1496 + Interview Questions in KINEMATICS IN GENERAL AWARENESS Page 16 InterviewSolution

751.	A ball is thrown up, it reaches a maximum height anf then comes down. IF ` t_1` and `t_2` (t_2 gt t_1)` ar the time that the ball takes to be at a particular height then the time taken by the ball to reach the highest point is .A. (a) ` (t_1+t_2)`B. (b) ` (t_2 -t_1)`C. (c ) ` (t_2 -t_1)//2`D. (d) ` (t_2 +t_1)//2`
Answer» Correct Answer - D Let (S) be the height of a particular point where the ball croosses in time ` t_1 ` and t_2` seconds while going upwards and coming downwards. If (u) is the initial velocity of projection of ball, then ` S= ut-1 - 1/2 gt_1^2 = ut^2 - 1/2 gt_2^@` or ` u ( t_1 -t_2) = 1/2 g (t_2^2 -t_1^2)` or ` u = 1/2 g (t- +t-10 . if (T) si the time taken by ball to reach to its highest then using the relation `v = u+ at, we have ` 0 = u + ( - g) T` ltbRgt or ` T= u /g = 1/2 g ( t_2- t_1) //g = 1/2 (t_2-t_1)`.

Discussion

752.	A man directly crosses a river in time `t_1` and swims down the current a distance equal to the width of the river I time `t_2`. If `u and v` be the speed of the current and the man respectively, show that : `t_1 :t_2 = sqrt(v + u) : sqrt(v - u))`.
Answer» We know that the time taken foa a man to cross the river directly is given by `t_1 = (d)/(sqrt(v^2 - u^2))`, where `d` is the width of the river. Now, if the man swims downstream then the resultant velocity is `v + u`. The corresponding time, `t_2 = (d)/(u + v)`. Now, `(t_1)/(t_2) = (v+u)/(sqrt(v^2 - u^2))= (sqrt(v + u) sqrt(v + u))/(sqrt(v + u)sqrt(v - u)) rArr (t_1)/(t_2) = (sqrt(v + u))/(sqrt(v - u))`.

Discussion

753.	A man `16 m` behind the door of a train when it starts it starts moving with an acceleration `a=2 ms^(-1)`.The man runs with a msximum constant speed to get into the train. How for does he have to run and after what time does he get the train ? What is his maximum constant speed ?
Answer» Let (v) be the maximum constant speed ot theman. He get into the train after time (t), of the man. He will get into tnto the trainaftre time (t), when his velocity becomes equal to the velocity of train. :.`v=0 + 2 xx t =t ` ….(i) Distance moved by man in time (t), ltbRgt `S=ut =2t xx t=2 t^(2)` Distance moved by train in time (t) is `S_(1) =0 +1/2 xx at^(2) =1/2 xx 2 xx t^(2)=1/2 xx2 xx t^(2) =t^(2)` Mam will get into the train, when `S=S_91) +16` :. 2 t^(2) =t +16 or t^(2) =16` or `t=4s` Distance moved by by man `=2 xx (4)^(2) =32 m` Maximum constant speed of man, `v= 2 t= 2 xx 4 =8 ms ^(-1)`.

Discussion

754.	Two particles `A` and `B` are initially ` 40 m`apart, `A` is behind `B`. Particle `A` is moving with uniform velocity of `10 m s^(-1)` towared `B`. Particle `B` starts moving away from `A` with constant acceleration of `2 m s^(-1)`. The minimum distance between the two is .A. `20 m`B. `15 m`C. `25 m`D. `30 m`
Answer» Correct Answer - b Distance travelled by `A` in time `5`, s `S_(1) =10 xx 5=50 m` Distance travelled by `B` in time `5 s` `S_(2) =(1)/(2) at^(2) =(1)/(2) xx 2 xx 5^(2) =25 m` Mimnimum distance `=40 +S^(2) -S_(1) =15 m`.

