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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.
| 251. |
“Don’t throw over the roads, they may cause slip” is applicable to A) bananas B) oils C) sand D) above all |
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Answer» Correct option is D) above all |
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| 252. |
You wanted to know effect of roughness on frictional force. Which of the following material is not needed for your experiment? A) Inclined plane B) Rough cloth C) Stop watch D) Ball |
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Answer» C) Stop watch |
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| 253. |
Why are Aeroplanes constructed as shown in figure?A) To overcome fluid friction B) To overcome rolling friction C) To overcome static friction D) To overcome sliding friction |
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Answer» A) To overcome fluid friction |
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| 254. |
Why is friction said to be a necessary evil? |
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Answer» Friction is all pervasive. Even a simple task; like writing on a page; cannot be done without friction. Walking on road won’t be possible because friction. But friction causes wear and tear of materials, and wastage of energy. Thus, friction is beneficial as well as harmful. Hence, friction is called a necessary evil. |
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| 255. |
Name the force which helps things to move and stop. |
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Answer» Frictional forces helps things to move and stop. |
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| 256. |
Cartilage is present in the joints of our body, which helps in their smooth movement. With advancing age, this cartilage wears off. How would this affect the movement of joints? |
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Answer» The wearing off of cartilage will increase the friction. As a result the movement of joints will become difficult which may lead to joint pains. |
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| 257. |
Which of the following play major role to avoid corrosion due to Motion? A) Paints B) Lubricants C) Alloys D) Galvanizing |
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Answer» Correct option is B) Lubricants |
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| 258. |
Which experiment do you perform on friction from the following equipments? Trolley, Weights, Weight hangers, Pulley, Inextensible string and table. A) Friction is proportional to the normal force B) To understand nature of friction and the concept of static friction C) Effect of roughness on frictional force D) Frictional force is independent of the area of contact |
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Answer» B) To understand nature of friction and the concept of static friction |
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| 259. |
“Friction is independent of weight, but depends on normal force between surfaces of contact where friction exists.” Do you agree with this statement? Discuss. |
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Answer» Yes, I agree with this statement. The above experiment helps us to come to this conclusion. In both cases we used the same brick hence the weight doesn’t change. But when its positions were changed the frictions at each case is changed. |
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| 260. |
A 60 kg weight is dragged on a horizontal surface by a rope upto 2 metres. If coefficient of friction is `mu=0.5` the angle of rope with the surface is `60^(@)`and `g=9.8m//sec^(2),` then work done isA. 294 joulesB. 315 joulesC. 588 joulesD. 197 joules |
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Answer» Correct Answer - B |
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| 261. |
Friction is said to be a Mend, because it helpsA) to walk on a road B) to hear a sound C) to sea a book D) to think a problem |
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Answer» A) to walk on a road |
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| 262. |
Directions of ‘a’, ‘b’ refers asA) weight (a), friction (b) B) friction (a), weight (b) C) friction (b), movement (a) D) movement (a), friction (b) |
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Answer» D) movement (a), friction (b) |
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| 263. |
The wrong in diagram isA) F B) W C) f D) f and W |
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Answer» Correct option is D) f and W |
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| 264. |
What is rolling friction? |
