total time")` |
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| 124. |
Asserion: Magnitude of the resultant of two vectors may be less than the magnitude of either vector. Reason: The resultant of two vectors is obtained by means of law of parallelogram of Vectors. |
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Answer» Correct Answer - B If `veca and vecb` are inclined at an angle greater than `90^(@)`, their resultant has magnitude. `R = sqrt(a^(2) + b^(2) + 2ab cos theta)` which is less than both a and b, when they are nearly equal. |
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| 125. |
Assertion: The difference of two vectors A and B can be treated as the sum of two vectors. Subtraction of vectors can be defined in terms of addition of vectors. |
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Answer» Correct Answer - A Let two vectors be `vecA and vecB`. Difference of vectors `vecA and vecB` can be taken as the sum of two vectors `vecA and -vecB`. i.e., `vecA - vecB = vecA = (-vecB)` |
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| 126. |
Initial speed of an `alpha` particle inside a tube of length 4m is `1 kms^(-1)`, if it is accelerated in the tube and comes out with a speed of `9 kms^(-1)` , then the time for which the particle remains inside the tube isA. `8xx10^(-3)s`B. `8xx10(-4)s`C. `80xx10^(-3)s`D. `800xx10^(-3)s` |
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Answer» Correct Answer - B Initial speed, `u=1 kms^(-1)=1000 ms^(-1)` Final speed, `upsilon=9 kms^(-1)=9000 ms^(-1)` By using the relation, `upsilon^(2)=u^(2)+2as` `(9000)^(2)=(1000)^(2)+2xx a xx4 rArr a=10^(7)ms^(-2)` `therefore` The time for which the particle remains in the tube `upsilon=u+at` `rArr t=(upsilon-u)/(a)=(9000-1000)/(10^(7))=8xx10^(-4)s` |
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| 127. |
A car covers the first half of the distance between two places at a speed of `40 kmh^(-1)` and second half at `60 kmh^(-1)` Calculate the average speed of the car. |
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Answer» Given, speed in first half, `v_(1)=40 kmh^(-1)`. Speed in second half, `v_(2) = 60 kmh^(-1)`. `because` Car covers equal distance with different speeds. `therefore` Average speed of car `v_(av)=(2v_(1)v_(2))/(v_(1)+v_(2))` `v_(av)=(2(40) (60))/(40+60)=48 kmh^(-1)` |
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| 128. |
The ceiling of a long hall is `20 m` high. What is the maximum horizontal distance that a ball thrown with a speed of `40 m` can go without hitting the ceiling of hall `(g=10 ms^(-2))` ? |
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Answer» : Here, `H=20m,u=40ms^(-1)` Suppose the ball is thrown at an angle `theta` with the horizontal. Now `H=(u^(2)sin^(2)theta)/2g rArr 20=((40)^(2)sin^(2)theta)/2xx10` or, `sintheta=0.5 rArr theta = 30^(@)` Now, `R=(u^(2)sin 2theta)/g= (40^(2)sin 60^(@))/10` `= ((40)^(2)xx0.866)/10=138.56m` |
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| 129. |
At the topmost point of its path, a projectile has acceleration of magntiude |
| Answer» Correct Answer - C | |
| 130. |
Statement -1 : A food packet is dropped from a rescue plane . Path of the food packed will be straight line for the pilot but parabolic for the person on the ground. Statement - 2: Food packet has initial velocity same as that of plane.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 -1C. Statement -1 is True, Statement -2 is FalseD. Statement -1 is False ,Statement -2 is True. |
| Answer» Correct Answer - A | |
| 131. |
Two bodies are thrown from the same point with the same velocity of `50ms^(-1)`.if their angles of projection are complimentary to each other and the difference of maximum heights is `30m`,the minimum and maximum heights are`(g=10 m//s^(2))`A. `50 m & 80m`B. `47.5 m & 77.5 m`C. `30 m & 60m`D. `25 m & 55m` |