Discussion

755.	A train `200 m` long is moving with a velocity of `72 km h^(-1)` Find the time taken by the trainto cross the bridge `1 km` logn .
Answer» Distance to be coverd`=1000+200=1200 m` Velocity `=72 km h^(-1) =20 ms^(-1)` `t=(Distance)/(Velocity)=(1200)/(20) =60 s`.

Discussion

756.	A particle is moving with constant speed in a circle as shown in (Fig. 5.163). Find the angular velocity of the particle A with respect to fixed point `B and C` if angular velocity with respect to `O is omega`. .
Answer» Angular velocity of `A` with respect to `O` is `omega_(A C) = ((v_(A O))_\|_)/(r_(A O)) = (v)/( r)= omega` `:. omega_(A B) = ((v_(A B))_\|_)/(r_(A B))= (v)/(2 r)=(omega)/(2)` and `omega_(A C) = ((v_(A C))_\|_)/(r_(A C)) = (v)/(3 r) = (omega)/(3)`.

Discussion

757.	Can we add a vector representing a force of ` 10 N` to a vector of force ` 2000 ` dune.
Answer» Force is a vector quantity. The two forces can be added by jaws of vectors , parallelogram law of vectors which is possible if we know the angle between the two forces, As the angle beteween the two given forces is bot known, hence they can not be added.

Discussion

758.	A particle moving along the x-axis has a pasition given by ` x= 10 t e^(-1)` metres where (t) is in seconds . How for is the partile from the origin when it momentarily stops ? (Do no consider its stop at infinity ).
Answer» ` x=10 t e^(-1)` Instantaneus veocity, ` v= (dx) /(dt) =10 t (-1)e^(-t) + 10 e^(-t)` `=e^(-t) (10 -10 t)` Foe the body to be at rest, ` v=0` S0, ` e^(-t) (10- 10 t) =0` :. T= prop or 1 s` But ` t= prop is not allowed , so ` t= 1s` Thus ` x =1= xx 1 xx e^(-t) = (10//e) ` metres.

Discussion

759.	When can an object be considered as a point object ? Explain in brief.
Answer» An object can be considered a point object if during motion in a given time it covers a distance much greater than its own size. For example, if a car of size ` 3` metres is moving a few hundred kilometre distance, then to study this motion, the car can be taken as a point object, since `3` m si very much less than few hundred kilometre distance.

Discussion

760.	Give ` vec A + vec B+ vec C + vec D` =0`, which of the following statements are correct ? (a) ` vec A, vec B ,vec C ` abd ` vec C` must each be a null vector. (b) The magnitude of ` ( vec A + vec C)` equals the magnitude of ` ( vec B + vec D0`. ltBrgt (c ) The magnitude of ` vec A` can never be greater than the sum of the magnitude of ` vec B , vec C ` and vec D`. ( d) ` vec B + vec C` must lie in the plance of ` vec A + vec D. if vec A and vec D` are not colliner and in the line of ` vec A` and `vec D`, if they are collinear.
Answer» (a) Not correct, because ` vec A + vec B+ vec C+ vec D` can be zero in may ways other than ` vec A, vec B, vec C` and ` vwc D` must each be a null vector. ltBrgt ( b) Correct. Since, ` vec A+ vec B+ vec C+ vec D=0 , (c ) Correct. Since ` vec A + vec B+ vec C + vec D=0` or ` vec A =- ( vec B+ vec C+ vec D)` It means the magnitude of ` vec A` is equal to the maniude of vector (vec B+ vec C + vec D). Since the sum of the magnitudes of ` vec B, vec C` and vec D` may be equal or greater than the magnitude of ` vec A`, hence the magnitude of `vec A` can never be greate than the sum of the maniude of ` vec B, vec C ` and vec D`. (d) Correct, Since ` vec A + vec B= vec D =0 , or vec A = 9 vec B= vec C) + vec D=0` and these three vectors are represented by the three sides of a triangle taken in one order. If ` vec A ` and ` vec B` are collinear, the ( vec B +vec C) must be in line of ` vec A` and vec D` , only then the vector sum of all the vectors will be zero.

