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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.
| 451. |
A block of mass 2kg rests on a rough inclined plane making an angle of `30^@` with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block isA. 10.3NB. 23.8NC. 11.9ND. 6.3N |
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Answer» Correct Answer - C Here, frictionless force, `f=muR` `=mu mg cos theta=0.7xx2xx9.8 cos 30^(@)` `=0.7xx2xx9.8xx0.866=11.9N` |
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| 452. |
A block of mass M rests on a rough horizontal surface as shown. Coefficient of friction between the block and the surface is `mu`. A force F=mg acting at angle `theta` with the vertical side of the block pulls it. In which of the following cases the block can be pulled along the surface? A. `tan theta ge mu`B. `tan ((theta)/(2) ge mu)`C. `cos theta ge mu`D. `cot ((theta)/(2)) ge mu` |
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Answer» Correct Answer - D (d) `N=Mg-F cos theta` `=Mg-Mg cos theta=2 Mg sin^(2)"" (theta)/(2)` Further block can be pulled if, `F sin theta ge mu N cos theta` `implies " " 2Mg sin""(theta)/(2).cos""( theta)/(2) ge 2 mu Mg sin^(2)"" (theta)/(2) implies cot"" (theta)/(2) ge mu` |
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| 453. |
A block rests on a rough plane whose inclination `theta` to the horizontal can be varied. Which of the following graphs indicates how the friction force `F` between the block and the plane varies as `theta` is increased?A. B. C. D. |
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Answer» Correct Answer - B Let `alpha =` angle of repose For `theta le alpha` Block is stationary and force of friction, `f = mg sin theta` or `f prop sin theta` i.e. it is sine graph For `theta ge alpha` Block sides downwards `f = mu mg sin theta` or `f prop cos theta` i.e. it is cosine graph the correct alternative is therefore (b). |
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| 454. |
A block of mass 1 kg is at rest elative to a smooth wedge moving leftwards with constant aasseleraton`a=5m//s^(2)` Let N be the noemal reaction between block and the wedge .Then `(g=10m//s^(2)`) A. `N=5sqrt(5)N`B. `n=15N`C. `tantheta=(1)/(2)`D. `tantheta =2` |
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Answer» Correct Answer - a,c |
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| 455. |
Block B has mass m and is released from rest when it is on top of wedge A, which has a mass 3m. Determine the tension in cord CD needed to hold the wedge from moving while B is sliding down A. Neglect friction. |
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Answer» Normal reation between `A` and `B` would be `N = mg cos theta` ,Its horizontal component is `N sin theta`. Therefore , tension in cord `CD` is equal to this horizontal component. `T = N sin theta = (mg cos theta) (sin theta)` `= (mg)/(2) sin 2 theta` |
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| 456. |
An inclined plane makes an angle `30^(@)` with the horizontal. A groove (OA) of length 5m cut in the plane makes an angle `30^(@)` with OX. A short smooth cylinder is free to slide down under the influence of gravity. The time taken by the cylinder to reach from A to O is `(g=10ms^(-2))`. A. `4 s`B. `2 s`C. `2sqrt(2) s`D. `1 s` |
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Answer» Correct Answer - B |
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| 457. |
The driver of a three wheeler moving with a speed of `36 km//h` sees a child standing in the middle of the road and brings his vehicle to rest in 4 s just in time to save the child What is the average retarding force on the vehicle ? The mass of three wheeler is 400 kg and mass of the driver is 65 kg. |
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Answer» Here mass of three-wheeler mi = 400 kg , mass of driver = `m_2` = 65 kg initial speed of auto , u = 36 km/h = `36 xx -m/s 10 ms^(-1)` , final speed , v - 0 and t = 4s . As acceleration , a = v-u / t = 0 - 10 / 4 = - 2.5 `ms^(-2)` Now F = `(m_1 + m_2)`a = ( 400 + 65 ) `xx` (- 2.5) = - 1162.5 N = `- 1.2 xx 10^(3) ` N . The -ve sign shows that the force is retarding force. |
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| 458. |
The driver of a three wheeler moving with a speed of `36 km//h` sees a child standing in the middle of the road and brings his vehicle to rest in 4 s just in time to save the child What is the average retarding force on the vechicle ? The mass of three wheeler is 400 kg and mass of the driver is 65 kg . |
