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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.
| 1351. |
Two particles start from rest simultaneously and are equally accelerated throughout the motion, the relative velocity ofone with respect to other is :A. ZeroB. Non zero and directed parallel to accelerationC. Non zero and directed opposite to accelerationD. Directed perpendicular to the acceleration |
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Answer» Correct Answer - A |
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| 1352. |
If ` ve A = 3 hat I + 4 hat j` and ` vec B = 7 hat I + 24 hat j`. Find vector haveng the same magnitude as ` vec B` and parallel to ` vec A` . |
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Answer» The magnitude of `vec A` is, ` A = sqrt( 3^2 + 4^2 ) = 5` : A unit vector parallel to `vec A` will be ` =hat A= hatA//A = ( 3 hat I + 4 hat j) //5` ltbRgt Now magnitude of ` vec B` is. ` B= sqrt ( 7^2 + ( 24) ^2) = 25` . Therefore a vector parallel to ` vec A` and haveing magnitude of ` vec B` will be ` = 25 hat A = 25 ( 3hat i + 4 hat j) //5 = ( 15 hat i + 20 hat j)`. |
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| 1353. |
Two particles , one with constant velocity ` 50 m//s` and the other with uniform acceleration ` 10 m//s^(-2)`. Start moving simultaneously from the same place in the same direction. They will be at a distance of `125m` other after.A. (a) ` 5 sec`B. (b) ` 5 (1 + sqrt 2) sec`C. (c ) 10 sec`D. (d) ` 10 (sqrt 2 + 1) sec` |
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Answer» Correct Answer - C For particle `1` moving with constant velocity , ` S_1 = 50 t`. For particle `2` moving with constant acceleration, ` S_2 = 1/2 at^2 = 1/2 xx 10t^2 = 5t^2 , As per question , ltbRgt ` S-2 - S_1 = 12 5` or ` 5 t^2 - 50 t= 125` or ` t^2 - 10 t- 25 =0` On solving we ger,` t= (5 +-5 sqrt 2 ) sec. ` t=(5+5 sqrt 2) sec`. |
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| 1354. |
A particle is moving in a circular orbit when a constant tangential acceleration. After `2s` from the beginning of motion, angle between the total acceleration vector and the radius E becomes `45^(@)`. What is the angular acceleration of the particle? |
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Answer» In the adjoining figure are shown the total acceleration vector `vec(a)` and its components the tangential accelerations `vec(a)_(tau)` and normal acceleration `vec(a)_(n)` are shown. These two components are always mutually perpendicular to each other and act along the tangent to the circle and radius respectively. Therefore, if the total acceleration vector makes an angle of `45^(@)` with the radius, both the tangential and the normal components must be equal in magnitudr. Now from equation and, we have `a_(tau)=a_(n) rarr alpha R=omega^(2)R rArr alpha=omega^(2)` ...(i) Since angular acceleration is uniform, form equation, we have `omega=omega_(o)+alphat` Substituting `omega_(0)=0` and `t=2 s`, we have `omega=2alpha` ...(ii) From equation (i) and (ii), we have `alpha=0.25 rad//s^(2)` |
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| 1355. |
A particle is moving on a circular path of radius `1.5 m` at a constant angular acceleration of `2 rad//s^(2)`. At the instant `t=0`, angular speed is `60//pi` rpm. What are its angular speed, angular displacement, linear velocity, tangential acceleration and normal acceleration and normal acceleration at the instant `t=2 s`. |
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Answer» Initial angular speed is given in rpm (revolution per minute). It is expressed in `rad//s` as `1 rp m=(2pi rad)/(60 s)` `omega_(o)=(60/pi)xx(2pi rad)/(60 s)=2 rad//s` At the instant `t=2s`, angular speed `omega_(2)` and angular displacement `theta_(2)` are calculated by using eq. `omega_(2)=omega_(0)+alphat` Substituting values `omega_(0)=2 rad//s, alpha=2 rad//s^(2), t=2 s`, we have `omega_(2)=6 rad//s` Substituting values `theta_(o)=0` rad, `omega_(o)=2 rad//s, omega=6 rad//s` `t=2 s`, we have `theta_(2)=8 rad` Linear velocity at `t=2 s`, can be calculated by using eq. `v_(2)=r omega_(2)` Substituting `r=1.5 m` and `omega_(2)=6 rad//s`, we have `v_(2)=9 m//s` Tangential acceleration `a_(tau)` and normal acceleration `a_(n)` can be calculate by using eq. and respectively. `a_(tau)=ralpha` Substituting `r=1.5 m` and `alpha=2 rad//s^(2)`, we have `a_(tau)=3 m//s^(2)` `a_(n)=omega^(2)r` substituting `omega_(2)=6 rad//s` and `r=1.5 m`, we have `a_(n)=54 m//s^(2)` |