Discussion

761.	` vec A , vec B and vec C` are three non-collinear, non co-planar vectors. What can you say about direction of ` vec A xx vec (B xx vec C`) ?
Answer» ` (vec Bxx vec C)` will act perpecndcular to the plane constaining ` vec B and vec C` and is derected as given by right hand rule. `(vec A xx (vec B xx vec C)` will lie in the plane of `vec B and vec C` and is perpwndicular to vector `vec A`.

Discussion

762.	Can a particle in one- dimensional motion have zero speed and a non-zero velocity ?
Answer» It is not possible because, velocity= speed + direction. If speed is zero velocity is also zero.

Discussion

763.	A particle is moving in a straight line with initial velocity `u` and uniform acceleration `f`. If the sum of the distances travelled in `t^(th) and (t + 1)^(th)` seconds is `100 cm`, then its velocity after `t` seconds, in `cm//s`, is.A. (a) `20`B. (b) 30`C. (c ) ` 50 `D. (d) ` 80`
Answer» Correct Answer - C Given , ` 100 = D_1 + D_(1-1)` ` `= [ u + f/2 (2t-1) ] + [u +f/2 {2 (t+1}]` or ` 100 =2 u + f/2 (4 t) = 2 u + 2 ft` or ` 50 =u + ft` Velocity after (t) seconds, ` v= u + ft 50 cm//s`.

Discussion

764.	Two particles, one with constant velocity 50 m/s and the other with uniform acceleration `10 m//s^(2)` start moving simultaneously from the same place in the same direction. They will be at a distance of 125m from each other after:A. 5 sec.B. `5(1+sqrt(2)) sec`C. 10 secD. `10(sqrt(2)+1)` sec
Answer» Correct Answer - A

Discussion

765.	Two particles A and B start moving simultaneously along the line joining them in the same direction with acceleration of `1 m//s^2 and 2 m//s^2` and speeds `3 m//s and 1 m//s` respectively. Initially, A is 10 m behind B. What is the minimum distance between them?
Answer» At minimum distance, their velocities are same, `:. v_A=v_B or u_A+a_At=u_B+a_Bt` or `(3+t)=(1+2t) or t=2s` Minimum distance, `d_(min)`=Initial distance- extra displacement of A upto this instant due to its greater speed `=10-(s_A-s_B)` `=10+s_B-s_A` `=10+(u_Bt+1/2a_Bt^2)-(u_At+1/2a_At^2)` `=10+[(1)(2)+1/2(2)(2)^2]-[(3)(2)+1/2(1)(2)^2]` `=8m`

Discussion

766.	The following graph shows the speed ofabody which is : A. Projected upwards with some velocity in vertical planeB. Having only constant accelerated motionC. Having only the constant retardationD. A perfectly elastic ball falling from a height on a fnction less and hard floor
Answer» Correct Answer - D

Discussion

767.	A plot of velocity versus time for a body in rectilinear motion is as shown in the figure.indicate whether acceleration is positive, negative , zero , variable or constant in the region OA, AB, BC and CD.Calculate the distance described up to the point C.What is the displacement up to point C ?
Answer» Correct Answer - In the region OA, acceleration is +ve and constant , in AB it is zero, in BC it is -ve and constant, and in CD it is again +ve but variable.Distance =18 m and displacement =15 m

Discussion

768.	If the same plot is of displacement (x) versus time (t), indicate whether the body is at rest or in motion and whether it has uniform velocity or variable velocity in the rigions OA, AB, BC and CD.Find the distance described in 8 s.What is the displacement of the body in 8 s ?
Answer» Correct Answer - In OA velocity is constant in AB it is zero , in BC it is constant and -ve and in CD it is variable . Distance =9 m .Displacement = -3 m

Discussion

769.	What happen when only magnitude of velocity changes?
Answer» Acceleration parallel or anti-parallel to velocity. e.g. Motion under gravity.

769.

What happen when only magnitude of velocity changes?

Answer»

Acceleration parallel or anti-parallel to velocity.

e.g. Motion under gravity.