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Answer» Here , ` u = 36 km//h = 10 m//s , upsilon = 0 , t = 4 s m = 400 + 65 = 465 kg ` Retarding force ` = F = ma = (m (upsilon - u ))/(t) = (465 (0 - 10 ))/(4) = - 1162 .5 N ` . |
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| 459. |
A sooterist moving with a speed of `36kmh^(-1)` sees a child standing in the middle of the road. He applies the brakes and brings the scooter to rest in 5 s just in time to save child. Calculate the average retarding force on the vehicle, if mass of the vehicle and driver is 300 kg. |
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Answer» Correct Answer - 600 N |
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| 460. |
A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is `5m//s^2`, the frictional force acting on the block is…………newtons. |
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Answer» Correct Answer - 5 N |
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| 461. |
A scooter weighs 120 kg f. Brakes are applied so that wheels stop rolling and start skidding. Find the force of friction if the coefficient of friction is `0.4`. |
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Answer» Correct Answer - 48 kg f |
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| 462. |
In the masses of A and B are 10 kg and 5 kg . Calculate the minimum mass of C which may stop A from slipping Coefficient of static friction between block A and table is 0.2 . |
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Answer» Correct Answer - 15 kg |
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| 463. |
In the masses of A and B are 10 kg and 5 kg . Calculate the minimum mass of C which may stop A from slipping Coefficient of static friction between block A and table is 0.2 . |
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Answer» Here , ` m_(A) = 10 kg , m_(B) = 5 kg `, `m_(C) = ? , mu = 0 . 2 ` To avoid slipping of A force of friction = T wt . Of B ` mu (m_(C) + m_(A)) g = m_(B) g` `0 .2 (m + 10 ) = 5 , m + 10 = (5) /(0.2) = 25 ` `m =25 - 10 = 15 kg` . |
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| 464. |
An aircraft executes a horizontal loop at a speed of `720 km h^(-1)` , with its wings banked at `15^(@)` What is the radiue of the loop ?A. 14.8kmB. 14.8mC. 29.6kmD. 29.6m |
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Answer» Correct Answer - A Here, `v=720 km h^(-1)=720xx(5)/(18)ms^(-1)=200 ms^(-1)` `theta=15^(@), g=10ms^(-2) "As" tan theta=(v^(2))/(rg)` `therefore r=(v^(2))/(tan thetag)=(200 ms^(-1))^(2)/(tan 15^(@)xx10ms^(-2))=14815m=14.8km` |
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| 465. |
For the situation shown in figure ,mark the correct options. A. Angle of friction is `tan ^(-1) (mu)`B. Angle of opposite is `tan ^(-1) (mu)`C. At `theta = tan ^(-1) (mu)`, minimum force will be required to move the blockD. minimum force required to move the block is `(mu mg)/(sqrt(1 + mu^(2))` |
| Answer» Correct Answer - A::B::C::D | |
| 466. |
For the situation shown in figure ,mark the correct options. A. At `t = 3 s`,pseudo force on `4 kg` block applied from `2 kg` is`4 N` in forward directionB. At `t = 3 s`,pseudo force on `2 kg` block applied from `4 kg` is `2 N` in forward directionC. Pseudo force does not make an equal pairsD. Pseudo force also makes a pair of equal and opposite forces |
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Answer» Correct Answer - B::C Maximum force of friction between `2 kg and 4 kg` `= 0.4 xx 2 xx 10 = 8 N` `2 kg` move due to friction. Therefore its maximum acceleration may be `a_(max) = (8)/(2) = 4 m//s^(2)` slip will start when their combined acceleration becomes `4 m//s^(2)` `:. a = (F)/(m)` or `4 = (2t)/(6)` or `t = 12 s` `A t = 3 s` `a_(2) = a_(4) = (F)/(m) = (2t)/(6) = (2 xx 3)/(6)` `= 1 m//s^(2)` Both `a_(2) and a_(4)` are towards right. Therefore pseudo force `F_(1)` (on `2 kg` from `4 kg`) and `F_(2)` (on `4 kg` from `2 kg)` are towards left `F_(1) = (2) (1) = 2 N` `F_(2) = (4) (1) = 4 N` From here we can see that `F_(1) and F_(2)` do not make a pair of equal and opposite force. |
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| 467. |
For the given situation shown in figure , choose the correct option `(g = 10 m//s^(-2))` A. At `t = 1 s`, force of friction between `2 kg` and `4 kg` is `2 N`B. At `t = 1 s`, force of friction between `2 kg` and `4 kg` is zero`C. At `t = 4 s`, force of friction between `4 kg` and ground is `8 N`D. At `t = 15 s`, acceleration of `2 kg 1ms^(-2)` |