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| 1356. |
Calculate the magnitude of linear acceleration of a particle moving in a circle of radius ` 0.5 m` at the instant when its angular velocity is ` 2/5 rads^(-1)` and its angular acceleration is `6 rad s^(-2)`. |
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Answer» Here, `r-0.5 m, xx 6=3.0 ms^(-1)` `alpha =6 rad s^(-2)` Tangential acceleration, `a_(T) =ralpha =0.5 xx 6= 3. 0 ms^(-2)` Centripetal acceleration, Magintude of total linear accr.rtation, `a = sqrt (a_(T)^(2) + a_(C)^(2) = sqrt ((3.0) ^(2) + (3.125 )^(2))` `=4. 33 ms^(-2)`. |
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| 1357. |
Particles `A and B` move with constant and equal speeds in a circle as shown in (Fig. 5.164). Find the angular velocity of the particle A with respect to `B`, if the angular velocity of particle `A w.r.t. O` is `omega`. . |
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Answer» Angular velocity of `A` with respect to `O` is `omega_(A O) = ((v_(A O))_|_)/(r_(A O))=(v)/(r) = omega` Now, `omega_(A B) = ((v_(A B))_|_)/(r_(A B)) rArr v_(A B) = 2 v` Since `v_(A B)` is perpendicular to `r_(A B) rArr (v_(A B)) _|_ = v_(A B) = 2 v , r_(A B) = 2 r` `rArr omega_(A B) = ((v_(A B))_|_)/(r_(A B)) = (2 v)/(2 r) = omega`. |
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| 1358. |
Study the following graph: The particle has negative acceletation.A. In graph (i)B. In graph (ii)C. In graph (iii)D. In graph (iv) |
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Answer» Correct Answer - c c. for the graph (iii) the partciles`s velocity first dereases and then increases in negative direction. It means neative acceleration is involved in this motion. |
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| 1359. |
Form a lift moving upwards with a uniform acceleration `a=2 m s^(-2), man throus a ball vertically upwards with a velcoity `v=`12 m s^(-1)` relative to the lift. The ball comes back to the man after a time `t`. Find the value of `t` in seconds. |
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Answer» Correct Answer - 2 Taking upward direction as positive, let us workin the frome of lift. Acceleration of ball relative to lift `=(g+a)` downwards, so `a_(real) =- (g+a)`, initial velocity: `u_(rel) =v`, final velcity : `v_(rel) =v`, as the ball will reach the man with same speed w.r.t. lift Apply `v_(rel) =u_(rel)+ a_(rel)t rArr - v+ (g-a)t rArr t =2 s` . |
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| 1360. |
Velocity of the river with respect to ground is given by `v_0.` Width of the river is d. A swimmer swims (with respect to water) perpendicular to the current with acceleration `a = 2t` (where t is time) starting from rest from the origin O at `t= 0.` The equation of trajectory of the path followed by the swimmer is A. `y= (x^3)/(3 v_0^3)`B. `y= (x^2)/(2 v_0^2)`C. `y= x/v_0`D. `y= sqrt(x/v_0)` |
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Answer» Correct Answer - A `(dv_Y)/(dt)=2t` `:. v_y=t^2 or (dy)/(dt)=t^2` or `y=t^3/3` and `x=v_0t rArr t=x/v` Substituting in Eq. (i) we have, `y=x^3/(3v_0^3)` |
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| 1361. |
What is the angle made by vector, ` vec A =2 hat I + 2 hat j` with x-axis ? |
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Answer» Here, `A_(x) =2, A_(y) =2 . Let ` theta` be the angle whith the ` vec A` make with x-axis then ` tan theta =A)_(y) //A_(x) =2//2 =1 or theta = 45^(@)`. |
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| 1362. |
A golfer standing on the ground hits a ball with a velocity of `52 m//s` at an angle `theta` above the horizontal if `tan theta=5/12` find the time for which the ball is at least `15m` above the ground? `(g=10m//s^(2))` |