Discussion

770.	A swimmer wishes to cross a `500 - m` river flowing at `5 km h^-1`. His speed with respect to water is `3 km h^-1`. The shortest possible time to cross the river is.A. `10 min`B. `20 min`C. `6 min`D. `7.5 min`
Answer» Correct Answer - A (a) Shortest time =`(d)/(v) = (1//2)/(3) = (1)/(6) h = 10 min`.

Discussion

771.	You are given the position-time graph of three deffernt bodies `A,B,` and `C`, Find which will have grater velocity an which will have least velocity. .
Answer» Slope for `C` is greatest, so `C` will have greater velocity and slope for `A` is least. So `A` will have least velocity.

Discussion

772.	Given below shows the desplacemen-time graph for a particle moving along a straight line path. . State true or false. a. Time during which the particle was at reat is `0` to `2s` b. Time maximum velocity of the particle is `-2.5 ms^(-1)` .
Answer» a. True, from graph, we see that for time `t=0` to `t=2 s`, displacement of the particle remains same as `3 m`. It means particle was at restfor this time interval. b.Ture velocity is maximum when the slope is maximum. And slope is maximum for time`t=5 s` to `t=7 s`. Velocty during this time =slope`=(0-5)/(7-5) =-2.5 ms^(-1)`.

Discussion

773.	Acceleration of a particle which is at rest at ` x=0 ` is `vec a = (4 -2 x) hat i`. Select the correct alternative (s).A. Particle further comes to rest at `x=4`B. Particle oscillates about `x=2`C. Maximum speed of particle is 4 unitsD. Maximum speed of particle is 2 units
Answer» Correct Answer - A::B

Discussion

774.	Acceleration of a particle which is at rest at ` x=0` is `vec a = (4 -2 x) hat i`. Select the correct alternative (s).A. Maximum speed of the particle is `4` unitsB. Particle further comes to rest at ` x =4`C. Particle oscillates about ` x=2`D. Particle will continuously acceleration along the x-axis`.
Answer» Correct Answer - B::C ` a= (dv)/(dt) = (dv)/(dx) xx (dx)/(dt) = (dv)/(dx) xx v` ` :. V (dv)/(dx) = ( 4- 2 x) ` or ` v dv = (4-2) dx` Intefrating it whithin the limits (when ` x=-0` to `x = 4`, the velocity changes ` 0` to `v`) :. ` int_0^v v dv = int_0^x (4-2 x) dx` ` v^2/2 = 4 x-x^2` or ` v^2 = 8 x- 2 x^2` or ` v= (8 - 2 x^2)^(1//2)` ....(i) ` `v` is zero at ` x=0` or ` x=4` Acceleration of the particles is zero , at ` x=2`. Hence the particcle will oscillate about ` x=2` Velocity will be maximum, when ` (dv)/(dx) =0` or ` d/(dx) [[8 x-2 x^2]^(1//2)]=0` or `1/2( 8 x- 2 x^2)^(-1//2) xx (8 - 4) =0` or ` 8-4 x= o` or ` x=2. It means the velocity of particle is maximum at mean position. From (i), ` v(max) = (xx 2 -2 xx 2^2) ^(1//2) =2 sqrt 2 units`.

