InterviewSolution
Saved Bookmarks
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.
| 801. |
Does it make a sense to call a physical quantity a vector, when its magnitude is zero ? |
| Answer» Yes, null vectro has a definite phusical significane. | |
| 802. |
An object is verically thrown upwards. The the dislacement-time graph for the motion is as shown in .A. .B. .C. .D. . |
|
Answer» Correct Answer - B Let the particle be thrown up with initial velocity `u`. Desplacement `s` at any time `t` is `S=ut -(1)/(2) g t^(2)`,. The graph should be parabolic downwards as shown in option `b` . |
|
| 803. |
A driver having a definite reaction time is capable of stopping his car over a distance of 30 m on seeing a red traffic signal, when the speed of the car is 72 km/hr andover a distance of 10 m when the speed is 36 km/hr. Find the distance over which he can stop the car if it were running at a speed of 54 km/hr. Assume that his reaction time and the deceleration of the car remains same in all the three cases. |
|
Answer» Correct Answer - [18.75 m] |
|
| 804. |
A point moving with constant acceleration from A to B in the straight line AB has velocities u and v at and B respectively. Find its velocity at C, the mid point of AB. Also show that if the time from A to C is twice that from C to B, then `v =7 u`. |
|
Answer» Correct Answer - `[sqrt((u^(2) + v^(2))/(2))]` |
|
| 805. |
A parachutist after bailing out falls freely for 50 metres.Then his parachute opens and he falls with deceleration of `2 ms^(-2)` When he reaches the ground his speed is reduced to only 3 metres per second.At what height did he bail out ? For how long was he in air ? |
|
Answer» Correct Answer - 293 m,17 s |
|
| 806. |
A block is released from rest at the top of a frictionless inclined plane, which is 10 m long. It reaches the bottom in 5 seconds.A second block is projected up the plane from the bottom at the instant the first block is released and the two return to the bottom simultaneously.(a)Find the acceleration of each block along the incline.(b)What is the initial velocity of the second block ? (c )How far up the inclined plane does it travel ? |
|
Answer» Correct Answer - (a)`0.8 m s^(-2)` , (b)`2 m s^(-1)` , ( c)2.5 m |
|
| 807. |
Given ` vec A+ vec B+ vec + vec D=0`, can the magnitude of ` vec A + vec B + vec C` be equal to the magnitude of vec D` Explain. |
|
Answer» Yes, Given ` vec A + vec B+ vec C + vec D =vec 0` or ` vwc A + vec B +vec C+ vec D =- vec D` Hence magitude of (vec A + vec B= vec C) must be equal to the magnitude of vec D`. |
|
| 808. |
A train is moving at a constant speed `V` when its driver observes another train in front of him on the same track and going in the same direction with constant speed `v`. If the distance between the trains is `x` then what should be the minimum retardation of the train so as to avoid collision?.A. ` (v + v_1^2) d`B. (v -v_1)^2` d`C. (v +v_1 )^(1//2)d`D. (v -v_1 ^(2//2)d` |
|
Answer» Correct Answer - D Relative velocity of a train with respect to another train, ` u= (v-v_1)` using , ` v^2 - u^2 = aS`, we have ` 0- (v-v_1)^2 = 2 a xx d` or ` a= - ((v-v_1)^2)/(2d)`. |
|
| 809. |
A boat can fo across a lake and retyrb ub tune ` T_0` at a speed ` v`. On a rough day there is a uniform currednt at speed ` v_1` to help the onward hourney and impede the return jourmey. If the time taken to go across and return on the same day be (T) then ` T//T_0` is`A. (1 -v_1^2 //v^2` `1/((1 -v_1^2//v^2))`(1 + v_1^2/v^2)`B. `1/((1 -v_1^2//v^2))`C. `(1 + v_1^2/v^2)`D. `1/ ((1 + v_1^2//v^2))` |
|
Answer» Correct Answer - B Let (d) be the width fo lake e to be crossed. Tehn ` T_0= (2d)/v` and ` T= d/(v=v_1) + d/(v-v_1)= (2 dv)/(v^2-v_1^2)= ( 2d)/(v(1- v_1^2 //v^2)` :. ` T/T_0 = 1/((1-v_1^2 //v^2))`. |