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Answer» Correct Answer - B::C `f_(1) rarr` force of friction between`2 kg` and` 4 kg` `f_(2) rarr` force of friction between 4 kg and ground `(fs_(1))_(max) = 0.4 xx 2 xx 10 = 8 N` `Fk_(1) = 0.2 xx 2 xx 10 = 4 N` `(fs_(2))_(max) = 0.6 xx 6 xx 10 = 36 N` `Fk_(2) = 0.4 xx 6 xx 10 = 24 N` At `t = 1 s`, `F = 2 N lt 36 N`, therefore system remains stationary and force of friction between `2 kg` and `4 kg` is zero At `t = 4 s`, `F = 8 N lt 36 N`, therefore system is again stationary and force of friction between `4 kg` from ground is `8 N` At `t = 15 s`, `F = 30 N lt 36 N`, therefore system remains stationary and force of friction between `2 kg` and `4 kg` is zero At `t = 1 s`, `F = 2 N lt 36 N`, and system is stationary. |
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| 468. |
The pulley arrangements of fig (a) and (b) are identical. The mass of the rope is negligible. In (a) the mass m is lifted up by attaching a mass 2 m to the other end of the rope. In (b) m is lifted up by pulling the other end of the rope with a constant downward force F = 2 mg. Which of the following is correct? |
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Answer» In pulling force , ` F = (2mg- T) = 2 mg - mg = mg ` total mass moved , ` = m + 2 m = 3 m ` `:.` Acceleration , ` a = a = ("force")/("mass")= (mg)/(3m )= (g)/(3) ` the pulling force , `F = (2mg-T) = 2 mg - mg = mg ` mass moved = m `:.` Acceleration `a (f)/(m) = (mg)/(m) = g ` , |
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| 469. |
A string breaks under a load of 4.8kg A mass of 0.5 kg is attached to one end of a string 2 m long and is rotated in a horizontal circle. Calculate the greatest number of revolutions that the mass can make without breaking the string. |
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Answer» Here , ` T = 4.8 kg wt . =4.8 xx 9.8 N ` ` m = 0. 5 , r = 2 m , v = ? ` As `T = m r omega^(2) = m r (2 pi v)^(2)` `= 4 pi^(2) m r v^(2)` `:. v^(2) = (T)/(4 pi^(2) mr )=(4.8 xx 9.8)/(4 xx 9.87 xx 0.5 xx 2)` ` = 1.215 ` `v^(2) = sqrt(1.215)=1.102 "rps"` Greatest number of revolutions that the mass can make per minute `= 1. 102 xx 60 = 66 . 12 "rpm"` . |
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| 470. |
Two masses `M_(1)=5kg,M_(2)=10`kg are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When masses are released, then acceleration of masses will be A. gB. `(g)/(2)`C. `(g)/(3)`D. `(g)/(4)` |
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Answer» Correct Answer - C |
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| 471. |
A bullet fired into a fixed target loses half of its velocity after penetrating 3cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?A. (a) 2.0cmB. (b) 3.0cmC. (c) 1.0cmD. (d) 1.5cm |
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Answer» Correct Answer - C Let K be the initial kinetic energy and F be the resistive force. Then according to work-energy theorem, `W=DeltaK` i.e., `3F=1/2mv^2-1/2m(v/2)^2` …(1) For B to C: `Fx=1/2m(v/2)^2-1/2m(0)^2` …(2) Dividing eqns. (1) and (2) we get `x/3=1/3` or `x=1cm` |
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| 472. |
A smooth inclined plane of length L, having an inclination `theta` with horizontal is inside a lift which is moving down with retardation a. The time taken by a block to slide down the inclined plane from rest will beA. `sqrt((2L)/(sqrt(a sin theta)))`B. `sqrt((2L)/(g sin theta))`C. `sqrt((2L)/((g-a) sin theta))`D. `sqrt((2L)/((g+a) sin theta))` |
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Answer» Correct Answer - D (d) Moving down with retardation a means, lift is accelerated upwards. With respect to lift, pseudo force on the block will be ma in downward direction, where m is the mass of block. So, downward force mg on the block will be replaced by m(g+a). Therefore, acceleration of block relative to plane will be, `a_(r)=(g+a) sin theta`(down the plane) From , `" " L=(1)/(2)a_(r)t^(2)" " (s=(1)/(2)at^(2))` `t=sqrt((2L)/(a_(r)))=sqrt((2L)/((g+a)sin theta))` |
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| 473. |
A block of mass m is placed on a smooth inclined plane of inclination `theta` with the horizontal. The force exerted by the plane on the block has a magnitudeA. mgB. ` mg sec theta`C. `mg cos theta`D. ` mg sin theta` |
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Answer» Correct Answer - D (d) Angle of plane is just equal to the angle of repose. or `" " mu =tan theta` |
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| 474. |