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Answer» When one body rolls over the surface of the other body the friction offered is called rolling friction. |
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| 265. |
What is meant by ‘rolling friction’? |
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Answer» Rolling friction is the friction or drag which opposes the motion when a body rolls. |
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| 266. |
Iqbal has to push a lighter box and Seema has to push a similar heavier box on the same floor. Who will have to apply a larger force and why? |
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Answer» Seema will have to apply larger force since she will experience more frictional force because heavy object will be pressed hard against the opposite surface and produces more friction. |
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| 267. |
Push a heavy box with a small force. It does not move. Now gradually increase the force, the box till does not move. From this experiment your conclusion is A) force increases, friction increases B) force increases, friction decreasesC) force increases, no change in friction D) none of the above |
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Answer» A) force increases, friction increases |
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| 268. |
Identifying forces acting on a body and effect of frictional force. Gently push a book on a horizontal floor. You may observe that the book acquires certain speed in the direction of push. However, the speed of the book gradually decreases and after sometime it stops.a) Why does the book stop after covering some distance?b) Is the book moving with uniform speed?c) Why does the speed of the book change gradually?d) How many forces act on the book when it is pushed on the floor?e) What are the forces acting on the book in the horizontal direction?f) What is the net force acting in the horizontal direction? |
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Answer» a) Some opposing force is acting to its motion. b) No, its speed decreases gradually c) Some net force is acting opposite direction to its motion. d) Two forces act on the vertical direction. They are 1. Weight of the book or gravitational force acting vertically down. 2. Normal force or reaction force applied by the floor vertically up. 3. Forces acting horizontal direction. e) 1. Muscular force applied by us. 2. Opposing force acting opposite to motion of book. f) The net force is the difference between muscular force and opposing force to the motion of book. |
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| 269. |
Why heavier box apply more frictional force than lighter box? |
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Answer» 1. We know that the force of friction is increased if the two surfaces are pressed harder. 2. A heavier box presses the ground harder to heavier box apply more frictional force. |
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| 270. |
Iqbal has to push a lighter box and Seema has to push a similar heavier box on the same floor. Who will have to apply a larger force and why? |
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Answer» Force of friction arises because of interlocking of irregularities on the two surfaces in contact. When a heavy object is placed on the floor, the interlocking of irregularities on the surfaces of box and floor become strong. This is because the two surfaces in contact are pressed harder. Hence, more force is required to overcome the interlocking. Thus, to push the heavier box, Seema has to apply a greater force than Iqbal. |
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| 271. |
Consider a car moving on a straight road with a speed of `100m//s`. The distance at which car can be stopped is `[mu_k=0.5]`A. `100 m`B. `400 m`C. `800 m`D. `1000 m` |
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Answer» Correct Answer - D `f = mu_(k) N = mu_(k) mg = ma rArr a = mu_(k) g = 0.5 xx 10 = 5 m//s^(2)` `v^(2) = u^(2) - 2as rArr 0 = (100)^(2) - 2 xx 5 s rArr s = 1000 m` |
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| 272. |
A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and table surface is 0.25, then the maximum fraction of length of the chain, that can hang over one edge of the table isA. 0.2B. 0.25C. 0.35D. 0.15 |
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Answer» Correct Answer - A |
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| 273. |
A block of mass `10kg` is placed on a rough horizontal surface having coefficient of friction `mu=0.5` . If a horizontal force of `100N` is acting on it, then acceleration of the will be.A. `0.5 m//s^(2)`B. `5m//s^(2)`C. `10m//s^(2)`D. `15m//s^(2)` |
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Answer» Correct Answer - B |
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| 274. |
