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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. |
A hunter tries to hunt a monkey with a small very poisonous arrow, blow form a pipe with initial speed `v_(0)`. The monkey is hanging on a branck of a tree at height `H` above the ground. The hunter is at a distance `L` form the the bottom of the tree. The monkey sees teh arrow leaving the blow pipe and immediately loses the grip on the tree, falling freeely down with zero initial velcoity. The minimum initial speed `v_(0)` of the arrow for hunter to succes while monkey is in air-A. `sqrt((g(H^(2)+L^(2)))/(2H))`B. `sqrt((gH^(2))/(H^(2)+L^(2)))`C. `sqrt((g(H^(2)+L^(2)))/(H))`D. `sqrt(((2gH^(2)))/(sqrt(H^(2)+L^(2))))` |
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Answer» Correct Answer - A To hit the monkey `L leR rArr L le (v_(0)^(2)sin 2 theta)/(g) rArr` `L le (v_(0)^(2)2 sin theta cos theta)/(g)` `(Lg)/(2sin theta cos theta) le V_(0)^(2) rArr sqrt((Lg)/(2sin theta cos theta)) = V_(0min)` `sqrt((Lg)/(2xx(H)/(sqrt(H^(2)+L^(2)))xx(L)/sqrt(H^(2)+L^(2))))V_(0(min))` `rArr sqrt(g(H^(2)+L^(2))/(2H)) =V_(0(min))` |
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| 152. |
A stone is thrown verticallyupwards and then it returns to the thrower. Is it a projectile ? Explain. |
| Answer» It is not a projectile, because a projectile should have two component velocities in two mutually perpendicular directions, but in this case the stone has velocity only in one direction wile going up or coming downwards. | |
| 153. |
A ball thrown up is caught by the thrower after ` 4 second`. How high did it go and with what velocity was it thrown ? How far below its highest point was in ` 3 second ` after starts ? Acceleration due to gravity is ` 9.8 ms^(-2)`. |
| Answer» The position of ball below the highest point in ` 3` secsn car be calculated by taking the motion of ball vertically downward grom its highest point for ` 1second` . | |
| 154. |
Points ` P, Q` and `R` are in vertical line such that ` PQ =QR`. A ball at (P) is allowed to fall freely. What is the ratio of the times of descent through PQ and QR ? |
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Answer» Let ` t_(1), t_(2) `be the times of descent through ` PQ` and `QR` respectively. Let ` PQ=QR=y`. Then ` y =1/2 gt_(1)^(2) and 2 y =1/2 g (t_(1) +t_(2))^(2)` or ` (2 y)/y =((t_(1) + t_(2))^(2)/t_(1)^(2)` or ` sqrt 2 =(t_(1) +t_(2))/t_(1) 1 + t_(2)/t_(1)` or ` t_(2)/t_(1) =sqrt 2-1 or t_(1)/t_(2) =1/(sqrt2 -1) =1 : (sqrt -1)`. |
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| 155. |
In the above figure, the portion AB of the curve shows:A. Accelerated motionB. Retarted motionC. Uniform motionD. State of rest |
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Answer» Correct Answer - C |
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| 156. |
In the above figure, the portion BC of the curve shows:A. Accelerated motionB. Retarted motionC. Uniform motionD. State of rest |
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Answer» Correct Answer - B |
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| 157. |
The distance-time curve of a moving motor-car is according to the following figure-1.101. The portion OA of the curve shows: A. Accelerated motionB. Retarted motionC. Uniform motionD. State of rest |
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Answer» Correct Answer - A |
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| 158. |
For a particle moving along a straight line, the displacement x depends on time t as `x = alpha t^(3) +beta t^(2) +gamma t +delta`. The ratio of its initial acceleration to its initial velocity dependsA. only on `alpha` and `beta`B. only on `beta` and `gamma`C. only on `alpha` and `gamma`D. only on `alpha` |
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Answer» Correct Answer - B |
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| 159. |
The displacement (x) of particle depends on time (t) as `x = alpha t^(2) - beta t^(3)`.A. The particle will return to its starting point after time `alpha//beta`B. The particle will come to rest after time `2 alpha//3beta`C. The initial velocity of the particle was zero but its initial acceleration was not zeroD. No net force will act on the particle at `t = alpha//3 beta` |
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Answer» Correct Answer - A::B::C::D |
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| 160. |
A projectile is thrown with a velocity of `10sqrt(2)m//s` at an angle of `45^(@)` with horizontal. The interval between the moments when speed is `sqrt(125)m//s` is `(g=10m//s^(2))`A. `1.0s`B. `1.5s`C. `2.0s`D. `0.5s` |