|
| 810. |
A car covers the first half of the distance between two places at 40 km/hr and another half at 60 km/hr. The average speed of the car isA. 40 km/hrB. 48 km/hrC. 50 km/hrD. 60 km/hr |
|
Answer» Correct Answer - 2 |
|
| 811. |
Two objects are projected with same velocity at different angles with the horizontal range and is same for both of them. If `t_1` and `t_2` are their time of flights, then `t_1t_2`=_____A. 2 RgB. `frac { R } {g}`C. `frac { 2 R } { g }`D. `frac { R } { 2g }` |
|
Answer» Correct Answer - 3 |
|
| 812. |
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors isA. `45 ^ { circ }`B. `180 ^ { circ }`C. `0 ^ { circ }`D. `90^ { circ }` |
|
Answer» Correct Answer - 4 |
|
| 813. |
If the displacement of a particle is `(2t^(2) +t +5)` meter then, what will be acceleration at `t = 5sec.` |
|
Answer» `v = (dx)/(dt) = (d)/(dt) (2t^(2)+t+5) = 4t +1m//s` and `a = (dv)/(dt) = (d)/(dt) (4t+1) a = 4m//s^(2)` |
|
| 814. |
Displacement of a particle is given by the expression `x = 3 t ^ { 2 } + 7 t - 9`, where x is in meter and t is in seconds. What is acceleration ?A. 10m/s²B. 15m/s²C. 8m/s²D. 6m/s² |
|
Answer» Correct Answer - 4 |
|
| 815. |
The velocity-time graph of a particle moving in a staight line is shown in the . Find the displacement and the distance trav elled by the particle in `6 s`. . |
|
Answer» Correct Answer - `8 m`; `16 m` We know that desplacement is area under `v-t` graph . Area from `0` to `2 s` is `4xx 2=8 m`, area form `2` to `4 s` is `-2 xx2 =-4 m`, area from `4` to `6` s `2xx 2=4 m` So displacement=`8-4+4=8 m` When we find distance, we consider all areas to be positive. So distance=`8+4+4=16 m`. |
|
| 816. |
A ball is projected horizontally with a velocity of `5ms^(-1)` from the top fo a building ` 19.6 m` high. How long will th ball take to hit the ground ? |
| Answer» Correct Answer - (b) | |
| 817. |
Path of the bomb released from an aeroplane moving with uniform velocity at certain height as observed by pilot is. |
| Answer» Correct Answer - (a) | |
| 818. |
A particle moves in straight line in same direction for 20 seconds with velocity `3 m//s` and the moves with velocity `4 m//s` for another 20 sec and finally moves with velocity `5 m//s` for next 20 seconds. What is the average velocity of the particle?A. `3 m//s`B. `4 m//s`C. `5 m//s`D. Zero |
|
Answer» Correct Answer - B Avg. velocity `=(20xx3+4xx20+5xx20)/(20+20+20)=4 m//s` |
|
| 819. |
For the displacement time graph shown in figure -1.100, the ratio of the speeds during the first two seconds and the next four second is : A. `1 : 1`B. `1 : 2`C. `2 : 1`D. `3 : 2` |
|
Answer» Correct Answer - A |
|
| 820. |
A particle A moves along a circle of radius `R=50cm` so that its radius vector r relative to the point O(figure) rotates with the constant angular velocity `omega=0.40rad//s`. Find the modulus of the velocity of the particle, and the modulus and direction of its total acceleration. A. `v=0.4 m//s, a=0.4 m//s^(2)`B. `v=0.32 m//s, a=0.32 m//s^(2)`C. `v=0.32 m//s, a=0.4 m//s^(2)`D. `v=0.4 m//s, a=0.32 m//s^(2)` |
|
Answer» Correct Answer - D Angular velocity `omega` about centre `=2omega` `=2xx0.40=0.30 rad//sec` `v=omega R` `=0.80xx1/2=0.40 m//s` `a=v^(2)/R=(0.40xx0.40xx100)/50=0.32 cm//s^(2)` |
|
| 821. |
Is dubruw eirRUIB cwxrie ? |
| Answer» The finite rotation about an axis is not a vector because its addition to another finite rotation about a different axis does bot obey commutative law of addition. | |
| 822. |
Consider a vector `vec(F)= 4hat(i)-3hat(j)`. Another vector that is perpendicular to `vec(F)` is |
| Answer» ` hat k` , It will be a unit vector along Z-axis as ` vec F` lines in X-Y` plance. | |
