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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.
| 1051. |
If at a height of ` 40 m`, the direction of moton of a projectile makes an angle `pi //4` with the orizontal, then its initial velocity and angle of projection are, respectively,A. ` 30 , 1/2 cos^(-1) (- 4/5)`B. ` 30 1/2 cos^(-1) (-1/2)`C. ` 50 1/2 cos^(1) (- 8 /(25)`D. 60 1/2 cos^(-1) (- 1/4)` |
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Answer» Correct Answer - C Let (u) be the initial velocity of projectile and ` theta` be the angel of projmection. Taking vertical upwarawd motion for hieght (h) , we have ` u =u sin theta, a=- g, S=h, v=v-y, As ` v^2 = u^2 + 2 aS, so ` v_y^2 = u^2 sin^2 theta- 2 gh` ...(i) At height ` h, tahn (pi)/4 = v_y/v_x ` or ` v_y =v-x = u cos theta` :. ` u^2 cos^2 theta = u ^2 sin^2 sin^2 theta - 2 gt` or ` u^2 (cos^2 theta- sin ^2 theta)=- 2 gt` ltBrgt or ` u^2 cos 2 theta =- 2 gt ` ....(ii) The equation (ii) is valid only is ` u= 50 m//s` and ` cos ` 2 theta =- 8//25` or ` theta 1/2 cos ^(-1) (- 8//25)`. |
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| 1052. |
A car is moving with uniform acceleration along a straight line between two stops X and Y. Its speed at X and Y are `2 ms^-1 and 14 ms^-1` , ThenA. its speed at mid-point of XY is `10 ms^-1`B. its speed at a point A such that `XA : AY = 1 :3` is `5 ms^-1`C. the time to go from X to the mid-point of XY is double of that to go from mid-point to YD. the distance travelled in first half of the total time is half of the distance travelled in the second half of the time |
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Answer» Correct Answer - A::C `(14)^2=(2)^2+2 as` `:. 2as=192` units At mid point, `v^2=(2)^2+2a(s/2)` `=4+192/2=100` `:. v=10 m//s` `XA:AY=1:3` `:. XA=1/4s` and `AY=3/4s` `v_1^2=(2)^2+2a(s/4)` `=4+192/4=52` `:. v_1=sqrt52!=5 m//s` `10=2+a t_1 (v=u+at)` `:. t_1=8/a 14=10+a t_2` `:. t_2=4/a` or `t_1=2t_2` `S_1=(2t)+1/2a(t^2)`=distance travelled in first half `S_2=2(2t)+1/2a(2t)^2` `S_3=S_2-S_1`=distance travelled in second half We can see that, `S_1!=(S_3)/2` |
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| 1053. |
A body wlaks to his school at a distance of `6 km ` with constant speed of ` 2-5 km h^(-1)` and walks bach with a constant speed of ` 4 km h^(-1)~ What is avetage speed for the round trip in km h^(-1)` ? |
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Answer» Total time in going and coming back, ` T= 6/(2.5) + 6 /4 `, total distance travelled, ` S = 6 + 6 12 km`, Average speed ` = S/T =(12)/((2//2 .5 )+ (6 //44))` `= (40)/(13) km h^(-1)`. |
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| 1054. |
If the magnitudes of two vectors are `2 and 3` and the magnitude of their scalar product is `3 sqrt 2`, then find the angle between the vectors. |
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Answer» Here, `A =2, B=3, vec A. vec B =3 sqrt2, As vec A. vec B=AB cos theta` So `cos theta =(vec A. vec B)/(AB) =(3sqrt2)/(2xx3) =1/9sqrt2) =cos^(0) , theta =45^(@)`. |
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| 1055. |
Calculate the area of a paralleleogram whose two adjacent sides are formed by the vectors ` vec A =3 hat I + 5 hat j and vec B =- 3 hat I + 7 hat j`. |
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Answer» `vec A xx vec B = |( hat I , gat j, hat k) ,(3, 5, 0), (-3, +7, 0)|` `= hat I [0 -0] + hat j [0 -0] + hat k [21 + 15]` `= +36 hat k` `Area of parallelogram `=| vec A xx vec B|` `= sqrt( 0^(2) + 0^(2) + (+36)^(2))` ` =36 sq. unt`. |
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| 1056. |
Calculate the area of the parallelogram whose two adjacent sides are formed by the vectors `vec A xx 4 hat I + 3 hat j and vec B =- 3 hati +6 hat j. |
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Answer» `vec A xx vecB =|(hat i,hat j,hatk,), (4,3,o),(-3, 6, 0)| =hati (3 xx0-6xx o) +hatj [0 xx(-3) -0xx 4 ] + hat k [4 xx 6-3 xx (-3)] =33 hatk` Area of parallelogram `=|vecA xx vec B| =sqrt (0^(2)0^(2) +(33)^(2) ) =33 sq a for vec F =(-3 hati +hat j [0xx (-3) -0 xx 4] + hat k [4 xx 6- 3 xx (-3) ] =33 hat k` Area of parallelogeam `=|vec A xx vec B| =sqrt(0^(2) +0^(2)+33)^(2)) =33 sq. units`. |
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| 1057. |