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Answer» Correct Answer - A |
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| 161. |
A bus starts movins with uniform acceleration from it position of rest. If moves 48 m in 4s. On applying the brakes, its stops after covering 24m. Find the deceleration of the bus |
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Answer» (i) What is the initial velocity of the bus? If the bus moves 48 m in 4,s find the velocity of the bus at the end fo 4s Use the formula, `syt=(1)/(2)at^(2)` and v+u at When brakes are applied, the bus stops after covering 24 m What is the final velocity fo the bus? Then , find the deceleration of the bus by using the formula, `v^(2)-u^(2)=2as`. (ii) `12 ms^(-1)` |
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| 162. |
A room has dimensions `3 m xx 4 m xx5 m. A fly starting at one cronet ends up at the diametrically opposite corner. (a) What is the magnitude of its displacement ? (a) What is the magnitude of its displacement ? (b) If the fly wer to walk, what is the length of the shortest pothe it cantake ? |
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Answer» (a) if the startong point of fly which is one corner of room is taken as origin of coordinates,then the coordinates of diametrically opposite corner of room are ( 3, 4, 5,). So displacement is ` vec r =3 hat I + 4 hat j + 5 hat k` and ` r = sqrt(3^(2) +4^(2) + 5^(2)) =sqrt 50 ~~7 m` (b) When the fly were to walk, then shortest distance travelled is ` =3 + sqrt ( 4^(2) + 5^(2)) =3 sqrt 41 =.4 m`. |
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| 163. |
A rope is lying on a table with one of its end at point O on the table. This end of the rope is pulled to the right with a constant acceleration starting from rest. It was observed that last 2 m length of the rope took 5 s in crossing the point O and the last 1m took 2 s in crossing the point O. (a) Find the time required by the complete rope to travel past point O. (b) Find length of the rope. |
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Answer» Correct Answer - (a) `8.5 s` (b) `2.41 m` |
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| 164. |
A meter stick AB is lying on a horizontal table. Its end A is pulled up so as to move it with a constant velocity `V_(A) = 4ms^(–1)` along a vertical line. End B slides along the floor. (a) After how much time `(t_(0))` speed `(V_(B))` of end B becomes equal to the speed `(V_(A))` of end A ? (b) Find distance travelled by the end B in time `t_(0)`. |
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Answer» Correct Answer - (a) `t_(0) = (1)/(4sqrt(2)) s` (b) `(1-(1)/(sqrt(2)))m` |
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| 165. |
One end of a rope is fixed at a point on the ceiling the other end is held close to the first end so that the rope is folded. The second end is released from this position. Find the speed at which the fold at F is descending at the instant the free end of the rope is going down at speed V. |
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Answer» Correct Answer - `V//2` |
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| 166. |
The acceleration of a particle, starting from rest, varies with time according to relation , `a =- r omega ^(2) sin ometa t`. Find the displacemnt of this particle at a time (t). |
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Answer» Acceleration, ` a = (dv)/(dt) =- r omega sin omega t`. or ` dv =- r omega ^(2) sin omega dt , ` Integratine it we have ` v=- r omega^(2) (-(cos omega t)/(ometa)) =r cos omega t`. Now, velocity ` v= (dx)/(dt) =r omega cos omega t` or ` dx= r ometa cos omega t dt` Integrating it, we have ` x= r omega ((sin omega t)/(omega)) =r sin omega t`. |
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| 167. |
A `0.098-kg` block slides down a frictionless track as shown in (Fig. 5.208). . The time taken by the block to move from `A` to `C` is.A. `sqrt((3)/(g))`B. `sqrt((2)/(g))`C. `sqrt((1)/(g))`D. `(1 + sqrt(3))/(sqrt(g))` |
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Answer» Correct Answer - D (d) `(1)/(2) mv^2 = mg(3 - 1) = 2 mg` [from conservation of energy] or `v = sqrt(4 g) = 2 sqrt(g)` Vertical component at A is `2 sqrt(g) sin 30^@ = sqrt(g)` Time of flight, `T = (2 v sin theta)/(g) = (2 sqrt(g))/(g) = (2)/(sqrt(g))` Using `S = ut + (1)/(2) at^2`, we get `-1 = sqrt(g) t - (1)/(2) "gt"^2 or (1)/(2) "gt"^2 - sqrt(g) t - 1 = 0` `t = (sqrt(g) +- sqrt(g + 4 xx (1)/(2) g))/(2 xx (g)/(2)) t = (1 +- sqrt(3))/(sqrt(g))` Neglecting negative time, `t = (1 +- sqrt(3))/(sqrt(g))` `x = 2 sqrt(g) cos 30^@ [(1 +- sqrt(3))/(sqrt(g))] or x = (sqrt(3) + 3) m`. |
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| 168. |
