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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 551. |
Find the velocity of the hanging block if the velocities of the free ends of the rope are as indicated in the figure. A. `3//2 m//s uarr`B. `3//2 m//s darr`C. `1//2 m//s uarr`D. `1//2 m//s darr` |
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Answer» Correct Answer - A |
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| 552. |
A flat car is given an acceleration `a_(0)=2m//s^(2)` starting from rest. A cable is connected to a crate of weight `50kg` as shown whose other end is attached to a fixed support on ground. Neglect friction between the floor and the car wheels and also the mass of the pulley. Calculate corresponding tension in th cable if `mu=0.30` between the crate and the floor or the car. A. 700 NB. 350 NC. 175 ND. 200 N |
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Answer» Correct Answer - B |
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| 553. |
A 2000 kg car has to go over a turn whose radius is `750 m` and the abgle of slopw is `5^(@)`. The coefficient of friction between the wheels and the road is 0.5. What should be the maximum speed of the car so that it may go over the turn without slipping ? |
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Answer» Correct Answer - `67.2m//s` . Here, `m = 2000kg, r = 750cm` `theta = 5^(@), mu_(s) = 0.5, upsilon_(max) = ?` `upsilon _(max) = [[rg(mu_(s) + tan 0))/((1 - mu _(s) tan theta))]^(1//2)` `[(750 xx 9.8 (0.5 + tan5^(@)))/(1 - 0.5 tan5^(5))]^(1//2)` `=[(750 xx 9.8 (0.5 + 0.087))/(1 - 0.5 xx 0.087)]^(1//2)` `upsilon_(max) =67.2 m//s` . |
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| 554. |
A package rest on a conveyor belt which is initially at rest . The belt is started and moves to the right for `1.3 s` with a constant acceleration of `2 m//s^(2)`. The belt then moves with a constant deceleration `a_(2)` and comes to a stop after a total displacement of `2.2 m` . Knowing that the coefficient of friction between the package and the belt are `mu_(s) = 0.35` and `mu_(k) = 0.25` , determine (a) the deceleration `a_(2)` of the belt , (b) the displacement of the package relative to the belt as the belt comes to a stop. `(g = 9.8 m//s^(2))` |
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Answer» (a) `nu = a_(1) t_(1) = 2.6 m//s` `s_(1) = (1)/(2) a_(1) t_(1)^(2) = (1)/(2) xx 2 xx (1.3)^(2) = 1.69 m` `s_(2) = (2.2 - 1.69) = 0.51 m` and `s_(2) = (nu^(2))/(2 a_(2))` `:. a_(2) = (nu^(2))/(2 s_(1)) = (2.6)^(2)/(2 xx 0.51) = 6.63 m//s^(2)` and `t_(2) = (nu)/(a^(2)) = 0.4 s` (b) Acceleration of package will be `2 m//s^(2)` while retardation will be `mu_(k)g or 2.5 m//s^(2) "not" 6.63 m//s^(2)` for the package. `nu = a_(1) t_(1) = 2.6 m//s implies s_(1) = (1)/(2) a_(1) t_(1) = 1.69m` ` s_(2) = nut_(2) - (1)/(2) d_(2) t_(2)^(2) = 2.6 xx 0.4 - (1)/(2) xx 2.5 xx (0.4)^(2)` `= 0.84 m` `:.` displacement of package w.r.t. belt `= (0.84 - 0.51) m = 0.33 m` |
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| 555. |
In case i wedge B is stationary and block A starts sliding when the angle of plane is greater than a certain value `theta`. While in case ii wedge B is moving upwards with an accelration `a_(0)` and the block A starts sliding when the angle of plane is greater than `alpha`. Coeffiecient of friction between A and B in both the cases is same. then . A. `theta=alpha`B. `thetagtalpha`C. `thetaltalpha`D. `(tantheta)/(tan alpha)=(g)/(a_(0))` |
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Answer» Correct Answer - a |
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| 556. |
In the diagram shown in figure wedge of mass M is stationary. Block of mass m=2 kg is slipping down. Force of friction on the wedge from ground is `(g=10 m//s^(2))` A. `5sqrt(3) N`B. `10 N`C. `50 N`D. `10 sqrt(3) N` |
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Answer» Correct Answer - A |
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| 557. |
