1496 + Interview Questions in KINEMATICS IN GENERAL AWARENESS Page 27 InterviewSolution

1301.	A body moving with uniform acceleration a stratght line describes `25 m` in the fifth second and `33 m`in the seventh second. Find its initial velocity an dacceleration.
Answer» `29 =u+(a)/(2)(2xx5-1)` and `33=u+(a)/(2) (2xx7-1)` Solving, we get `u=7 m s^(-1), a=4 ms^(-2)`.

Discussion

1302.	A body covera `10 m` in the seconds second and `25 m` in finfth second of its motion. If the motion is uniformly accelerated, how farwill it go in the the seventh second?
Answer» `10=u+(a)/(2)(2xx2-1)` and `25=u+(a)/(2) (2xx5-1)` Solving, we get `u=(5)/(2)m s^(-1), a= ms^(-2)` `D_(7)=(5)/(2)+(5)/(2)[2xx7-1] =35 m`.

Discussion

1303.	A body has an initial velocity of `3 ms^-1` and has an acceleration of `1 ms^-2` normal to the direction of the initial velocity. Then its velocity `4 s` after the start is.A. `7 m s^-1` along the direction of initial velocity.B. `7 m s^-1` along the normal to the direction of initial velocity.C. `7 m s^-1` midway between the two directions.D. `5 m s^-1` at an angle `tan^-1(4//3)` with the direction of initial velocity.
Answer» Correct Answer - D (d) `Let u_x = 3 ms^-1, a_x = 0` `v_y = u_y + a_y t = 0 + 1 xx 4 = 4 ms^-1` `v = (sqrt(v_x^2 + v_y^2)) = (sqrt(3^2 + 4^2))` Angle made by the resultant velocity w.r.t. direction of initial velocity, i.e., x - axis, is `beta = tan^-1 ((v_y)/(v_x)) = tan^-1 ((4)/(3))`.

Discussion

1304.	The distance betweenn two place A and B is 180 km by road adn 120 km by air. An aeroplane takes 20 min to go from A to B whereas a deluxe bus strarting at A takes 5h to reach B. Find the average speed and average velocity by both means of transport)(Ignore the vertical ascent of aeroplane)
Answer» (i) Calculate the total time and total distance to find the average speed . (ii) `36 km h^(-1), 360 km h^(-1)` `24 km h^(-1), 360 km h^(-1)`

Discussion

1305.	A projectile is moving at `60 m//s` at its highest point, where it breaks into two equal parts due to an internal explosion. One part moves vertically up at `50m//s` with respect to the ground. The other part will move atA. `110m//s`B. `120 m//s`C. `130m//s`D. `10 sqrt(6) m//s`
Answer» Correct Answer - C

Discussion

1306.	Block A is placed on block B, whose mass is greater than that of A. There is friction between the blocks, while the ground is smooth. A horizontal force P, increasing linearly with time, begins to act on A. The accelerations a1 and a2 of A and B respectively are plotted against time (t). Choose the correct graph.
Answer» Correct answer is: (C) The two blocks will move together with the same acceleration as long as the force of friction between them is less than the limiting friction, as the only force on the lower block B is the force of friction. Once limiting friction is reached, the acceleration of B becomes constant (= F_lim / mass of B), and the acceleration of A continues to increase at a faster rate.

1306.

Block A is placed on block B, whose mass is greater than that of A. There is friction between the blocks, while the ground is smooth. A horizontal force P, increasing linearly with time, begins to act on A. The accelerations a1 and a2 of A and B respectively are plotted against time (t). Choose the correct graph.

Answer»

Correct answer is: (C)

The two blocks will move together with the same acceleration as long as the force of friction between them is less than the limiting friction, as the only force on the lower block B is the force of friction. Once limiting friction is reached, the acceleration of B becomes constant (= F_lim / mass of B), and the acceleration of A continues to increase at a faster rate.

Discussion

1307.	A variable force F acts on a body which is free to move. The displacement of the body is proportional to t 3, where t = time. The power delivered by F to the body will be proportional to (a) t (b) t2 (c) t3 (d) t4
Answer» Correct Answer is: (c) t³ Displacement = s = kt³ where k = constant. v = s = 3kt² a = v = 6kt F = ma = 6mkt Power = Fv = 18mk²t³ ∝ t³.