A `0.098-kg` block slides down a frictionless track as shown in (Fig. 5.208). . The time taken by the block to move from `A` to `B` is.A. `(1)/(sqrt(g))`B. `(2)/(sqrt(g))`C. `(3)/(sqrt(g))`D. `(4)/(sqrt(g))` |
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Answer» Correct Answer - B (b) `(1)/(2) mv^2 = mg(3 - 1) = 2 mg` [from conservation of energy] or `v = sqrt(4 g) = 2 sqrt(g)` Vertical component at A is `2 sqrt(g) sin 30^@ = sqrt(g)` Time of flight, `T = (2 v sin theta)/(g) = (2 sqrt(g))/(g) = (2)/(sqrt(g))` Using `S = ut + (1)/(2) at^2`, we get `-1 = sqrt(g) t - (1)/(2) "gt"^2 or (1)/(2) "gt"^2 - sqrt(g) t - 1 = 0` `t = (sqrt(g) +- sqrt(g + 4 xx (1)/(2) g))/(2 xx (g)/(2)) t = (1 +- sqrt(3))/(sqrt(g))` Neglecting negative time, `t = (1 +- sqrt(3))/(sqrt(g))` `x = 2 sqrt(g) cos 30^@ [(1 +- sqrt(3))/(sqrt(g))] or x = (sqrt(3) + 3) m`. |
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| 169. |
Acceleration of a particle moving along the x-axis is defined by the law `a=-4x`, where a is in `m//s^(2)` and x is in meters. At the instant `t=0`, the particle passes the origin with a velocity of `2 m//s` moving in the positive x-direction. (a) Find its velocity v as function of its position coordinates. (b) find its position x as function of time t. (c) Find the maximum distance it can go away from the origin. |
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Answer» (a) By substituting given expression in the equation `a=v dv//dx` and rearranging, we have `vdv=-4xdxrArr underset(2)overset(v)(int)vdv=-4 underset(0)overset(x)(int)xdx rArr v= +-2sqrt(1-x^(2)) rarr v=2sqrt(1-x^(2))` since the particle passes the origin with positive velocity of `2 m//s`, so the minus sign in the eq. (i) has been dropped. (b) By substituting above obtained expression of velocity in the equation `v=dx//dt` and rearranging, we have `(dx)/sqrt(1-x^(2))=2dtrArr underset(0)overset(x)(int)(dx)/sqrt(1-x^(2))=2underset(0)overset(t)(int)dtrArr sin^(-1)(x)=2t rarr x=sin 2t` (c) The maximum distance it can go away from the origin is `1m` because maximum magnitude of fine function is unity. |
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| 170. |
Acceleration of particle moving along the x-axis varies according to the law `a=-2v`, where a is in `m//s^(2)` and v is in `m//s`. At the instant `t=0`, the particle passes the origin with a velocity of `2 m//s` moving in the positive x-direction. (a) Find its velocity v as function of time t. (b) Find its position x as function of time t. (c) Find its velocity v as function of its position coordinates. (d) find the maximum distance it can go away from the origin. (e) Will it reach the above-mentioned maximum distance? |
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Answer» (a) By substituting the given relation `a=dv//dt`, we have `(dv)/v=-2dtrArr underset(2)overset(v)(int)(dv)/v=-2underset(0)overset(t)(int)dt rarr v=2e^(-2t)`…(i) (b) By substituting the above equation in `v=dx//dt`, we have `dx=2e^(-2t)dt rArr underset(0)overset(x)(int)dx=2underset(0)overset(t)(int)e^(-2t)dt rarr x=1-e^(-2t)`...(ii) (c) Substituting given expression a in the equation `a=vdv//dx` and rearranging, we have `dv=-2dxrArr underset(2)overset(v)(int)dv=-2underset(0)overset(x)(int)dx rarr v=2(1-x)` ...(iii) (d) Eq. (iii) suggest that to cover `1m` it will take whose value tends to infinity. Therefore, it can never cover this distance. |
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| 171. |
Every car, you know , is fitted with an odometer, whichn indicates the actual distance travelled by the car, In going from Ambala to Delhi taken was 9 hours, ltbRgt Read the above passage and answer the following questins: (i) What are the values of avetage speed and avetage velocity over the jouney ? (ii) What is more relevant : average speed or avetage velocity ? (iii) What are the paracticla mplications of this study ? |
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Answer» Average speed= (total distance travelled )/( time taken ) ` = (( 16124 - 15 674) km)/( 9 hr) = ( 450 ) /4 km //hr = 50 km // hr` Average velocity = `( displacement /(time )` As journey starting from Ambala eneds at Ambals, therefore, displacement = 0` :. Average velocity = zero (ii) In a journey, average velocity is calculated only theoretically . Average speed is what matters actually. (ii) As averave speed = ( distance travelled)/(time taken), therefore, to increase average speed, giben distance should be travelled in least posssibe time. It implies that if tow starions (A) and (B) between which we have to travel, are connected by different routs, we should choose the rout of monimum distance along which road condition are really good so that tiem taken is small. |