A lift is going up. The total mass of the lift and the passengers is 1500 kg . The variation in the speed of the lift is given by the graph shown in Fig . What will be the tension in the rope pulling the lift at (i) `t = 1 s` , (ii) `t =6 s` (iii) `t = 11 s` ? What is the height to which the lift takes the passengers ? During the course of entirs motion What is the average velocity and average acceleration of the lift ? Taken `g = 9.8 m//s^(2) ` . |
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Answer» Here , ` m = 1500 kg , T = R = ? ` (i) From 0 to 2 s , u = 0, `upsilon = 3.6 m //s`, `a = (upsilon-u)/t =(3.6 - 0 )/0 = 1.8 m// s^(2) ` As acceleration is unifrom therefore at t = 1 s , ` a = 1.8 m//s^(2) ` `:. R= m (g+a) = 1500 (9.8 + 1.8) ` `1500 xx 11.6 = 17400 N` (ii) From the graph we find that at t = 6 s,a=0 . `:. R = mg = 1500 xx 9.8 N = 14700 N ` (iii) At `t = 11 s` : From the graph the velocity of body is decreasing at a constant rate , From `t = 10 s` to `t = 12 s` `a = (0-3.6)/(2) = - 1.8 m//s^(2)` This is the retardation at `t = 11 s ` `:. R = m (g+a) = 1500 (9.8 - 1.8)` `12000 N` The height to which the lift takes the passengers = area of trapezium oABC `=(( AB + OC )/(2))AD = (8+2)/2 xx 3.6 = 36 m` ` =(( AB + OC )/(2))AD = (8+2)/2 xx 3.6 = 36 m` As initial velocity of lift =0 and final velocity of lift is also zero , thereforce , change in velocity of lift , Delta upsilon = 0 . Average acceleration of lift , `a_(av) = (Delta upsilon)/(t) = (0)/(2)` = Zero Further in 12 secound, net displacement is zero . Therefore , average velocity is zero. |
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| 558. |
A lift starts from rest with a constant upward acceleration It moves 1.5 m in the first 0.4 A person standing in the lift holds a packet of 2 kg by a string Calculate the tension in the string during the motion . |
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Answer» Here , ` u = 0 , s = 1.5 m , t = 0.4 s ` From ` s = ut + (1)/(2) at^(2) ` ` 1.5 = 0 + (1)/(2) a (0.4)^(2) , ` ` a = (1.5 xx 2 )/(0.4 xx 0.4 )= 18 . 75 m // s^(2) ` As the string is moving upwards with this acceleration ` :. T = m (g+a) = 2 (9.8 + 18 . 75) = 57 .1 N ` . |
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| 559. |
A lift starts from rest with a constant upward acceleration It moves 1.5 m in the first 0.4 A person standing in the lift holds a packet of 2 kg by a string Calculate the tension in the string during the motion .A. 5.89 NB. 57.1 NC. 6.71 ND. none of these |
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Answer» Correct Answer - B (b) Given, `u=0ms=15 m,t=0.4s` From `s=ut(1)/(2) at^(2) implies 1.5 =0+(1)/(2)a(0.4)^(2)` `implies a=(1.5xx2)/((0.4)^(2))=18.75 ms^(-2)` As the string is moving upwards with in acceleration `therefore T=m(g+a)=2(9.8+18.75)=57.1 N` |
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| 560. |
Assertion : The work done in bringing a body down from the top to the base along a frictionless inclined plane is the same as the work done in bringing it down along the vertical side . Reason : The gravitational force on the body along the inclined plane is the same as that along the vertical side.A. If both, Assertion and Reason are true and the Reason is the correct explanation of the AssertionB. If both, Assertion and Reason are true but Reason is not a correct explanation of the AssertionC. If Assertion is ture but the Reason is falseD. If both, Assertion and Reason are false |
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Answer» Correct Answer - c Work done is same because gravitational forces are conservative forces Gravitational force on the body along the inclined plane is not same as that along the vertical The reason is false . |
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| 561. |