1307.

A variable force F acts on a body which is free to move. The displacement of the body is proportional to t 3, where t = time. The power delivered by F to the body will be proportional to (a) t (b) t2 (c) t3 (d) t4

Answer»

Correct Answer is: (c) t³

Displacement = s = kt³ where k = constant.

v = s = 3kt² a = v = 6kt F = ma = 6mkt

Power = Fv = 18mk²t³ ∝ t³.

Discussion

1308.	A cannot shell is fired to hit a target at a horizontal distance R. However, it breaks into two equal parts at its highest point. One part (A) returns to the cannon. The other partA. will fall at a distance of R beyond the targetB. will fall at a distance of 3R beyond the targetC. will hit the targetD. have nine times the kinetic energy of A
Answer» Correct Answer - A::D

Discussion

1309.	Two blocks A and B are placed on a table and joined by a string. The limiting friction for both blocks is F. The tension in the string is T. The forces of friction acting on the blocks are FA and FB. An external horizontal force P = 3F/2 acts on A, directed away from B.(a) FA = FB = T = 3F/4 (b) FA = F/2, FB = F, T = F (c) FA = FB = 3F/4, T = 0 (d) FA = F, FB = T = F/2
Answer» Correct Answer is: (d) F_A = F, F_B = T = F/2 Tension will appear in the string only when its length tends to change, i.e., when A tends to move. This can occur only when F_A reaches its limiting value.

1309.

Two blocks A and B are placed on a table and joined by a string. The limiting friction for both blocks is F. The tension in the string is T. The forces of friction acting on the blocks are FA and FB. An external horizontal force P = 3F/2 acts on A, directed away from B.(a) FA = FB = T = 3F/4 (b) FA = F/2, FB = F, T = F (c) FA = FB = 3F/4, T = 0 (d) FA = F, FB = T = F/2

Answer»

Correct Answer is: (d) F_A = F, F_B = T = F/2

Tension will appear in the string only when its length tends to change, i.e., when A tends to move. This can occur only when F_A reaches its limiting value.

Discussion

1310.	A train of length 200 m switches on its headlight when it starts moving with acceleration 0.5 m/s2. Some time later, its tail light is switched on. An observer on the ground notices that the two events occur at the same place. The time interval between the two events is(a) 10√2s (b) 20 s (c) 20√2 s (d) 40 s
Answer» Correct Answer is: (c) 20√2 s The train has covered a distance of 200 m in the time interval. 200 m = 0 + 1/2 (0.5 m/s²)t². t = 20√2 s.

1310.

A train of length 200 m switches on its headlight when it starts moving with acceleration 0.5 m/s2. Some time later, its tail light is switched on. An observer on the ground notices that the two events occur at the same place. The time interval between the two events is(a) 10√2s (b) 20 s (c) 20√2 s (d) 40 s

Answer»

Correct Answer is: (c) 20√2 s

The train has covered a distance of 200 m in the time interval.

200 m = 0 + 1/2 (0.5 m/s²)t².

t = 20√2 s.

Discussion

1311.	A cannon shell lands 2 km away from the cannon. A second shell, fired identically, breaks into two equal parts at the highest point. One part falls vertically. How far from the cannon will the other land? (a) 2 km (b) 3 km (c) 4 km (d) 5 km
Answer» Correct Answer is: (b) 3 km Let v = horizontal component of velocity of shell, and T = total time of flight ∴ vT = 2 km. When the shell breaks, the part which falls vertically does not have any horizontal velocity, and hence the other part acquires horizontal velocity 2v, by conservation of momentum. Also, its remaining time of flight will be T/2. It will travel a further distance (2v) (T/2) = vT = 2 km. ∴ it will land 3 km away from the cannon.

1311.

A cannon shell lands 2 km away from the cannon. A second shell, fired identically, breaks into two equal parts at the highest point. One part falls vertically. How far from the cannon will the other land? (a) 2 km (b) 3 km (c) 4 km (d) 5 km

Answer»

Correct Answer is: (b) 3 km

Let v = horizontal component of velocity of shell,

and T = total time of flight

∴ vT = 2 km.