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Answer» Correct Answer - `(2)` `y = u sin theta t - (1)/(2) "gt"^2` `15 = 52 xx (5)/(13) t - (1)/(2) xx 10 t^2` `5t^2 + -20t + 15 = 0` `t^2 - 4t + 3 = 0` `t^2 - 3t xx t + 3 = 0 rArr t (t - 3)-1(t - 3) = 0` Thus, `t_2 - t_1 = 3 - 1 = 2 s`. |
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| 1363. |
A ball is released from rest. If it takes 1 second to cross the last 20 m before hitting the ground, find the height from which it was dropped. |
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Answer» Correct Answer - [31.25 m] |
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| 1364. |
A block of weight 9.8N is placed on a table. The table surface exerts an upward force of 10 N on the block. Assume `g = 9.8 m//s^(2)`.A. The block exerts a force of 10N on the tableB. The block exerts a force of 19.8 N on the tableC. The block exerts a force of 9.8N on the tableD. The block has an upward acceleration |
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Answer» Correct Answer - A::D |
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| 1365. |
A particle starts sliding down a frictionless inclined plane. If ` S_(n) ` is the distance travelled by it from time ` t = n-1 sec `, to `t = n sec`, the ratio ` (S_(n))/(s_(n+1))` is |
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Answer» Correct Answer - C `S_(n) = u +(a)/(2) (2n-1) rArr S_(n) = (a)/(2) (2n-1)` `rArr S_(n+1) = (a)/(2) [2(n+1)-1]` `{:(S_(n)=(a)/(2)[2n+1),,rArr,,(S_(n))/(S_(n+1))=(2n-1)/(2n+1)):}` |
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| 1366. |
Can a particle accelerate if its speed is constant ? Can it accelerate if its velocity is constant ? Explain. |
| Answer» When the particle describes a uniform cirular motion, its speed is constant but it has centripetal acceleration action along the radius directed towards the centre of the circular path. When the particle is moving with a constant velcity, there is no change in velcity with time and hence its acceleration is zero. | |
| 1367. |
In a journey, a car travels at the rate of `20 km h^(-1)` for minutes and then at `30 km h^(-1)` for `20 minutes`. Find (i) the total distance traveled by the car and (ii) the average speed of the car during the journey. |
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Answer» Here, `v_(1) =20 km h^(-1) , `t_(1) =(10)/(60) h 1/6 h. Distance travelled, `S_(1) =v_(1) v_(1) t_(1) =20 xx 1/6 =(10)/3 km` Distance travelled, `S_(2) =v_(2) t_(2) =30 xx 1/3 =10 km` (i) Total distance travelled , `S=S_(1) +s_(2) =(10)/3 + 10 =13.33 km` (ii) Average speed `=S/(t_(1) +t_(2)) =((10//3) +10)/((1//6) +(1//3))` `=26.66 km h^(-1)` |
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| 1368. |
A river is flowing from north to south at a speed of 0.3 kph. A man on the west bank of the river, capable of swimming 1 kph in still water, wants to swim across the river in the shortest time. He should in a direction:A. Due eastB. `30^(@)` north of eastC. `30^(@)` weat of northD. `60^(@)` north of east |
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Answer» Correct Answer - A |
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| 1369. |
When a projectile is projected with velocity (v) making an angle ` theta` with ground.then its velocity at the highest point is ……………………….. . |
| Answer» Correct Answer - ` v cos theta` | |
| 1370. |
A river flowing from east to west at a speed of `5 m//"min"`. A man on south bank river, capable of swimming `10 m//"min"` in still water, wants to swim across the river in shorter time, he should swim:-A. Due northB. Due north-eastC. Due north-east with double the speed of riverD. None of the above |
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Answer» Correct Answer - A For shortest time then maximum velocity is in the direction of displacement. |
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| 1371. |
The magnitude of two vectors are ` 3` and `4` units and their dot product is ` 6 units`. The angle between the vectors is. |
| Answer» Correct Answer - (b) | |
