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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.
| 51. |
A block of mass 5 kg is kept on a horizontal , floor having cofficient of friction 0.09 Two mutually perpendicular horizontal forces of 3 n and block is `(0.x)m//s^(2) `Find value of `x.(g=10m//s^(2))` |
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Answer» Correct Answer - 1 |
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| 52. |
A small objective placed on a rotating horizontal turn table just slips when it is placed at a distance 4cm from the axis of rotation. If the angular velocity of the trun-table doubled, the objective slip when its distance from the axis of ratation is.A. 1cmB. 2cmC. 4cmD. 8cm |
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Answer» Correct Answer - A The object will slip if centripal force `le` "force of friciton" `mromega^(2) ge mu mg` `r omega^(2) ge mug` `r omega^(2) ge "constant", or ((r_(1))/(r_(2)))=((omega_(2))/(omega_(1)))^(2)` `(4cm)/(r_(2))=((2omega)/(omega))^(2) therefore r_(2)=1cm` |
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| 53. |
If an electron is subjected to a force of `10^(-25)` N is an X-ray machine, then find out the time taken by the electron to cover a distance of 0.2m. Take mass of the electron `10^(-30)` kg. |
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Answer» The acceleration of the electron `a=(F)/(m)=(10^(-25))/(10^(-30))=10^(5) ms^(-2)` The time taken by the elecron (t) to cover the distance (s) of 0.2m can be given by `s=ut+1/(2)at^(2)` `implies 0.2=0+(1)/(2)xx10^(5)xxt^(2)` `implies t^(2)=0.4xx10^(-5)=4xx10^(-6)implies t=2xx10^(-3)s` |
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| 54. |
Assertion:On a merry-go-around, all parts of our body are subjected to an inward force. Reason: We have a feeling of being pushed outward the direction of impending motion.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 The object concept of force in physics should not be confused with the subjective concept of the feeling of force. Both statements are true. |
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| 55. |
A body of mass 0.40 kg moving initially with a constant speed of `10 m//s` to the north is subjected to a constant force of 8.0 N directed towards the south for 30 s Take the instant the force is applied to be t = 0, and the position of the particle at that time to be x = 0, predict its position at `t = -5 s , 25 s, 100 s` ? |
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Answer» Here, m = 0.40 kg , u = 10 `ms^(-1)` , F = -8 N (retarding force) As a = F / m = 8 / 0.4 = - 20 `ms^(-2)` Also , S = `ut + 1/2 aT^(2)` (i) Position at t = - 5s S = 10 (-5) + 1/2 `xx 0 xx (-5)^(2)` = - 50 m (ii) Position at t = 25 s `S_1` = `10 xx 25 + 1/2 xx (-20) xx (25)^(2)` = - 6000 m = - 6 km (iii) Position at t = 100 s `S_2` = ` 10 xx 30 + 1/2 xx (-20) xx (30)^(2)` = - 8700 m At t = 30 s , v = u + at v = 10 - 20 `xx` 30 = -590 `ms^(-1)` Now , for motion from 30 s to 100 s `S_3` = - 590 `xx 70 + 1/2(0) xx (70)^(2)` = - 41300 m Total distance = `S_2 + S_3` = - 8700 - 41300 = - 50000 m = - 50 km . |
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| 56. |
A body of mass 0 .40 kg moving intially with a constant speed of `10 m//s` to the north is subjected to a constant force of 8.0 N directed towards the south for 30 s Take the instant the force is applied to be t = 0 , and the position of the particle at that time to be x = 0 , predict its position at `t = -5 s , 25 s , 100 s` ? |
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Answer» Here , ` m = 0.4 kg u = 10 m//s "due" N,F = - 8.0 N` `a = (F)/(m) = (-8.0)/(0.40) = - 20 ms^(-2)` for `0 le t le 30 s ` (i) At `t = - 5 s , x = ut = 10 xx (-5) = - 50 m ` (ii) At `t = 25 s , x = ut + (1)/(2) at^(2) = 10 xx 25 (1)/(2) (-20) (25)^(2) = - 600 m` (iii) At `t = 100 s`, The problem is divided into two parts Upto 30 s there is force/acc. `:.` from `x_(1) = ut + (1)/(2) at^(2) = 10 xx 30 (1)/(2) (-20) (30)^2) = - 8700 m` At `t = 30 s,upsilon = u + "at" = 10 - 20 xx 30 = - 590 m//s, :.` for motion from 30 s to 100 s ` x_(2) = vt = - 590 xx 70 = - 41300 m. :. x = x_(1) + x_(2) = - 8700 - 41300 = - 50000 m = - 50 km`. |
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| 57. |
A rocket with a lift off mass 20000 kg is blasted upwards with a net initial acceleration of `5 ms^(-2)` Calculate the initial thrust (force) of the blast. |