A metal block of mass `0.5` kg is placed on a plane inclined to the horizontal at an angle of `30^(@)`. If the coefficient of friction is `0.2`, what force must be applied (i) to just prevent the block from sliding down the inclined plane (ii) to just move the block up the inclined plane and (iii) to move it up the inclined plane with an acceleration of `20cms^(-2)`? |
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Answer» Correct Answer - (i) `1.6N` (ii) `3.2999N` (iii) 3.399 N` |
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| 562. |
A block `B` is pushed momentarily along a horizontal surface with an initial velocity `v` . If mu is the coefficient of sliding friction between `B` and the surface, block `B` will come to rest after a time: A. `(v)/(mug)`B. `(vg)/(g)`C. `(v mu)/(g)`D. `(mu g)/(v)` |
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Answer» Correct Answer - A (a) Given, initial velocity =v Final velocity =0 coefficient of friction =`mu` For a straight line motion , v-u=at Here, `0-v=-(mu g)t" "(as,a=(mu mg)/(m))` Negative sign indicates retardation. `therefore" " t=(v)/(mu g)` |
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| 563. |
A mass of `3kg` descending vertically downwards supports a mass of `2kg` by means of a light string passing over a pulley At the end of `5s` the stringbreaks How much high from now the `2kg` mass will go `(g = 9.8m//^(2))`.A. `9.8m`B. `19.6m`C. `2.45m`D. `4.9m` |
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Answer» Correct Answer - d Acceleration of the system before the string breaks `a = ("net pulling force")/("total mass") = (3 g -2g)/((3 + 2)) = (g)/(5)` After 5s, velocity of the system, `upsilon = "at" = (g)/(5) xx 5 = gm//s` Required height `h = (upsilon^(2))/(2 g) = (g xx g)/(2 g) = (g)/(2)` ` = (9.8)/(2) = 4.9m` . |
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| 564. |
A block of mass 10 kg is sliding on a surface inclined at `30^(@)` with horizontal . If coefficient of friction between the block and the surface is 0.5 , find acceleration produced in the block . Take ` g = 9.8 m //^(2)` . |
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Answer» Here , ` 10 kg , theta = 30^(@) , mu = 0.5 , ` ` a = ? ` ` a = g ( sin theta - mu cos theta) ` ` = 9.8 (sin 30^(@)) = 0.657 m // s^( 2 ) ` . |
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| 565. |
Assertion : No force is required to move a body unifromly along a straight line Reason : Because `F = ma = m (0) =0` .A. If both, Assertion and Reason are true and the Reason is the correct explanation of the AssertionB. If both, Assertion and Reason are true but Reason is not a correct explanation of the AssertionC. If Assertion is ture but the Reason is falseD. If both, Assertion and Reason are false |
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Answer» Correct Answer - a Both the assertion and reason are true only when there are no forces opposing the motion of the body . |
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| 566. |
Find the force required to move a train of mass `10^(5)` kg up an incline of 1 in 50 with an acceleration of `2 ms^(-2)`. Coefficient of friction between the train and rails is 0.005. Take `g = 10^(2)`. |
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Answer» ` F = ? , m = 10^(5) kg , sin theta (1)/(50) , ` ` a = 2 m // s^(2) , mu = 0.005 , g = 10 ms^(-2) ` Now `cos theta = sqrt ( 1 - sin^(2) theta) = sqrt (1 - ((1)/(50))^(2))= sqrt((2499)/(2500))=1` ` F = mg (sin theta + mu cos theta ) + ma ` ` = 10^(5) xx 10 ((1)/(50) + 0.005 xx 1 ) + 10^(5) xx 2 ` ` 10^(6) xx 0 .025 + 2 xx 10^(5) = 2.25 xx 10^(5) N ` . |
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| 567. |
A block is placed on an inclined plane whose angle `theta` can be varied as shown. Given `mu_(0)=3//4` and `mu_(k)=1//2`. The angle at which block starts moving and minimum acceleration of the block once it starts moving is A. `37^(@),2m//s^(2)`B. `53^(@),4m//s^(2)`C. `37^(@),4m//s^(2)`D. `53^(@),2m//s^(2)` |
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Answer» Correct Answer - a |
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| 568. |
