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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.
| 951. |
Statement I: Distance and displacement are different physical quantities. Statement II : Distance and displacement have same dismension.A. Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I.B. Statemnt I is true, Statement II is true, Statement II is true, Statement II is false.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
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Answer» Correct Answer - B b. Two different physical quantities may have samedimension. |
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| 952. |
The displacement of a particle as a function of time is shown in . It indicates .A. ainvelocity, but the motion is retarded and finally the particle stops.B. The velocity of the particle dereases.C. The accleration of the particle is in opposits direction to the velocty.D. The particle stares with a constant velocity, the motion is accelerated and finaly pparticle moves with another constant velocity. |
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Answer» Correct Answer - A::B::C Initially at origin, slope is not zero, so the pariticle has some initialvelcoty but with time we see that slope is decreasing and finally the slope becomes zero. So the particle stops finally. As the magnitude of velocityis decreasing, velocity and acceleratioin will be in oppsite derections. |
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| 953. |
The velcity of a paarticle is given by v=u_(0) + gt+ 1/2 ft^(2). If its position is `x =0` at `t=0`, then what is its displacement after `t=1 s` ?A. `v_(0) +g//2+f`B. `v_(0) +2 g+3f`C. `v_(0) +g//2 +f//3`D. `v_(0) +g +f` |
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Answer» Correct Answer - C `v = v_(0) +g t +ft^(2)` or `(dx)/(dt) = v_(0) +g t +ft^(2)` `rArr dx = (v_(0) +gt +ft^(2)) dt` So, `int_(0)^(x) dx = int_(0)^(1) (v_(0) +gt +ft^(2))dt` `rArr x = v_(0) +(g)/(2) +(f)/(3)` |
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| 954. |
STATEMENT-1: The maixmum range along the inclined plane, when thrown downward is greater than that when thrown upward along the same inclined plane with constant velocity. STATEMENT-2: The maximum range along inclined plane is independent of angle of inclination.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. |
| Answer» Correct Answer - C | |
| 955. |
Statement I: The displacement of a body may be zero, though its distance can be finite. Statement II: If the bodt moves such that finally it arrives at the initial point, then displacement is zero while distance is finite.A. Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I.B. Statemnt I is true, Statement II is true, Statement II is true, Statement II is false.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
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Answer» Correct Answer - A When the body returns to its initial point, point, its displacement is zero but distance travelled is bot zero, |
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| 956. |
STATEMENT-1: When a body is dropped or thrown horizontally form the same height, it reaches the ground at the same time. STATEMENT-2: They have same acceleration and same initial speed in vertical direction.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. |
| Answer» Correct Answer - A | |
| 957. |
STATEMENT-1: When a body is dropped and another body is thrown horizontally from the same height, it reaches the ground at the same time. STATEMENT-2: They have same acceleration and same initial speed in vertical direction.A. Statement-I is true, Statement-II is true, Statement-II is correct explanation for Statement-IB. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanation for Statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true. |
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Answer» Correct Answer - A Because initial vertical velocity component is zero in both cases. |
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| 958. |