The upper half of an inclined plane with inclination `phi` is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given byA. `mu=sintheta`B. `mu=costheta`C. `mu=2costheta`D. `mu=2tantheta` |
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Answer» Correct Answer - D |
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| 275. |
A long chain of mass of mass `M` length `L` is being pulled with constant velocity on a rough incline with coefficient of friction `mu`. What rate frictional force on chain is increasing A. `(3)/(10)mu(M)/(L)gv`B. `(5)/(6)mu(M)/(L)gv`C. zeroD. `mu(M)/(L)gv` |
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Answer» Correct Answer - A In a small time let `dx` length crossing from lower incline to upper incline `df=mu(M(dx))/(L)g cos 37^(@)-mu (M)/(L)dx cos 60^(@)` `df=mu(M)/(L)g (cos 37-cos 60^(@))dx` `(df)/(dt)=(muM)/(L)g((4)/(5)-(1)/(2))(dx)/(dt)=(3)/(10)(muM)/(L)gv` |
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| 276. |
A particle is projected up a `37^(@)` rough incline with velocity `v_(0)`. If `mu = 1//2`, the speed with which it returns back to the starting point is `v`. Then `v//v_(0)` isA. `(1)/(2)`B. `(1)/(sqrt(3))`C. `(1)/(sqrt(5))`D. `(1)/(sqrt(2))` |
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Answer» Correct Answer - C `0 = v_(0)^(2) -2 (g sin 37^(@) + mu g cos 37^(@))d` (i) `v^(2) = 2(g sin 37^(@) - mu g cos 37^(@))d` (ii) `(v)/(v_(0)) = (1)/(sqrt(5))` |
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| 277. |
A particle is projected up a rough inclined plane of inclination `theta` and friction coefficient `mu`. If the time of ascent is half of the time of descent, the friction coefficient `mu` is equal toA. `(3)/(4) tan theta`B. `(3)/(5) tan theta`C. `(4)/(5) tan theta`D. `(1)/(2) tan theta` |
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Answer» Correct Answer - B Upward motion : Retardation : `a_(1) = g sin theta + mu g cos theta , t_(1) = t` Downward motion : Acceleration : `a_(2) = g sin theta - mu g cos theta , t_(2) = 2t` `s = (1)/(2) a_(1) t_(2)^(2) = (1)/(2) a_(2) t_(2)^(2) rArr a_(1) (t)^(2) = a_(2) ( 2t)^(2)` `a_(1) = 4 a_(2)` `g sin theta + mu g cos theta = 4 ( g sin theta - mu g cos theta)` `mu = (3)/(5) tan theta` |
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| 278. |
A small object is kept on a groove on rough incline plane of inclination `theta`. Groove wakes an angle `alpha` as shown in diagram. `mu` coefficient of friction. Which of the following is correct A. Normal force by incline plane. `N=mg cos theta`B. Normal force by incline is `N=mgsqrt(cos^(2)theta+sin^(2)theta sin^(2) alpha)`C. maximum frictional force that can develop is `f_(max)=mumg cos theta`D. If `mu=0` then acceleration of blocks is `g sin theta cos alpha` |
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Answer» Correct Answer - B::D `N_(1)= mg cos theta` perpendicular to the plane `N_(2)= mg sin theta sin alpha` `N_(2)` is in the plane of incline perpendicular to groove. `N=sqrt(N_(1)^(2)+N_(2)^(2))` If `mu=0` Net force along groove. `mg sin theta cos alpha`. Hence acceleration is `g sin theta cos alpha`. |
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| 279. |
In order to stop a car in shortest distance on a horizontal road, one should(a) apply the brakes very hard so that the wheels st rotating(b) apply the brakes hard enough to just prevent slipprn(c) pump the brakes (press and release)(d) shut the engine off and not apply brakes. |
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Answer» (b) apply the brakes hard enough to just prevent slipprn Explanation: In order to stop a car in shortest distance on a horizontal road, Maximum possible force of friction by applying brakes is needed. Maximum force of friction can only be achieved when the surfaces in contact are just on the verge of slipping but do not slip. In this situation the limiting/maximum static force of friction come into action. So apply the brakes hard enough to just prevent slipping. |
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| 280. |
Two carts of masses `200 kg` and `300 kg` on horizontal rails are pushed apart. Suppose the coefficient of friction between the carts and the rails are same. If the `200 kg` cart travels a distance of `36 m` and stops. Then the distance travelled by the cart weighing `300 kg` is : (if both have same initial speed) .A. 32 mB. 24 mC. 16 mD. 12 m |
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Answer» Correct Answer - C |
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| 281. |