If `vec A=hat I + 2 hat j -3 hat k`, vec B =2 hat I -hat j + hat k` and ` vec Chat I -3 hat j + 2 hat k`, then find ` vec A xx ( vec B xx vec C). |
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Answer» `vec B xx vec C = |(hat I hat j hat k),(2, -1, 1), (1, -3, 2)|` `=hat I (-2 +3) + hat j (1 -4 ) + hat k (-6 +1)` `=hat I -3 hat j -5 hat k` ` vec A xx ( vec B xx vec C) =|(hat I, hat j, hatk), (1, 2, -3) ,(1, -3, -5)|` `= hat I (-10 -90 + hatj (- 3 +5 ) + hat k (- 3 -2 0` =- 19 hat i + 2 hat j- 5 hat k`. |
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| 1058. |
A fan is rotating with angular velocity `100 rev s^-1`. Then is switched off. It takes `5 min`to stop. (a) Find the total number of revolution made before the fan stops. (assume uniform angular retardation). (b) Find the value of angular retardation. ( c) Find the average angular velocity during this interval. |
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Answer» (a) `theta = ((omega + omega_0)/(2)) t = ((100 + 0)/(2)) xx 5 xx 60 = 15000 rev`. (b) `omega = omega_0 + prop t` `rArr 0 = 100 - prop (5 xx 60) rArr prop = (1)/(3) rev s^-2` ( c) `omega_(a v) = ("Total angle of rotation")/("Total time taken")` =`(15000)/(50 xx 60) = 50 rev s^-1`. |
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| 1059. |
The tangential acceleration of a particle moving in a circular path of radius `5 cm is 2 m s^-2`. The angular velocity of the particle increases from `10 rad s^-1 "to" 20 rad s^-1` during some time. Find (a) this duration od time and (b) the number of revolutions completed during this time. |
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Answer» `r = 5 cm = 0.05 m, a_t = 2 ms^-2, omega_1 = 10 rad s^-1` `omega_2 = 20 rad s^-1, prop = a_t//r = 2//0.05 = 40 rad s^-2` (a) `omega_2 = omega_1 + omega_t rArr 20 = 10 + 40 t rArr t = 0.25 s` (b) `theta = omega_1 t + (1)/(2) prop t^2 = 10 xx (1)/(4) + (1)/(2) xx 40 xx ((1)/(4))^2 = (15)/(4) rad` Number of revolutions = `(theta)/( 2pi) = (15)/( 4 xx 2 pi) = 0.6`. |
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| 1060. |
A particle moves in a circular path such that its speed `1v` varies with distance `s` as `v = sqrt(s)`, where `prop` is a positive constant. Find the acceleration of the particle after traversing a distance `s`. |
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Answer» Total acceleration, `a = sqrt(a_t^2 + a_r^2) = sqrt(((d v)/(d t))^2 + ((v^2)/(R))^2)` where `v = prop sqrt(s)` Differentiating `v = prop sqrt(s)` with respect to time, we have `( dv)/(d t)= prop (s^(-1//2))/(2) (d s)/(d t)=` Substituting `(d s)/(d t) = prop sqrt(s)`, we have `(d v)/(d t) = (prop^2)/(2)` Now substituting `dv//dt` and `v` in the expression of `a`, we have `a = sqrt(((prop^2)/(2))^2 + [(prop sqrt(s))/(R)]^2` This gives `a = prop^2 sqrt((1)/(4) + (s^2)/(R^2))`. |
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| 1061. |
Two straight lines drawn on the same displacement-time graph make anles `30^(@)` and`60^(@)` with time-axis respectiv ely Fig. 2 (a) .36, Which line repersents greater veloc ity ? What is the ratio of two velocities? . |
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Answer» Since the slpe of the displacement-time graph of uniform motion in one dimension represents the velocity of the object, hence the line showing grearer slope in graph corresponds to greater velocity of the object. Therfore the line making angle `60^(@)` with time axis represents greater velocity: Ratio of two velocities , `= v_(A)/v_(B) = (tan 30^(@))/(tan 60^(@)) = (1 //sqrt 3)/(sqrt3 ) = 1/2`/ |
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| 1062. |
Two swimmers start a race. One who reaches the point `C` first on the other bank wins the race.`A` makes his strokes in a direction of `37^(0)` to the river flow with velocity `5km//hr` relative to water. `B` makes his strokes in a direction `127^(0)` to the river flow with same relative velocity.River is flowing with speed of `2km//hr` and is `100m` wide.speeds of `A` and `B` on the ground are `8km//hr` and `6km//hr` respectively. |
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Answer» Correct Answer - [B wins, time of A = 165 s, time of 5 = 150 s] |
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| 1063. |