STATEMENT-1: A man can cross river of width `d` in minimum time `t`. On increasing river velocity, minimum time to cross the river by man will remain unchanged. STATEMENT-2: Velocity of river is perpendicular to width of river. So time to cross the river is independent of velcoity of river.A. Statement-I is true, Statement-II is true, Statement-II is correct explanation for Statement-IB. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanation for Statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true. |
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Answer» Correct Answer - A Yes, river velocity does not any help to cross the river in minimum time. |
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| 959. |
Two trains are moving with velocities `v_1 = 10 ms^-1 and v_2 = 20 ms^-1` on the same track in opposite directions. After the application of brakes if their retarding rates are `a_1 = 2 ms^-2 and a_2 = 1 ms^-2` respectively, then the minimum distance of separation between the trains to avoid collision isA. 150 mB. 225 mC. 450 mD. 300 m |
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Answer» Correct Answer - B `d=d_1+d_2` `=(v_1^2)/(2a_1)+(v_2^2)/(2a_2)` `((10)^2)/(2xx2)+((20)^2)/(2xx1)` `=225m` |
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| 960. |
Two particles `A and B` are moving with constant velocities `v_1 and v_2`. At `t = 0`, `v_1` makes an angle `theta_0` with the line joining `A and B and v_2` makes an angle `theta_2` with the line joining `A and B`. (a) Find the condition for `A and B` to collide. (b) Find the time after which `A and B` will collide if separation between them is `d at t = 0`. . |
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Answer» (a) For `A and B` to collide, their relative velocity must be directed along the line joining them. Therefore their relative velocity along the perpendicular to this line must be zero. Thus `v_1 sin theta_1 = v_2 sin theta_2`. (b) `v_(A P P) = v_1 cos theta_1 + v_2 cos theta_2` `t = (d)/(v_(a P P)) = (d)/(v_1 cos theta_1 + v_2 cos theta_2)`. |
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| 961. |
Statement-I : If two particles, moving with constant velocities are to meet, the relative velocity must be along the line joining the two particles. Statement-II : Relative velocity means motion of one particle as viewed from the other.A. Statement-I is true, Statement-II is true, Statement-II is correct explanation for Statement-IB. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanation for Statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true. |
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Answer» Correct Answer - B To meet, co-ordinates must be same. So in frame of one particle, second particle should approach it. |
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| 962. |
Particles `A` and `B` are moving with constant velocities along `x` and `y` axis respectively, the graph of separation between them with time is A. B. C. D. |
| Answer» Correct Answer - D | |
| 963. |
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. `(pi)/(2)(v^(4))/(g^(2))`B. `pi(v^(2))/(g^(2))`C. `pi(v^(2))/(g)`D. `pi(v^(4))/(g^(2))` |
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Answer» Correct Answer - D `A = piR_(max)^(2)` `= pi xx ((U^(2)sin 2 xx 45^(@))/(g))^(2)` `= (piv^(4))/(g^(2))` |
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| 964. |
A particle is projected at an angle of θ with respect to the horizontal direction. Match the following for the above motion. (a) vx – decreases and increases (b) vy – remains constant (c) Acceleration – varies (d) Position vector – remains downward |
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Answer» (a) vx = remains constant (b) vy = decreases and increases (c) a = remains downward (d) r = varies |
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| 965. |
A compartment of an uniformly moving train is suddenly detached from the train and stops after covering some distance. The distance covered by the compartment and distance covered by the train in the given time –(a) both will be equal(b) second will be half of first (c) first will be half of second (d) none |
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Answer» (c) first will be half of second |
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| 966. |