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Answer» Here , m = 20000 kg = `2.0 xx 10^(4)` kg initial acceleration = `5 ms^(-2)` Clearly, the thrust should be such that it overcome the force of gravity besides giving it an upward acceleration of `5 ms^(-2)` Thus the force should produce a net acceleration of 9.8 + 5.0 = 14.8 `ms^(-2)` . Since, thrust = force mass `xx` acceleration f = `2.0 xx 10^(4) xx 14.8 ` = ` 2.96 xx 10^(5)` N . |
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| 58. |
A truck starts from rest and accelerate uniformly with `2 ms^(-2)`. At t = 10 s a stone is dropped by a person standing on the top of the truck (6 m high from ground). What are the (a) velocity and (b) acceleration of the stone at t = 11 s ? Neglect air resistance. |
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Answer» (a) (b) The moment the stone is dropped from the car, horizontal force on the stone is zero. The only acceleration of the stone is that due to gravity. This gives a vertically downward acceleration of `10 ms^(-2)` . This is also the net acceleration of the stone. |
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| 59. |
All surfaces shown in figure are smooth system is released with the spring unstretched In equilibrium, compression in the spring will be A. `(mg)/(sqrt(2)K)`B. `(2mg)/(k)`C. `((M+m)g)/(sqrt(2)k)`D. `(mg)/(k)` |
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Answer» Correct Answer - d |
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| 60. |
Which of the following statements is incorrect?A. A cricketer moves his hands backwards while holding a catch.B. A person from Hing from a certain height receives more iniuries when he falls on a cemented floor than when he falls on a heap of sand.C. It is easier lo push a lawn mower than to pull it.D. Mountain roads are generally made winding upwards rather than going strainght up. |
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Answer» Correct Answer - C It is easier to pull lawn mowe than to push it. All other statements are true. |
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| 61. |
A rocket is going upward with acceleration motion. A man string in it feels his weight increased 5 time his own weight. If the mass of the rocket including that of the man is `1.0xx10^(4)kg` , how much force is being applied by rocket engine? `(Take g=10ms^(-2))` .A. `5xx10^(4)N`B. `5xx10^(5)N`C. `5xx10^(8)N`D. `2xx10^(4)N` |
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Answer» Correct Answer - B As the weight of man increased by 5 times, so acceleration of the rockets, `a=5g=5 xx 10=50ms^(-2)` Force applied by roccket engine is `F=ma=1.0xx10^(4)xx50=5xx10^(4)N` |
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| 62. |
A cyclist speeding at `18 km// h` on a level road takes a sharp circular turn of radius 3 m without reducing the speed . The coefficient of static friction between the tyres and the road is 0.1 Will the cyclist slip while taking the turn ? |
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Answer» Correct Answer - `Yes` . On a level road centripetal force is provided only by force of friction Therefore, max safe speed `upsilon_(max) = sqrt(mu r g) = sqrt( 0.1 xx 3 xx 9.8) = sqrt(2.94` Actual speed, `upsilon, upsilon = 18km//h= 5m//s` Hence the cyclist will slip . |
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| 63. |
A sphere is rotatiing between a a rough wall and a smooth wedge as shown in figure If `mu` is the cofficent of frication between wall and sphere then normal reation between wall and sphere will A. increase if `mu` increasesB. decreases if `mu` increasesC. increases if `theta` increasesD. decreses if `theta` increases |
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Answer» Correct Answer - b.c |
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| 64. |
A cyclist speeding at `6 m//s` in a circle of 36 m diameter makes an angle 0 with the verticl . What is the value of 0 ? Also , deter - mine the minimum possible value of the coefficient of friction between the tyres and the road . |
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Answer» Here , ` upsilon = 6 m//s , r = (36)/(2) = 8 m` ` theta = ? mu = ? ` As ` tan theta = upsilon^(2)/(rg) = (6 xx 6 )/(18 xx 9.8 )=(2)/(9.8) = 0.2040 ` ` theta = tan^(-1) (0.2040) = 11^(@) 32 ` Also ` mu = tan theta = 0.2040` . |