When the shell breaks, the part which falls vertically does not have any horizontal velocity, and hence the other part acquires horizontal velocity 2v, by conservation of momentum. Also, its remaining time of flight will be T/2. It will travel a further distance (2v) (T/2) = vT = 2 km.

∴ it will land 3 km away from the cannon.

Discussion

1312.	STATEMENT-1: When a particle is thrown obliquely form the surfcae of the Earth, ity always moves in a parabolic path, provided the air resistance is negligible. STATEMENT-2: A projectile motion is a two dimensional motion.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 - D The particle will move in a parabolic path till acceleration (due to gravity) is constant. For this the particles should be near the surface of the Earth and air resistance should be negligible.

Discussion

1313.	A stick is thrown in the air and lands at some distance from the thrower. The centre of mass of the stick will move along a parabolic path (a) in all cases (b) only if the stick is uniform (c) only if the stick does not have any rotational motion (d) only if the centre of mass of the stick lies at some point on it and not outside it
Answer» Correct Answer is: (a) in all cases

Discussion

1314.	A person sitting in the rear end of the compartment throws a ball towards the front end. The ball follows a parabolic path. The train is moving with uniform velocity of `20ms^(-1)`. A person standing outside on the ground also observers the ball. How will the maximum heights `(h_(m))` attained and the ranges (R) seen by thrower and the outside observer compare each other?A. Same `h_m`, differect RB. same `h_m`, and RC. different `h_m`, same RD. different `h_m`, and R
Answer» Correct Answer - A (a) The motion of the train will affect only the horizontal component of the velocity of the ball. Since, vertical component is same for both observers, `h_m`, will be same, but but `R` will be different.

Discussion

1315.	STATEMENT-1: When a particle is thrown obliquely form the surfcae of the Earth, ity always moves in a parabolic path, provided the air resistance is negligible. STATEMENT-2: A projectile motion is a two dimensional motion.A. Statement-I is true, Statement-II is true, Statement-II is correct explanation for Statement-IB. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanation for Statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true.
Answer» Correct Answer - D

Discussion

1316.	The motion of an in secton at able is given as `x = 4t-2 sin t` and `y = A-1 cos t`, where x and y are in metres and t is in seconds. Find the magnitude of minimum and maximum velocities attained by the insect.
Answer» Correct Answer - [2 m/s, 6 m/s]

Discussion

1317.	A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by `(4t^3-2t)`, where t is in sec and velocity in m/s. what is the acceleration of the particle when it is 2 m from the origin?A. `28m//s^(2)`B. `22 m//s^(2)`C. `12 m//s^(2)`D. `10 m//s^(2)`
Answer» Correct Answer - A

Discussion

1318.	A particle is moving along x-axis. Its X-coordinate varies with time as, `X=2t^2+4t-6` Here, X is in metres and t in seconds. Find average velocity between the time interval `t=0 to t=2s.`
Answer» In 1-D motion, average velocity can be written as `v_(av)=(Delta s)/(Delta t)=(X_f-X_i)/(Delta t)=(X_(2 sec)-X_(0 sec))/(2-0)` `=([2(2)^2+4(2)-6]-[2(0)^2+4(0)-6])/2` `=8m//s`

Discussion

1319.	A stone is dropped from an aeroplane which is rising with acceleration `5 ms^(-2)`. If the acceleration of the stone relative to the aeroplane be `f`, then the following is (are) true :A. `f=5 ms^(-2)` downwardB. `f=5 ms^(-2)` upwardsC. `f=15 ms^(-2)` upwardD. `f=15 ms^(-2)` downward
Answer» Correct Answer - D

Discussion

1320.	`x` and `y` co-ordinates of a particle moving in `x-y` plane at some instant of time are `x=2t` and `y=4t`.Here `x` and `y` are in metre and `t` in second. Then The distance travelled by the particle in a time from `t=0` to `t=2s` is ………`m`A. `2sqrt(3)`B. `4sqrt(5)`C. `sqrt(2)`D. `3sqrt(40)`
Answer» Correct Answer - B