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Answer» Correct Answer - B `H_(2)-H_(1)=30,H_(2)-H_(1)=u/(2g)=125` |
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| 132. |
The horizontal and vertical displacements of a projectile are given as `x=at` & `y=bt-ct^(2)`.Then velocity of projection isA. `sqrt(a^(2)+b^(2))`B. `sqrt(b^(2)+c^(2))`C. `sqrt(a^(2)+c^(2))`D. `sqrt(b^(2)-c^(2))` |
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Answer» Correct Answer - A `u_(x)=(dx)/dt,u_(y)=(dx)/dt` and `u=sqrt(u_(x)^(2)+u_(y)^(2))` |
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| 133. |
Motion of a particle is governed by following relations `y=x/alpha,V_(x)=b-ct.(a,b,c are + ve "const")` The displacement `(S)` verson from `(t)` graph osA. B. C. D. |
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Answer» Correct Answer - D `y=x/alpha,V_(x)=b-ct,y=x/alpha.rArrx=bt-1/2ct^(2)` `s=sqrt(x^(2)+y^(2))=sqrt(x^(2)+x^(2)/alpha^(2)),xsqrt(a+a/a^(2))` `S=sqrt(1+1/a^(2))(bt-1/2ct^(2))` `=kbt-(kc)/2.t^(2)(k=sqrt(1+1/alpha^(2)))` Parabolic curve |
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| 134. |
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. `4ms^(-2)`D. `8ms^(-2)` |
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Answer» Correct Answer - D Centripetal acceleration `a=v^(2)/r` |
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| 135. |
During the first `18 min` of a `60 min` trip, a car has an average speed of `11 ms^-1.` What should be the average speed for remaining `42 min` so that car is having an average speed of `21 ms^-1` for the entire trip?A. `25.3 ms^(-1)`B. `29.2 ms^(-1)`C. `31 ms^(-1)`D. `35.6 ms^(-1)` |
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Answer» Correct Answer - A `21=(11xx18+42xx upsilon)/(60)` `therefore upsilon = 25.3 ms^(-1)` |
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| 136. |
A ball projected with a velocity of `10m//s` at angle of `30^(@)` with horizontal just clears two vertical poles each of height `1m`. Find separation between the poles. |
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Answer» `h=u_(y)t+1/2g t^(2)=(10sin 30^(@))t+1/2(-10)t^(2)` `1=5t-5t^(2) rArr t=0.72s,2.76s` are the instants at which projectile crosses the poles `:.` Separation between poles `=OS-OQ` `=u cos theta (t_(2)-t_(1))` `=10cos 30^(@)(2.72-0.72)=17.7m` |
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| 137. |
A particle projected from the level ground just clears in its ascent a wall `30 m` high and `120sqrt3` away measured horizontally.The time since projection to clear the wall is two second.It will strike the ground in the same horizontal plane from the wall on the other side of a distance of (in metres)A. `150sqrt3`B. `180sqrt3`C. `120sqrt3`D. `210sqrt3` |
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Answer» Correct Answer - B `x=u cos theta.t, y=u sin thetat-1/2"gt"^(2)` `R=(2u cos theta.u sin theta)/g,x^(1)R-x` |
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| 138. |
Which one of the following statements is true?A. A scalar quantity is the one that is conserved in a processB. A scalar quantity is the one that can never take negative valuesC. A scalar quantity is the one that does not vary from one point to another in spaceD. A scalar quantity has the same value for observers with different orientation of the axes |
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Answer» Correct Answer - D A scalar quantity is independent of direction hence has the same value for observers with different orientations of the axes. |
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| 139. |
The angle between `A=hat(i)+hat(j) " and " B=hat(i)-hat(j)` isA. `45^(@)`B. `90^(@)`C. `-45^(@)`D. `180^(@)` |
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Answer» Correct Answer - B Given, `A=hat(i)+hat(j)` `B=hat(i)-hat(j)` We know that `A.B=|A||B|cos theta` `rArr " " (hat(i)+hat(j)).(hat(i)-hat(j))=(sqrt(1+1))(sqrt(1+1)) cos theta` where `theta` is the angle between A and B `rArr " " cos theta=(1-0+0-1)/(sqrt(2)sqrt(2))=(0)/(2)=0` `rArr " " theta=90^(@)` |