Two cars (A) and (B) travel in straight line . The distance of (A) from the starting point is given as a function of time be ` a_A (t) = pt + qt^2 `, with ` p = 2 . 60 ms^(-1)` and ` q= 1.20 ms^(-2)`. The distance of (B) from the starting pint is ` x_B (t) = rt^2 - st^3 ` are ` r= 2. 80 ms^(-2)` and ` s = 0.20 ms^(-3)`. Answer the following questions , Which car is ahead just after they have the starting point ?A. ` Car (A) moves ahead `B. Car (B) moves aheadC. Cars (A) and (B) move simultaneouslyD. Data is insufficient to decide. |
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Answer» Correct Answer - A ` x_A= pt + qt^2 , x_b =rt^2 - st^3` Velocity, ` v_A= (dx)_A)/(dt) = p+ 2 qt` ` v_B = (dx_B)/(dt) = 2 rt - 3 st^2` Acceleration, ` a_A = (dv_A)/(dt) = 2 q , a_B= 2 r- 6 st` The car that initially moves ahead is the one which bas largr velocity at ` t= 0. at ` t=0 , v= p` and ` v_B =0`. So, initially car (A) moves ahead. |
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| 967. |
Two cars (A) and (B) travel in straight line. The distance of (A) from the starting point is given as a function of them be ` a_A (t) = pt + qt^2 `, with ` p = 2 . 60 ms^(-1)` and ` q= 1.20 ms^(-2)`. The distance of (B) from the starting point is ` x_B (t) = rt^2 - st^3 ` are ` r= 2. 80 ms^(-2)` and ` s = 0.20 ms^(-3)`. Answer the following questions, At what time (s) are the cars at the same point ?A. ` 2. 60`B. ` 2. 27 s`C. ` 5. 7 3 s`D. ` both ` 2. 27 `and ` 5 . 73 s` |
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Answer» Correct Answer - D ` x_A=A_b` :. Pt+qt^2 =rt^2 -st^3` One solution is ` t= 0`, which means both the cars start from the same point. To find the other solution , divede it by ` 9t), we have ` p + qt =rt -st^2` or ` st^2 + (q-2) t + p=0` or ` t= 1/(2 s) [ q- r) =- sqrt((q -r)^2 -4sp)]` ` =1/(2 xx 0.20) [1 . 20 - 2 . 89 )` ` +- sqrt ((1.20 - 2 .80)^2 - 4 xx 0.24 xx 2.60)]` `= 4. 00 +- 73 = 2.27 s ` and ` 5. 73 s`. Hence ` x_A= x_B` for ` t= 0, t= 2.27 s` and ` t= 5. 73 s`. |
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| 968. |
A body is dropped from the top of a tower. If its average velcoity when it reaches the ground, is ` 10 ms^(-1)` and the time taken by it to reach the ground is 2s, the height of the tower is ______ m.A. 20B. 10C. 40D. 25 |
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Answer» Correct Answer - A Average velocity =` ("total distance")/("total time ") ` ` 10 h/2` h =20 m, |
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| 969. |
A train moves from one station to another in two hours time. Its speed during the motion is shown in the graph Calculate (i) Maximum acceleration during the journey. (ii) Distance covered during the time interval from 0.75 hour to 1 hour |
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Answer» Correct Answer - (i) `160 km//hr^(2)` (ii) 10 km From given situation : (i) `a_(avg)=(60-20)/(1.00-0.75)=4000/25=160 km//hr^(2)` (ii) Area `=1/2xx[20+60]xx0.25` `=40xx25/100=10 km` |
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| 970. |
A particle is projected vertically upward with velocity `u` from a point `A`, when it returns to the point of projection .A. Its average velocity is zeroB. Its displacement is zeroC. Its average speed is `u//2`D. Its average speed is `u` |
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Answer» Correct Answer - A, B, C |
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| 971. |
Find the time `t_0` when x-coordinate of the particle is zero. |
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Answer» Correct Answer - B::C::D `8.25 s` (b) Total displacement =`4+4-16-32-16=-56 m` This is also equal to `x_f-x_i=-46-10=-56 m` Total distance=`4+4+16+32+16=72 m` Total time =`16s` Now, average velocity=total displacement/total time ` = -56/16=-3.5 m//s` average speed =`("total distance")/("total time")` `=72/16=4.5 m//s` (c) Average acceleration=`(Delta v)/(Delta t)` `=(v_f-v_i)/(Delta t)=(v_(8 sec)-v_(2 sec))/(8-2)` `=(-8-4)/6=-2 m//s^2` |
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| 972. |
A particle is moving with initial velocity `bar(u)=hati-hatj+hatk`. What should be its acceleration so that it can remain moving in the same straight line?A. `vec(a)=2hati-2hatj+2hatk`B. `vec(a)=-2hati+2hatj+2hatk`C. `vec(a)=3hati+3hatj+3hatk`D. `vec(a)=1hati-1hatj` |