A plank with a box on it at one end is gradully raised about the other end. As the angle of inclination with the horizntal reaches `30^(@)` , the box starts to slip and slide `4.0m` down the plank in `4.0s` . The coefficients of static and knitic friction between the box and the plank will be, respectively. A. `0.4` and `0.3`B. `0.6`and `0.6`C. `0.6` and `0.5`D. `0.5` and `0.6` |
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Answer» Correct Answer - c When box just slips then Coefficient of static friction `mu_(s) = tan30^(@) = (1)/sqrt3 ~~0.6` When box slides down the plane, coefficient of kinetic friction will be involved Let a be the acceleration of the box sliding down the plane then using `S = ut + (1)/(2)at^(2)` we have `4 = 0 + (1)/(2) xxa xx 4^(2)` or `a=(2)/(4) = 0.5 ms^(-2)` Here, `a =g sin theta -mu_(k) g cos theta` `0.5 =9.8xx sin30^(@) -mu_(k) xx 9.8 cos 30^(@)` `0.5 =9.8 xx(1)/(2) -mu_(k) xx 9.8 xx sqrt3/2` `0.5 =4.9 -mu_(k) xx4.9sqrt3` `mu_(k) =(4.9 -0.5)/(4.9sqrt3) =(4.4)/(4.9 xx1.73) ~~0.5` . |
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| 569. |
A wooden block is kept on a polished wooden plank whose inclination is increased gradully . The block starts slipping when the plank makes an angle of `25^(@)` with the horizontal . However , once started , the block can continue with unifrom speed , if the inclination is reduced to 21 Calculate coefficient of static and dynamic friction between the block and the plank . |
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Answer» Here angle of repose , `alpha = 25^(@)` `:.` Coefficient of static friction , `mu_(s) = tan alpha = tan 25^(@) = 0.4663` As the block continues to move uniformly , when `theta = 21^(@)` , `:.` Coefficient of kinetic friction , `mu_(k) = tan theta = tan 21^(@) = 0.3839 ` . |
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| 570. |
A `0.5 kg` ball moving with a speed of `12 m//s` strikes a hard wall at an angle of `30^(@)` with the wall. It is reflected with the same speed and at the same angle . If the ball is in contact with the wall for `0.25 s`, the average force acting on the wall is A. `96N`B. `48N`C. `24N`D. `12N` |
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Answer» Correct Answer - c Here, `m = 0.5kg, upsilon = 12m//s, theta = 30^(@)` `t = 0.25 s, F = ?` As Impulse = change in linear momentum `:. F xx t = 2 m upsilon sin 30^(@)` (perpendicular to the wall) `F = (2m upsilon sin 30^(@))/(t) = (2xx0.5 xx12)/(0.25) xx (1)/(2) = 24N` . |
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| 571. |
Assertion: For applying the second law of motion, there is no conceptual distinction between inanimate and animate objects. Reason: An animate object requires an external force to acceleration.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason is the not correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - B Second law of motion can be applied to both inanimate and animate objects. An animate object such as a human also requires an external force to accelerate. For example, without the external force of friction, we cannot walk on the ground. |
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| 572. |
A suitcase is gently dropped on a conveyor belt moving at ` 3 m//s`. If the coefficient of friction between the belt and the suitcase is 0.5. Find the displacement of suitcase relative to conveyor belt before the slipping between the two is stopped `(g=10 m//s^(2))`A. `2.7 m `B. `1.8 m`C. `0.9 m`D. `1.2 m` |
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Answer» Correct Answer - C |
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| 573. |
A 75 kg man stands in a lift. What force does the floor does the floor exert on him when the elevator starts moving upwards with an acceleration of `2.0ms^(-2)`. ? Take g = 10`ms^(-2)`. |
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Answer» Correct Answer - 90 kg f |
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| 574. |