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| 140. |
Assertion: If `hat(i)` and `hat(j)` are unit Vectors along x-axis and y-axis respectively, the magnitude of Vector `hat(i)+hat(j)` will be `sqrt(2)` Reason: Unit vectors are used to indicate a direction only. |
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Answer» Correct Answer - B Since `hati and hatj` are unit vectors, their magnitude are `|hati| = 1 and |hatj| = 1` . Magnitude of resultant vector is equal to `sqrt(|hati|^(2) + |hatj|^(2)) = sqrt((1)^(2) + (1)^(2)) = sqrt(2)` |
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| 141. |
Three vectors `vecA,vecB` and `vecC` add up to zero.Find which is false.A. `(AxxB)xxC` is not zero unless B, C are parallelB. `(AxxB).C` is not zero unless B, C are parallelC. If A, B, C define a plane, `(AxxB)xxC` is in that planeD. `(AxxB).C=|A||B||C| rarr C^(2)=A^(2)+B^(2)` |
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Answer» Correct Answer - B::D Given A+B+C=0 Hence, we can say that A, B and C are in one plane and are represented by the three sides of a triangle taken in one order, Now consider the options one by one. (a) We can write `Bxx(A+B+C)=Bxx0=0` `rArr " " BxxA+BxxB+BxxC=0` `rArr " " BxxA+0+BxxC=0` `rArr" "BxxA=-BxxC` `rArr " " AxxB=BxxC` `:. " " (AxxB)xxC=(BxxC)xxC` It cannot be zero. If B||C, then `BxxC=0`, than `(BxxC)xxC=0`. (b) `(AxxB).C=(BxxAC).C=0` whatever be the positions of A, B and C. If B||C, then `BxxC=0`, then `(BxxC)xxC=0`. (c) `(AxxB)=X=AB sin theta X`. The direction of X is perpendicular to the plane containing A and B. `(AxxB)xxC=XxxC`. Its direction is in the plane of A, B, and C. (d) If `C^(2)=A^(2)+B^(2)`, then angle between A and B is `90^(@)` `:. " " (AxxB).C`=(AB `sin 90^(@)` X).C=AB (X.C) `" " ABC cos 90^(@)=0` |
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| 142. |
For tow vectros ` vec A` and `vec B` ` | vec A + vec B| = | vec A- vec B|` is always true when.A. `|A|=|B|!= 0`B. `A_|_B`C. `|A|=|B|!= 0` and A and B are parallel or anti-parallelD. when either |A| or |B| is zero |
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Answer» Correct Answer - B::D Given, |A+B|=|A-B| `rArr " " sqrt(|A|^(2)+|B|^(2)+2|A||B| cos theta)=sqrt(|A|^(2)+|B|^(2)-2|A||B|cos theta)` `rArr " " |A|^(2)+|B|^(2)+2|A||B|cos theta = |A|^(2)+|B|^(2)-2|A||B|cos theta` `rArr " " 4|A||B| cos theta =0` `rArr " " |A||B|cos theta=0` `rArr " " |A|=0 or |B| = 0 or cos theta =0` `rArr " " theta=90^(@)` When `theta=90^(@)`, we can say that `A_|_B` |
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| 143. |
For a particle performing uniform circular motion, choose the incorrect statement form the following.A. Magnitude of particle velocity (speed) remains constant.B. Particle velocity remains directed perpendicular to radius vector.C. Direction of acceleration keeps changing as particle moves.D. Magnitudes of acceleration does not remain constant. |
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Answer» Correct Answer - D For a praticle performing uniform cirular motion, magnitude of the acceleration remains constant. |
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| 144. |
Which of the following statements is incorrect ?A. In one dimension motion, the velocity and the acceleration of an object are always along the same line.B. In two or three dimensions, the angle between velocity and acceleration vectors may have any value between `0^(@)` and `180^(@)`C. The kinematic equations for uniform acceleration can be applied in case of a uniform circular motion.D. The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant. |
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Answer» Correct Answer - C The kinematic equations for uniform acceleration do not apply in case of uniform circular motion because in this case the magnitude of acceleration is constant but its direction is changing. |