Discussion

1321.	A particle is projected up the inclined such that its component of velocity along the incline is `10 m//s`. Time of flight is `2` sec and maximum height above the incline is `5m`. Then velocity of projection will beA. `10 m//s`B. `10 sqrt(2) m//s`C. `5sqrt(5) m//s`D. none
Answer» Correct Answer - B Using the given data in the formulae for projection up the inclined plane. (`theta` is angle of projectile with inclined plane, `beta` is angle of inclined with horizontal) `T=2 sec =(2v sin theta)/(g cos beta)` `v cos theta=10.....(1)` `h=5 =(v sin theta)/2xx1` `v sin theta =10` from (i) and (ii) `v^(2)=200 ` `v=10sqrt(2) m//s`

Discussion

1322.	Under the action of force P, the constant acceleration of block B is `6 m//sec^(2)` up the incline. For the instant when the velocity of B is `3 m//sec` up the incline. Choose the correct option(s) A. Velocity of B relative to A is 1 m/sB. Acceleration of B relative to A is `2 m//s^(2)`C. The velocity of point C of the cable (in ground frame) is 4 m/sD. Velocity of B relative to A is 2 m/s
Answer» Correct Answer - A, B, C

Discussion

1323.	A point moves along an arc of a circle of radius R. Its velocity depends on the distance covered s as `v=asqrts`, where a is a constant. Find the angle `alpha` between the vector of the total acceleration and the vector of velocity as a function of s.A. `tan alpha=R/(2s)`B. `tan alpha=(2a)/R`C. `tan alpha=(2R)/s`D. `tan alpha=s/(2R)`
Answer» Correct Answer - B `tan alpha=v^(2)/Rxx1/(dv//dt)=(a^(2)s)/(Rav//2sqrt(s))=(2s)/R`

Discussion

1324.	A projectile is thrown with an initial velocity of `v = a hat i + b hat j`. If the range of projectile is double the maximum height reached by it. Find the ratio `(b)/(a)`.
Answer» Angle of projection, `theta = tan^-1 ((v_y)/(v_x)) = tan^-1 ((b)/(a)) :. Tan theta = (b)/(a)` …(i) From formula : `R = 4 H cot theta = 2 H` `rArr cot theta = (1)/(2) :. Tan theta = 2` `[As R = H given]`…(ii) From Eqs (i) and (ii), `b = 2a`.

Discussion

1325.	The maximum height reached by projectile is `4 m`. The horizontal range is `12 m`. The velocity of projection in `m s^-1` is (g is acceleration due to gravity)A. `5 sqrt(g//2)`B. `3 sqrt(g//2)`C. `(1)/(3) sqrt(g//2)`D. `(1)/(5) sqrt(g//2)`
Answer» Correct Answer - A (a) `tan theta = (4 H)/(R) =(4 xx 4)/(12) = (4)/(3) rArr sin theta = (4)/(5)` `H = (u^2 sin^2 theta)/(2 g) rArr u = (sqrt(2 g H))/(sin theta) = (sqrt(2 xx g xx 4))/(4//5) = 5 sqrt((g)/(2))`.

Discussion

1326.	The maximum height attain by a projectile is increased by ` 10%` by increasing its speed of projection, without changing the angle of projection. What will the percentage increase in the horizontal range.
Answer» As max. height , `H=u^(2)/ 2g) sin ^(2) theta` …(i)` Let `Delta H` be the increase in (H) when (u) changes by `Delta u` . Differentiating (i), we get `DeltaH` =(2 u Delta u sin^(2) theta)/(2g)` :. (DeltaH)/H =(2 Delta u)/u` Given, % increase in (H) is `10%, so `(Delta H) /H (10)/(1000) =0.1. Therfore, (2 Deltau)/u =0.1 As `R= (u^(2) sin2 theta )/g` Therfore, Delta R =(2 u Delta u)/g xx sin 2 theta` And `(Delta R)/R (2 Delta u)/u =0.1` :. `% increase in horizontal tange `(Delta R)/R xx 100 =0.1 xx 100 =10%`.

