InterviewSolution
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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.
| 1. |
A particle has a velocity u towards east at `t=0`. Its acceleration is towards west and is constant. Let `x_A` and `x_B` be the magnitude of displacements in the first 10 seconds and the next 10 secondsA. `x_Altx_B`B. `x_A=x_B`C. `x_Agtx_B`D. the information is insufficient to decide the relation of `x_A with x_B`. |
| Answer» Correct Answer - D | |
| 2. |
A man at a speed of 6 km/hr for 1 km and 8 km/hr for the next 1 km. What is t his average speed for the walk of 2 km? |
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Answer» Distance traveled is 2 km Time taken ` = (1km)/(6km/hr)+(1km)/(8km/hr)` `=(1/6+1/8)hr = 7/24 hr` `Averge speed = (2kmx24)/(7hr)=48/7 km/hr` =7 km/hr` |
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| 3. |
A particle moves along the X-axis as `x=u(t-2s)=at(t-2)^2`.A. the initial velocity of the particle is uB. the acceleration of the particle is aC. the acceleration of the particle is 2aD. at t=2 s particle is at the origin. |
| Answer» Correct Answer - C::D | |
| 4. |
A particle starting from rest moves with constant acceleration. If it takes 5.0 s to reash the speed 18.0 km/h find a. the average velocity during this period, and b. the distance travelled by the particle during this period. |
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Answer» Correct Answer - A::B Given, `u=0 v=18km/hr =5m/s t=5 sec a=v-ut =5-0xx5 =5m/s^2 S=ut+1/2at^2ltbrge1/2xx1xx(5xx5) =12.5m a. Average velocity V_(ve)=(12.5)/5 =2.5 m/s b. Distance travelled = 12.5 m`. |
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| 5. |
An athelete takes 2.0 s to reach his maximum speed of 18.0 km/h. What is the magnitude of his average acceleration? |
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Answer» Correct Answer - B Initial velocity u=0 (Since starts from rest) FiN/Al velocity v=18km/hr=5sec (i.e. maximum velocity) Time inteval t=2 sec. `:. Acceleration = a_(av)=((v-u))/2 =5/2= 2.5m/s^2` |
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| 6. |
A ball is projected vertically upward with speed of 50 m/s. Find a. the maximum height, b. the time to reach themaximum height, c. the speed at half the maximum height. Take g=`10 m/s^2`. |
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Answer» Correct Answer - B::C Given, `u=50 m/s g=-10 m/s^2` when moving upward ` v=0` t highest point a. `S=(v^2-u^2)/(2a) =((0.50)^2)/)2(-10))=125m ` maximum height reached = 125m b. `t= (v-u)/a=(0-50)/(-10)=5 sec c. S= 125/2= 62.5 m v=? u= 50 m/s a = -10 m/s^2 From v^2-u^2=2aS v=sqrt((u^2+2as)) =sqrt((50)^2+2(-10)(62.5)) =sqrt((2500-1250)) sqrt(1250)=35 m/s`. |
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| 7. |
Six particles situated at the corners of a regular hexagon of side a move at a constant speed v. Each particle maintains a direction towards the particle at the next corner. Calculate the time the particles will take to meet each other. |
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Answer» Correct Answer - A::B A approaches B, B approaches C and so on. Relative velocity between A and E, `vecV_(AB)= vecV_A-vecV_B=vecV-vecV cos 60^0` `=vecV-vecV/2= vecV/2` ` Time, t= (Displacement) /(Velocity) ` `= a/(vecV/2)= (2a)/vecV`. |
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| 8. |
A car travelling ast 60 km/h overtakes another car travellign at 42 km/h. Assuming each car to be 5.0 m long, find the time taken during the overtake aned the total road distance used for the overtake. |
