1496 + Interview Questions in KINEMATICS IN GENERAL AWARENESS Page 30 InterviewSolution

1451.	STATEMENT-1: During a turn, the value of centripetal force should be less than the limiting frictional force. STATEMENT-2: The centripetal force is provided by the frictional force between the tyre and the road.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 If the value of frictional force `mu mg` is less than centripetal froce, than it is not possible for a vehicle to take twin and the bicycle would take a over turns. Thus condition for no over turning is `mu mg ge (mv^(2))/(r )` So statement -`I` and statement -`II` are correct and statement - `II` si the correct explanantion of statement - `I`. Hence `(a)` is correct.

Discussion

1452.	Can two equal vects `vec a and vec b` at different locations in space necessarilly have identical physical effects ?
Answer» Yes they may have same thusical effects. For example two bodies while falling under gravity, will have same acceleration though their locations may be different with respect ot a common origin .

Discussion

1453.	The acceleration. If ` v_(0)` is themagnitude the engine is cut off, is given by ` (dv)/(dt) =- k v^(2)`, wher (k) is a constant. If `v_(0)` is the magnitude of the velocity at cut off, find the magnitude of the velocity at time (t0 after the cut off.
Answer» Given, ` (d v)/(dt) =- k v^(3) dv =- k dt` Integration it wighin the conditions of motion i.e. as time changes from ` 0 to t`, velocity changes from ` v_(0) to v`, we have ` int _(v0)^(v) v^(-3) d v=- k int _(0)^(t) dt or [v^(-2)/(-2)](v_(0)^(v) =- k (t) _(0)^(t)` or -1/2 [1/v^(2) -1/v_(0)^(2)] =- kt` or 1/v^(2) =2 kt + 1/v_(0)^(2) =(2 kt v_(0)^(2) +1)/v_(0)^(2)` or ` v=v_(0)/(sqrt 1 + 2 kt v_(0)^(2)`.

Discussion

1454.	A particle slides down a frictionless paraboli `(y=x^(2))` track `(A-B-C)` starting from rest at point `A`.Point `B` is at the vertex of parabola and point `C` is at a height less than that of point `A`.After `C`,the particle moves freely in air as a projectile. If the particle reaches highest point at `P`,then A. (a) KE at ` P = KE at B`B. (b) height at ` P = height at (A)C. (c ) total eneragy at ` P = total eneragy at (A)`D. (d) time of travel from ` A` oto ` B = time of travel from (B) tp (P).
Answer» Correct Answer - C Since ` y= x^2` , the motion is in two dimensions as shown in figure. Velocity at (B) greater than at (P) . In the given motion of a particle, the law of conservation of energy . Therefore, total energy at ` P=` total eneragy at (A). As vertical distance ` ABgt BP`, time of travel from (A) to (B) is greater than that from (B) to (P).

Discussion

1455.	Whe a projectile is projeected with velcity (v0 making an angle ` theta` with the horizontal direction, then maximum horizontal range is…………………………… .
Answer» Correct Answer - ` ve^2//g`

Discussion

1456.	A player throws a ball upwards with an initial speed of `29.4 ms^(-1).` What are the velcity and acceleration of the hall at the highest point os its motion ?
Answer» At the highest position of the ball projected verticall upwards the velocity is zero and accelration is equal to acceleration due to grav ity (=9.8 ms^(p-1)) acting vertically downward.

Discussion

1457.	Two balls of equal masses are thrown upwards along the same vecticla direction at an interavel of ` 2 seconsds`. With the same intial velcity of ` 39.2 m//s` . The these collede at a height of .A. (a) ` 44.1 m`B. (b) ` 73.5 m`C. (c ) ` 11 .6 m`D. (d) 1 9 .0 m`
Answer» Correct Answer - B Here `u = 39.2 m//s` Let two balls collide at a height (S) from the groun after (t) seconds when second ball is thrown upwards. :. `the time taken by first ball to teach the point of collision = ( t+ 2) sec`. Taking motion of the first ball, we have ` S= 39.2 ( t + 2) + 1/2 (- 9.8 ) ( t+ 2)^@` `= 39.2 (t + 2) - 4 .9 ( t+ 2)^2` ...(i) Taking motion of second ball, we have ` S= 39.2 t = 1/2 (- 9.8 0 T^2 = 39.2 t- 4.9 t^@` ...(ii) From (i) and (ii), we have ltbRgt ` 39.2 (t+2) - 4.9 (t + 2)^2 = 39. 2 t - 4 .9 t^2` On solving we get ` t= 3 sec ` From (ii), ` S = 39.2 xx 2 - 4.9 (3)^2 = 117. 6 - 44.1` `= 73.5 m`.