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Answer» Correct Answer - A To move in a straight line `vec(epsilon)||vec(u)` |
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| 973. |
Assertion : Velocity and acceleration of a particle are given as, `v=hati-hatj and a=-2 hati+2 hatj` This is a two dimensional motion with constant acceleration. Reason : Velocity and acceleration are two constant vectors.A. If the both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D `a=-2(hati-hatj)=-2v` i.e. a and v are constant vectors and antiparallel to each other. So motion is one dimensional. First retarded then accelerated in opposite directions. |
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| 974. |
The maximum range of a rifle bullet on the horizontal ground is `6 km`. Find its maximum range on an inclined of `30^@`. |
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Answer» Maximum range on horizontal plane, `R = (u^2)/(g) = 6 km` (given) Maximum range on a inclined plane, `R_(max) = (u^2)/(g(1 + sin prop))` Putting `prop = 30^@` `R_(max) = (u^2)/(g(1 + sin 30^@)) = (2)/(3) ((u^2)/(g)) = (2)/(3) xx 6 = 4 km`. |
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| 975. |
Unit vector `hat(P)` and `hat(Q)` are inclined at an angle `theta`. Prove that `|hat(P)-hat(Q)|= 2 sin(theta//2)`. |
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Answer» We know that `|vec A - vec B|^(2) = ( vec A - vec B). ( vec A - vec B)` or `|vec A - vec B|^(2) = vec A . Vec A + vec B. vec B -2 vec A . Vec B` `= 1 +1 -2 (1) 91) cos theta =2 91 - cos theta)` ltvbrgt `= [1 - (1-2 sin ^(2) theta //2 )] =4 sin ^(2) theta //2 ` `(:. cos 2 theta =1-2 sin ^(1) theta )` `:. | vec A -vec B =2 sin theta //2`. |
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| 976. |
Find the angle vetween force ` vec F =(3 hat I + 4 hat j -5 hat k)` and displacement ` vec d =(5 hat I + 4 hat j + 3 hat k)` unit. Also find the projection of ` vec F and vec d`. |
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Answer» Here, `vec F = (3 hat I + 4 hat j - 5 khat k) ,` `verc d = (5 hat I + 4 hat j + 3 hat k) ` ` vec F . Vec D =(3 hat I + 4 hat j -5 hat k) . (5 hat I + 4 hat j + 3 hat k)` `= 3 (5) + 4 (4) -5 93) = 16 units `F= sqrt ( F_(_(x)^(2) + F_(y)^(2) + F_(z)^(2)) ` ` = sqrt (3^(2) + 4^(2) + (-5)^(2) ) =sqrt 50` `d= sqrt (d_(x)^(2) + d_(y) ^(2) + d_(z)^(2) ) = sqrt (5^(2) + 4^(2) + 3^(3) ) =sqrt 50)` Now , cos theta = (vec F. vec d) /(F d) = (16)/(sqrt 50 dqrt 50 )) =(16) /(50) =0.32` `thets = cos ^(-1) (0. 32) = 71 .3^^(@)` Unit vector along `vec d ` is `vec d = (5 hat i + 4 hat j _ 3 hat k)/(sqrt ( 5^(2) +4 %^(2) + 3^(2))) =(5 hat i + 4 hat j + 3 hatk)/(sqrt 50 ` ` vec F . vec d = (3 hat i _ 4 hat j - 5 hat k) . (( 5 hat i + 4 hat j + 3 hat k)/(sqrt 50 )` `=(3 (5) + (4) -5 (3)) /(sqrt 50 ) = (16) /(sqrt)` :. Projection of `vec F on vec d` = component vector of `vec F ` along vedc d` In is` ` =( vec F .vec d) vec d = (160 /(sqrt 50 ) ((5 hat i + 4 hat j + 3 hat k)) /(sqrt50)` ` =0. 32 (5 hat i +4 hat j+ 3 hat k)`. |
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| 977. |
A car A has just overtaken car B which is moving in the same direction as A. Gien below are few cases explanainig the motion of the cars at that instance. Identify th given cases as cases that are necessarily true, cases that can be true only under certain conditions and cases that are necessarily false. (i) The car is accelerating and car B isb decleariting (ii) The car A is decelearting and car B accelerating (iii) Both car accelerating (iv) Both cars are decelerating (v) Both cars moving with uniform velocity (vi) Car A is moving with uniform velocity and car B is accelerationg. (vii) Instantaneous spee of car A is less than that of car B (viii) Instaneous speed of car B is less thatn that of car A. |
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Answer» What is instantaneous velocity? When car A just overtaken car B, which one has higher insaneous velocity? If the instaneous velocity of A is greater than that of B, then is it a must that it should be accelerated or decelerated? Is it not possible that both the cars have acceleration or deceleration or more with uniform velocity? |