A hose lying on the ground shoots a stream of water upward at an angle of `60^@` to the horizontal with the velocity of `16 m s^-1`. The height at which the water strikes the wall `8 m` away is.A. 8.9 mB. 10.9 mC. 12.9 mD. 6.9 m |
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Answer» Correct Answer - A (a) `u_x = 16 cos 60^@ = 8 ms^-1` time taken to reach the wall `= 8//8 = 1 s` Now `u_y = 16 sin 60^@ = 8 sqrt(3) ms^-1` `h = 8 sqrt(3) xx 1 - (1)/(2) xx 10 xx 1 = 13.86 - 5 = 8.9 m`. |
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| 1064. |
A man can swim at a speed of `3 km h^-1` in still water. He wants to cross a `500-m` wide river flowing at `2 km h^-1`. He keeps himself always at an angle to `120^@` with the river flow while swimming. The drift of the man along the direction of flow, when he arrives at the opposite bank is.A. `(1)/(6 sqrt(3)) km`B. ` 6 sqrt(3) cm`C. `3 sqrt(3) km`D. `(1)/(3 sqrt(3)) km` |
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Answer» Correct Answer - A (a) `t = (d)/( v sin theta) =(0.5 km)/(3 sin 120^@ km h^-1)=(1)/(3 sqrt(3)) h` `x = (u + v cos theta) t = (2 + 3 cos 120^@)(1)/(3 sqrt(3)) = (1)/(6 sqrt(3))`. |
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| 1065. |
A constant force `(2 hat i+ 3 hat j+ 4 hat k)` mewton pewton produces a desplacement of `(2 hati +3 hat j+ 4 hat k)` metre. What is the word done? |
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Answer» Work done, `W=vec F.vec S =(2 hat I +3 hatj +4 hat k).(2 hati +3 hat j +4 hat k) `=2xx2 +3xx 3+4xx4=4 +9 16 =29 J` |
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| 1066. |
Two particles A and B are placed in gravity free space at `(0, 0, 0) m and (30, 0, 0) m` respectively. Particle A is projected with a velocity `(5hat i + 10 hat j + 5 hat k) ms^-1,` while particle B is projected with a velocity `(10 hat i + 5 hat j + 5 hat k) ms^-1` simultaneously. Then,A. they will collide at `(10, 20, 10) m`B. they will collide at `(10, 10, 10) m`C. they will never collideD. they will collide at 2 s |
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Answer» Correct Answer - C For collision, `r_A=r_b` (at same instant) `:. (r_i+v_t)_A=(r_i+v_t)_B` `rArr (5 hati+10 hatj+5 hatk)t=30 hati+(10 hati+5 hatj+5 hatk)t` Equating the coefficients of x `5t=30+10t` `rArr t=-ve` So, they will never collide. |
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| 1067. |
The diagram shows the velocity-time graph for a particle moving in a straight line. The sum of the two shaded areas represents : A. The increase in displacement of the particleB. The average velocity of the particleC. The average acceleration ofthe particleD. The distance moved by the particle |
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Answer» Correct Answer - D |
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| 1068. |
The distance travelled by the moving body is :A. The area between the speed time graph and time axis.B. The area between the speed time graph and speed axisC. The area between the distance time graph and time axisD. The area between the distance time graph and distance axis |
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Answer» Correct Answer - A |
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| 1069. |
A particle is moving with velocity ` vecv = k( y hat(i) + x hat(j)) `, where `k` is a constant . The genergal equation for its path isA. (a) `y^1 = x + constantB. (b) ` xy = cosntant`C. (c ) y^2 = x^2 + constatnD. (d) ` y =x^2 + constant |
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Answer» Correct Answer - C ` vec v = K ( y hat I +x hat j) ` or ` v_x hat i+ v_y hat j = K (y hat i+ x hat j)` :. ` v_x = Ky ` and ` v_y = K x` ` (dx) (dt) = Ky` or ` dx = Ky dt` ,brgt or ` x= Kyt` ….(i) ` (dy)/(dt) = Kx` or ` dy = Kx dt` or ` y= Kxt` ….(ii) From (i0, ` t= x/(Ky)` From (ii), ` y = Kx (x/(Ky)) + a constant ltbRgt or ` y^2 = x^2 + a constant`. |
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| 1070. |
A particle starts from the origin of coordinates at time t = 0 and moves in the xy plane with a constant acceleration `alpha` in the y-direction. Its equation of motion is `y= betax^2`. Its velocity component in the x-direction isA. (a) sqrt (2 b)/a`B. (b) sqrta/(2b)`C. (c ) sqrt a/b`D. (d ) sqrt b/a` |