A car of mass `1000kg` negotiates a banked curve of radius `90m` on a fictionless road. If the banking angle is `45^(@)` the speed of the car is:A. 20 `ms^(-1)`B. 30 `ms^(-1)`C. 5 `ms^(-1)`D. 10 `ms^(-1)` |
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Answer» Correct Answer - B |
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| 575. |
A car of mass `1000kg` negotiates a banked curve of radius `90m` on a fictionless road. If the banking angle is `45^(@)` the speed of the car is:A. `20ms^(-1)`B. `30ms^(-1)`C. `ms^(-1)`D. `10ms^(-1)` |
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Answer» Correct Answer - b Here, `m = 1000kg, r = 90m, theta = 45^(@) , upsilon = ` ? For banking, `tan theta= (upsilon^(2))/(rg)` `upsilon^(2) = rg tan theta = 90 xx 10 tan 45^(@) = 900` `upsilon = sqrt(900) = 30m//s` . |
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| 576. |
For traffic moving at `60km//h` if the radius of the curve is `0.1km` what is the correct angle of banking of the road Given `g = 10m//s^(2)` . |
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Answer» Correct Answer - `15.5^(@)`. Here, `upsilon = 60 km//h = (60 xx 1000)/(60 xx 60) m//s = (50)/(3) m//s` `r = 0.1 km = 100 m, theta = ? g = 10m//s^(2)` `tan theta = (upsilon^(2))/(rg) = (50)/(3) xx (1)/(100 xx 10) = 0.2777` `theta = tan^(-1) (0.2777) =15.5^(@)` . |
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| 577. |
Calculate the accelerations of the pulleys B and C and the tension in the string passing over the pulley A of the figure Figure 5.22. |
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Answer» Correct Answer - `g//11`,12.46 kgf |
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| 578. |
A ball moving with a speed of `9m//s` strikes an identical ball at rest, such that after the collision, the direction of each ball makes an angle of `30^(@)` with the original line of motion. Find the speeds of the two balls after collision. |
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Answer» Correct Answer - `3 sqrt3 m s^(-1)`, No |
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| 579. |
A cord is tied at one end to a sprind balance of weight 3 kg supporting a mass of 5 kg . The cord passes over a smooth fixed pulley carrying a load of 10 kg at the other end . What will be the reading of the spring balance when the system moves ? How will the reading vary if the load of 10 kg be replaced by a load of 6 kg ? [Hint : Consider free-body diagrams of 5 kg , 10 kg and the system (5 kg body + 3 kg spring ).] |
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Answer» Correct Answer - `5.55` kg , `4.3` kg . |
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| 580. |
A small body slides from a height h down a smooth plane inclined at `45^(@)` to the horizontal. How will the body move if at the end of the plane it encounters (a) a completely elaatic plane, (b) an inelastic but smooth plane. |
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Answer» Correct Answer - It leaves the plane with velocity `sqrt(2gh)` at `45^(@)` with the verticle, it moves along the plane with velocity `sqrt(gh)` |
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| 581. |
Two identical buggies 1 and 2 with one man in each move without friction due to inertia along the parallel rails toward each other. When the buggies get opposite each other, the men exchange their places by jumping in the direction perpendicular to the motion direction. As a consequence, buggy 1 stops and buggy 2 keeps moving in the same direction, with its velocity becoming equal to v. Find the initial velocities of the buggies `v_1` and `v_2` if the mass of each buggy (without a man) equals M and the mass of each man m. |
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Answer» Correct Answer - `v_(1)=(mv)/(M-m),v_(2)=(Mv)/(M-m)` |
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| 582. |