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Answer» Correct Answer - A `V_1=60` km/hr=16.6 m/s `V_2`= 42 km/h=11.6m/s Before crossing` `Relative velocity between te cars `=(16.6-11.6)=5 m/s` `By first car =`5+5=10 m Time t=S/v=10/5 sec =2 sec to cross the 2nd car`.` He also covered distasnce which is own length equal its 5m`. `:.` Total road distance used for the overtake =33.2+5=38 m. |
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| 9. |
A person travelling on a straight line moves with a uniform velocity `v_1 for a distance x and with a uniform velocity `v_2` for the next equal distance. The average velocity v is given byA. `v=(v_1+v_2)/2`B. `v=sqrt(v_1v_2)`C. `2/v=1/v_1+1/v_2`D. `1/v=1/v_1+1/v_2` |
| Answer» Correct Answer - C | |
| 10. |
A person travelling at 43.2 km/h applies the brake giving a deceleration of `6.0 m/ss^2` to his scooter. How far will it travel before stopping? |
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Answer» Correct Answer - A::B Initial velocity `u=43.2 km/hr= 12 m/s u=12 m/s v=0 a=6m/s^2 (Deceleration) Distance S= (v^2-u^2)/(2(-a)) =(0-(12)^2)/(2)-6)) =(12xx12)/12=12m` |
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| 11. |
When a person leaves his home for sightseeing by his car, the meter reads 12352 km. When he returns home after two hours the reding is 12416 km. a. What is the average speed of the car during this period? B. What is the averge velocity? |
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Answer» Correct Answer - A::B::C a. Total distance covered `=12416-12352 = 64 km, in 2 hours ` :. Speed = 64/2=32km/h` b. As he returns to his house, the displacement is zero., `:. Velocity =(Displacement/Time) =0` |
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| 12. |
A police is chasing a culprit going n a motorbike. The motorbike crosses a turning at a speed of 72 km/h. The jeep follows it at a speed of 90 km/h, crossing the turning tenseconds latert than the bike. Assuming that they tavel at constant speeds, how far from the turning will the jeep catch up with the bike? |
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Answer» Correct Answer - A::B::C::D `V_p=0` km/h=25 m/s `V_e`= 72 km/h=20 m/s In 10 sec culprit reaches at point B from A. Distance covered by culprit `S=vt=20xx10=200m` At time t=10 sec the police jeepis 200 m behind the Culprit. Relative velocity between jeep and culprit `25-20`=`5m/s` Time =S/v= 200/5=40s (Relative velocity is considered).` In 40 s the police jeep will move from A to a distance S where ` S=vt=25xx40 = 1000 m = 1.0 km` away `:.` The jeep will catch up with the bike, 1 km far from the turning. |
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| 13. |
A ball is dropped from a ballon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take in reaching the ground? |
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Answer» Correct Answer - A::B::C Initially the ball is going upward `u=-7 m/s S=60 a= g= 10 m/s^2 From S=ut+1/2 at^2 60=-7t+1/2 10 t^2 rarr 60=-7t+5t^2 rarr5t^2-7t-60= :. T=(7+-(sqrt(49+1200)))/(2xx5) =(7+-(sqrt(49+1200)))/1 =(-7+-35.34)/10 Taking +ve sign t= (7+35.34)/10=4.2 sec` Therefore, the ball will take 4.2 sec, to reach the ground. |
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| 14. |
Mark the correct statements for a particle going on a straight line:A. If the velocity and acceleration have opposite sign, the object is slowing down.B. If the position andvelocity have opposite sign, the particle is moving towards the origin.C. If the velocity is ero at an instant, the acceleration should also be zero at that instasnt.D. If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval. |
| Answer» Correct Answer - A::B::D | |
| 15. |
A person standing near the edge of the top of a building throws two balls A and B. The ball A is thrown vertically upward and B is thrown vertically downward with the same speed. The ball A hits the ground with speed `v_A` and the ball B hits the ground wiht a speed `v_B`. We haveA. `v_Altv_B`B. `v_Altv_B`C. `v_A=v_B`D. the relation between `v_A and v_B` depends on height of the building above the ground. |