A fixed pulley carries a weightless thread with masses `m_(2) and m_(1)` at its ends `(m_(2) gt m_(1))`. There is friction between the thread and the pulley. If it is such that the thread starts slipping when the ratio of `m_(2)` to `m_(1)` is `n_(0)`. Find : (a) the friction coefficient (b) the acceleration of the masses when `(m_(2))/(m_(1))=n gt n_(0)` [Hint : Consider an element of the thread of width `d theta` at angular distance `theta `from the horzontal diameter of the pulley. The forces at the ends are T and T = dT. If dN is the normal reaction of the pulley on the element than df (frictional force on the element) `= mu dN` Resolving the tensions at the ends along and perpendicular to the radius vector of the element, we have for equilibrium of the element `dT = df and dN = T d theta` Integrate from `T_(1)` to `T_(2).T_(1)=m_(1)g and T_(2)=m_(2)g`. These are obtained by considering equilibrium of `m_(1) and m_(2)`] |
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Answer» Correct Answer - `mu-(ln n_(0))/(pi)a=(g(n-n_(0)))/((n+n_(0)))` |
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| 282. |
A block of mass 2kg is placed on a trolley of mass 20 kg free to move on a frictionless floor. What are the frictional force, accelerations of the block and trolley if the block is pulled by a force of 3 N ? The coefficient of friction between block and trolley is 0.25. What will be the values of all three if the block is pulled by 8 N force ? At what value of the applied force will the bodies just tend to separate ? [Hint : So long as the force applied does not produce the limiting frictional force the two will move with common acceleration. When the force applied exceeds that limiting value the block will move on the trolley with a different acceleration] |
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Answer» Correct Answer - `2.73 N,0.136 ms^(-2),0.136 ms^(-2)`,4.9 N,1.55 ms^(-2),0.245 ms^(-2),5.39 N` |
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| 283. |
The handle of a floor mop makes an angle `theta ` with the vertical. If `mu_(k) and mu_(s)` be the coefficients of kinetic and static friction between floor and mop, show that if `theta` is smaller than certain value `theta_(0)`, the mop cannot be made to slide over across the floor, no matter how great a force is directed along the handle towards the centre. What is the angle `theta_(0)` ? |
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Answer» Correct Answer - `theta_(0)=tan^(-1) mu_(s)` |
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| 284. |
A bullet of mass 20 kg travelling horizontally at `100 ms^(-1)` gets embedded at the centre of a block of wood of mass 1kg, suspended by a light vertical string of 1m in length. Calculate the maximum inclination of the string to the vertical. |
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Answer» Correct Answer - `37^(@)` |
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| 285. |
`M_(A) = 3 kg ,M_(B) = 4 kg ,M_(C) = 8 kg`. Coefficient of friction between any two surface is `0.25` pulley is frictionless and string is massless .A is connected to wall through a massless right rod. |
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Answer» Correct Answer - 8 kg.wt |
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| 286. |
Masses `M_(1),M_(2) and M_(3)` are connected by string of negligible mass which pass over massless and frictionless pulleys `P_(1) and P_(2)` as shown in figure 7.15. The masses move such that the string between `P_(1) and P_(2)` is parallel to the incline and the portion of the string between `P_(2) and M_(3)` is horizontal. The masses `M_(2) and M_(3)` are 4.0 kg each and the coefficient of kinetic friction between the masses and the surfces is 0.25. The inclined plane makes an angle of `37^(@)` with the horizontal. If the mass `M_(1)`moves downwards with a uniform velocity, find (i) the mass of `M_(1)`, (ii) the tension in the horizontal portion. `(g=9.8 ms^(-2) , sin 37^(@)=(3)/(5))` |
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Answer» Correct Answer - 4.2 kg (ii) 9.8 N |
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| 287. |
A 60 kg man leaps vertically from a crouching position when his centre of mass is 40 cm above the ground. As his feet leave the floor his centre of mass is 90 cm above the ground and rises to 120 cm at the top of his leap. Calculate up upward froce, assumed constant, that the ground exerts on the man. [Hint : Human body is a system of particles] |
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Answer» Correct Answer - 96 kgf |
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| 288. |
A ball of mass `m` is projected horizontally with velocity `v` into the barrel of a spring gun of mass M initially at rest on a frictionless table. The mass `m` sticks in the barrel at the point of maximum compression of the spring. What fraction of the initial energy of the ball is stored in the spring ? [Hint : Apply energy conservation principal to the system to obtain the maximum work done on the spring which is stored as potential energy of the spring. The spring will be progressively compressed till the two acquire a common velocity] |