Discussion

1327.	In a motorcycle race, a rider A is leading another rider B by 36 m and both riders are travelling at a constant speed of 170 kph. At t=0 both starts accelerating at a constantrate. It is given that after 8s, So vertakes And at this instant speed of A is 220 kph. Find the accelerations of the two riders
Answer» Correct Answer - `[1.74 m//s^(2), 2.86 m//s^(2)]`

Discussion

1328.	A car starts moving along a line, first with acceleration a=5 `ms^(-2)` starting from rest then uniformly and finally decelerating at the same rate a, comes to rest.The total time of motion is `tau=25 s`. The average velocity during the time is equal to lt v gt =72 km/hr.How long does the partial move uniformly ?
Answer» Correct Answer - [15 sec]

Discussion

1329.	A swimmer capable of swimming with velocity `v` relative to water jumps in a flowing friver having velocity `u`. The man swims a distance `d` down stream and returns back to the priginal position. Find out the time taken in complete motion.
Answer» Tatal time =Time of swinning downstream +Time of swinming upstream Velocity of the man during swimming downstream`=v+u` Velovity ot the man during swimming sustream`=v-u` `t=t_(down)+t_(up)=t_(down)+t_(up)=(d)/(u)+(d)/(v-u)(2dv)/(v^(2)-u j^(2)`.

Discussion

1330.	A jet airplane travelling at the speed of ` 500 km h^(-1)` , ejects the burnt agses at the speed of `1100 km h^(-1)`, relative to the jet airplance. Find the speed of the burnt gases .w.r.t. a stationary onserver on earth.
Answer» Let downward motion be be taken positive and upward motion negative Speed of jet air plane ` v_j =- 5 00 km h^(-1)` Relatve velocity velocity of burnt gases .wr.t. jet airplacne is ` v_(bj) = 1100 km h^(-1)` Let ` v_(b)` be the velocit of the burnt gases w.r.t. observer on ground. Then ltbRgt 1 v_(bj) = v_b -v-j ` or 1100 = v_b - (- 500)` or ` v_b = 100 - 500 = 6p00 km h^(-1)`.

Discussion

1331.	Ball `A` is dropped from the top of a building. At the same instant ball `B` is thrown vertically upwards from the ground. When the balls collide, they are moving in opposite direction and the speed of `A` is twice the speed of `B`. At what fraction of the height of the building did the collision occurs ?A. `1//3`B. `2//3`C. `1//4`D. `2//5`
Answer» Correct Answer - B

Discussion

1332.	A football is kicked vertically upward from the ground and a student gazing out of the window sees it moving upwards past her at `5.00 m//s.` The window is `15.0 m` above the ground. Air resistance may be ignored. Take `g=10 m//s^2.` (a) How high does the football go above ground? (b) How much time does it take to go from the ground to its highest point?
Answer» Correct Answer - A::B (a) `h=15+u^2/(2g)` `=15+(5)^2/(2xx10)` `=16.25m` (b) `t=sqrt((2h)/g)=sqrt((2xx16.25)/10)=1.8s`

Discussion

1333.	A ball is thrown vertically upwards from the ground and a student gazing out of the window sees it moving upward past him at `10ms^-1.` The window is at 15 m above the ground level. The velocity of ball 3 s after it was projected from the ground is [Take `g= 10 ms^-2`]A. `10 m//s,` upB. `20 m//s,` upC. `20 ms^-1,`downD. `10 ms^-1,` down
Answer» Correct Answer - A Total height=`15+(u^2)/(2g)` `=15+((10)^2)/(2xx10)` or `h=20m` initial velocity `u=sqrt2gh` `sqrt(2xx10xx20)` `=20 m//s` Now applying `v=u+at,` we have `v=(+20)+(-10)(3)` `:.` Velocity is `10m//s,` downwards.