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| 65. |
A 60 kg body is pushed with just enough force to start it moving across a floor and the same force continues to act afterwards. The coefficient of static friction and sliding friction are 0.5 and 0.4 respectively. The acceleration of the body isA. `1 ms^(-2)`B. `3.9 ms^(-2)`C. `4.9 ms^(-2)`D. `6 ms^(-2)` |
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Answer» Correct Answer - A (a) Kinetic friction `-mu_(k)R=mu_(k) mg=0.4xx60xx10=240N` and the limiting friction `=mu_(s)R=mu_(s) mg=0.5xx60xx10=300N` So, the force applied on the body is 300N and if the body is moving afterwards with the same force, then Net accelerating force =Applied force- Kinetic force `ma=300-240` `a=(60)/(m)=(60)/(60)=1 ms^(-1)` |
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| 66. |
A 60kg is pushed horizontaly with just enough force to start it moving across a floor and the same force continues to act afterwards. The coefficient of static friction and sliding friction are 0.5and 0.4 respectively the accleration of the body isA. `6m//s^(2)`B. `2m//s^(2)`C. `3m//s^(2)`D. `1m//s^(2)` |
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Answer» Correct Answer - d |
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| 67. |
A body continues to move along the same straight line some external force comples it to change its direction of motion Rather the body moving along a straight line opposes the force that tries to change its straight line the path This is well known property of inertia of direction Read the above passage and answer the following questions: (i) Mud gurads over the wheels of an auto save us from mud .How ? (ii) What are the implication of this study in day to day life? |
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Answer» (i) The function of mud guards over the wheels of an auto is based on inertia of direction When the auto happens to pass through mud the mud particles fly along the tangent to the weels Theyare stoped by the mud gurads before they spoil our cloths. (ii) In day to day life we find that some external effort in the form of motivation is always required to change any of our usal routine/habits If we are used to some sort of seting/convenience we ourselves resist the effort/motivation for change Consistent efforts and motivation are essential for bringing about any change . |
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| 68. |
When a ball is thrown upwards . Its momen tum first decreases and the increases Is conservation of linear momentum violated in this process? |
| Answer» No the momentum conservtion principle applies only when no external forces act on the system . | |
| 69. |
In the arrangement shown, the mass m will ascend with an acceleration ( Pulley and rope are massless) A. zeroB. `(g)/(5)`C. gD. 2g |
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Answer» Correct Answer - B (b) Acceleration `a=[(m_(2)-m_(1))/(m_(1)+m_(2))]" " g=([(3)/(2)m-m]g)/([(3)/(2)m+m])=(g)/(5)` |
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| 70. |
Two blocks are in contact on a frictionless table. One has mass m and the other 2m. A force F is applied on 2m as shown in the figure. Now the same force F is applied from the right on m. In the two cases the ratio of force of contact between the two blocks will be A. sameB. `1:2`C. `2:1`D. `1:3` |
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Answer» Correct Answer - B (b) Acceleration in both the case will be same.l `N_(1)=ma " but " N_(2)=(2m)a` `therefore " " (N_(1))/(N_(2))=(1)/(2)` |
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| 71. |
Three blocks are placed at rest on a smooth inclined plane with force acting on `m_(1)` parallel to the inclined plane. Find the contace force between `m_(2) " and " m_(3)` A. `((m_(1)+m_(2)+m_(3))F)/(m_(3))`B. `(m_(3)F)/(m_(1)+m_(2)+m_(3))`C. `F-(m_(1)+m_(2))g`D. None of the above |
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Answer» Correct Answer - B (b) Acceleration of system, `a=(" Net pushing force")/(" Total mass")` or `a=(F-(m_(1)+m_(2)+m_(3))g sin theta)/((m_(1)+m_(2)+m_(3)))` Equation of motion for `m_(3)` `N-m_(3)g sin theta=m_(3)a` or `N=m_(3)g sin theta +m_(3)[(F-(m_(1)+m_(2)+(m_(3))/((m_(1)+m_(2)+m_(3)))]` `=(m_(3)F)/(m_(1)+m_(2)+m_(3))` |
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| 72. |