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Answer» Correct Answer - B Here, ` y = bx^2`. Differentiating it w.r.t. t`, we get ,br. ` (dy)/(dt) = b 2 x (dx)/(dt)` or v_y = 2 bx v_x` Again differentiating it w.r.t.(t).we get ` (dv_y)/(dt) = 2 b v_x 9dx)/(dt) + 2 b x 9dv_x)/(dt) = 2 b v_x^2 + 0` ` [ (dv _x)/(dt) =0`. becaude the particle has constant acceleration along y-direction only]` ltbRgt As per question, ` (d v_y)/(dt) =a = 2 b v_x^2` or ` v_x^2 = a/(2 b) ` or v_x = sqrt a/(2 b)`. |
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| 1071. |
Can the direction of velcity of a body change, when accleration is constant? |
| Answer» Yes, when the body moves vertically upwards, after reaching the highest point, the body starts falling down, i.e. the direction of velocity is reversed, wheras the acceleration is constat. | |
| 1072. |
A particle is projected from a point P with a velocity v at an angle `theta` with horizontal. At a certain point Q it moves at right angles to its initial direction. ThenA. velocity of particle at `Q` is `v sin theta`B. velocity of particle at `Q` is `v cot theta`C. time of flight from `P` to `Q` is `(v//g) cosec theta`D. time of flight from `P` to `Q` is `(v//g)sectheta` |
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Answer» Correct Answer - B::C |
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| 1073. |
Velocity of a particle at any time t is `v=(2 hati+2t hatj) m//s.` Find acceleration and displacement of particle at `t=1s.` Can we apply `v=u+at` or not? |
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Answer» Correct Answer - A::B `v=(2 hati+2t hatj)` (i) `a=(dv)/(dt)=(2hatj)m//s^2`=constant `:. v=u+at` can be applied. (ii) `s=int_0^1vdt=int_0^1(2hati+2thatj)dt=[2t hati+t^2 hatj]_0^1` `=(2hati+hatj)m` |
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| 1074. |
An open elevator is ascending with constant speed `v=10m//s.` A ball is thrown vertically up by a boy on the lift when he is at a height `h=10m` from the ground. The velocity of projection is `v=30 m//s` with respect to elevator. Find (a) the maximum height attained by the ball. (b) the time taken by the ball to meet the elevator again. (c) time taken by the ball to reach the ground after crossing the elevator. |
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Answer» Correct Answer - A (a) Absolute velocity of ball=`40 m//s` (upwards) `:. h_(max)=h_i+h_f` Here, `h_i`=initial height =`10m` and `h_f`=further height attained by ball `=u^2/2g=(40)^2/(2xx10)=80m` `:. h_(max) =(10+80)m=90m` (b) The ball will meet the elevator again when displacement of lift=displacement of ball or `10xxt=40xxt-1/2xx10xxt^2` or `t=6s` (c) Let `t_0` be the total time taken by the ball to each the ground. Then, `-10=40xxt_0-1/2xx10xxt_0^2` Solving this equation we get, `t_0=8.24s` Therefore, time taken by the ball to reach the ground after crossing the elevator, `=(t_0-t)=2.24s` |
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| 1075. |
The velocity of a particle moving in a straight line is decreasing at the rate of `3 m//s` per metre of displacement at an instant when the velocity is `10 m//s.` Determine the acceleration of the particle at this instant. |
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Answer» Correct Answer - B::C `(dv)/(ds)=-(3 m//s)/m` `a=v.(dv)/(ds)` `=(10)(-3)` `=-30 m//s^2` |
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| 1076. |
"A lift is ascending with decreasing speed". What are the directions of velocity and acceleration of the lift at the given instant. |
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Answer» Correct Answer - A (i) Direction of motion is the direction of velocity. Lift is ascending (means it is moving upwards). So, direction of velocity is upwards. Speed of lift is decreasing. So, direction of acceleration should be in opposite direction or it should be downwards. |
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| 1077. |
A man is sitting inside a moving train and observes the stationary objects outside of the train. Then choose the single correct choice from the following statements`-`A. all stationary objects outside the train will move with same velocity in opposite direction of the train with repsect to the man.B. stationary objects near the train will move with greater velocity & objects far from train will move with lesser velocity with respect to the man.C. large objects like moon or mountains will move with same velocity as that of the train.D. all of these |