Discussion

1458.	A body is projectd with the velcity ` (U_1)` from the point (A) as shown in Fig. 2. (d). 37. At the same tiem another body uis projected vertcally upwards with the velcoity ` u_2` from the point (B) .What should be the value of ` u_1)//u_1` for both the bodies to collide ? .
Answer» The two bodies will collide, if they reach at a point covering the same vertical distaance in the same time. Therefore ` y=u_1 sin 60^2 xx t -1/2 t^@ =u_2 t - 1/2 g t^2` or ` u_1 sin 60^@ xx t = ut_2 t` or ` u_1/u_2 1/(sin 60^2) = 2/(sqrt3)`.

Discussion

1459.	A particle moves with a tangential acceleration `a_(t)=vec(a).hat(v)` where `vec(a)=(5hat(i)) m//s^(2)`. If the speed of the particle is zero at `x=0`, then find v (in `m//s`) at `x=4.9 m`.
Answer» Correct Answer - 7 As `dvd=vec(a).vec(d)r=adx=5dxrArr int_(0)^(v)vdv=5 int_(0)^(4.9)dxrArr v^(2)/2=5(4.9)rArr v^(2)=49rArrv=7 m//s`

Discussion

1460.	A particle moves along a circle if radius (`20 //pi`) m with constant tangential acceleration. If the velocity of the particle is ` 80 m//s` at the end of the second revolution after motion has begun the tangential acceleration is .A. `40 m//s^2`B. ` 640 pi //s^@`C. 1 60 mss^(2)`D. 40 m//s^2`
Answer» Correct Answer - A ` r= ( 20 // pi ) m , v= 80 m//s , ` ` theta = 2 rev = 4 pi rad , omega _0=0` From the equatin ` omega^2 =omega_0^2 + 2 alpha theta , we have ` moega^2 = 2 alpha theta ` or v^2/r^2 = 2 . A/ r theta` ` or ` a = u^2 /(2 r theta) = ( 80^2)/(2 xx (20 //pi ) xx 4 pi) = 40 m//s^2`.

Discussion

1461.	A particle moves with deceleration along the circle of radius R so that at any moment of time its tangential and normal acceleration are equal in moduli. At the initial moment `t=0` the speed of the particle equals `v_(0)`, then th speed of the particle as a function of the distance covered S will beA. `v=v_(0) e^(-S//R)`B. `v=v_(0)e^(S//R)`C. `v=v_(0) e^(-R//S)`D. `v=v_(0) e^(R//S)`
Answer» Correct Answer - A Given `(dv)/(dt)=v^(2)/rrArr(dv)/(ds)=v^(2)/r, -underset(v_(0))overset(v)(int)1/vdv=underset(o)overset(s)(int)(ds)/r` `rArr ln[v_(0)/v]=S/rrArr v_(0)/v=e^(S//r)` `rArr v_(0)=ve^(S//r)rArrv=v_(0)e^(-S//r)`

Discussion

1462.	A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, the magnitude of acceleration is :A. `20 ms^(-2)`B. `12 m//s^(-2)`C. `9.9 ms^(-2)`D. `8 ms^(-2)`
Answer» Correct Answer - C `omega=(14xx2pi)/25` `:.` magnitude of acceleration `=omega^(2)r=((14xx2pi)/25)^(2)xx80/100=9.9 m//s^(2)`