Discussion

775.	The motion of a body is given by the equation `dv//dt=6-3v`, where v is in m//s. If the body was at rest at `t=0` (i) the terminal speed is `2 m//s` (ii) the magnitude of the initial acceleration is `6 m//s^(2)` (iii) The speed varies with time as `v=2(1-e^(-3t)) m//s` (iv) The speed is `1 m//s`, when the acceleration is half initial valueA. the terminal speed is `2.0 ms^(-1)`B. the manitude of initial accelration is ` 6.0 ms^(-2)`C. the speed varies with time as ` v(t) = 2 (1 -e^(-3)) m//s`D. the speed is ` 1.0 ms^(-1)` when the acceleration is half the initial value.
Answer» Correct Answer - B::C::D Acceleration, ` (d v(t))/(dt) = 6.0- 3 v(t)` …(t) At ` t=0`, acceleration `= 6.0 ms^(-2)` Thus option (b) is true Form (i), ` (dv)/(6-3 v) = dt` Integrating it within the conditions of motion, we have ` int_0^v = (dv)/(6-3 v)= int_0^t dt` lrt ` 6 - 3 v = y ` or ` -3 dv = dy ` or ` dv =- dy//3` :. ` int_6^(6-3 v) (-dy //3)/y = int_0^t dt or int_6^(6-3 v) (dt)/y =- int_0^t 3 dt` `(log_ey)_6^(6-3 v) =- 3 t` ltBrgt or ` log_e (6-3 v) - log_e 6=- 3 t` or ` log_e (6-3 v)-log_e 6 =- 3 t` or ` log_e (6 - 3v)/6 =- 3t` or ` (6 - 3v0/6 e^(-3t)` or ` 6- v =6 e^(-t) ` or ` 3 v= 6-6 e^(3t)` or ` v=2 (1 -e^(-3t))` ,brgt Thes, option (c ) is true When ` (ev(t))/(dt) = 1/2 xx 6 = 3.0 ` theta from (i) we have ` 3= 6.-3 v (t) or ` v (t) = (6.0-3)//3 = 1. 0 ms^(-1)` Thus , option (d) is true.

Discussion

776.	How many coordinates are required there in one, two and three dimensional motion?
Answer» One dimensional motion has 1 coordinate, two dimensional motion has 2 coordinates and three dimensional motion has three coordinates.

Discussion

777.	In which type of motion a particle or a system of particles move around a fixed axis.
Answer» Rotational motion.

Discussion

778.	What is projectile motion?
Answer» Projectile motion: Projectile motion is a form of motion in which object or particle (called a projectile, is thrown near the earth’s surface; and it moves along a curved path under the action of gravity only. To study such kind of motion the velocity working on the object and the acceleration components are divided in both the horizontal and vertical directions and analyzed independently.

Discussion

779.	A runner completes a circular path of 1000 m in 2 min 5 s. What is its average speed and average velocity?
Answer» Total distance travelled by the runner in 2 min 5 s = 1000 m Total time of the whole journey t = 2 min 5 s or t = (2 × 60) s + 5 s = 120 s + 5 s = 125 s Therefore average speed of the runner, \(\vec{v}=\frac{s}{t}=\frac{1000}{125}\) = 8m/ s v or \(\vec{v}\) = 8 m/s Since, the runner runs for one complete one cycle, therefore its initial and final positions coincide with each other. Thus, the displacement becomes zero i. e, \(\overrightarrow{\Delta x}\) = 0 (zero) Hence, average velocity \(\vec{v}=\frac{\overrightarrow{\Delta x}}{\Delta t}=\frac{0}{125}=0\) or \(\vec{v}\) = 0 (zero)

779.

A runner completes a circular path of 1000 m in 2 min 5 s. What is its average speed and average velocity?

Answer»

Total distance travelled by the runner in 2 min 5 s = 1000 m
Total time of the whole journey t = 2 min 5 s
or t = (2 × 60) s + 5 s
= 120 s + 5 s = 125 s
Therefore average speed of the runner,
\(\vec{v}=\frac{s}{t}=\frac{1000}{125}\) = 8m/ s v
or \(\vec{v}\) = 8 m/s
Since, the runner runs for one complete one cycle, therefore its initial and final positions coincide with each other.
Thus, the displacement becomes zero i. e,
\(\overrightarrow{\Delta x}\) = 0 (zero)
Hence, average velocity

\(\vec{v}=\frac{\overrightarrow{\Delta x}}{\Delta t}=\frac{0}{125}=0\)
or \(\vec{v}\) = 0 (zero)

Discussion

780.	A one km long train is travelling with 2 km/min velocity. How much time will it take to cross a tunnel of 1 km in length?
Answer» Total distance travelled by the engine of the train in crossing the tunnel completely S = 1 + 1 = 2 km The average speed of the train v = 2 km/min ∴ The time of crossing the tunnel, t = \(\frac{S}{v}=\frac{2 \mathrm{km}}{2 \mathrm{km} / \mathrm{min}}\) or t = 1 min

Discussion

781.	Name the physical quantity which obtained by the slope of velocity time graph.
Answer» Acceleration is the physical quantity obtained by the slope of velocity-time graph.