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| 978. |
A particle is parojected vertically upwards from grund with initial velocity `u`. a. Find the maximum height `H` the particle will attain and time `T` that it will attain and time `T` that it will take to return to the ground . . b. What is the velocity when the particle returns to the ground? c. What is the displacement and distance travelled by the particle during this time of whole motion. |
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Answer» Cosnider the motion from `A` to `B`: `s=+H` (final point lies above the initial point), initial velocity `=u`, final velocity `v=0`. Let the time taken to go from `A` to `B` be `t_(1)`. Using `v=u-g t`, we get `0=u-g t_(1) rArrt_(1)=(u)/(g)` Using `s=ut-(1)/(2)g t^(2)`, `rArr h=u ((u)/(g))-(1)/(2)g((u)/(g))^(2)=(u^(2))/(2g)` So the maximum height attained is `H=(u^(2))/(2g)`. Cosnsider the return motion from `B` to `A`: `s=H` (final point lies below the initial point) `u=0` (at point `B`, velocty is zero) Let time taken to go from `B` to `A`be `t_(2)`. We have `t_(2)=sqrt((2H)/(g))=sqrt((2)/(g)(u^(2))/(2g))(u)/(g)` Hence `t_(1)` is known as the time of ascent and `t_(2)` is known as the time of descent. We can see that time of ascent=Time of descent`=(u)/(g)` Total time flight `T=t_(1)+t_(2)=(2u)/(g)` Hence, time of flight is the time for which the particle remains in air. Alternative method to find the time of flight: a. Consider the motion from `A` to `B`: `s=0` (initial and final points are same ) Initial velocity `=u`, time taken`=T` Using `s=ut-(1)/(2)g t^(2)`, we have `0=uT-(1)/(2)g t^(2)` `rArr T =(2u)/(g)` b. Magnitude of velocity on returning the fround will be same as that of inital velocity but directionwill be opposite. Proof:Let `v` be the velocity on reaching the ground. Then from previous formulae, we get ` v=- sqrt(2gH)=-sqrt(2g(u^(2))/(2g))rArrv=-u`, hence proved, c. Displacement`=0`, distance travelled `=2H=u^(2)/(g)`. |
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| 979. |
If the displacement of a particle varies with time as `sqrt x = t+ 3`A. velocity of the particle is inversely proportional to tB. velocity of particle varies linearly with tC. velocity of particle is proportional to `sqrt t`D. initial velocity of the particle is zero |
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Answer» Correct Answer - B `sqrtx=t+3` `:. x=(t+3)^2` or `v=(dx)/(dt)` `=2(t+3)` `:.` v-t equation is linear |
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| 980. |
Can the speed of a body will be negative ? |
| Answer» No, because the speed of an object is the distantce travelled by onject in unit time and distance can never be begative. | |
| 981. |
What is the magniude and direction of ` ( hat I + hat j) ?` |
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Answer» Magnitude of ` (hat v + hat j) =| hat I + hat j|` `=sqrt ((1)^(2) + (1)^(2))=sqrt 2` Let ` (hat I + hat j)` make an angle ` beta` with the direction of ` hat i` then ` tan ` beta =1//1=1 tan 45^(@) , :. Beta =45^(@)`. |
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| 982. |
Which speed is measured by the speedometer of your scooter? |
| Answer» The speedometer measures the instantaneous speed of the scooter at a given instant of time. | |
| 983. |
Why is the speed in general, greater than the magniude of the velcity ? |
| Answer» If ther is a chage in the direction of motion of a body, then the actual path langth in a given time is greater than the displacement in that time. Therefore speed (=path length//time taken) is greatre than the magnitude of velocity (= displacement // time). | |
| 984. |
A particle moves along the curve `(x^(2))/(9) +(y^(2))/(4) =1`, with constant speed `v`. Express its "velocity vectorially" as a function of `x,y`. |
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Answer» Given `(x^(2))/(9)+(v^(2))/(4)=1` `rArr (2x)/(9) [(dx)/(dt)]+(2y)/(4)[(dy)/(dt)]=1 [(dx)/(dt)=v_(x),(dy)/(dt)=v_(y)]` `rArr v_(x) =-(9)/(4)(y)/(x)v_(y)` Also `v_(x)^(2)+v_(y)^(2) =v^(2) rArr (-(9)/(4)(y)/(x)v_(y))^(2) +v_(y)^(2)=v^(2)` `rArr v_(y)^(2) =(16x^(2)v^(2))/(16x^(2)+81y^(2)) rArr v_(y)= (+-4xv)/(sqrt(16x^(2)+81y^(2))` From (i) and (ii), we get ,`v_(x)=(+-yv)/(sqrt(16x^(2)+81y^(2))` So, velocity is given by `v_(x) jat i+ v_(y) hat j = ((+- 9 y hat i+- 4 x hat j)v)/(sqrt(16x^(2)+81y^(2)))`. |