A body of mass `0.5kg` is wirled in a circle with a velocity of `2ms^(-1)` using `0.5m` length of a string which can withstand a tension of `15N` Neglecting the force of gravity on the body predict whether or not the string will break Give reasons for your answer . |
| Answer» Correct Answer - The string will not break as centrifugal force is 4 N which is less than 15N` . | |
| 583. |
Impulse of a force is a measure of total effect of the force It is given by the product of everage force and the time for which the force acts on the body . Impulse of a force is measured by the total change in linear momentum produed during impact Impulse ` vec(I) = vec(F)_("avg") xxt = vec (p_(2)) - vec(p_(1)) ` A given change in linear momentum can be produced by applying a larger force for a shorter time or by applying a smaller force for a longer time Read the above passage and answer the following questions (i) What is the function of shockers in autos (ii) What values of life do you learn from this study ? |
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Answer» (i) When autos ply over an uneven road , impulsive forces are exerted by the road The funcation of shockers is to increase the time of impact As `F xx t` = constant , therefore when `t` increases `F` decreases the force/jerk experienced by the rider of the vehicle raduces (ii) In day life the relation `F xx t ` = change in linear momentum = constant implies that to achieve any goal or to affect any change the product of force and time shall be constant when your efforts are weak you require longer time to achieve the goal . |
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| 584. |
A particle is projected along the line of greatest slope up a rough plane inclined at an angle of `45^(@)` with the horizontal. If the coefficient of friction is `1//2` . Their retardation is:A. `(g)/(2sqrt(2))`B. `(g)/(sqrt(2))`C. `(3g)/(2 sqrt(2))`D. `(g)/(2)` |
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Answer» Correct Answer - C ( c) Acceleration,`a=(g sin theta+ mu g cos theta)` `=(g)/(sqrt(2))+0.5xxgxx(1)/(sqrt(2)=(3g)/(2sqrt(2))` |
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| 585. |
Which one is real force - centripetal or centrifugal force ? |
| Answer» Centripetal force is real . | |
| 586. |
Can centripetal force produe rotation ? |
| Answer» No Centripetal force can move a body along a circle . But it cannot produce rotation . | |
| 587. |
A bomb at rest explodes into three parts of the same mass. The linear momenta of the two parts are `-2phati` and `p hati` Calculate the magnitude of momentum of third part . |
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Answer» Correct Answer - `psqrt5` . Here, `vec(p_1) = -2 phati = 2p` along negative X-axis `vec(p_(2)) = p hat j = p` along Y-axis Resultant momentum of the two parts `p = sqrt(p_(1)^(2) + p_(2)^(2)) = sqrt((2 p)^(2) + p^(2)) = p sqrt5` As the bomb was initially at rest final momentum of all the three parts must be zero. `vec(p_(3)) + vec(p) = 0 ` `vec(p_(3)) = - vec(p)= - p sqrt(5) = p sqrt(5)` (in magnitude). |
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| 588. |
An object of mass ` 3 kg ` is at rest. Now a force of ` vec F = 6 t^2 hat I + 4 t hat j` is applied on the object, the velocity of object at `t= 3 s` is.A. `18hat(i)+3hat(j)`B. `18hat(i)+6hat(j)`C. `3hat(i)+18hat(j)`D. `19hat(i)+4hat(j)` |
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Answer» Correct Answer - B |
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| 589. |
A body of mass `1kg` initially at rest explodes and breaks into threee fragments of masses in the ration breaks into three fragments of masses in the ration `1: 1: 3`. The two pieces of equal masses fly off perpendicular to each other with a speed of `30m//s` each What is the velocity of heavier fragments ? . |