| 823. |
The velocity of a particle moving along x-axis is given as `v=x^(2)-5x+4` (in m`//`s) where x denotes the x-coordinate of the particle in metres. Find the magnitude of acceleration of the particle when the velocity of particle is zero?A. `0 m//s^(2)`B. `2 m//s^(2)`C. `3 m//s^(2)`D. None of these |
|
Answer» Correct Answer - A For `v=0, x=1, 4` and `a=v(dv)/(dx)` so `a|._(x=1)=0xx(dv)/(dx)=0, a|_(x=4)=0xx(dv)/(dx)=0` |
|
| 824. |
The displacement of a particle si given by `y = a+bt +ct^(2) +dt^(4)`.find the acceleration of a particle. |
|
Answer» `v = (dy)/(dt) = (d)/(dt) (a+bt +ct^(2) +dt^(4)) =b +2ct +4dt^(3)` `a = (dv)/(dt) = 2c +12dt^(2)` |
|
| 825. |
a particle is moving in x-y-plane at `2 m//s` along x-axis. `2` seconds later, its velocity is `4 m//s` in a direction making `60^(@)` with positive x-axis. Its average acceleration for the period of motion is:-A. `sqrt(5) m//s^(2)`, along y-axisB. `sqrt(3) m//s^(2)`, along y-axisC. `sqrt(5) m//s^(2)`, along at `60^(@)` with positive x-axisD. `3m//s^(2)` at `60^(@)` with positive x-axis. |
|
Answer» Correct Answer - B `v_(i)=2hat(i)` `v_(f)=4 cos 60^(@) hat(i)+4 sin 60^(@) hat(j)` `=4/2hat(i)+(4sqrt(3))/2hat(i)` `=2hat(i)+2sqrt(3) hat(j)` `Deltavec(v)=vec(v)_(f)-vec(v)_(i)=2hat(i)+2sqrt(3)hat(j)-2hat(i)=2sqrt(3) hat(j)` `lt lt vec(a)gt gt =(2sqrt(3)hat(j))/2=sqrt(3)hat(j) m//s^(2)` |
|
| 826. |
For a particle moving along a straight line, the displacement x depends on time `t` as `x=At^(3)+Bt^(2)+Ct+D`. The ratio of its initial velocity to its initial acceleration depends on:A. A & CB. B & CC. CD. C and D |
|
Answer» Correct Answer - B |
|
| 827. |
A particle has initial velocity of `(3hat(i)+4hat(j)) m//s` and a acceleration of `(4hat(i)-3hat(j)) m//s^(2)`. Its speed after one second will be equal to :-A. `0`B. `10 m//s`C. `5sqrt(2) m//s`D. `25 m//s` |
|
Answer» Correct Answer - C `vec(v)_((1))=(3+4xx1)hat(i)+(4+(-3)xx1) hat(i)+(4+(-3)xx1)hat(j)=7hat(i)+hat(j)` `|vec(v)_((1))|=sqrt(49+1)=5sqrt(2) m//s` |
|
| 828. |
The co-ordinates of a moving particle at any time t are given by `x=ct^(2) and y=bt^(2)`The speed of the particle isA. `2t(c+b)`B. `2tsqrt(c^(2)-b^(2))`C. `tsqrt(c^(2)+b^(2))`D. `2t sqrt(c^(2)+b^(2))` |
|
Answer» Correct Answer - D `|vec(v)|=sqrt(v_(x)^(2)+v_(y)^(2))` here `v_(x)=(dx)/(dt)=2ct, v_(y)=(dy)/(dt)=2bt` therefore `|vec(v)|=sqrt(4t^(2)(c^(2)+b^(2)))=2tsqrt((c^(2)+b^(2)))` |
|
| 829. |
When a `2` kg car driven at `20 m//s` on a level road is suddenly put into netural gear (i.e. allowed to coast), the velocity decreases in the following manner: `V=20/(1+(t/20))m//s` Where t is time in sec. The deceleration of car at the instant its speed is `10 m//s` is.A. `1//4 m//s^(2)`B. `1//2 m//s^(2)`C. `1 m//^(2)`D. `3//4 m//s^(2)` |
|
Answer» Correct Answer - A Given that initial velocity `=u=20 m//s` `v=20/(1+t/20)` at `u=10 m//s` `10/1=20/(1+t/20)` `2=1+t/20` `t=t/20` `t=20` sec From `u=20/(1+t/20)=400/(20+t)` acceleration `a=(dv)/(dt)=400xx1/((20+t)^(2))` acceleration `a=(-400)/(40xx40)` acceleration `a=(-1)/4 m//s^(2)` so deceleration `a=1/4 m//s^(2)` |
|
| 830. |
Velocity and acceleration of a particle are `v=(2 hati) m/s` and `a = (4t hati+t^2 hatj) m /s^2` where, t is the time. Which type of motion is this ? |
|
Answer» Correct Answer - A::C::D a is function of time and v and a are neither parallel nor antiparallel. |
|
| 831. |
Average speed is always equal to magnitude of average velocity. Is this statement true or false ? |
|
Answer» Correct Answer - A Distance may be greater than or equal to magnitude of displacement. |
|
| 832. |
When a particle moves with constant velocity its average velocity, its instantaneous velocity and its speed all are equal. Is this statement true or false ? |