| Answer» Correct Answer - C | |
| 16. |
A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the positions of the 3rd, 4th and 5th ball when the 6th ball is being dropped. |
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Answer» Correct Answer - A::B::C::D For every ball `u=0, a=g=9.8 m/s^2 `4th ball moves for 2 sec. 5th ball for 1 sec and 3rd ball for 3 sec when 6th ball is being dropped. For 3rd ball `t=3 sec :. S_1 = ut+ 1/2 atI^2ltbrge0+1/2 9.8(3)^2 =44.1 m` below the top. For 4th ball t=2 sec ` :. S_2 = 0+1/2 g^2=1/2xx9.8 xx(2)^2 =19.6`m below the top [u=0] for 5th ball t=1 sec `:. S_3 = ut+ 1/2 at^2 =0+ 1/2 xx 9.8x t^2 =4.9` m below the top. |
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| 17. |
The accleration of a particle as seen from two frames `S_(1)` and `S_(2)` have equal magnitudes 4 `m//s^(2)`A. The frames must be at rest with respect to each other.B. The frames may be moves with respect to each other but neigther should be accelerates with respect to the other.C. The acceleration of `S_2` with respect to `S_1` may either be zero or `8 m/s^2.D. The acceleration of `S_2` with respect to `S_1` may be anything between zero and `8 m/s_2` |
| Answer» Correct Answer - D | |
| 18. |
A ball is thrown up at a speed of 4.0 m/s. Find the maximum height reached by the ball. Take `g=10 m//s^2`. |
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Answer» Let us take vertically upward direction as the positive Y-axis. We have `u=4.0 m/s and a=-10m/s^2`. At the highest point the velocity becomes zero.Using the formula. `v^2=u^2+2ay`, `0=(4.0 m/s)^2+2(-10 m/s^2)y` ltbrlt or, `y=(16 m^2/s^2)/(30 m/s^2)=0.80 m`. |
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| 19. |
Figure shows a 11.7 ft wide ditch with the approach roads at and angle of `15^0` with the horizontal. With what minimum speed should a mororbike be moving on the road so that it safely croses the ditch? Assume that the length of thebike is 5 ft, and it leaves the road when the front part runs out of the approch road. |
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Answer» Correct Answer - A Given Horizontal range, `= 11.7+5` = 16.7 ft covered by the bike `g= 9.8m/s = 32.2 ft/s^2 Now, R= (u^2 sin2alpha)/g `u^2`= (Rg)/(sin2alpha)= (16.7xx32.2)/(1/2) ` u= 32 ft/s` |
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| 20. |
A ball is thrown from a field with a speed of 12.0 m/s at an angle of `45^0` with the horizontal. At what distance will it hit the field again ? Take `g=10.0 m/s^2` |
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Answer» The horizontal range `= (u^2sin 2 theta)/g` `=((12 m/s)^2xxsin(2xx45^0))` `=(144 m^2/s^2)/(10.0m/s^2) = 14.4 m. Thus, the ball hits the field at 14.4 m from the point of projection. |
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| 21. |
A healthy youngman standing at a distance of 7 m from 11.8 m high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms hieght (1.8m)? |
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Answer» Correct Answer - A::B::D At point B (i.e. over 1.8 m from ground), the kid should be caught. For kid: Initial velocity `u=0` Acceleration = 9.8 m/s^2 Distance S= 11.8-1.8=10 m No S= ut+ 1/2 `at^2` 10=0+ 1/2 (9.8)`t^2` ` t^2`= 10/4.9=2.04 t= 1.42 `` In this time the man has to reach at the bottom of the building `:. Velocity =S/t=7/1.42= 4.9 m/s` |
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| 22. |
A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he ils driving a car at a speed of 54 km/h and the brakes cause a deceleration of `6.0 m/s^2`, find the distance travelled by the car after he sees the need to put the brakes on. |