| 1372. |
A projectile is projected from a level ground making an angle `theta` with the horizontal (x direction). The vertical (y) component of its velocity changes with its x co-ordinate according to the graph shown in figure. Calculate `theta`. Take `g = 10 ms^(-2)`. |
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Answer» Correct Answer - `theta = 45^(@)` |
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| 1373. |
A projectile is projected from the ground making an angle of `30^(@)` with the horizontal.Air exerts a drag which is proportional to the velocity of the projectileA. at highest point velocity will be horizontalB. the time of ascent will be equal to the time of descentC. the time of ascent will be greater than the time of descentD. the time of descent will be greater than the time of ascent |
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Answer» Correct Answer - A::D |
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| 1374. |
Find the component of a vector ` vec A = 3 hat I + 4 hat j` along the direction of ` 2hat I -3 hat j`. |
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Answer» Here, `vec A =(3 hat I + 4 hat j ), vec B =( 2hat I - 3 hat j)` Unit vector ob ` vec B` , ` hat B (vec B) /B = (2 hai -3 hat j)/( sqrt( (2)^(2) + (-3)^(2)) =(2 hat - 3 hat j)/(sqrt 13)` Let `theta` be the anle between ` vec A and vec B`. The componet of `vec A` along the direction of ` vec B` is `=(A cos rhea ) hat B = (vec A . hat B) hat B` `=[(3 hat i + 4 hat j). ((2 hati -3 hat j)/(sqrt 13))] ((2 hat i -3 hatj )/(sqrt 13 ))` `((6 -12))/(13) (2 hat i -3 hat j) =- 6/(13) (2 hat i -3 hat j)`. |
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| 1375. |
If ` 9 3hat i-2 hat j+ 2 hat k+ 2hat k).(2 hat i-x hat j+3 hat k) =- 12`, the value of (x) is . |
| Answer» Correct Answer - (d) | |
| 1376. |
From a motorboat moving downstream with a velocity `2 m//s` with respect to river, a stone is thrown. The stone falls on an ordinary boat at the instant when the motorboat collides with the ordinary boat. The velocity of the ordinary boat with repsect to the river is equal to zero. The river flow velocity is given to be `1 m//s`. The initial velocity vector of the stone with respect to earth is :- A. `2i+20j`B. `3i+40j`C. `3i+50j`D. `2i+50j` |
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Answer» Correct Answer - C Time of collision of two boat `=20//2=10` sec. As given in question i.e. the time of flight of stone is also equal to `10` sec. so vertical component of stone initially is `50 m//s` and the horizontal component w.r.t. motorboat equals to `2 m//s`. Hence `vec(v)_(BG)=3hat(i)+50hat(j)` |
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| 1377. |
A train, travelling at `20 km//hr` is approaching a platform. A bird is sitting on a pole on the platform. When the train is at a distance of 2 km from pole, breakes are applied which produce auniform deceleration in it. At that instant the bird flies towards the train at `60 km//hr` and after touching the nearest point on the train flies back to the pole and then flies towards the train and continues repeating itself. Calculate how much distance will the bird have flown before the train stops? |
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Answer» Correct Answer - 12 km Deaceleration of train, `a=|(v^(2)-u^(2))/(2s)|=(20xx20)/(2xx2)=100 km//hr^(2)` Time to reach platform `=20/100=1/5` hr `:.` Total distance travelled by the bird`=vt=60xx1/5=12 km` |
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| 1378. |
A fighter jet plane is flying at a height of ` 500 m` with a velocity ` 450 km h^(-1)` . If release a bob when ` 500 m` away from the enemy post, Will the bomb hit post ? Take g=10 ms^(-2)`. |
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Answer» Here, ` y= 500 m`, u_x = 450 km h^(-1)` = 450 xx 5 ( 18) = 125 ms^(-1)` Let (t) be the time taken by bomb to reach the ground. ` Taking vertical vertical downward motion of the bomb, we have ` ` u_y =0 , a-y = 10 ms^(-2), y= 500 m, t=?` y= u_y t + 1/2 a_y t^2` or ` 500 =0 1/2 xx 10 xx t^2 ` or t= 10 s` ` Horizontal distance coverd by the bobm ` u_x xx t= 125 xx 10 = 1250 m lt 600 m` Thus the bomb will not hit the enemy post but will fall faar ahead of the target. |