Discussion

782.	Name the physical quantity which obtained by the slope of displacement time graph.
Answer» Velocity is physical quantity obtained by the slope of displacement time graph.

Discussion

783.	A particle is moving with a definite velocity then how much would be its acceleration?
Answer» Zero acceleration.

Discussion

784.	Which physical quantity obtained by the area of velocity time graph.
Answer» Distance physical quantity obtained by the area of velocity time graph.

Discussion

785.	Assrtion o (A) : A body may have velocity and displacement in the oppostie direction Reason (R) : Velocity is distance travellled by the body in a unit timeA. A and R are correct and R is the correct explanation for AB. A and R are correct but R is not rhe correct explanation for AC. A is correct but R is incorrectD. Both A and R are incorrect
Answer» Velocity= `("displaement")/("time")` The displacement and the velocity need not be in the same direction. Example : SHM Hence , A is correct, R is incorrect.

Discussion

786.	What are the Two dimensional motion and Three dimensional motion ?
Answer» Two dimensional motion : The motion of an object is said to be two dimensional motion if two out of three coordinates specifying the position of the object change with time. In such motion the object moves in a plane. Three dimensional motion : The motion is said to be three dimensional motion if all the three coordinates specifying the position of an object change with respect to time ,in such a motion an object moves in space.

786.

What are the Two dimensional motion and Three dimensional motion ?

Answer»

Two dimensional motion :

The motion of an object is said to be two dimensional motion if two out of three coordinates specifying the position of the object change with time. In such motion the object moves in a plane.

Three dimensional motion :

The motion is said to be three dimensional motion if all the three coordinates specifying the position of an object change with respect to time ,in such a motion an object moves in space.

Discussion

787.	What is meant by Position-time graph ?
Answer» It is the graph between the time t and position x of a particle relative to a fixed origin. Its slope at any point gives the instantaneous velocity at the point. (i) For a stationary object, the position –time graph is a straight line parallel to the time-axis. (ii) For a body in a uniform motion, the position-time graph is a straight line inclined to the time-axis. (iii) For uniformly accelerated motion, the position time graph is a parabola.

787.

What is meant by Position-time graph ?

Answer»

It is the graph between the time t and position x of a particle relative to a fixed origin. Its slope at any point gives the instantaneous velocity at the point.

(i) For a stationary object, the position –time graph is a straight line parallel to the time-axis.

(ii) For a body in a uniform motion, the position-time graph is a straight line inclined to the time-axis.

(iii) For uniformly accelerated motion, the position time graph is a parabola.

Discussion

788.	Define the term Speed and Velocity.
Answer» Speed: It is rate of change of distance covered by the body with respect to time Speed = Distance travelled /time taken Speed is a scalar quantity .Its unit is meter /sec. and dimensional formula is[M⁰L¹ T^-1]. It is positive or zero but never negative. Velocity: The rate of change of position of an object in a particular direction with respect to time is called velocity. It is equal to the displacement covered by an object per unit time. Velocity =Displacement /Time Velocity is a vector quantity, its SI unit is meter per sec. Its dimensional formula is [M⁰L¹T^-1]. It may be negative, positive or zero.

788.

Define the term Speed and Velocity.

Answer»

Speed: It is rate of change of distance covered by the body with respect to time

Speed = Distance travelled /time taken

Speed is a scalar quantity .Its unit is meter /sec. and dimensional formula is[M⁰L¹ T^-1]. It is positive or zero but never negative.

Velocity: The rate of change of position of an object in a particular direction with respect to time is called velocity. It is equal to the displacement covered by an object per unit time.

Velocity =Displacement /Time

Velocity is a vector quantity, its SI unit is meter per sec. Its dimensional formula is [M⁰L¹T^-1]. It may be negative, positive or zero.