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| 172. |
The sependometerof a car indicatesA. its instantaneous speedB. average speedC. averae velocityD. Both (1) and (3) |
| Answer» The speedometer of a vechicle indicates the instantaneous speed of the vehicle. | |
| 173. |
Two trains ` 120m ` and ` 100 m` in length are tunning in opposite directions with velocityes ` 42 km h^(-1) ` and ` 30 km h^(-1)` In what tiem they will completely cross each other ? |
| Answer» Correct Answer - ` 10 secs` | |
| 174. |
A man sitting in a train in motion is facing the engine . He tosses a coin up, the coin falls………………………………………..him. |
| Answer» Correct Answer - behind | |
| 175. |
When is the magnitude of (\(\bar A\) + \(\bar B\)) equal to the magnitude of (\(\bar A\) – \(\bar B\)) ? |
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Answer» When \(\bar A\) is perpendicular to \(\bar B\). |
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| 176. |
What is the maximum number of component into which a vector can be resolved? |
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Answer» The maximum number of component is Infinite. |
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| 177. |
A body projected horizontally moves with the same horizontal velocity although it moves under gravity. Why ? |
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Answer» Because horizontal component of gravity is zero along horizontal direction. |
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| 178. |
What is the maximum number of component into which a vector can be resolved ? |
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Answer» The maximum number of component Infinite. |
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| 179. |
A car acceletrates uniformaly at `4 m s^(-2)` froms rest. Find its velocity at the end of 5 seconds. |
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Answer» Given acceleration of the car, `a=4 m s^(-2)` Initial velocitt of the car,` u=0 m s^(-1)` Time interval, t=5s Final velocity of the car,` v=u+at` `=0+(4)(5)` `=20 m s^(-1)` |
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| 180. |
A Car accelerates uniformaly form rest. If the reading on the speedometer after 5 s is `km h^(-1)` |
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Answer» `u=0, v=72 km h^(-1) 72 35 m s^(-1) 20 m s^(-1)` `:.` acceleration `=(v-u)/(t)=(20-0)/(5)=4 m s^(-1)` |
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| 181. |
A particle exeriences constant acceleration form ` 20 seconds after strating from gest. If it travels a distance ` S_(1)` in the first ` 10 seconds` and distance ` S_(2)` in the next 10 seconds`. Find the relation between ` S_(1) and S_(@)`. |
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Answer» Hence, ` u =0 , t= 10 s , a=a (say ) , S=S_(1)` As, S=ut + 1/2 at ^(2)` :. S_(1) =0 xx 10 + 1/2 a xx 10^(2) =50 a ` ….(i) Taking motion of particle for ` 10 s + 10 s = 20 s`, we have `(S_(1) + S_(2) ) =0 xx 20 + 1/2 a xx 20^(2) = 200 a ` ...(ii) :. S_(2) =(S_(1) + S_(2)) - S_(1) =200 a- 50 a` :. ` S_(2)/S_(1) = (150 a)/(50 A) =3 or S_(2) =3 S_(1)`. |
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| 182. |
A particle is projected up an inclined plane with initial speed `v=20m//s` at an angle `theta=30^(@)` with plane. The component of its velocity perpendicular to plane when it strikes the plane isA. `10sqrt(3)m//s`B. `10m//s`C. `5sqrt(3)m//s`D. Data is insufficient |
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Answer» Correct Answer - B |
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| 183. |
To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0 ms, what is its average acceleration during that contact? Take `g= 98 m//s^2.` |
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Answer» Correct Answer - A::B::C::D `a_(av)=(Deltav)/(Deltat)=(v_f-v_i)/(Deltat)` `=(sqrt(2gh_f)+sqrt(2gh_i))/(Deltat)` `=(sqrt(2xx9.8xx2)+sqrt(2xx9.8xx4))/(12xx10^-3)` `=1.26xx10^3 m//s^2` |
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| 184. |
Velocity-time graph for a car is semicircle as shown here. Which of the following is correct: A. Car must move in circular path.B. Acceleration of car is never zero.C. Mean spped of the aprticel is `pi//4 m//s`.D. The car makes a turn once during its motion. |
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Answer» Correct Answer - C Car is moving in straight line which changing velcoity. `V_(av) = (distance)/(time) rArr V_(av) = (pia.b)/(2xx2) rArr V_(av) = (pi)/(4)` |
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| 185. |