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| 1379. |
(a) Prove that bodies starting at the same time `t = 0` from the same point, and following frictionless slopes in different directions in the same vertical plane, all lie in a circle at any subsequent time. (b) Using the above result do the following problem. A point P lies above an inclined plane of inclination angle `alpha`. P is joined to the plane at number of points by smooth wires, running in all possible directions. Small bodies (in shape of beads) are released from P along all the wires simultaneously. Which body will take least time to reach the plane. |
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Answer» Correct Answer - (b) Body travelling along a line making an angle `(alpha)/(2)` with vertical |
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| 1380. |
A particle has an initial velocity of 3i + 4j and an acceleration of 0.4i + 0.3j. Its speed after 10 s is(a) 10 units (b) 7 units (c) 7√2 units (d) 8.5 units |
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Answer» Correct answer is: (c) 7√2 units Vector u = 3i +4j ∴ ux = 3, uy = 4 Vector a = 0.4i + 0.3j ∴ ax = 0.4, ay =0.3 vx = ux + axt = 3 +0.4 x 10 =7. cy = uy + ayt = 7. v = √(vx2 + vy2). |
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| 1381. |
A food packet is released from a helicopter which is trising steadily at ` 3 ms^(-1)`. After ` 3 seconds`, (i) what is the velocity of the packet? (ii) how far is it below the helicopter ? Take ` g= 9.8 m//s^2`. |
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Answer» Taking downwad motion of fodd parcet dropped from helicopter. ` Her, u =- 3 ms^(-1) , t = 3 s, a= + 9.8 m//s(-2)` (i) ` v=u + at =- 3 + (+ 9/8 ) 3 = + 26.4 ms^(-1)` This velocity is directed downwards. (ii) Downward distance coverd by packet in ` 3` seconds ` S= ut 1/2 at^2` ` =- 3 xx3 + 1/2 xx 9.8 xx3^2 = 35 .1 m` The heilcopter rises up in `3` seconds ` = 3 ms^(-1)` xx s= 9m` therfore, the distance of the food packet from the helecopter ` =35.1 + 9 44.1 m```. |
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| 1382. |
A particle of mass m, initially at rest, is acted upon by a variable force F for a brief interval of time T. It begins to move with a velocity u after the force stops acting. F is shown in the graph as a function of time. The curve is a semicircle.(a) u = πF02 / 2m(b) u = πT2 / 8m(c) u =πF0T / 4m(d) u = F0T / 2m |
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Answer» Correct answer is: (c) u =πF0T / 4m |
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| 1383. |
A particle of mas 70g, moving at `50cm//s`is acted upon by a variable force opposite to its direction of motion. The force F is shown as a function of time t. A. Its speed will be `50 cm//s` after the force stops actingB. Its direction of motion will reverseC. Its average acceleration will be `1m//s^(2)` during the interval in which the force actsD. Its average acceleration will be `10 m//s^(2)` during the interval in which the force acts. |
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Answer» Correct Answer - A::B |
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| 1384. |
A body of mass 10 kg is being acted upon by a force `3t^2` and an opposing constant force of 32 N. The initial speed is `10 ms^-1.`The velocity of body after 5 s isA. `14.5 ms^-1`B. `6.5 ms^-1`C. `3.5 ms^-1`D. `4.5 ms^-1` |
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Answer» Correct Answer - B `F=3t^2-32` `a=F/m=(0.3t^2-3.2)` `int_0^vdv=int_0^5adt=int_0^5(0.3t^2-3.2)dt` `v-10=-3.5` `:. v=6.5 m//s` |
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| 1385. |
The variation of quantity A with quantity B is plotted in the fig. Describes the motion of a particle in a straight line. (a) Quantity B may represent time. (b) Quantity A is velocity if motion is uniform. (c) Quantity A is displacement if motion is uniform (d) Quantity A is velocity if motion is uniformly accelerated.A. (a) Quantity (B) may represent time.B. (b) Quantity (A) is velocity if motion is uniform.C. (c ) Quantity (A) is displance if motion is uniform.D. (d) Quantity (A) is velocity if motion is uniformly accelerted. |