Discussion

1463.	A particle starts moving along a circle of radius `(20//pi)m` with constant tangential acceleration. If the velocity of the parthcle is `50 m//s` at the end of the second revolution after motion has began, the tangential acceleration in `m//s^(2)` is :A. `1.6`B. `4`C. `15.6`D. `13.2`
Answer» Correct Answer - C Given `r=20/pi m` Angular velocity after second revolution `omega=v/r=(50pi)/20=(5pi)/2` `omega_("final")^(2)=omega_("initial")^(2)+2alpha theta` `25/4 pi^(2)=2alpha(4pi) rArr alpha=(25 pi)/32` `a_(t)=alphar=(25pi)/32xx20/pi=15.6`

Discussion

1464.	The speed of river current close to banks is nearly zero. The current speed increases linearly from the banks to become maximum `(= V_(0))` in the middle of the river. A boat has speed ‘u’ in still water. It starts from one bank and crosses the river. Its velocity relative to water is always kept perpendicular to the current. Find the distance through which the boat will get carried away by the current (along the direction of flow) while it crosses the river. Width of the river is l.
Answer» Correct Answer - `(V_(0)l)/(2u)`

Discussion

1465.	Two trains are headed towaads each other on the same straight rrach, each vaving a speed of ` 30 km h^(-1)`. A bird that can fly at ` 60 km h^(-1)` flies off one train when they are ` 60 km` apart and leads firectly form the otjer train, On reaching the other train, tg flies back to the first train and so on. (a) How many trips can the bird make from one trin to the other train befre they meet ? (b) What is the total distance the bird travels ?
Answer» (a) Relative velocit of train (A) relative to train (B) ` = 30 + 30 = 60 km h^(-1)` Distansce between train ` A and B=60 km` Time taken by the two trains to meet ` T = (60)/(60) =1 h` Relative velocity of bird w.r.t. train (B) ` ` 60 + 30 -90 km h^(-1)` Time taken for first trip, t_1 = ( 60 km )/(90 km h^(-1))` = 2/3 h` ` Now, separation betwwen the two trains just after first trip of bird ` =60 - 60 xx 2//3 =20 km` Time taken for the second trip ` t_2 + (20)/(90) = 2/9 h = 2/(3^2) h` Similarly taken for the nth trip ` t_n = 2/ 3^n h` and time taken for the nth trip ` t-n = 2/3^n h` and time taken for the nth trip ` t_n = 2/3^n h` Total time of flight of bird will be ` T= t_1 = t_ 2 + t_3 + .........+ t_n` `= 1/2 + 2 /3^2 + 2/3^3 + ...........+ 2/3^n` ` = 2/3 [1 + 1/3 + 1/3^2 + ......... + 1/(3^n -1)]` ` 2/3 [ (1- (1//3)^n)/( 1- (1//3 ))] =1 1- (1/3)^n` [Formula for sum of G.P. series is , ` S= (a (1 -r^n)/(1 -r) , Here, a = 1 , r = 1//3 ]` But ` T =1 h, so 1=1 - (1/3)^n or 1/3^n or 1/3^n =0 or n= prop` (b) Total distance travelled by the bird ` = constant velocity xx time = 60 xx 1= 60 60 km`.

Discussion

1466.	An aeroplane at a constant speed releases a bomb.As the bomb drops away from the aeroplane,A. it will always be vertically below the aeroplaneB. it will always be vertically below the aeroplane only if the aeroplane is flying horizontallyC. it will always be vertically below the aeroplane only if the aeroplane is flying at an angle of `45^(@)` to the horizontalD. it will gradually fall behind the aeroplane if the aeroplane is flying horizontally
Answer» Correct Answer - A

Discussion

1467.	An aeroplane flying at a constant velocity releases a bomb. As the bomb drops down from the aeroplane, (a) it will always be vertically below the aeroplane (b) it will always be vertically below the aeroplane only if the aeroplane is flying horizontally (c) it will always be vertically below the aeroplane only if the aeroplane is flying at an angle of 45° to the horizontal (d) it will gradually fall behind the aeroplane if the aeroplane is flying horizontally
Answer» Correct answer is: (a) it will always be vertically below the aeroplane The horizontal component of the velocity of the bomb is the same as the horizontal component of the velocity of the aeroplane in all cases. Hence, in all cases, the two have the same horizontal displacement in the same time.