A steel is suspended in an accelerating cabin by two cords A and B If the accelerration of the frame is `a=(g)/(3sqrt(3))` then the tesion in A is x times the tension in B Find x. |
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Answer» Correct Answer - 2 |
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| 475. |
A block of mass `m` is placed at rest on an inclination `theta` to the horizontal.If the coefficient of friction between the block and the plane is `mu`, then the total force the inclined plane exerts on the block isA. `mg`B. `mu mg cos theta`C. ` mg sin theta`D. `mu mg tan theta` |
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Answer» Correct Answer - A Only two force are acting, `mg` and net connect force (resultant of fricting and normal reation) from the inclined plane. Since the body is at rest. Therefore these two force should be equal and opposite. :. Net contact force `= mg` |
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| 476. |
These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following four responses Assertion Two forces are acting on a rope lying on a smooth table as shown in figure. In moving from A to B, tension on string decrease from 2F to F. Reason Situation will becomes in determinant, if we take it a massless string. |
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Answer» Correct Answer - B (b) In case of massless string. `a=(2F-F)/(0)=oo` |
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| 477. |
These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following four responses Assertion In the shown figure, tension T connected to the ceiling is `10 N lt T lt 20 N` Reason System is not stationary. |
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Answer» Correct Answer - B (b) Let `T_(0)` is the tension in the string connecting the two blocks. Then `T_(0)-10=1xxa` `therefore T_(0) gt 10N, 20-T_(0)=2xxa` `therefore" " T_(0) lt20N` |
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| 478. |
These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following four responses Assertion In the system of two blocks of equla masses as shown, the coefficient of friction between the blocks `(mu_(2))` is less than coefficient of friction `(mu_(1))` between lower block and ground. For all values of force F applied on upper block, lower block remains at rest. Reason Frictional force on lower block due to upper block is not sufficient to overcome the frictional force on lower block due to ground. |
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Answer» Correct Answer - A (a) `(f_(2))_("max")=mu_(2) mg` `(f_(1))_("max")-mu_(1)(2m)(g)` Since `(f_(1))_("max")gt (f_(2))_("max")` Lower block will not move at all. |
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| 479. |
These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following four responses Assertion Block A is resting on one corner of a box as shown in figure. Acceleration of box is `(2hati+2 hatj)ms^(-2)`. Let `N_(1)` is the normal reaction on block from vertical wall and `N_(2)` from ground of box. Then `(N_(1))/(N_(2))=(1)/(6)`.(Neglect friction) Reason `N_(1)=0`, if lift is stationary. |
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Answer» Correct Answer - A (a) `N_(1)=ma_(x)=2m` `N_(2)=m(g+a_(y))=12m` `therefore" " (N_(1))/(N_(2))=(1)/(6)` |
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| 480. |
In the diagram shown in figure, match the following columns (take `g=10 ms^(-2)`) `{:("Column I","Column II"),("(A) Normal reaction","(p) 12 SI unit"),("(B) Force of friction","(q) 20 SI unit"),("(C) Acceleration of block","(r) zero"),(,"(s) 2 SI unit"):}` |
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Answer» Correct Answer - A`rarr`q,B`rarr`p,C`rarr`s `N=mg-20sqrt(2) sin 45^(@) =20N` `f mu_(k)N=12N` Since, `20sqrt(2) cos 45^(@) gt muN` block will move and kinetic friction will act, `a=(20sqrt(2) cos 45^(@)-mu_(k)N)/(m)=(20-12)/(4)=2 ms^(-2)` |
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| 481. |
It is easier to roll a barrel than to pull it along the road. Why ? |
| Answer» This is because rolling friction is less than the sliding friction | |
| 482. |
The displacement vector of a mass m is given by `r (t) = hati A cos omegat + hatj B sin omegat ` (a) Show that the trajectory is an ellipse (b) Show that `F = - m omega^(2) r` . |
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Answer» Here (a) ` vec(r) (t) = hati A cos omegat + hatj B sin omegat, :. x = A cos omegat, y = B sin omegat ` ` x^(2)/(A^(2)) + y^(2)/(B^(2)) cos^(2) omegat + sin^(2) omegat = 1` which is the equation of an ellipse `:.` The trajectory of the particle is elliptical (b) Now ` vec(upsilon) =vec (dr)/(dt) = -hatiomega A sin omegat + hatjomega Bcos omegat ` ` vec(a) = vec (d upsilon)/(dt) = hati omega^(2) A cos omegat - hatj omega^(2) B sin omegat = - omega^(2) [hati A cos omegat + hatjBsin omegat] = - omega^(2) vec(r) ` ` vec(F) = m vec (a) = - m omega^(2) vec(r) ` . |