Discussion

789.	Differentiate between Uniform Speed and Instantaneous speed.
Answer» Uniform Speed: If an object covers equal distances in equal intervals of time than the speed of the moving object is called uniform speed. In this type of motion, position –time graph is always a straight line. Instantaneous speed: The speed of an object at any particular instant of time is called instantaneous speed. In this measurement, the time ∆t→0. When a body is moving with uniform speed its instantaneous speed = Average speed = uniform speed.

789.

Differentiate between Uniform Speed and Instantaneous speed.

Answer»

Uniform Speed: If an object covers equal distances in equal intervals of time than the speed of the moving object is called uniform speed. In this type of motion, position –time graph is always a straight line.

Instantaneous speed: The speed of an object at any particular instant of time is called instantaneous speed. In this measurement, the time ∆t→0.

When a body is moving with uniform speed its instantaneous speed = Average speed = uniform speed.

Discussion

790.	It is found that `\|A+B\|=\|A\|`,This necessarily implies.A. (a) ` vec B= 0`B. (b) vec A, vec B` are antiiparallelC. (c ) ` vec A, vec B` are perpendicularD. (d) ` vec A. vec B-lt0`
Answer» Correct Answer - A::B If ` \| vec A +vec B\|=\| vec A\| `, then either ` \| vec B\| = 0 ` or vec A` and vec B` will be antiparallel, where \| vec B\| = 2 \| vec A\|`.

Discussion

791.	Block A rests on inclined surface of wedge B which rests on a horizontal surface. The block A is connected to a string, which passes over a pulley P (fixed rigidly to the wedge B) and its other end is securely fixed to a wall at Q. Segment PQ of the string is horizontal and Q is at a large distance from P. The system is let go from rest and the wedge slides to right as A moves on its inclined face. Find the distance travelled by A by the time it reaches the bottom of the inclined surface.
Answer» Correct Answer - `10 sin 15^(@)`

Discussion

792.	Two frictionless ropes connect points A & B in vertical plane. Bead 1 is allowed to slide along the straight rope AB and bead 2 slides along the curved rope ACB. Which bead will reach B in less time?
Answer» Correct Answer - Bead 2

Discussion

793.	A particle is projected from a point on the level ground and its height is `h` when at horizontal distances `a and 2 a` from its point of projection. Find the velocity of projection.
Answer» Correct Answer - `(1)/(2) sqrt(((4 a^2)/(h) + 9 h)) g`. If `v_0` is the velocity of projection and `prop` the angle of projection, the equation of trajectory is `y = x tan prop - (1)/(2) (gx^2)/(v_0^2 cos^2 prop)`…(i) With origin at the point of projection, `gx^2 -m 2 v_0^2 sin prop cos prop. x + 2 v_0^2 cos^2 prop.y = 0` ...(ii) Since the projectile passes through two points `(a,h) and (2a, h)`, then `a and 2a` must be roots of Eq. (ii), `a + 2 a = (2 v_0^2 sin prop cos prop)/(g)`...(iii) and `a xx 2a = (2 v_0^2 cos^2 prop h)/(g)`...(iv) Dividing Eqs (iii) by (iv), we get `(3a)/(2a^2) = (tan prop)/(h) or tan prop = (3 h)/(2a)` From Eq. (iv), `v_0^2 = (ga^2)/(h) sec^2 prop = (ga^2)/(h) (1 + tan^2 prop) = (ga^2)/(h) (1 + (9 h^2)/(4 a^2))` =`(g)/(4) ((4a^2)/(h) + 9h) or v_0 = (1)/(2)sqrt((4a^2)/(h) + 9 h)) g`.

Discussion

794.	Two equal forces have their resultant equal to either. At what angle are they inclined ?
Answer» Here, ` A = B = R=x (say)` As, ` R = sqrt (A^2 + B^2 + 2 AB cos theta` )` :. ` x^2=x^2 +x^2 + 2 xx cos theta` or ` 1=1 + 1 =2 cos theta` or ` cos theta =- 1/2 + 2 cos 120^@` or ` theta = 120^@ s` .