1467.

An aeroplane flying at a constant velocity releases a bomb. As the bomb drops down from the aeroplane, (a) it will always be vertically below the aeroplane (b) it will always be vertically below the aeroplane only if the aeroplane is flying horizontally (c) it will always be vertically below the aeroplane only if the aeroplane is flying at an angle of 45° to the horizontal (d) it will gradually fall behind the aeroplane if the aeroplane is flying horizontally

Answer»

Correct answer is: (a) it will always be vertically below the aeroplane

The horizontal component of the velocity of the bomb is the same as the horizontal component of the velocity of the aeroplane in all cases. Hence, in all cases, the two have the same horizontal displacement in the same time.

Discussion

1468.	A particle falls from rest under gravity. Its potential energy with respect to the ground (PE) and its kinetic energy (KE) are plotted against time (t). Choose the correct graph.
Answer» Correct answer is: (b)

Discussion

1469.	A particle is moving eastwards with a velocity of ` 5 m//s`. In `10 s` the velocity changes to `5 m//s` nothwards. The average acceleration in this time isA. ` 1 /(sqrt 2) ms^(-2)` , 45 West of NothB. (b) ` 1/2 ms^(-2), 60^@ west of NorthC. ` 2 ms^(-2) 60^@` East of soutD. `1/(sqrt 2) ms^(-2) `, 30^@` West of south
Answer» Correct Answer - A Here, initial velocity ` v_1 = 5 ms^(-1)` along East ` = 5 hat i` ` vec _1 = 5 ms^(-1)` along North `= 5 hat j` Change in velocity, ` Deltav= vec v_2 = vec v_1 = 5 hat j - 5 hat i=- 5 hat I + 5 hat j` Magnitude of ` Delta v= sqrt ((-5) ^2 + 5^2 ) = 5 sqrt 2 m//s` Acceleration, ` \|vec a\| = (Delta v0/t = (5 sqrt 2)/(10) = 1/(sqrt 2) ms^(-2)` ` tan theta = (BD)/(OB) = 5/ 5 = 1 tan 45^@` ` theta 45^2` West of Noth`.

Discussion

1470.	A cyclist starts form centre ` O` of a circular park oa radius ` 1 km` and moves along the path ` OPRQO` as whown Fig. 2 (EP).15. If he maintais constant speed of ` 10 ms^(-1)`, what is his acceleration at point (R )in magnitude and direction ? .
Answer» Centripetal acceleration, `a_(c) =v^(2) /r =(10^(2) /(1000) =0.1 m//s^(2)` along RO`.

Discussion

1471.	A cyclist starts from centre O of a circular park of radius 1 km and moves along the path OPRQO as shown. If he maintains constant speed of 10 ms-1 . What is his acceleration at point R in magnitude & direction?
Answer» Centripetal acceleration, a_c = \(\frac{v^2}{r}\) = \(\frac{10^2}{1000}\) = 0.1 m/s² along RO.

Discussion

1472.	The velocity of a moving particle is given by V=6+18t+9t2 (x in meter, t in seconds) what is it’s acceleration at t=2s.
Answer» Differentiation of the given equation eq. w.r.t. time We get a = 18 + 18t At t = 2 sec. a = 54 m/sec².

1472.

The velocity of a moving particle is given by V=6+18t+9t2 (x in meter, t in seconds) what is it’s acceleration at t=2s.

Answer»

Differentiation of the given equation eq. w.r.t. time

We get a = 18 + 18t

At t = 2 sec.

a = 54 m/sec².

Discussion

1473.	If ` vec R = (vec A = vec B), show that ` R^(2) =A^(2) +B^(20 +2 AB cos theta` wher, ` theta` is the smaller angle between `vec A and vec B`.
Answer» Given ` vec `R = (vec A + vec B) , Taking dot product of ` R` with itself , we have, ` vec R .vec R = ( vec A+ vec B). ( vec A+ vec B)` or ` R^2 = vec A. vec A + 2 vec A.vec B + vec B. vec B` or ` R^2 = A^2 + 2 AB cos theta + B^2`.