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| 483. |
A person driving a car suddenly applies the brakes on seeing a child on the road ahead . If he is not wearing seat belt, he falls forward and hits his head against the steering weel. Why ? |
| Answer» When the car is brought to a sudden halt upper part of driver s body continues to move forward on account of inertia of motion If he is not wearing seat belt the only force that tries to stop him is frictional force exerted by the seat which is too small That is why he hits his head against the steering wheel . | |
| 484. |
A block of mass 100 kg is set into motion on a frictionless horizontal surface with the help of a frictionless pulley and rope system shown in What horizontal force should be applied on the rope to produce an acceleration of `0.1 m//s^(2)` . |
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Answer» Here , m = 100 kg If is force applied on the free end ogf the rope , then as is clear from ` F + F = T = ma ` ` F = (ma)/(2) = (100xx0.1)/(2)= 5N ` . |
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| 485. |
A body is moving along a rough horizontal surface with an initial velocity 6m/s m If the body comes to rest after travelling 9 m , then the coefficient of sliding friction will beA. 0.5B. 0.4C. 0.6D. 0.2 |
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Answer» Correct Answer - D (d) Retardation,`a=(mu mg)/(m)=mu g=10 mu` Now, `0=u^(2)-2as" or " 0=(6)^(2)-2(10mu)(9)` `therefore" " mu=0.2` |
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| 486. |
A block `A` with mass 100kg is resting on another block `B` of mass 200kg. As shown in figure a horizontal rope tied to a wall hold it. The coefficient of friction between `A` and `B` is `0.2` while coefficient of friction between `B` and the ground is `0.3` . the minimum required force `F` to start moving `B` will be. A. 900 NB. 200 NC. 1100 ND. 700 N |
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Answer» Correct Answer - C (c ) Friction force between block A and block B and between bock B and surface will oppose the F ` therefore F=F_(AB)+F_(BS)` `=mu_(AB) m_(A)g+ mu _(BS) (m_(A)+m_(B))g` `=0.2xx100xx10+0.3(100+200)10` `=200+900=1100N` This is the required minimum force to move the block B. |
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| 487. |
Find the force required to move a train of mass `5000` quintal up an incline of `1` in `50` with an acceleration of `2m//s^(2)` Take force of friction = 0.2 N quintal and `g = 10ms^(-2)`. |
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Answer» Correct Answer - `1.101 xx 10^(6)` . Here, `m = 5000 "quintals" = 5 xx 10^(5)kg` `sin theta = 1//50, a = 2m//s^(2) F = 0.2N//quintal` `0.2 xx 5000 = 10^(3) N` Total force required `f = mg sin 0 + F + ma` `f = 5 xx 10^(5) xx 10 xx (1)/(5) + 10^(3) + 5 xx 10^(5) xx 2` `= 10^(5) + 10^(3) + 10^(6) = 10^(6) (0.1 + 0.001 + 1)` `f = 1.101 xx 10^(6)N` . |
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| 488. |
Calculate the power of an engine which can pull a train of mass `25000` quintal up an incline of 1 in 100 at the rate of `10.8km//h` Resistance due to friction is `2N//"quintal"`. |
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Answer» Correct Answer - 885 kW |
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| 489. |
Why are the passengers thrown outwards when a car in which they are travelling suddenly takes a circular turn? |
| Answer» This is because of inertia of direction of the passengers They tend to maintain their direction of motion while the direction of car changes on turning . | |
| 490. |
What is the relation between coefficient of friction and angle of repose ? |
| Answer» Correct Answer - `mu = tan alpha` | |
| 491. |
In the figure shown, angle of repose is `45^(@)`.Find force of friction, net force and acceleration of the block when (a) `theta = 30^(@)` (b)`theta = 45^(@)` (c)`theta = 60^(@)` |
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Answer» If `theta le ` angle of repose , the block is stationary , `a = 0, F_("net") = 0` and `f = mg sin theta`,if `theta gt` angle of repose , the block will move , `f = mu mg cos theta` `a = (m g sin theta - mu m g cos theta)/(m)` `g sin theta - mu g cos theta` and `F_("net") = ma` further `mu = tan alpha` `(alpha = "angle of repose" = 45^(@)` `= tan 45^(@) = 1` |