A stone is tied to a weightless string and revolved in a vertical circle of radius 5 m . What should be the minimum speed of the stone at the highest point of the circle so that the string does not slack ? What should be the speed of the stone at the lowest point of vertical circle ? Take ` g = 9.8 ms^(2) ` . |
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Answer» Here, ` r = 5 m , upsilon_(H) = ? ` ` upsilon_(L) = ? G = 9.8 m//s^(2) ` The string will not slack at the highest point when ` upsilon_(H) = sqrt(gr )= sqrt(9.8 xx 5 ) = sqrt( 49) = 7 m//s ` In that case , ` upsilon_(L) = sqrt(5 gr )= sqrt( 5) sqrtgr = 7 sqrt(5) m // s ` . |
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| 73. |
Assertion : The maximum speed with which a vehicle can go round a level curve of diameter `20m` without skidding is `sqrt10m//s` given `mu =0.1` Reason : If follows from `upsilon le sqrt(mu r g) .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 `upsilon= sqrt(mu rg) = sqrt(0.1 xx 10xx10) =sqrt10m//s` Both the statements are ture and statement-2 is correct explantion of statement- 1. |
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| 74. |
Assertion : A ball of mass `100g` hits a bat with a speed of `72km//hr` and bounces back with the same speed in one second The force exerted by the bat on the ball is `4N` Reason : It follows from `F =ma` .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 - b Here, `m = 100 g = 10^(-1) kg, u = 72km//hr =20m//s` `= (10^(-1) (20 + 20))/(1) = 4N` The statement -1 is true Statement-2 is also true but it does not explaine the assertion properly . |
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| 75. |
The mass of a bicycle rider along with the bicycle is 100 kg. he wants to cross over a circular turn of radius 100 m with a speed of `10 ms^(-1)`. If the coefficient of friction between the tyres and the road is 0.6, will the rider be able to cross the turn? Take `g = 10 ms^(-2)`.A. 300NB. 600NC. 1200ND. 150N |
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Answer» Correct Answer - B Centripetal force `=(mv)^(2)/(r)=(100xx10xx10)/(100)=100N` Required frictional force to cross the turn `=mumg=0.6xx100xx10=600N` As the friciton force is greater than the centripetal force, so the rider will be able to cross the turn. |
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| 76. |
A bird is sitting on the floor of a closed glass cage and the cage is in the hand of a girl. Will the girl experience any change in the weight of the cage when the bird (i) starts flying in the cage with a constant velocity (ii) flies upwards with acceleration (iii) flies downwards with acceleration ? |
| Answer» In a closed glass cage , air inside is bound with the cage . Therefore . (i) there would be no change in weight of the cage if the bird flies with a constant velocity . (ii) the cage becomses heavier , when bird flies upwards with an accelertion . (iii) the cage appears lighter when bird flies downwards with an acceleration . | |
| 77. |
Why buffers are provided between the bogies of a train ? |
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Answer» When a train starts or stops , buffer springs increase the time of impact between the bogies ` F = ("impulse")/(t)` decreases . Thereforce , strongjerks to the passengers are avoided . |
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| 78. |
A bob of mass 0.1 kg hung from the ceilling of a room by a string 2 m long is set into osillation The speed of the bob at its mean position ` 1 m//s ` What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme position (b) at is mean position ? |
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Answer» (a) We shall study in unit 10 that at each extreme position velocity of the bob is zero If the string isc cut at the extreme position bob is only under the action of g Hence the bob will fall vertically downwards (b) At the mean position velocity of the bob is `1 ms^(-1)` along the tangent to the arc which is in the horizontal direction If the string is cut at mean position the bob will behave as a horizontal projectile Hence it will follow a parabolic path . |
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| 79. |
A ball is suspended by a cord from the ceilling of a motor car What will be the effect on the position of the ball , if (i) the car is moving with unifrom velocity (ii) car is accelerated (iii) car is turning towards left ? |