Discussion

1474.	The position of a particle is given by ` vec r= 3.0 t hat I - 2 2- 0 t^@ hat j + 4.0 hat k m`, wher 9t) in seconds and the coefficits have the proper units for ` vec r` to be in metres. (a) Fing the ` vec v` and ` ve a` of the particle ? (b) What is the magnitude and direction fo velocity of the particle at ` t= 2 s`?
Answer» (a) Velocity, ` vec v = (d vec r0/ (dt) = d/(dt0 ( 3.0 t hat I - 2.0 t hat j + 4.0 hat k) = [ 3. 0 hat I - 4.0 t hat j] ms^(-1)` Acceleration , ` vec a = (d vec v)/(dt0 = d/(dt) ( 3.0 hat I - 4 .0 t hat j ) = 0 -4.0 hat =- 4.0 hat j ms^(-2)` (b) At time ` t 2 s, vec v = 3.0 hat i - 4. 0 xx 2 hat j = 3. 0 hat i - 8.0 hat j ` :. ` vec = sqrt (93.0)^2+ (-8)^2) = sqrt 72 = 8. 54 ms^(-10` If ` theta` is the angle which ` vec v` makes with x-axis, then ` atn = v_y/v_x - (-8)/3 =- 2.667 =- tan 69. 5^@` ` theta = 69.5^@` below teh x-axis`.

Discussion

1475.	Block A and C starts from rest and move to the right with acceleration `a_(A)=12tms^(-2)` and `a_(C )=3ms^(-2)`. Here t is in second. The time when block B again comes to rest is A. `2s`B. `1s`C. `3/2s`D. `1/2 s`
Answer» Correct Answer - D Block B will again comes to rest if `v_(A)=v_(c)` i.e. `3t=(12t)trArr t=.^(1)//_(2) s`

Discussion

1476.	Particles `P` and `Q` of masses `20g` and `40g`, respectively, are projected from positions `A` and `B` on the ground. The initial velocities of `P` and `Q` make angles of `45^(circ)` and `135^(circ)`, respectively with the horizontal as shown in the fig. Each particle has an initial speed of `49m//s`. The separation `AB` is `245m`. Both particles travel in the same vertical plane and undergo a collision. After the collision `P` retraces its path. The separation of `Q` from its initial position when it hits the ground is
Answer» Correct Answer - [122.5 cm, 3.53 sec]

Discussion

1477.	A trolley was moving horizontally on a smooth ground with velocity `v` with respect to the earth. Suddenly a man starts running from rear end of the trolley with a velocity `(3//2)v` with respect to the trolley. After reaching the other end, the man turns back and continues running with a velocity `(3//2)v` with respect to trolley in opposite direction. If the length of the trolley is L, find the displacement of the man with respect to earth when he reaches the starting point on the trolley. Mass of the trolley is equal to the mass of the man.A. `2.5L`B. `1.5 L`C. `(5L)/3`D. `(4L)/3`
Answer» Correct Answer - D For man on trolley `3/2 vt=LrArr t=(2L)/(3v)` with respect to ground : `vt+3/2vt=L+(2L)/3=(5L)/3` `:. 3/2 vt-vt=L-(2L)/3=L/3 :. DeltaS=(5L)/3-L/3=(4L)/3`

Discussion

1478.	In an elastic collision between spheres A and B of equal mass but unequal radii, A moves along the x-axis and B is stationary before impact. Which of the following is possible after impact? (a) A comes to rest. (b) The velocity of B relative to A remains the same in magnitude but reverses in direction. (c) A and B move with equal speeds, making an angle of 45° each with the x-axis. (d) A and B move with unequal speeds, making angles of 30°and 60° with the x-axis respectively.
Answer» Correct answer is: (a, b, c, & d)

Discussion

1479.	A particle of mass `m` moving with kinetic energy `K`, makes a head - on elastic collision with a stationary particle of mass `eta m`. The maximum potential energy stored in the system during the collision isA. `nE//(n+1)`B. `(n+1)E//n`C. `(n-1)E//n`D. `E//n`
Answer» Correct Answer - A