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Answer» Correct Answer - A::C::D Quantity (B) may represent time. Thus, option (a0 is true, If motion is uniform, the velcoity-time graph showul be a straight line parallel to tiem axis. Theus, quantiy (A) cannot be a velocity Thus, Option (b) is wrong. If motion os uniform , the velocity is constant. The displacement-time graph is a straight line with positive slope. theus, quantity ()A is displacement. If motion is uniformly accelerated, the accelerationis constant. The velocity-tieme graph is a straight line with positive slope. Thus, quanitiy (A) is vellocity. |
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| 1386. |
The figure shows a velocity-time graph of a particle moving along a straight line The total distance4 travelled by the particel isA. `66.6 m`B. `51.6 m`C. zeroD. `36.6 m` |
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Answer» Correct Answer - A Distance `=` Area convered by `v-t` graph on time axis `rArr (1)/(2) (+(14)/(3)) xx 10 +(1)/(2) xx (8-(14)/(3)) xx 20` `=(100)/(3)+(100)/(3) = (200)/(3) rArr 66.6 m` |
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| 1387. |
In the given ` v-t` graph, the distance travelled by the body in `5` second will be .A. B. C. D. |
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Answer» Correct Answer - C For the give velociyt-position graph, intercept`= v_0 ` and slope `=- v_0//x-0` The equation of given line is , ` v=- v_0/x_0 x = v_0` Differentiatng it w.tr.t. t, we have ` (dv)/(dt)=- v-0/x_0 (dx)/(dt)` or ` a=- v_0 /x_0 v` ` =- v-0 /x_0 ( - v_0/x_0 x + v_0 ) = v_0^2/x_0^2 x - v_0^@/x_0` It is an equation of straight line between (a) and (x) with positive slope ` (=(v_0^2)/x_0^20` and negative intercept ` ( = - ( v_0^2/x_0)`. |
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| 1388. |
The acceleration time graph of a particle is shown in the Fig. 2 (CF). 11. At time ` t= 10 s` is the particle is ` 8 ms^(-1)`. Its velocity ` t= 10 s` is. .A. (a) (50)/3 ma^(-1)`B. (b) (70)/3 ma^(-1)`C. (c ) ` (74)/3 ms^(-1)`D. (d) (144)/3 ms^(-1)` |
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Answer» Correct Answer - C As, acceleration = (change in velocity)/(time taken)` ltbRgt :. Change in velocity=acceleration xx time taken Hence, change in velocity in ` 10 s=area OAB` `+area BCD` ` = 1/2 xx 6 xx 10 +1/2 xx (10 -6 ) xx (- (20)/3)` `= 30 - (40)/3 = (50)/3 ms^(-1)` :. Final velocity at ` 10 s=initail veloicty+change in velocity = 8 + (50)/3 = (74)/3 ma^(-1)`. |
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| 1389. |
What will be the (a) ve (x) graph for the graph shown in Fig. 2 (CF). 18 .A. ` 20 m`B. ` 40 m`C. 80 m`D. ` 100m` |
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Answer» Correct Answer - D Distance travelled by hody in `5` second is the toatl area which the velocity-time graph encloses with time axis , given by ` S= area OABB_1 + area BCC_1 B_1` ltBrgt `+ area CDC_1 + area DEF` ` = (( 400 + 20)/2) + (2 -0) + 20 xx ( 3-2)` ` + 1/2 xx ( 4 - 3) xx 20 + 1/2 xx 5- 4 ) xx 20` ` = 60 + 20 + 10 + 10` `= 100 m`. |
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| 1390. |
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vector, the angle between these Vector isA. ` 0^@`B. (b) ` 90^@`C. (45^@`D. ` 180^@` |
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Answer» Correct Answer - B Give , `| vec A+vec B| =| vec A-vec B|` or ` sqrt(A^2 +B^2 + 2 B cos theta)` ` = sqrt A^2 +B^2 = 2 AB cos theta)` ltbRgt or ` A^2 + B^2 + 2 AB cos theta = A^2 + B^2 - 2 AB cos theta` or ` 4 AB cos theta = 0 ` or ` cos theta = 0` or `theta = 90^@`. |
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| 1391. |
The motion of a projectile as seen from other projectile isA. Accelerated motionB. Uniform motionC. Motion with uniform distanceD. None of these |
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Answer» Correct Answer - B Uniform motion |