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Answer» (i) The ball will remain suspended vertically only . (ii) The ball will move in the backward direction (iii) The ball will move towards right . |
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| 80. |
In the diagram shown in figure, match the following `(g=10 m//s^(2))` `{:(,"Column-I",,,"Coloumn-II"),((A),"Acceleration of 2kg block",,(P), "8 Si unit"),((B),"Net force on 3kg block",,(Q),"25 SI unit"),((C),"Normal reaction between 2 kg and 1kg",,(R),"2 SI unit"),((D),"Normal reaction between 3 kg and 2kg " ,,(S)," 45 SI unit"),(,,,(T)," None"):}` |
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Answer» Correct Answer - A`rarr`r,B`rarr`t,C`rarr`q,D `rarr`t Acceleration of system, `a=(60-18-(m_(1)+m_(2)+m_(3))g sin 30^(@))/((m_(1)+m_(2)+m_(3))=2 ms^(-2)` Net force on 3 kg block=`m_(3)a=6N` From free body diagram of 3 kg block, we have `N_(12)-m_(1)g sin 30^(@)-18=m_(1)a implies N_(12)=25N` From free body diagram of 3 kg block, we have `60-m_(3)g sin 30^(@)-N_(32)=m_(3)a` `therefore" " N_(12)=39N` |
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| 81. |
Why is it difficult to move a bike with its brakes on ? |
| Answer» When the brakes of a bik are on its wheels cannot rotate They will simply skid Thus rolling friction will be converted into sliding friction which is comparatively larger . Hence it friction which is comparatively larger Hence it becomes difficult to move the bike . | |
| 82. |
How do you account for the function of mud guards ? |
| Answer» Due to directional inertia , the rotating wheels of any vehicle throw out mud , if any tangentially . The mud guards over the wheels stop this mud protecting the clothes etc . Of the driver of the bike or motor bike . | |
| 83. |
Assertion : A body droped from a given height and another body projected horizontal from the same height strike the ground simultaneously Reason : Because horizontal velocity has no effect in the vertical direction .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 Component of horizontal velocity `(u)` along the verticle ` =cos 90^(@) =0` Therefore vertical motion is not affected Time taken by the two bodies to strike the ground is the same . |
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| 84. |
A thin circular wire of radius `R` rotates about its vertical diameter with an angular frequency `omega`. Show that a small bead on the wire remain at its lowermost point for `omegalesqrt(g//R)` . What is angle made by the radius vector joining the center to the bead with the vertical downward direction for `omega=sqrt(2g//R)` ? Neglect friction. |
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Answer» Let the radius vector joining the bead to the centre of the wire make an angle `theta` with the vertical downward direction. If N is normal reaction . Then from fig. ltbRgt mg = Ncos `theta` … (i) mr`omega^(2)` = N sin`theta` … (ii) or m(R sin`theta` )`omega^(2)` = N sin`theta` or mR`omega^(2)` = N From equation (i), mg = mR`omega^(2)`cos`theta` or cos`theta` = g/ R`omega^(2)` ... (iii) As `` `le` 1 , therefore, bead will remain at its lowermost point for g/R`omega^(2)` `le` 1 or `omega` `le` `sqrt g/R` When `omega` = `sqrt 2g/R` from equation (iii), cos`theta` = g/R(R/2g) = 1/2 `therefore` `theta` = `60^(@)`. |
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| 85. |
Two block of masses `2kg` and `4kg` are released from rest over a smooth inclined place of inclination `30^(@)` as shown in figure what is the normal force between the two blocks? |
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Answer» Since surface is smooth acceleration of both is `g sin theta = 10 sin 30^(@) = 5 m//s^(2)` down the plane. The component of `mg` down the plane `(= mg sin theta)` provides this acceleration. So normal reation will be zero. |
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| 86. |
Three forces acting on a body are shown in figure. To have the resultant force only along the y-directon, the magnitude of the minimum additional force needed si A. 0.5NB. 1.5NC. `(sqrt(3))/(4)N`D. `sqrt(3)N` |
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Answer» Correct Answer - A (a) Minimum additional force needed `F=-(F_("resultant"))_(x)` `F_("resultant")=[(4-2)(cos 30 jhatj-sin 30 hati)+( cos 60 hati +sin 60 hatj)]` `=[2((sqrt(3))/(2)hatj-(1)/(2)hati)+((1)/(2) hati+(sqrt(3))/(2) hatj)]` `=[(sqrt(3)+(sqrt(3))/(2))hatj+(- hati+(hati)/(2))]` `=[-(1)/(2)hati+(3sqrt(3))/(2) hatj]=-(1)/(2)+(3sqrt(3))/(2)hatj` `implies F_(x)=(-( hati)/(2))` Hence, `|F|=0.5N` |