Discussion

1334.	A particle travels with constant speed along a circle of radius `3 m` and completes one revolution in `20 s`. Starting with the lowest point as origin, find (a) the magnitude and direction of the displacement vectors 5 s, 7.5s and 10 s later , (b) the magnitude and direction of the displacement in the 5 s interval from the fifth to the lenth second, (c) average velocity in this interval, (d) the instantaneous velocity at the begining and at the end of the interval.
Answer» Correct Answer - (a)4.2 m at `45^@` , 5.5 m at `68^@` and 6 m at `90^@` , (b)4.2 m at `135^@` , (c )`0.85 ms^(-1)` at `135^@` , (d)`0.94 ms^(-1)` at `90^@`, `0.94 ms^(-1)` at `180^@`

Discussion

1335.	A dog sees a flowerpot sail upand then back down past a window 5 ft high. If the total time the pot is in sight is 1.0 sec, find the height above the window that the pot rises. Take `g = 32 ft//s^(2)`
Answer» Correct Answer - `[1//16ft]`

Discussion

1336.	A man traversed half the distance with a velocity `v_(0)`. The remaining part of the distance was covered with velocity `V^(1)`. For half the time and with velocity `v_(2)` for the other half of the time . Find the average speed of the man over the whole time of motion. .
Answer» Time to cover the distance `AB`, `T_(AB)=(s)/(v_(0))` …(i) Let the time taken to cover the distance `BC` be`t_(BC)` Given: `(t_(AB))/(2)=(s_(1))/(v_(1))` …(ii) `S_(1)`= dismilarly, covered during first halt of time From (ii) and (iv), `s_(1)=v_(2) (t_(AB))/(2)` ..(iii) Similarly, `s_(2)=v_(2)(T_(AB))/(2)` ...(iv) From (iii) and (iv), `s_(1)+s_(2)=s=(v_(1)+v_(2))/(2) t_(BC)` Hence, `t_(BC)=(2s)/(v_(1)+v_(2))` Hence, average velocity `v_(AC)=(2s)/(t_(AB)+t_(BC))` `v_(AC)=(2s)/((s)/(v_(0))+(2s)/((v_(1)+v_(2)))) rArr v_(AC)=(2v_(0)(v_(1)+v_(2)))/((v_(1)+v_(2)+2v_(0)))`.

Discussion

1337.	A shell is fired from a point O at an angle of `60^(@)` with a speed of `40 m//s` & strikes a horizontal plane throught O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed v. If it hits the target which starts to rise vertically from A with a constant speed `9sqrt(3) m//s` at the same instant as the shell is fired, find v.
Answer» Correct Answer - `50 ms^(-1)` Range `(OA)=(u^(2) sin 2 theta)/(g)=(1600 xx sqrt(3))/(10 xx 2)=80 sqrt(3)` `h = 80 sqrt(3) xx tan 60^(@) = (10 xx 80 xx 80 xx 3)/(2 xx v^(2) cos 60^(@))` Time to strike `rArr v cos 60^(@) xx t = 80 sqrt(3)` `rArr t =(80 sqrt(3) xx2)/(v xx1) =(10 sqrt(3))/(v)` `h=9 sqrt(3) xx (160 sqrt(3))/(v) (480 xx 9)/(v) = (240^(2) - 38400)/(v^(2))` `v^(2) -1600 - 18v =0` `v =(18 pm sqrt(324)+6400)/(2)` `rArr v =50 m//s`.

Discussion

1338.	A body cover 20 m in 2 s and another 20 m in next 4s. Arrange the following steps in sequential order to find th average velocity of the body . (A) Find the displacements of the body in first 2s and next 4 s from the given data. (B) find the displacements of the body . (C) find the total time taken by the body to complete the total displacement . (D) Use the formula, average velocity = `("total displacement")/(" Total time taken")`A. ABCDB. ADCBC. DCBAD. DBCA
Answer» Correct Answer - A

Discussion

1339.	A particle is moving in a circle of radius `R`. a. What is its displacement when it covers (i) half the circle, (ii) full circle? b. What is its distance when it comers (i) half the the circle and (ii) full circle ?. .
Answer» a. Displacemect is the vector drawn from the initial position to the final position, and its magnitude is equal to the shortest distance between the initial and final positions. i. Hence, displacemect is equal to the diameter of the circle `=2R` ii. As initial and final positions are same, i. e., desplacement will be zero. Distance is the length of the path travelled by the particlae. distance tramelled in case (ii) will be equal to `2pi R`.