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| 1392. |
In a projectile motion the velocity(a) is always perpendicular to the acceleration(b) is never perpendicular to the acceleration(c) is perpendicular to the acceleration for one instant only(d) is perpendicular to the acceleration for two instants. |
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Answer» (c) is perpendicular to the acceleration for one instant only Explanation: In projectile motion, velocity is perpendicular to acceleration only at the highest point. Here, velocity is along the horizontal direction and acceleration is along the vertically downward direction. |
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| 1393. |
Two tal buldings are situated ` 200 m` apart, With what speed muat a ballbe thrown horixontally fromt eh winow ` 550` m above the ground in one building so that it will enter a window ` 50 m` above the ground in the other building ? Take ` g= 10 ms^(-2)`. |
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Answer» Vertcal distanace to be travelled by ball, ` y= 550 - 50 = 500 m` Horizontal distance to be ravelled by ball, ` x = 200 m` Let (u) be the horizontal velocity of projection of ball and it reaches in time (t) in the other building. ` Then, ` y = 1/2 gt^2` or ` t= sqrt (2 y)/g = sqrt ( 2 xx 500) /(10) = 10 s` Now ` x=ut ` or 200 =u xx 10 or ` u = 20 ms^(-1)`. |
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| 1394. |
A car moving with constant acceleration covered the distance between two points `60.0 m` apart in `6.00 s.` Its speed as it passes the second point was `15.0 m//s.` (a) What is the speed at the first point? (b) What is the acceleration? (c) At what prior distance from the first was the car at rest? |
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Answer» Correct Answer - A::B::C (a) and (b) `60=u(6)+1/2xxaxx(6)^2` `15=u+(a)(6)` Solving these two equations,we get `u=5 m//s` and `a=5/3m//s^2` (C) `(5)^2=(2)(5/3)s (v^2=2as)` `:. s=7.5m` |
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| 1395. |
Calculate the angular velcity of the minute`s hand of a clock. |
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Answer» The minute`s hand of a clock completes one rotation in ` 60`mimutes. i.e. ` T= 60 min = 60 xx 60 s` and ` theta = 2 pi ` rad, :. ` omega= theta//T = 2 pi // ( 60 xx 60) rad. ` s^(-1)`. |
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| 1396. |
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 displacement in the first 10 seconds and the next 10 seconds:A. `x_(A)ltx_(B)`B. `x_(A)=x_(B)`C. `x_(A)gtx_(B)`D. the information is insufficient to decide the relation of `x_(A)` with `x_(B)` |
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Answer» Correct Answer - D |
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| 1397. |
The motion of a simple pendulum is and example for oscillatory motion. |
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Answer» Correct Answer - 1 The motion of a simple pendulum is an oscillatory motion. |
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| 1398. |
One hour is equal to 360 second. |
| Answer» One hour is equal to 3600 second. | |
| 1399. |
A particle movig with a uniformacceleration travels ` 24 ` metre and ` 64` metre in first two successive intervals of ` 4` seconds each. Its initial velocity is.A. (a) ` 1 m//s`B. (b) 2 m//s`C. (c ) ` 5 m//s`D. (d)` 10 m//s`. |
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Answer» Correct Answer - A ` 24 = u xx 4 + 1/2 a xx 4^2 = 4 u + 8 a` ,brgt or ` 6= u = 2 a` …(i) ltbr. ( 24 + 64) = u + 8 + 1/2 a xx 8^2 =8 + 32 a` On ` 11 = u + 4 a` …(ii) on solving (i) and (ii) , we ger ` u= 1 m//s`. |
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| 1400. |
Two balls are dropped from the same point after an interval of 1s. If acceleration due to gravity is `10 m//s^(2)`, what will be their separation 3 seconds after the release of first ball?A. 5 mB. 10 mC. 25 mD. 30 m |
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Answer» Correct Answer - C |
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