Discussion

795.	Two equal vector have a resultant equal to either of them, then the angle between them will be:
Answer» Hence ` A =F , B =F, R=F, theta= ?` As ` R^(2) =A^(@) + B^(2) + 2 AV cos theta` :. ` F^(2) =F^(2) + F^(2) + 2 FF cos theta` or ` cos theta =- 1/2 =cos 120^(@) or theta = 120^(@)`.

Discussion

796.	What are the maximum muber of (i) rectangular component vectors (ii) componet vectrs, component vectors, into which a vector can ve resolved in a plane ?
Answer» (i) Two only in a plane (ii) Any number of cpmponent vecctors in a plane.

Discussion

797.	Mention the condition when on object in motion (a) can be considered point object (b) cab bot considered point object.
Answer» As object can be considerd as a point object if (a) its sixze is negligble in comparison to the distance travelled by onject (b) its size is not negligible in comparison to the distance travelled by object.

Discussion

798.	Find the scalr and vector products of two vectors ` vec a =(3 hat I - 4 hat j + 5 hat k) and `vec b = (-2 hat I + hat j- 3 hat k).
Answer» Here, ` vec a =3 hat I -4 hat j + 5 hat k , hat b = - 2 hat I + hat j -3 hat k` ` vec a ,. Vec b =(3 hat I -4 hat j+ 5 hat k) .(-2 hat I + hat j - 3 hat k)` `=3 (-2 ) -4 (1) + 5 (-3)` =- 6- 4 -15 =-25` `vec a xx vec b = ( 3hat I - 4 hat j + 5 hat k ) xx (-2 hat i + hat j -3 hat k)` ` =\|(vec i, vec i, vec k) , (3, -4, 5 ) ,( -2 , 1, -3 )\|` `= vec i (12 -5 ) + hat j (-10 + 9) + hat k (3 -8)` `= 7 hat i - hat j - 5 hat k`.

Discussion

799.	Can a vector be multiplied with both dimensional and non-dimensional scalr ?
Answer» Yes, when a vectro is multiplied with a dimensional scalar, the resultant vector will have different diemensions. For example, if acceleration vector is multant vector has the dimnsions ot forec. When a vector is multplied with non-dimensional scalr, it will be a vector havubg dunebsions as that of the gi en vector.

Discussion

800.	A particle moves from position vector ` vecr_1 = ( 23 hat I + 2 hat j - 6 hat k)` to position vector, ` vec r-2 = ( 14 hat I + 13 hat j+ 9 hat k)` in metre under the action of a constant force of ` vec F = 9 14 hat i+ hat j + 3 hat k) N`. Calculat word done by the force.
Answer» Here, ` vec r_1 = ( 3 hat i+ 2 aht - 6 hat k)` ltBrgt ` vecr_2 = ( 14 hat i+ 13 hat j + 9 hat k)` ` vec F ( 4 hat i+ hat j+ 3 aht k) N , W=?` Edsplacement ` vec r_(12) = vec 2 - vec _1` ` = ( 14 hat i+ 13 hat j + 9 hat k) - ( 3 hat i+ 2 aht j -6 aht k)` ltbRgt ` = 11 hat i+ 11 aht j + 15 aht k` ` W= vec F. vecr_(12) = (4 hat i +hat j = 3 aht k) . 11 hat j + 11 hat i+ 11 hat j= 15 aht k)` ltbRgt ` =344 hat i hat i+ 11 hat j hat j= 45 hathat k. hat k = 100 j`.

Discussion

Explore topic-wise InterviewSolutions in .

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Find the scalr and vector products of two vectors ` vec a =(3 hat I - 4 hat j + 5 hat k) and `vec b = (-2 hat I + hat j- 3 hat k).

Can a vector be multiplied with both dimensional and non-dimensional scalr ?

A particle moves from position vector ` vecr_1 = ( 23 hat I + 2 hat j - 6 hat k)` to position vector, ` vec r-2 = ( 14 hat I + 13 hat j+ 9 hat k)` in metre under the action of a constant force of ` vec F = 9 14 hat i+ hat j + 3 hat k) N`. Calculat word done by the force.