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| 87. |
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth horizontal surface are tied to the ends of a light string A horizontal force F = 600 N is applied to (i) A and (ii) B along the direction of string. What is the tension in the string in each case ? |
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Answer» Acceleration = 600 N / 10 kg + 20 kg = 20 `ms^(-2)` (i) When force is applied on 10 kg mass 600 - T = `10 xx 20` or T = 400 N (ii) When force is applied on 20 kg mass, 600 - T = `20 xx 20` or T = 200 N |
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| 88. |
Two blocks are connected over a massless pulley as shown in figure. The mass of block A is 10 kg and the coefficient of kinetic friction is 0.2 Block A sliders down the incline at constant speed. The mass of block B in kg is A. 5.4B. 3.3C. 4.2D. 6.8 |
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Answer» Correct Answer - B (b) Net pulling force on the system should be zero, as velocity is constant. Hence, `m_(A)g sin 30^(@)=mu m_(A)g cos 30^(@)+m_(B)g` `therefore" " m_(B)=((m_(A))/(2))-((mu m_(A)sqrt(3))/(2))` `=10[(1)/(2)-0.2xx(sqrt(3))/(2)]=3.3 kg ` |
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| 89. |
Air is thrown on a sail attached to a boat from an electric fan placed on the boat . Will the boat start moving ? |
| Answer» No when the fan pushes the sail by air also pushes the fan in the opposite direction . Since fan is placed on the boat , the vector sum of linear momenta of fan and boat is zero . The boat can move only under the force of reaction from some external agency . | |
| 90. |
A stone of mass 0.05 kg is thrown vertically upwards. What is the direction and magnitude of net force on the stone during its upward motion?A. 0.98 N vertically downwardsB. 0.49 N vertically downwardsC. 9.8 N vertically downwardsD. 0.49 N vertically upwards |
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Answer» Correct Answer - B (b) Given, m=0.05 kg We know that, net force acting on a stone, `f_(g)=mg=0.05xx(-9.8)="-0.49N"` =0.49 N vertically downwards |
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| 91. |
An engine of `100H.P` draws a train of mass `100` metic ton with a velocity of `36km h^(-1)` Find the coefficient of friction. |
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Answer» Correct Answer - `0.0076` Here `P = 100 HP = 100 xx 746 "watt"` `upsilon =36km h^(-1) = 10ms^(-1)` If `F` is frictional force overcome by the engine then `P = F xx upsilon F = (P)/(upsilon) = (100 xx 746)/(10) = 7460 N` `R = mg = 100 xx 1000 xx 9.8 N = 9.8 xx 10^(5)` `mu = (F)/(R) = (7460)/(9.8 xx 10^(5)) = 0.0076` . |
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| 92. |
In the given figurethe wedge is fixed, pulley is frictionless and string is light. Surface AB is frictionless whereas AC is rough. If the block of mass 3m. Slides down. With conttant velocity, then the coefficient of friction between surface AC and the block is A. `(1)/(3)`B. `(2)/(3)`C. `(1)/(2)`D. `(4)/(3)` |
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Answer» Correct Answer - b |
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| 93. |
Two block of mass `1 kg` and `2 kg` are connected by a string `AB` of mass `1 kg`. The blocks are placed on a smooth horizental surfacec. Block of mass `1 kg` is pulled by a horizental force `F` of magnitude `8 N`. Find the tension in the string at points `A` and `B` |
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Answer» `a = ("Net pulling force")/("Total mass") = (8)/(1 + 2 + 1)` `= 2 m//s^(2)` `T_(B)` `8 - T_(B) = m_(1)a = (1) (2)` `:. T_(B) = 6 N` `T_(A) T_(A) = m_(1)a = (2) (2)` ` :. = 4 N` |
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| 94. |
A block of mass `4kg` is placed on a rough horizontal plane A time dependent force `F = kt^(2)` acts on the block where `k = 2N//s^(2)` Coefficient of friction `mu = 0.8` force of friction between the block and the plane at `t = 2s` isA. `32N`B. `4N`C. `2N`D. `8N` |
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Answer» Correct Answer - d Maximum force of friction `f_(max) = mu mg = 0.8 xx 4 xx 10 =32N` and applied force at `t =2 s is ` `F = kt^(2) = 2 (2)^(2) = 8 N` The body fails to move Hence the force of friction at `t =2s` is equal to applied force `=8N` . |