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Answer» Correct Answer - A `vec(V)_(O,M)=vec(V)_(O)-vec(V)_(M)` `vec(V)_(O,M)=vec(V)_(O)-vec(V)_("Train")` `V_(O,M)` =velocity of object with respect to man `V_(O)`= velocity of object `V_(M)` =velocity of man Here velocity of object is zero. So, `vec(V)_(O,M)=-vec(V)_(M)` |
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| 1078. |
Figure shows the displacement-time graph of a particle moving in a straight line. Find the signs of velocity and acceleration of particle at time `t = t_1 and t = t_2.` |
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Answer» Correct Answer - A::B::D At time `t=t_1,` slope of s-t graph (=velocity) is positive and increasing. Therefore velocity is positive and accleration is also positive. At time `t=t_2` slope is negative but decreasing (in magnitude). Therefore velocity is negative but decreasing in magnitude. Hence, acceleration is positive. |
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| 1079. |
A particle projected atv a definite angle `prop` to the horizontal passes through points `(a,b) and (b,a)`, referred to horizontal and vertical axes through the points of projection. Show that : (a) The horizontal range `R = (a^2 + ab + b^2)/(a + b)` . (b) The angle of projection `prop` is given by `tan^-1 [(a^2 + ab + b^2)/(ab)]`. |
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Answer» (a) The equation of the trajectory of the particle `y = x tan prop (1 - (x)/( R))`….(i) Since the particle passes through the points with coordinates `(a, b) and (b,a)`, they should satisfy the equation of the curve. `b = a tan prop (1 - (a)/(R )) and (ii) a = b tan prop (1 - (b)/( R))` ...(ii) Dividing `(b^2)/(a^2) = ((1 - (a)/(R)))/((1 - (b)/(R)))` `rArr b^2 - (b^3)/(R) = a^2 - (a^3)/(R)` `rArr (1)/(R) [a^3 - b^3] = a^2 -b^2` `rArr R = (a^3 - b^3)/(a^2 - b^2) = ((a- b)(a^2 + ab + b^2))/((a - b)(a + b)) = (a^2 + ab + b^2)/(a + b)` Hence, proved. `[:. a != b]` (b) Substituting the expression for `R` in (ii), `(b)/(a) = tan prop [1 - (a( + b))/(a^2 + ab + b^2)]` =`tan prop [(a^2 + ab + b^2 - a^2 - ab)/(a^2 + ab + b^2)]` `tan prop = (a^2 + ab + b^2)/(ab) rArr prop = tan^-1 [(a^2 + ab + b^2)/(ab)]` Hence proved. |
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| 1080. |
How is thevlocity-time graph of uniformly accelerated motion helpful in studying the motion of the object in one dimension? |
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Answer» This graph is useful in following ways: (a) To determine the velocity of the object at the given instant. (b) To determine the acceleration of the object. (c ) To determine the total distance travelled by the object in a given time. |
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| 1081. |
How is the position-time graph of uniformly accelerated motion id one dimension helpful in studying the motion of the object ? |
| Answer» This graph helps us to determine the distance travelled by object during any interval of time and also the velocity of the object at any instant of time. | |
| 1082. |
A car moving along a straight highway with a speed of `72 km h^(-1)` is brought to a stop within the distance of `100 m`. What is the retardation of the car and how long does it take for the car to stop ? |
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Answer» Here, `u=72km h^(-1) =72 xx 5/(18)` `=20 ms^(-1)`: `v=0, S=100 m: a=? Acceleration, `a(v^(2)-u^(2))/(2 S) =(0-20^(2))/(2xx 100) =- 2 ms^(2)` i.e. retardation `=- a=2 ms^(-2)` ltbtgt Time `t= (v-u)/a = (0-20)/(-2) =10 s`. |
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| 1083. |
What is the nature of the displacement time curve of a body moving with constant acceleration ? |
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Answer» If a body starts form rest and moves along a straight line with a uniform acceleration ` a`, then the diplacement (x) of the body in time (t) is ` x =1/2 at^(2)` which is an equation of a parabola. Thus the displacement-time graph for a body moving with constant acceleration is a parabola. |