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Answer» Correct Answer - B In reactiion time the body moves with the speed of 54 km/hr = 15m/sec (constant speed). Distance travelled in this time, `S_1=15xx0.2 =3m When breades are applied u=15 m/s v=0 a=-6m/s^2(deceleration) S_2=(v^2-u^2)/(2a) =(0-15^2)/(2(-6)) =18.75m Total distance = S_1+S_2 =3-18.75+21.75m = 22m`. |
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| 23. |
A football ils kicked with a velocity of 20 m/s at an angle of `45^0` wilth the horizontal. A. Find the time taken by the ball to strike the ground. b.Find the maximum height it reaches. C. How far away from the kick dows it hit the ground? Take `g=10 m/s^2`. |
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Answer» a. Take the origin at the point where theball is kicked, vertically upward as the Y-axis and the horizontl iln the plane of motion as the X-axis. The initial velocity has the components `u_x=(20 m/s)cos45^0=10 sqrt2m/s` and `u_y=(20 m/s)sin45^0=10 sqrt2 m/s` When the ball reaches the ground, y=0. Using `y=u_yt-1/2 gt^2, `0=(10sqrt2m/s)t-1/2xx(10 m/s^2)xxt^2` or, `t=2sqrt 2s=2.8 s.` Thus, it takes 2.8s for the football to fall on the ground. a. At the highest point `v_y ^2=u_y^2-2gy,` `=0(10sqrt2m/s)^2-2xx(10 m/s^2)H` `or H=10 m` Thus, the maximum height reached is 10 m. c. The horizontal distance travelled before falling to the ground is `x=u_xt` `=(10 sqrt2 m/s)(2sqrt2s)=40m` |
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| 24. |
A bullet travellling with a velocity of 16 m/s penetrates a tree trunk and comes to rest in 0.4 m. Find the time taken during the retardation. |
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Answer» `u=16m/s (initial) v=0 S=0.4m Deceleration a= (v^2-u^2)/(2S) (0-16^2)/(2xx0.4) =320m/s^2 time= (v-u)/a= (0-16).(-320) =0.05 sec` |
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| 25. |
A particle starts from rest with a constant aceleration. At a time t secon, the speed is found tobe 100 m/s and one second later the speed becomes 150 m/s. Find a. the acceleratiuon and b. the distance travelled during the `(t+1)^th` second. |
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Answer» a. Velocity at time t is `100 m/s=a.(t second)`…………i and velocity at time (t+1) second `150 m/s=a.(t+1) ………..2 Subtracting 1 from 2, `a=50 m/s^2` b. consider the interval t second to `(t+1)` second, time elapsed = 1s initial velocity = 100 m/s fiN/Al velocity = 150 m/s Thus `(150 m/s)^2=(100 m/s)^2+2(50+m/s^2)x` or, `x=125m`. |
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| 26. |
Two projectiles A and B are projected with angle of projection B. If `R_A and R_B` be the horizontal range for the two projectile then.A. `R_AltR_B`B. `R_A=R_B`C. `R_AgtR_B`D. the information is insufficient to decide the relastion of `R_A with R_B` |
| Answer» Correct Answer - D | |
| 27. |
An elevator is descending with uniform acceleration.To measure the acceleration, a person in the elevator drops a coin at momen the elevator strts. The coin is 6 ft asbove the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate these dta the acceleration of the elevator. |
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Answer» Correct Answer - B::D For elevator and coin u=0 as the elevator descends down ward with acceleratiion a (say) the coin has to move more distance than 1.8 cm to strike the floor. Time takes t=1 sec `S_e=ut+ 1/2 at^2 =0+ 1/2 g(1)^2=1/2 g`, S_3 = ut+ 1/2 a`t^2` = 0+ 1/2 a`(1)^2`= 1/2 a `` Total distance covered by coin given by `1.8+1/2 a= 1/2 g rarr 1.8 + a/2= 9.8/2=4.9 rarr a/2= 4.9=1.8=3.1m rarr a= 6.2 m/s^2 = 6.2xx3.28=20.34 ft/s^2`. |
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