Discussion

1340.	A teain tramels from city `A` to city `B` with constant speed of `10 m s^(-1)` and returns back to city `A` with a constant spiid of `20 m s^(-1)`. Finde its average speed during its entire journey.
Answer» Let the distance between the two cities `A` and `B` be `x` m. Time taken by the train to travel from `A` to `B=(x)/(10)=t_(1)` (say) Time taken to come back from `B` to `A =(x)/(10)=t_(2)` (say) Average speed `=(Total distance)/(Total time) (x+x)/(t_(1)+t_(2))=(2x)/((x)/(10)+(x)/(10))=(40)/(3) m s^(-1)`

Discussion

1341.	The acceleration will be positive in . A. `(I) and (III)`B. `(I) and (IV)`C. ` (II) and (IV)`D. None of these
Answer» Correct Answer - B In I, slope is negative and its magnitude is decreasing with time. It means slope is increasing mumerically, So velocity is inreasing towards right and so acceleration is positve. In IV slope is positive and its magnitude is increasing with time, So velocity is increasing towards right, and so acceleration is positive.

Discussion

1342.	If a vector is added or subtravted from a vector, the resultant is a vector . Is this also true in case of multiplication to two vectros ?
Answer» May or may bot be true. If the multiplcation of two vectors is a scalar product or dot product, then the new phusical wuantity is called scalar. If the multiplication of two vectors is a vector product or cross product then the bew phuiscal quantity is called vector.

Discussion

1343.	The magnitude of vectros vec A, vec B` and vec C` are 12, 5 and 13 units respectively and ` vec A + vec B= vec C`, find the angle between ` vec A and vec B` .
Answer» Since, ` (12)^(2) + (5)^(2) =(13)^(2). So A^(2) + b^(2) =C^(2)`. It means the angle between `vec A` and vec B` is `90^(@)`.

Discussion

1344.	Two bodies are projected vertically upwards from one point with the same initial velocity `v_0.` The second body is projected `t_0 s` after the first. How long after will the bodies meet?
Answer» Correct Answer - B At the time of collision, `s_1=s_2` `:. v_0t-1/2g t^2=v_0(t-t_0)-1/2g(t-t_0)^2` Solving this equation we get, `t=(v_0)/g+(t_0)/2`

Discussion

1345.	A bead is free to slide down on a smooth wire rightly stretched between points `A and B` on a vertical circle of radius `10 m`. Find the time taken by the bead to reach point `B`, if the bead slides from rest from the highest point `A` on the circle. .
Answer» Correct Answer - `(2)` `AB = 2R cos theta` `AB = (1)/(2) g cos theta t^2 rArr 2 R cos theta = (1)/(2) g cos theta t^2` `2 sqrt((R )/(g)) = t rArr 2 sqrt((10)/(10)) = t = 2 s`.

Discussion

1346.	What is the name of branch of Physics related with the study of motion of particles?
Answer» The name of branch of Physics-related with the study of motion of particles Dynamics.

Discussion

1347.	What happen When only direction of velocity changes?
Answer» Acceleration perpendicular to velocity. e.g. Uniform circular motion

1347.

What happen When only direction of velocity changes?

Answer»

Acceleration perpendicular to velocity.

e.g. Uniform circular motion

Discussion

1348.	What happen when both magnitude and direction of velocity changes ?
Answer» Acceleration has two components one is perpendicular to velocity and another parallel or antiparallel to velocity. e.g. Projectile motion

1348.

What happen when both magnitude and direction of velocity changes ?

Answer»

Acceleration has two components one is perpendicular to velocity and another parallel or antiparallel to velocity.

e.g. Projectile motion

Discussion

1349.	How many types of acceleration ?
Answer» Types of acceleration : (i) Uniform acceleration (ii) Non-uniform acceleration (iii) Average acceleration (iv) Instantaneous acceleration

1349.

How many types of acceleration ?