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| 1084. |
A river is flowing from west to east at a speed of 5 metres per minute. A man on the south bank of the river, capable of swimming at 10 metres per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction.(a) due north (b) 30° east of north(c) 30° north of west (d) 60° east of north. |
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Answer» (a) due north Explanation: If the man swims at any angle east to the north direction, although his relative speed will increase, he will have to travel a larger distance. So, he will take more time. If the man swims at any angle west to the north direction, his relative speed will decrease. So, he will take more time. |
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| 1085. |
The resultant of \(\vec A\,+\,\vec B\) acts along x - axis. if A = 2\(\hat i\) – 3\(\hat j\) + 2 \(\hat k\) then B is -(a) -2\(\hat i\) +\(\hat j\) + \(\hat k\) (b) \(3\hat j\) – 2 \(\hat k\) (c) -2 \(\hat i\) -3\(\hat j\) (d) -2\(\hat i\) – 2 \(\hat k\) |
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Answer» Correct answer is (b) 3 \(\hat j\) – 2 \(\hat k\) |
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| 1086. |
The angle between (\(\vec A\)+ \(\vec B\) ) and ( \(\vec A\) – \(\vec B\) ) can be –(a) only 0° (b) only 90° (c) between 0° and 90° (d) between 0° and 180° |
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Answer» Correct answer is (d) between 0° and 180° |
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| 1087. |
A body is projected with a velocity ` 30 ms^(-1)` at an angle of `30^@` with the vertical. Find (i) the maximum height (ii) time of flight and (iii) the horizontal range of the projectile. Take ` g=10 m//s^@`. |
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Answer» Here, ` u= 30 ms^(-1)` , Angle of projection with horizontal, ` theta = 90^@ = 60^@. (i) Max. height, ` H= (u^2 sin^2 theta)/(2 g) = ((30)^2 sin^2 60^@)/ ( 2xx 10) = 33 . 75 m` (ii) Time of flight, `T = ( 2 u sin theta )/ g = ( 2 xx 30 2 xx 60^2)/(10) = 5.2 s` (iii) Horizontal range, ` R= (u^2 sin theta0/g = ((30^sin 2 xx 60^2) /(10)` `= ( 900) /(10) sin 120^@ = 90 sin 60^@ = 77 . 94 m`. |
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| 1088. |
Two balls A and B are projected vertically upwards with different velocities. What is the relative acceleration between them? |
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Answer» `a_4=a_B=g (downwards)` `:. a_(AB)=a_A-a_B=0` |
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| 1089. |
A ball is thrown vertically upwards from the ground. It crosses a point at the height of 25 m twice at an interval of 4 secs. The ball was thrown with the velocity ofA. 20 m/secB. 25 m/secC. 30 m/secD. 35 m/sec |
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Answer» Correct Answer - C |
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| 1090. |
Three particles are projected in air with the minimum possible speeds, such that the first goes from Ato B, the second goes from 5 to C and the third goes from C to A. Points A and Care at the same vertical level. The two inclines make the same angle a with the horizontal as shown. Thenthe relation among the projection speeds of the three particles is :A. `u_(3)=u_(1)+u_(2)`B. `u_(3)^(2)=2u_(1)u_(2)`C. `1/u_(3)=1/u_(1)+1/u_(2)`D. `u_(3)^(2)=u_(1)^(2)+u_(2)^(2)` |
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Answer» Correct Answer - B |
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| 1091. |
A particle is projected on an inclined plane with a speed `u` as shown in (Fig. 5.61). Find the range of the particle on the inclined plane. . |
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Answer» Correct Answer - `[2R//3]` |
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| 1092. |
A particle is projected on an inclined plane with a speed `u` as shown in (Fig. 5.61). Find the range of the particle on the inclined plane. . |
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Answer» `R = (u^2)/(g cos^2 theta_0) [sin(2 prop + theta_0) - sin theta_0]` where `theta_0 = 30^@ and prop = 45^@` Then, `R = (u^2)/(g cos^2 30^@) [sin(2 xx 45^@ + 30^@)- sin30^(@)]` =`(2 u^2sqrt(3 - 1))/(3 g)`. |