Assertion: The instantaneous velocity does not depend on instantaneous position vector. Reason: The instantaneous velocity and average velocity of a particle are always same.A. Statement -1 is True, Statement-2 is Ture, Statement-2 is a correct explanantion for Statement-1.B. Statement-1 is Ture, Statement-2 is Ture, Statement-2 is Not a correct explanantion for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - C As `vecv = (dvecs)/(dt)`, hence statement `1` is correct. Instantaneous velcoity is same always to average velocity if the particle is moving with constant velocity. |
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| 186. |
What does the slope of v-t graph indicate? |
| Answer» Acceleration. | |
| 187. |
Under what condition the average velocity equal to instantaneous velocity? |
| Answer» For a uniform velocity. | |
| 188. |
Give an example when a body moving with uniform speed has acceleration. |
| Answer» In the uniform circular motion. | |
| 189. |
State the essential condition for the addition of the vector. |
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Answer» They must represent the physical quantities of same nature. |
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| 190. |
Two balls of different masses are thrown vertically upward with same initial velocity. Height attained by them are h1 and h2 respectively what is h1/h2. |
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Answer» 1/1, because the height attained by the projectile is not depend on the masses. |
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| 191. |
Two balls of different masses (one lightre an other heavier) are thrown vertically upward with same initial speed. Which one will rise to greater height ? |
| Answer» Wheb a ball is projected veritially upward with speed ` u`, the maximum height attained, ` h_(max) =u^(2)//2g`, which is independent of mass of ball. Thus both the balls ( of different masses) will attain same height when projected vertically upward wigh the same velocity. | |
| 192. |
Give any two examples for parallelogram law of vectors. |
Answer»
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| 193. |
Two balls of different masses are thrown vertically upward with same initial velocity. Maximum heights attained by them are h and h respectively what is h1 /h2 ? |
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Answer» Same height, ∴ h1 /h2 = 1 |
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| 194. |
Which of the two – linear velocity or the linear acceleration gives the direction of motion of a body? |
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Answer» Linear velocity. |
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| 195. |
A car moving with velocity of 50 kmh–1 on a straight road is ahead of a jeep moving with velocity 75 kmh–1 . How would the relative velocity be altered if jeep is ahead of car ? |
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Answer» The relative velocity : No change |
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| 196. |
If two objects A and B are moving along a straight line in the opposite direction with the velocities VA and VB respectively, then relative velocity is-(a) VA + VB(b) VA -VB(c) VA VB(d) VA/VB |
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Answer» Correct answer is (a) VA+ VB |
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| 197. |
A car moving with velocity of 50 kmh-1 on a straight road is ahead of a jeep moving with velocity 75 kmh-1 would the relative velocity be altered if jeep is ahead of car? |
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Answer» No change in the relative velocity. |
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| 198. |
A person moving horizontally with velocity\(\vec V_m\) . Rain falls vertically with velocity \(\vec V_R\) To save himself from the rain, he should hold an umbrella with vertical at an angle of – |
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Answer» Correct answer is (b) tan -1 (\(\frac{V_m}{V_R}\)) |
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| 199. |
The distance travelled by a body, falling freely from rest in t = 1s, t = 2s and t = 3s are in the ratio of –(a) 1 : 2 : 3 (b) 1 : 3 : 5 (c) 1 : 4 : 9(d) 9 : 4 : 1 |
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Answer» (c) The distance travelled by a free falling body S = \(\frac{1}{2}\) gt2 ∴ S α t2 ∴ S1 : S2 : S3 : 12 : 22 : 32 = 1 : 4 : 9. |
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| 200. |
An object falling freely under gravity close to earth is –(a) one dimensional (b) circular motion (c) rotational motion (d) spinning motion |
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Answer» (a) one dimensional |
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