|
Answer» Constant velocity means constant speed in same direction. Further if any physical quantity has a constant value (here, the velocity) then its average value in any interval of time is equal to that constant value |
|
| 833. |
On a cricket field, the batsman is at the origin of coordination and a fielder stands in position given as `(46 hat(i) + 28 hat(j))m`. The batsman hits the ball so that it rolls along the ground with constant velocity given `(7.5 hat(i) + 10 hat(j))m//s`. The fielder can run with a speed of `5m//s`. If the starts to run immediately the ball is hit, what is the shortest time in which he could intercept the ball. |
|
Answer» Correct Answer - [4s] |
|
| 834. |
The displacement ` x ` of the body is motion is given by ` x=A sin (omega t+ theta) , Determine the time at which the displacement is maximum. |
|
Answer» The value of displacement (x) will be maximum, when the value of sin ` (omega t + theta)` is maximum. If will be so if ` sin (omega t + theta ) =1 = sin pi //2` or omega t+ theta (pi)/2 or ometa t = (pi)/2 -theta` :. ` t= ((pi)/(2 omega) -(theta)/(omega)`. |
|
| 835. |
A ball is projected as shown in figure. The ball will return to point: A. OB. left to point OC. right to point OD. none of these |
|
Answer» Correct Answer - A Here `a_(x)/a_(y)=(g cot theta)/g=1/(tan theta)=u_(x)/u_(y)rArr` Initial velocity & acceleration are opposite to each other. `rArr` Ball will return to point O. |
|
| 836. |
Throughout a time interval, while the speed of a particle increases as it moves along x-axis, its velocity and acceleration might beA. positive and positive respectivelyB. positive and negative respectivelyC. negativetive and negative respectivelyD. negative and positive respectively |
|
Answer» Correct Answer - A, C Spped increases if both velocity & acceleration have same signs. |
|
| 837. |
Three point particle A, B and C are projected from same point with same speed at `t=0` as shown in figure. For this situation select correct statement(s). A. All of them reach the ground at same timeB. All of then reach the ground at different time.C. All of them reach the ground with same speedD. All of them have same horizontal displacement when they reach the ground |
|
Answer» Correct Answer - B, C Vertical component of initial velocities are different `rArr` reach the ground at different time. |
|
| 838. |
A particle is thrown at time `t=0` with a velocity of 10 m/s at an angle of `60^(@)` with the horizontal from a point on an incline plane, making an angle of `30^(@)` with the horizontal. The time when the velocity of the projectile becomes parallel to the incline is : A. `2/sqrt(3)` secB. `1/sqrt(3)` secC. `sqrt(3)` secD. `1/(2sqrt(3))` sec |
|
Answer» Correct Answer - B |
|
| 839. |
A particle moves along x-axis with acceleration `a = a0 (1 – t// T)` where `a_(0)` and T are constants if velocity at t = 0 is zero then find the average velocity from t = 0 to the time when a = 0. |
|
Answer» Correct Answer - `a_(0)T//3` `a=a_(0)(1-t/T)` where `a_(0)` & `T` are constant `underset(0)overset(v)(int)dv=a_(0)underset(t=0)overset(t)(int)(1-1/T)dt rArr v=a_(0)[t-t^(2)/(2T)]` `rArr int dx=a_(0) underset(t=0)overset(t)(int)[t-t^(2)/(2T)]dt` For `a=0rArr 1-t/T=0" "t=T" "a_(0)[t^(2)/2-t^(3)/(6T)]` `:. lt v gt =(underset(0)overset(T)(int)v dt)/(underset(0)overset(T)(int)dt)=(a_(0)[T^(2)/2-T^(3)/(6T)])/T=(a_(0)T)/3` |
|
| 840. |
Velocity of a particle in x-y plane at any time t is `v=(2t hati+3t^2 hatj) m//s` At `t=0,` particle starts from the co-ordinates `(2m,4m).` Find (a) acceleration of the particle at `t=1s.` (b) position vector and co-ordinates of the particle at `t=2s.` |