Answer»

Types of acceleration :

(i) Uniform acceleration

(ii) Non-uniform acceleration

(iii) Average acceleration

(iv) Instantaneous acceleration

Discussion

1350.	The initial velocity of a body moving along a straight lines is `7m//s`. It has a uniform acceleration of `4 m//s^(2)` the distance covered by the body in the `5^(th)` second of its motion is-A. `25 m`B. `35 m`C. `50 m`D. `85 m`
Answer» Correct Answer - A `S=7+4/2(2xx5-1)=25 m`

Discussion

Explore topic-wise InterviewSolutions in .

A body moving with uniform acceleration a stratght line describes `25 m` in the fifth second and `33 m`in the seventh second. Find its initial velocity an dacceleration.

A body covera `10 m` in the seconds second and `25 m` in finfth second of its motion. If the motion is uniformly accelerated, how farwill it go in the the seventh second?

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A particle is projected up the inclined such that its component of velocity along the incline is `10 m//s`. Time of flight is `2` sec and maximum height above the incline is `5m`. Then velocity of projection will beA. `10 m//s`B. `10 sqrt(2) m//s`C. `5sqrt(5) m//s`D. none

A projectile is thrown with an initial velocity of `v = a hat i + b hat j`. If the range of projectile is double the maximum height reached by it. Find the ratio `(b)/(a)`.

The maximum height reached by projectile is `4 m`. The horizontal range is `12 m`. The velocity of projection in `m s^-1` is (g is acceleration due to gravity)A. `5 sqrt(g//2)`B. `3 sqrt(g//2)`C. `(1)/(3) sqrt(g//2)`D. `(1)/(5) sqrt(g//2)`

The maximum height attain by a projectile is increased by ` 10%` by increasing its speed of projection, without changing the angle of projection. What will the percentage increase in the horizontal range.

A swimmer capable of swimming with velocity `v` relative to water jumps in a flowing friver having velocity `u`. The man swims a distance `d` down stream and returns back to the priginal position. Find out the time taken in complete motion.

A jet airplane travelling at the speed of ` 500 km h^(-1)` , ejects the burnt agses at the speed of `1100 km h^(-1)`, relative to the jet airplance. Find the speed of the burnt gases .w.r.t. a stationary onserver on earth.

A dog sees a flowerpot sail upand then back down past a window 5 ft high. If the total time the pot is in sight is 1.0 sec, find the height above the window that the pot rises. Take `g = 32 ft//s^(2)`

A man traversed half the distance with a velocity `v_(0)`. The remaining part of the distance was covered with velocity `V^(1)`. For half the time and with velocity `v_(2)` for the other half of the time . Find the average speed of the man over the whole time of motion. .

A particle is moving in a circle of radius `R`. a. What is its displacement when it covers (i) half the circle, (ii) full circle? b. What is its distance when it comers (i) half the the circle and (ii) full circle ?. .

A teain tramels from city `A` to city `B` with constant speed of `10 m s^(-1)` and returns back to city `A` with a constant spiid of `20 m s^(-1)`. Finde its average speed during its entire journey.

The acceleration will be positive in . A. `(I) and (III)`B. `(I) and (IV)`C. ` (II) and (IV)`D. None of these

If a vector is added or subtravted from a vector, the resultant is a vector . Is this also true in case of multiplication to two vectros ?

The magnitude of vectros vec A, vec B` and vec C` are 12, 5 and 13 units respectively and ` vec A + vec B= vec C`, find the angle between ` vec A and vec B` .

Two bodies are projected vertically upwards from one point with the same initial velocity `v_0.` The second body is projected `t_0 s` after the first. How long after will the bodies meet?

A bead is free to slide down on a smooth wire rightly stretched between points `A and B` on a vertical circle of radius `10 m`. Find the time taken by the bead to reach point `B`, if the bead slides from rest from the highest point `A` on the circle. .

What is the name of branch of Physics related with the study of motion of particles?

What happen When only direction of velocity changes?

What happen when both magnitude and direction of velocity changes ?

How many types of acceleration ?

The initial velocity of a body moving along a straight lines is `7m//s`. It has a uniform acceleration of `4 m//s^(2)` the distance covered by the body in the `5^(th)` second of its motion is-A. `25 m`B. `35 m`C. `50 m`D. `85 m`