Discussion

1480.	In an elastic collision between spheres A and B of equal mass but unequal radii, A moves along the x-axis and B is stationary before impact. Which of the following is possible after impact ?A. A comes to restB. The velocity of B relative to A remains the same in magnitude but reverses in directionC. A and B move with equal speeds, making an angle of `45^(@)` each will the x-axisD. A and B moves with unequal speeds, making angles of `30^(@)` and `60^(@)` with the x-axis respectively.
Answer» Correct Answer - A::B::C::D

Discussion

1481.	Two identical spheres move in opposite directions with speeds `v_(1)` and `v_(2)` and pass behind an opaque screen, where they may either cross without touching (Even 1) or make an elastic hand-on collision (Event 2).A. We can never make out which event has occurredB. We cannot make out which event has occurred only if `v_(1) = v_(2)`C. We can always make out which event has occurredD. We can make out which event has occurred only if `v_(1) = v_(2)`
Answer» Correct Answer - A

Discussion

1482.	Find the magnitude of the linear acceleration of a particle moving in a circle of radius 10 cm with uniform speed completing the circle in 4s.
Answer» The distance covered in completing the circle is `2r = 2 xx 10 cm`. The linear speed is `v = (2r)/(t) = (2 pi xx 10 cm)/(4 s) = 5 cm s^-1` The accleration is `a = (v^2)/r = ((5 pi cm s^-1)^2)/(10 cm) = 2.5 cm s^-2`.

Discussion

1483.	Two diamonds begin a free fall from rest from the same height, 1.0 s apart. How long after the first diamond begins to fall will the two diamonds be 10 m apart? Take `g = 10 m//s^2.`
Answer» Let the first body for (t) sec and second body for time ( t-10 second when their separation becomes ` 10 m`. Then ` 10 = 1/2 gt^2 - 1/2 g ( t=12) ^2` On solving , ` t= 1.5 s`.

Discussion

1484.	Two cars, A and B, are traveling in the same direction with velocities `v_(a)` and `v_(b)` respectively. When car is at a distance d behind car B, the brakes on A are applied, causing a deceleration at a rate a. Show that to prevent a collision between A and B it is necessary that:
Answer» Correct Answer - `v_(A) - v_(B) = sqrt(2ad)`

Discussion

1485.	Two cars are moving in the same direction with the same speed `30 km//hr`. They are separated by a distance of `5 km`, the speed of a car moving in the opposite direction of it meets these two cars at an interval of `4` minutes, will be.A. ` 40 km h^(-1)`B. 45 km h^(-1)`C. ` 50 km h^(-1)`D. ` 35 km h^(-1)`
Answer» Correct Answer - B Let car ` C` be moving in opposite direction to (A) and (B) with velocity ` vec_C` relave to ground. Then relative velocity of ` C` relative to ` A` and ` B` will be ` vec v (rel) = vec v _C - vec v`, As ` vec v` is opposite to ` vec v_C` Hence, time taken by car ` C` to cross the cars `A` and ` B` will be ` t= S/v_(rel)` or ` 4 /( 60) = 5/( v_C+30)` or` v_c = 45 km h^(-1)`.

Discussion

1486.	Two cars are moving in the same direction with the same speed `30 km//hr`. They are separated by a distance of `5 km`, the speed of a car moving in the opposite direction of it meets these two cars at an interval of `4` minutes, will be.
Answer» Since the two cars (A) and (B) are moving with same velocity, the separation between then will remain the same (=5 km). Let car (c ) be moving in opposite derection to cars ` A and B`, with velocity ` vec v_c` relative to ground, Then relative velocuty of car ` c` relative to car (A) and (B) wil be ` vec v _(rel) = vec c_C-vec V` As 1 vec V` is opposite ot ` vec v_C`, so ` v_(rel) =v_C - (-v)` ` =v_C + v= (v_C + 30 ) km h^(-1)` Time taken by car (C ) to cross the cars (A) and (B) will be ` t= S/(v_(rel)) or 4/(60) = 5 /(v_C + 30)` On solving , ` v_C= 45 km h^(-1)`.