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| 145. |
A cyclist starts from the centre O of a circular park of radius 1km, reaches the edge P of the park, then cycles along the PQ cicumference and returns to the centre along OQ as shown in fig. If the round trip taken ten minute, the net displacement and average speed of the cylists (in kilometer and kinetic per hour) is A. 0, 1B. `(pi+4)/(2),0`C. `21.4, (pi+4)/(2)`D. `0, 21.4` |
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Answer» Correct Answer - D Net displacement = 0 and total distance = OP + PQ + QO `=1+(2pixx1)/(4)+1=(14.28)/(4)km` Averagespeed `= (14.28)/(4xx10//60)=(6xx14.28)/(4)=21.42 km//h` |
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| 146. |
A cyclist starts from the centre O of a circular park of radius 1km, reaches the edge P of the park, then cycles along the PQ cicumference and returns to the centre along OQ as shown in fig. If the round trip taken ten minute, the net displacement and average speed of the cylists (in kilometer and kinetic per hour) is A. 0.1B. `(pi + 4)/(2), 0 `C. `21.4, ""(pi+4)/(2)`D. 0.21.4 |
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Answer» Correct Answer - D Since the initial position coincides with the final position. So, net displancement of the cyclist =zero `"Aveerage speed of the cyclist"=("Total distance travelled")/("Total time taken")` `=(OP+PQ+QO)/(10)kmmin""^(-1)=(1+(pi)/(2)xx1+1)/(10)kmmin""^(-1)` `=(pi+4)/(20)kmmin""^(-1)=(pi+4)/(20)xx60kmh^(-1)=21.4kmh^(-1)` |
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| 147. |
The equations of motion of a projectile are given by `x=36tm and 2y =96t-9.8t^(2)m`. The angle of projection isA. `sin^(-1)((4)/(5))`B. `sin^(-1)((3)/(5))`C. `sin^(-1)((4)/(3))`D. `sin^(-1)((3)/(4))` |
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Answer» Correct Answer - A Given : `x = 36t, 2y = 96t - 9.8 t^(2)` or `y = 48t - 4.9t^(2)` Let the initial velocity of projectile be u and angle of projection is `theta`. Then, Initial horizontal component of velocity, `u_(x) = u cos theta = ((dx)/(dt))_(t=0) = 36 or u cos theta = 36 " "...(i)` Initial vertical component of velocity, `u_(y) = u sin theta = ((dy)/(dt))_(t = 0) = 48 or u sin theta = 48 " "...(ii)` Dividing (ii) by (i), we get ` tan theta = (48)/(36) = (4)/(3) therefore sin theta = (4)/(5) or theta = sin^(-1) ((4)/(5))` |
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| 148. |
Which of the following is true regarding projectile motion ?A. Horizontal velocity of projectile is constant.B. Vertical velocity of projectile is constant.C. Acceleration is not constant.D. Momentum is constant. |
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Answer» Correct Answer - A Horizontal velocity of a projectile is not affected by gravity. |
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| 149. |
A bomb is released by a horizontal flying aeroplane. The trajectory of the bomb isA. a parabolaB. a straight lineC. a circleD. a hyperbola |
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Answer» Correct Answer - A It is an example of projectile motion. Therefore, the trajectory of the bomb is parabola. |
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| 150. |
In a two dimensional motion,instantaneous speed `v_(0)` is a positive constant.Then which of the following are necessarily true?A. The average velocity is not zero at any timeB. Average acceleration must always vanish.C. Displacements in equal time intervals are equalD. Equal path lengths are traversed in equal intervals. |
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Answer» Correct Answer - D As speed is a sclar quantity, hence is will be related with path length (scalar quantity only) |
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