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| 95. |
A small stone of mass 0.2 kg tied to a massless , inextensible string is rotated in a vertical circle of radius 2 m If the particle is just able to complete the vertical circle what is its speed at the highest point of the circular path ? How would the speed get affected if the mass of the stone is increased by 50% ? Take ` g = 10 m//s^(2) ` . |
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Answer» Here , ` m = 0.2 kg , r = 2 m` When the particle is just able to complete the vertical circle , `upsilon_(L) = sqrt(5 g r ) ` and `upsilon_(H) = sqrt(g r ) = sqrt(10 xx 2 ) = 4 .47 ms^(-1) ` As `upsilon_(H)` does not depend upon mass , m of the stone , therefore , value of `upsilon_(H)` will not be affected by any change in mass of the stone . |
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| 96. |
A man wants to slide down a block of mass `m` which is kept on a fixed inclined plane of inclination `30^(@)` as shown in the figure .Initially the block is not sliding . To just start sliding the man pushed the block down the inclined with a force `F`. Now , the block starts accelerating. To move it downwards with conatant speed the man starts pulling the block with same force. Surface are such that ratio of maximum static friction to kinetic friction is `2` Now, answer the following questions. What minimum force is required to move it up the incline with constant speed?A. `(2)/(3) mg`B. `(mg)/(2)`C. `(7 mg)/(6)`D. `(5 mg)/(6)` |
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Answer» Correct Answer - D `F' = mg sin theta + mu_(k) mg cos theta` `= (mg)/(2) + ((2)/(3 sqrt(3))) mg ((sqrt3)/(2))` `= (5 mg)/(6)` |
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| 97. |
Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a constant speed (b) a crok of mass 10 g floating on water (c) a kit skilfully held stationary in the sky (d) a car moving with a constant velocity of `30 kh//h` on a rough road (e) a high speed electron in space for from all gravitational obects and free of electric and magnetic fields. |
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Answer» (a) As the rain drop I falling with a constant speed its acceleration a = 0 Hence net force F = ma = 0 (b) As the cork is floating on water its weight is being balanced by the upthrust (equal to weight of water displaced) Hence net force on the cork is zero (c) As the kite is held stationary net force on the kite is zero in accordance with Newton s first law (d) Force is being applied to overcome the force of friction But as velocity of the car is constant its acc a = 0 Hence net force on the car F = ma = 0 (e) As no fied (gravitational/electric/magnetic ) is action on the electron net force on it is zero. |
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| 98. |
A cork of mass 10 g is floating on water. The net force acting on the cork isA. 10NB. `10^(-3)N`C. `10^(-2)N`D. zero |
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Answer» Correct Answer - D When the cork is floating its weight is balanced by the upthrust. Therefore net force on the cork is zero. |
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| 99. |
Detrmine the maximum acceleration of the train in which a box lying on the floor will remain stationary given that the coefficient of static friction between the box and the train s floor is 0.15 given `g = 10 m//s^(2)` . |
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Answer» As acceleration (a) of box is due to static friction , therefore, `ma = f le mu_(s) R le mu_(s) mg` `a le mu_(s) g ` or ` a_(max) = mu_(s) g = 0. 15 xx 10 = 1.5 m// s^(2)` . |
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| 100. |
A stone of mass l kg is lying on the floor of a train which is accelerating with `1ms^(-2)` The net force acting on the stone isA. zeroB. 1NC. 5ND. 10N |
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Answer» Correct Answer - B Here, mass of the stone, m=1kg As the stone is lying on the floor of the train, its acceleration is same as that of train `therefore "Force acting on the stone," F=ma=(1kg)(1ms^(-2))=1N` NOTE: Here weight of the stone is balanced by the normal reaction of hte floor. |
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