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| 985. |
A particle is projected with a certain velocity at an angle `alpha` above the horizontal from the foot of an inclined plane of inclination `30^(@)`. If the particle strikes the plane normally then `alpha` is equal toA. `30^(@)+"tan"^(-1)((sqrt(3))/2)`B. `45^(@)`C. `60^(@)`D. `30^(@)+tan^(-1)(2sqrt(3))` |
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Answer» Correct Answer - A |
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| 986. |
If the didplacement is given by , ` x =1 + 2 t= 3 t^2 `, the value of instantaneous acceleration is ……………………………….... |
| Answer» Correct Answer - units | |
| 987. |
A cae is travelling along a straight line. It covers one-half of the total, distance with a velocity ` 10 km//h` . The remaining part of the distance was covered with velocity `12 ms^(-1)`. For half the time and with velocity ` 16 ms^(-1)` for the other half the tiem. Find the average speed over the whole time of motion. |
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Answer» Let (S) be total distance to be travelled by car. The taken by car to cover the distance ` S //2 ` with velocity ` Then S/2 = 12 xx t-2 /2 + 16 xx t_2 /2 = 14 t-2` or ` t_2 = S/928)` Total time taken , `T= t_1 = t_2 = S/ ( 20) = S/( 28) = ( 12S)/( 140)` ltbRgt Average speed ` = S /T = S/ ( 12 S// 140)` ` = ( 140)/( 12) = 11. 6 ms^(-1)`. |
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| 988. |
A truck has to carry a load in the shortest time from one point to another at a distance L from the first.It can only start up or slow down at the same acceleration a. What maximum velocity must the truck attain to satisfy this condition ? |
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Answer» Correct Answer - `sqrt(La)` |
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| 989. |
Velocity of a projectile at height `15 m` from ground is `v=(20hati+10hatj)m//s`. Here `hati` is in horizontal direction and `hatj` is vertically upwards. Then Speed with which particle is projected from ground is……… m/sA. 30B. `20sqrt(2)`C. `sqrt(20)`D. `3sqrt(40)` |
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Answer» Correct Answer - B |
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| 990. |
A police car B is chasing a culprit’s car A. Car A and B are moving at constant speed `V_(1) = 108 km//hr` and `V_(2) = 90 km//hr` respectively along a straight line. The police decides to open fire and a policeman starts firing with his machine gun directly aiming at car A. The bullets have a velocity `u = 305 m//s` relative to the gun. The policeman keeps firing for an interval of `T_(0) = 20 s`. The Culprit experiences that the time gap between the first and the last bullet hitting his car is `Deltat`. Find `Deltat`. |
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Answer» Correct Answer - `Deltat = 23.33 s` |
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| 991. |
A chain of length L is supported at one end and is hanging vertically when it is released. All of the chain falls freely with acceleration g. The moment, the chain is released a ball is projected up with speed u from a point 2 L below the point of support. Find the interval of time in which the ball will cross through the entire chain. |
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Answer» Correct Answer - `(L)/(u)` |
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| 992. |
A valley has two walls inclined at `37°` and `53°` to the horizontal. A particle is projected from point P with a velocity of `u = 20 m//s` along a direction perpendicular to the incline wall OA. The Particle hits the incline surface RB perpendicularly at Q. Take `g = 10 m//s^(2)` and find: (a) The time of flight of the particle. (b) Vertical height h of the point P from horizontal surface OR. `[tan 37^(@) = (3)/(4)]` |
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Answer» Correct Answer - (a) `2.5 s` (b) `4.05 m` |
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| 993. |