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| 492. |
In the diagram shown coefficient of friction between 2 kg block and 4 kg block is `mu =0.1 ` and between 4 kg and ground is `mu =0.2 ` then answer the following quwstions .(tis time) The graph of frication force acting on 4 kg block and ground as a function of timeA. B. C. D. |
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Answer» Correct Answer - a |
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| 493. |
Angle of repose for a rough inclined plan is `60^(@)` The coefficient of friction is.A. `sqrt2`B. `1//sqrt3`C. `1`D. zero |
| Answer» Correct Answer - a | |
| 494. |
A massless spring balance is attached to 2 kg trolley and is used to pull the trolley along a flat smooth surface as shown in the figure. The reading on the spring balance is 10kg during the motion. The acceleration of the trolley is (Use g=9.8 `ms^(-2)`) A. `4.9ms^(-2)`B. `9.8ms^(-2)`C. `49ms^(-2)`D. `98ms^(-2)` |
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Answer» Correct Answer - c |
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| 495. |
A block of mass 3kg is at rest on a rough inclined plan as shown in the Figure. The magnitude of net force exerted by the surface on the block will be A. `15sqrt(3)N`B. 15 NC. 10 ND. 30 N |
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Answer» Correct Answer - D (d) Since, the block is at rest under tow forces (i) weight of block. (ii) contact force from the plane (resultant of force of friction and normal reaction). Contact force should be equal to weight (or 30N) in upward direction. Because under the action of two forces, a body remains in equilibrium when both the forces are equal in magnitude but opposite in direction. |
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| 496. |
A time varying force is applied on a block placed over a rough surface as shown in figure. Let `theta` be the angle between net contact force on the block and the normal reaction, then with time, `theta` will A. remain constantB. first increase to a maximum value (say `theta _(max))` and then becomes constant in a value less then `theta_(max)`C. first decrease to a minimum value (say `theta_(min)`) and then becomes constant in a value more then `theta_(min)`D. None of the above |
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Answer» Correct Answer - B |
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| 497. |
In figure (i) surface is rough and net force on the block has magnitude `F_(1)` . In figure (ii) surface is smooth and net force on the block has magnitude `F_(2)`. In figure (iii) again surface is rough and net force has magnitude `F_(3)`. In cases (i) block is moving up and in cases (ii) and (iii) block is moving down. `[theta gt tan^(-1) (mu)` for (i) and (iii)] Then, A. `F_(1) gt F_(2) gt F_(3)`B. `F_(2) gt F_(3) gt F_(1)`C. `F_(3) gt F_(2) gt F_(1)`D. `F_(1) gt f_(3) gt F_(2)` |
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Answer» Correct Answer - A |
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| 498. |
A 2 kg mass placed on a level table and is attached by a string passing over the edge of a table as illustrated in the diagram. (i) Calculate the magnitude of acceleration of the system. (ii) Calculate the tension in the string. |
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Answer» From FBD in `m_(1)` `T-m_(1)a=0` `implies T-2a=0`……(i) FBD in `m_(2)` `m_(2)g-T=m_(2)a` `5g-T=5a` Putting T=2a from Eq. (i) 5g-2a=5a 7a=5g `a=(5)/(7)xx10=(50)/(7) ms^(-2)` Again Tension in the string `T=2a=2xx(50)/(7)=(10)/(7) N` |
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| 499. |
A wedge of mass 2m and a cube of mass m are shown in figure. Between cube and wedge, there is no friction. The minimum coefficient of friction between wedge and ground so that wedge does not move is A. `0.10`B. `0.20`C. 0.25D. `0.50` |
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Answer» Correct Answer - b |
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| 500. |
a Block of mass m is placed on a wedge of mass M there is no friction between the block and the wedge .The minmum coefficient of friction between the wedge and the gound so that wedge does not move is, A. `(mcos^(2)theta)/(M+mcos^(2) theta)`B. `(msin^(2)theta)/(M+mcos^(2) theta)`C. `(mcos thetasintheta)/(M+mcos^(2) theta)`D. `(mcos thetasintheta)/(M+msin^(2) theta)` |
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Answer» Correct Answer - C |
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