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Answer» Correct Answer - `10 sqrt2 m//s` . Here, `m = 1kg` `m_(1) : m_(2) : m_(3) = 1 : 1: 3` `:. M_(1) = (1)/(5)kg, m_(2) = (1)/(5)kg, m_(3) =(3)/(5)kg` `upsilon_(1) = upsilon_(2) = 30 m//s, upsilon_(3) =` ? `p_(1) = p_(2) = m_(1) upsilon_(1) = m_(2) upsilon_(2)` `= (1)/(5) xx 30 = 6 kg m//s` As these two pieces fly off perpendicular to each other their resultant momentum `sqrt(p_(1)^(2) + p_(2)^(2)) = sqrt(6^(2) + 6^(2)) = 6sqrt2kgm//s` As the body was initially at rest therefore from the principle of conservation of linear momentum `p_(3) =m_(3) upsilon_(3) =6sqrt2 kg m//s` `upsilon_(3) = (6sqrt2)/(m_(3))= (6sqrt2)/(3//5) = 10 sqrt2 m//s` . |
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| 590. |
A player takes 0.1s I catching a ball of mass 150 g moving with velocity of 20 m/s. The force imparted by the bal on the hands of the player isA. 0.3 NB. 3 NC. 30 ND. 300 N |
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Answer» Correct Answer - C |
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| 591. |
A body of mass `1kg` initially at rest explodes and breaks into threee fragments of masses in the ration breaks into three fragments of masses in the ration `1: 1: 3`. The two pieces of equal masses fly off perpendicular to each other with a speed of `30m//s` each What is the velocity of heavier fragments ? .A. `(10)/(sqrt(2))m//s`B. `10sqrt(2)m//s`C. `20sqrt(2)m//s`D. `30sqrt(2)m//s` |
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Answer» Correct Answer - B |
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| 592. |
A circular race track of radius 400 m is banked at an angle of `10^(@)` . If the coefficient of friction between the wheels of a race car and the road is 0.2 , what is the (i) optimum speed of the race car to aviod wear and tear on its tyres . maximum permissible speed to aviod slipping ? |
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Answer» Here, ` r = 400 m , theta = 10^(@) , mu = 0.2 ` (i) optimum speed of the car to aviod wear and tear ` upsilon = sqrt(rd tan ) = sqrt(400 xx 9.8 tan 10^(@) )= ` ` = sqrt(400 xx 9.8 xx 0 .1763) = sqrt(69. 1.1)` ` = 26 . 29 m//s ` (ii) maximum permissible speed is ` upsilon_(max) = sqrt((r g(mu + tan theta))/(1- mu tan theta) ` ` = sqrt((400 xx 9.8 (0.2 + 0.1763))/(1 - 0 .2 xx 0. 1763 ` ` = 39 . 10 m // s ` . |
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| 593. |
A circular racetrack of radius 300 m is banked at an angle of `15^(@)` If the coefficient of friction between the wheels of a race car and the road is 0.2 what is the (a) optimum speed of the race car to avoid wear and tear on its tyres , and (b) maximum permissible speed to aviod slipping ?A. `10sqrt3 "ms"^(-1)`B. `9sqrt10 "ms"^(-1)`C. `sqrt10 "ms"^(-1)`D. `2sqrt10 "ms"^(-1)` |
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Answer» Correct Answer - B Here, R=300m, `theta=15^(@), g=10ms^(-2), mu=0.2` The optimum speed of the car to avoid wear and tear is given by `v=sqrt(Rg tan theta)=sqrt(300xx10xxtan 45^(2))=sqrt(810)=9sqrt10ms^(-1)` |
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| 594. |
Assertion : Force of friction depends on the actual area of contact Reason : Smoother the surfaces of contact smaller is opposition to motion .A. If both, Assertion and Reason are true and the Reason is the correct explanation of the AssertionB. If both, Assertion and Reason are true but Reason is not a correct explanation of the AssertionC. If Assertion is ture but the Reason is falseD. If both, Assertion and Reason are false |
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Answer» Correct Answer - d Force of friction is independent of the actual area of contact Statement `1` is false but Statement -2 is true . |
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| 595. |
What are the factors on which coefficient of friction depends ? |
| Answer» Coefficient of friction between any two surfaces in contact depends on nature of surfaces and nature of material of the surfaces in contact . | |
| 596. |
Why do we easily slip on a rainy day ? |
| Answer» Whater acts as a lubricant reducing the friction between the road our feet . | |