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| 1093. |
Two particles A and B are moving along x-axis. Their x-coordinate versus time graphs are as shown below (a) Find the time when the particles start their journey and the x-coordinate at that time. (b) Find velocities of the two particles. (c) When and where the particles strike with each other. |
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Answer» Correct Answer - A::B::C::D (b) v=Slope of x-t graph |
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| 1094. |
Which of the following statements are true for a moving body ?A. If its speed changes, its velocity must change and it must have some accelerationB. If its velocity changes, its speed must change and it must have some acceleration.C. If its velocity changes, its speed may or may not change, and it must have some acceleration.D. If its speed changes but direction of motion does not change, its velocity may remain constant |
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Answer» Correct Answer - A::C |
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| 1095. |
Two small pegs (A and B) are at horizontal and vertical separation b and h respectively. A small block of mass M is suspended with the help of two light strings passing over A and B as shown in fig. The two string are always kept at right angles (i.e., `lt APB = 90^(@))`. Find the minimum possible gravitation potential energy of the mass assuming the reference level at location of peg A. [Hint: the potential energy is minimum when the block is at its lowest position] |
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Answer» Correct Answer - `U_(min) = - (1)/(2)Mg [sqrt(h^(2)+b^(2))-h]` |
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| 1096. |
Balls are thrown vertically upwards in such a way that the next ball is thorwn when the previous one is at the maximum height. If the maximum hieght is `5m`, the number of balls thrown per minute will beA. `40`B. `50`C. `60`D. `120` |
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Answer» Correct Answer - C `h_("max")=u^(2)/(2g)rArr u=12xx10xx5=10 m//s` `t_(H)=sqrt((2xx5)/10)=1 s` no. of balls in one min. `=1xx60=60` |
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| 1097. |
In the figure shown the acceleration of A is, `vec(a)_(A) = 15 hat (i) + 15 hat (j)` then the acceleration of B is (A remains in contact in B) A. `6 hat(i)`B. `-15 hat(i)`C. `10 hat(i)`D. `-5 hat(i)` |
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Answer» Correct Answer - D Let acceleration of B `vec(a)_(B)=a_(B) hat(i)` Then acceleration of A w.r.t. `B=vec(a)_(A)-vec(a)_(R)=(15-a_(B))hat(i)+15 hat(j)` This acceleration must be along the inclined pplane so `tan 37^(@)=15/(15-a_(B))rArr 3/4=15/(15-a_(B)) a_(B)=-5` `rArr vec(a)_(R)=-5hat(i)` |
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| 1098. |
A bullet is fired in a horizontal direction from a tower while a stone is simultaneously dropped from the same point then :A. The bullet and the stone will reach the ground simultaneouslyB. The stone will reach earlierC. The bullet will reach earlierD. None of these |
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Answer» Correct Answer - A |
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| 1099. |
A very broad elevator is going up vertically with a constant acceleration of `2m//s^(2)`. At the instant when its velocity is `4m//s`, a ball is projected from the floor of the lift wht as of `4m//s` relative to the floor at an elevation of `30^(@)`. The time taken by the ball to return the floor is `(g=10m//s^(2))`A. `1/2s`B. `1/3s`C. `1/4s`D. `1s` |
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Answer» Correct Answer - B |
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| 1100. |
An elavator is going up vertically with a constant accelerartion of `2m//s^(2)`. At the instant when its velocity is `4m//s` a ball is projected from the floor of the elavator with a speed of `4m//s` relative to the floor with an angular eleavation of `60^(@)`. The time taken by the ball to return the floor is `(g=10m//s^(2))`A. `1/3s`B. `1/(sqrt(3))s`C. `2/(sqrt(3))s`D. `sqrt(3)s` |
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Answer» Correct Answer - B |
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