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Answer» Correct Answer - `M//(m+M)` |
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| 289. |
An ideal massless spring is compressed by 1 m by a force of 100 N. The same spring is placed at the bottom of a frictionless inclined plane `(theta = 30^(@))`. A 10-kg mass is released from the top of the incline and is brought to rest momentarily after compressing the spring by 2m. Through what distance does the mass slide before coming to rest ? [Hint : Apply work-energy theorem] |
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Answer» Correct Answer - 4.08 m |
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| 290. |
1 kg block collides with a horizontal massless spring of force constant `2 N//m`. The block compresses the spring by 4m. If the coefficient of kinetic friction between the block and the· surface is 0.25, what was the speed of the block at the instant of collision? (take `g = 10 m//s^(2)`) |
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Answer» Correct Answer - `7.2 ms^(-1)` |
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| 291. |
A light spring is hung vertically from a fixed support and a heavy mass is attached to its. lower end. The mass is then slowly lowered to its equilibrium position: This stretches the spring by an amount d. If the same body is permitted to fall instead, through what distance does it stretch the string? |
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Answer» Correct Answer - 2d |
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| 292. |
A 2-kg block is dropped from a height of 0.4 m on a spring of force constant `k=1960 N`. Find the maximum distance the spring will be compressed. (Take `g = 9.8 ms^(-2)`) [Hint : Consider conservation of energy] |
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Answer» Correct Answer - 0.1 m |
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| 293. |
A uniform rod is made to lean between rough vertical wall and the ground. Friction coefficient between rod and walls is `mu_(1)=(1)/(2)` and between the rod and the ground is `mu_(2)=(1)/(4)`. The rod is about to slip at contact surfaces. The correct options are: A. The normal reaction between rod and wall is `(mu_(2)W)/(1+mu_(1)mu_(2))`B. Normal reaction between rod and ground is `(W)/(1+mu_(1)mu_(2))`C. `N_(2) gt N_(1)`D. `N_(1) gt N_(2)` |
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Answer» Correct Answer - A::B::C For equilibrium of rod `sumF_(x)=0 , N_(1)=mu_(2)N_(2)sumF_(y)=0, N_(1)=(mu_(2)W)/(1+mu_(1)mu_(2))` So, choice (a) is correct. and `N_(2)=(W)/(1+mu_(1)mu_(2))` So, choice (b) is correct. by subsutituting the values `N_(1)=(2w)/(9)& N_(2)=(8w)/(9):. N_(2) gtN_(1)` |
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| 294. |
A body of mass 2 kg is kept by pressing to a vertical wall by a force of 100 N . The coefficient of friction between wall and body is 0.3. Then the frictional force is equal toA. 2 NB. 20 NC. 50 ND. 100 N |
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Answer» Correct Answer - B |
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| 295. |
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. 0.4B. 0.2C. 0.6D. 0.8 |
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Answer» Correct Answer - B |
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| 296. |
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. 100 NC. 1100 ND. 1200 N |
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Answer» Correct Answer - C |
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| 297. |
……………. and ……………. are called fluids. A) Liquids, gases B) Solids, liquids C) Solids, gases D) Solids, moltens |
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Answer» A) Liquids, gases |
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| 298. |
Explain why, objects moving in fluids should have streamlined shape. |
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Answer» The objects moving in fluids have streamlined shape so as to reduce the air drag or fluid friction that opposes their motion in any fluid. |
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| 299. |
Define fluids. |
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Answer» Fluids: Liquids and gases are called fluids. |
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| 300. |
How does life history of Leonardo Davinci inspire students? |
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Answer» Leonardo Davinci was concerned with laws of science and nature which greatly informed his work as a painter, sculptor, inventor and draft men. His versatile personality inspire students. |
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