|
Answer» Correct Answer - A::B::C::D (a) `a=(dv)/dt=d/dt(2t hati+3t^2hatj)` `=(2hati++6thatj) m//s^2` At `t=1s,` `a=(2hati+6 hatj) m//s^2` (b) intds=intv dt or `s=intv dt=int(2 thati+3t^2 hatj)dt` `:. r_f-r_i=int_(i ntitial)^(fi nal) (2t hati+3t^2 hatj)dt` `or r_(2 sec)-r_(0 sec)=int_0^2(2t hati+3t^2 hatj)dt` `:. r_(2 sec)=r_(0 sec)+[t^2 hati+t^3 hatj]_0^2` `=(2 hati+4 hatj)+(4 hati+8 hatj)` `=(6 hati+12 hatj) m` Therefore, coordinates of the particle at `t=2s` are `(6m,12m)` |
|
| 841. |
What does the area under acceleration-time graph for any interal of time represents when the accleration of the moving body is varing with time ? |
| Answer» The area under acceleration-time graph for a given interval of time represents the change in speed of the body during that interval of time . | |
| 842. |
Accleration-time graph of a particle moving in a straight line is as shown in Fig. Velocity of particle at time `t=0` is `2 m//s.` Find the velocity at the end of fourth second. |
|
Answer» Correct Answer - A `int dv=inta dt` or change in velocity=area under a-t graph Hence, `v_f-v_i=1/2(4)(4)` `=8 m//s` `:. v_f=v_i+8=(2+8) m//s` `= 10 m//s` |
|
| 843. |
A bike startingh from rest picks up velocity of 72 `km h^(-1)` over distance of 40m. Calculate its accleration |
|
Answer» Given `u=0` `v=72km h^(-1)=72xx(5)/(18)=20 m s^(-1)` `s=40 m` using `v^(2)-u^(2)=2as` `(20)^(2)-0=2axx40` `a=(400)/(2xx40)=5 m s^(-2)` |
|
| 844. |
Two particles 1 and 2 start simultaneously from origin and move along the positive X direction. Initial velocity of both particles is zero. The acceleration of the two particles depends on their displacement (x) as shown in fig. (a) Particles 1 and 2 take `t_(1)` and `t_(2)` time respectively for their displacement to become `x_(0)`. Find `(t_(2))/(t_(1))`. (b) Which particle will cover `2x_(0)` distance in least time? Which particle will cross the point `x = 2x_(0)` with greater speed? (c) The two particles have same speed at a certain time after the start. Calculate this common speed in terms of `a_(0)` and `x_(0)`. |
|
Answer» Correct Answer - (a) `sqrt(2)` (b) particle 1 will cover `2x_(0)` in lesser time. Both will cross `2x_(0)` with same speed. (c) `v = (2 +sqrt(2)) sqrt(a_(0)x_(0))` |
|
| 845. |
The speed -time graph of the ball in the above situation is.A. B. C. D. |
|
Answer» Correct Answer - B |
|
| 846. |
Two particles `A` and `B` start from the origin along `x`-axis. Velocity time graph of both particles are shown in the figure. During the given time interval, the maximum separation between the particles is A. `4m`B. `1m`C. `2m`D. `3m` |
|
Answer» Correct Answer - C |
|
| 847. |
A ball is dropped from a certain height on a horizontal floor. The coefficient of restitution between the ball and the floor is `(1)/(2)`. The displacement time graph of the ball will be.A. B. C. D. |
|
Answer» Correct Answer - C |
|
| 848. |
A particle is moving in `x-y` plane with `y=x/2` and `V_(x)=4-2t`. The displacement versus time graph of the particle would beA. B. C. D. |
|
Answer» Correct Answer - C |
|
| 849. |
A particle is moving in a straight line. Particle was initially at rest. Acceleration versus time graph is shown in figure. Acceleration of particle is given by `a=3sin pi t` in `m//s^(2)`. The time (in s) when the particle comes to rest is A. `t=0,1,2,3,4`B. `t=1,3`C. `t=0,2,4`D. at `t=0.5,1.5, 2.5` |
|
Answer» Correct Answer - C |
|
| 850. |
Graph of velocity versus displacement of a particle moving in a straight line is as shown in figure. The acceleration of the particle is. A. constantB. increases linearly with `x`C. increases parabolically with `x`D. zero |
|
Answer» Correct Answer - B |
|