AB is a pipe fixed to the ground at an inclination of `37°`. A ball is projected from point O at a speed of `u = 20m//s` at an angle of `53°` to the horizontal and it smoothly enters into the pipe with its velocity parallel to the axis of the pipe. [Take`g = 10 ms^(–2)]` (a) Find the length L of the pipe (b) Find the distance of end B of the pipe from point O. |
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Answer» Correct Answer - (a) `L = 14.58 m` (b) `OB = 41.66m` |
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| 994. |
What are the angles made by vector ` vec A =hat I + sqer 3 hat j` with x-axis and y-axis ? |
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Answer» Comparing the given vector ` vec A= hat i + sqrt 3 hat j` with vector, ` vec A= A_(x) hat I + A_(x) hat j`, we have ` A_x =1 and A_x =sqrt` :. A = sqrt( A_x ^2 + A_y^2) = sqrt( 1^2 + (sqrt3)^2 )=2` Let ` alpha, beta` are the angles which `vec A` make with x-axis and y-axis respectively. Then ` cos alpha = A_x/X = 1/2 = cos 60^@ or alpha =60^@` ` cos beta = A_y /A = (sqrt3)/2 = cos 30^@ or beta = 30^@`. |
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| 995. |
Two persons are pulling, the ends of a strong in such a way so that the string is stretched horizontally. When a weight of ` 10 kg ` is suspended in the middle of the string. The string does not remanin horizontal. Can the persons maken it horizontal again by pulling it with a greater force ? |
| Answer» No, because the vetical weight suspended cannot be balanced by horizonatl forces howsoever large these may be. [Refer to Concepetual Problem 21] . | |
| 996. |
The revolution of earth aroud the sun is…………………………….motion. |
| Answer» Correct Answer - two dimensional | |
| 997. |
The windshield of a truck is inclined at `37^(@)` to the horizontal. The truck is moving horizontally with a constant acceleration of `a = 5 m//s^(2)`. At the instant the velocity of the truck is `v_(0) = 0.77 m//s`, an insect jumps from point A on the windshield, with a velocity `u = 2.64 m//s` (relative to ground) in vertically upward direction. It falls back at point B on the windshield. Calculate distance AB. Assume that the insect moves freely under gravity and `g = 10 m//s^(2)`. |
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Answer» Correct Answer - `AB = 0.57 m` |
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| 998. |
There is a large wedge placed on a horizontal surface with its incline face making an angle of `37^(@)` to the horizontal. A particle is projected in vertically upward direction with a velocity of `u = 6.5 m//s` from a point O on the inclined surface. At the instant the particle is projected, the wedge begins to move horizontally with a constant acceleration of `a = 4 m//s^(2)`. At what distance from point O will the particle hit the incline surface if (i) direction of a is along BC? (ii) direction of a is along AB? |
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Answer» Correct Answer - (i) `3.38 m` (ii) `2.5 m` |
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| 999. |
A cylinder of radius R has been placed in a corner as shown in the fig. A wedge is pressed against the cylinder such that its inclined surfaces touches the cylinder at a height of `(2R)/(5)` from thw ground. Now the wedge is pushed to the left at a constant speed `V = 15 m//s`. With what speed will the cylinder move? |
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Answer» Correct Answer - `20 m//s` |
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| 1000. |
The entrance to a harbour consists of `50 m` gap between two points A and B such that B is due east of A. Outside the harbour there is a `8 km//hr` current flowing due east. A motor boat is located 300 m due south of A. Neglect size of the boat for answering following questions- (a) Calculate the least speed `(V_(min))` that the motor boat must maintain to enter the harbour. (b) Show that the course it must steer when moving at Vmin does not depend on the speed of the current. |
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Answer» Correct Answer - (a) `sqrt((48)/(37)) km//hr` |
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