Discussion

1487.	A particle is projected upwards with a velocity of 110m/sec at an angle of `60^(@)` with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g `=10m//sec^(2)`A. 10 secondsB. 22 secondsC. 5 secomdsD. `10 sqrt(3)` seconds
Answer» Correct Answer - B

Discussion

1488.	River is flowing with a velocity `v_(BR)=4hatim//s`. A boat is moving with a velocity of `v_(BR)=(-2hati+4hatj)m//s` relative to river. The width of the river is `100m` along `y`-direction. Choose the correct alternative(s)A. The boatman will cross the river in `25s`B. Absolute velocity of boatman is `2sqrt(5)m//s`C. Drift of the boatman along the river current is `50m`D. The boatman can never cross the river
Answer» Correct Answer - A::B::C

Discussion

1489.	The following figure-1.114 shows the linear motion velocity-time graph of a body.The body will be displaced in 5 seconds by: A. 2 mB. 3 mC. 4 mD. 5 m
Answer» Correct Answer - B

Discussion

1490.	A man in a river boat must get from point `A` to point `B` on the opposite bank of the river (see figure).The distance `BC=a`.The width of the river `AC=b`.At what minimum speed `u` relative to the still water should the boat travel to reach the point `B`?The velocity of flow of the river is `v_(0)`
Answer» Correct Answer - `[(v_(0)b)/(sqrt(a^(2)+b^(2)))]`

Discussion

1491.	The horizontal range of a projectile is R and the maximum height at tained by it is H. A strong windnow begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now be:A. `R+H/2`B. `R+H`C. `R+(3H)/2`D. `R+2H`
Answer» Correct Answer - D

Discussion

1492.	Particle is dropped form the height of `20m` form horizontal ground. There is wind blowing due to which horizontal acceleration of the particles becomes `6 ms^(-2)`. Find the horizontal displacement of the particle till it reaches ground.A. `6 m`B. `10 m`C. `12 m`D. `24 m`
Answer» Correct Answer - C

Discussion

1493.	The sum of two forces at a point is 16N. if their resultant is normal to the smaller force and has a magnitude of 8N, then two forces areA. ` 6 newton and 10 newton `B. ` 8 newton and `8` newton `C. `4 newton and `12` newton `D. `2` newton and `14` newton
Answer» Correct Answer - A Here, ` A+ B= 16` …..(i) ` sqrt (a^2 + b^2 + 2 AB cos theta ) = 8 ` …(ii) and ` tan 90^@ = ( B sin theta)/(A+ B cos theta)` ` or ` A= B cos theta = (B sin theta) /( tan 90^@) =0` or ` B cos theta =- A` or cos theta =- B//A` From ` (ii) , A^2 + B^2 + 1 AB (- B//A) = 64` or ` A^2 - B^2 = 64` ....(iii)` solving (i) and (iii) , we get ` ` A= 10 N ` and ` B = 6 N`.

Discussion

1494.	The position time graph for a particle travelling along x axis has been shown in the figure. State whether following statements are true of false. (a) Particle starts from rest at `t = 0`. (b) Particle is retarding in the interval 0 to `t_(1)` and accelerating in the interval `t_(1)` to `t_(2)`. (c) The direction of acceleration has changed once during the interval 0 to `t_(3 )`
Answer» Correct Answer - (a) F (b) T (c) T

Discussion

1495.	Two particles are projected simultaneously in the same vertical plane from the same point, with different speeds `u_(1)` and `u_(2)`, making angles `theta_(1)` and `theta_(2)` respectively with the horizontal, such that `u_(1) cos theta_(1) = u_(2) cos theta_(2)`. The path followed by one, as seen by the other (as long as both are in flight), isA. A horizontal straight lineB. A vertical straight lineC. A parabolaD. A straight line making an angle `\|theta_(1)-theta_(2)\|` with the horizontal.
Answer» Correct Answer - B

Discussion

1496.	At time `t=0`, a particle is at `(-1m, 2m)` and at `t=2s` it is at `(-4m,6m)`. From this we can conclude that in the given time interval.A. particle may be accelerateB. particle may be acceleratedC. average speed of the particle is `2.5m//s`D. average velocity of the particle is `2.5m//s`
Answer» Correct Answer - B::D

Discussion

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