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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 201. |
A block is placed on an inclined plane moving towards right horizontally with an acceleration `a_(0)=g`. The length of the plane `AC=1m`. Friction is absent everywhere. Find the time taken (in seconds) by the block to reach from C to A. A. `1.2 s`B. `0.74 s`C. `2.56 s`D. `0.42 s` |
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Answer» Correct Answer - B |
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| 202. |
For oscillation of a simple pendulum of length L what is the maximum possible velocity at the lowest position What happens to the motion if velocity exceeds this value ? |
| Answer» For oscillation of a simple pendulum max possible velocity at the lowest point `= sqrt (3 g l )` When the velocity exceeds `sqrt(3 g L )` but is less than `sqrt(5 g L)` , the bob leaves the vertical circle When upsilon `= sqrt(5 g L )` the bob will complete the vertical circle . | |
| 203. |
The driver of a truck travelling with a velocity upsilon suddenly notices a brick wall in front of him at a distance d It is better for him to apply brakes or to make a circular turn without applying brakes in order to just avoid crashing into the wall ? Why ? |
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Answer» In applying brakes suppose `F_(B)` is the force required to stop the truck in distance `:. F_(B) xx d = (1)/(2) m upsilon^(2) or F_(B) = (m upsilon^(2))/(2 d ) ` In taking a turn of radius d the force required is ` F_(T) = (m upsilon^(2))/(d) = 2 F_(B) or F_(B) = (1)/(2) F_(T) ` Hence it is better to apply brakes . |
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| 204. |
The linear momentum of a body can change only in the direction of applied force Comment . |
| Answer» The statement is corrent It is in accordance with Newton s secound law of motion . | |
| 205. |
A force is always required to move a body uniformly . Comment . |
| Answer» The statement is wrong when we assume that there is no force of friction and air resistance etc . In that case , force is required only for producing acceleration in the body . | |
| 206. |
A person sitting in the compartment of a tarin moving with unifrom speed throws a ball in the upward direction . What path of the ball will appear to him ? What to a person standing outside ? |
| Answer» No change in speed , but change to a person standing outside . | |
| 207. |
Consider, a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of static friction between the distance in which the car can be stopped is (Take `g=10 //s^(2)`)A. 30mB. 40mC. 72mD. 20m |
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Answer» Correct Answer - B |
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| 208. |
lnertia is that property of a body by virtue of which the body isA. unable to change by itself the state of restB. unable to change by itself the state of uniform motion.C. unable to change by itself the direction of motion.D. unable to change by itself the state of rest or of uniform motion |
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Answer» Correct Answer - D Inertia means resistance to change. It is the property of the body by virtue of which it cannot change by itself its state of rest or of uniform motion. |
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| 209. |
If a force is acting on a moving body in a direction perpendicular to the direction of motion what will be its effect on speed and direction of the body ? |
| Answer» No change in speed but change in direction is possible . | |
| 210. |
A passenger getting down from a moving bus, falls in the direction of the motion of the bus. This is is an example forA. second law of motionB. second law of motionC. second law of motionD. inertia of motion |
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Answer» Correct Answer - D A passenger getting down from a moving bus, falls in the direction of the motion of the bus. This is because his feet come to rest on touching the ground and the remaining body continues to move due to inertia of motion. |
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| 211. |
A heavy uniform chain lies on horizontal table top. If the coeficient of friction 0.25, then the maximum friction of length of the chain that can hang over on edge of the table isA. `20%`B. `25%`C. `35%`D. `15%` |
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Answer» Correct Answer - A |
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| 212. |
Physical independence of force is a consequence ofA. Third law of motionB. secon law of motionC. first law of motionD. All of these |
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Answer» Correct Answer - C |
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| 213. |
A ball is travelling with uniform translatory motion . This means thatA. it is at rest.B. the path can be a straight line or circular and the ball travels with uniform speed.C. all parts of the ball have the same velocity (magnitude and direction) and the velocity is constant.D. the centre of the ball moves with constant velocity and the ball spins about its centre uniformly. |
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Answer» Correct Answer - C In uniform translatory motion, all parts of the ball have the same velocity in magnitude and direction and this velocity is constant. |
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| 214. |
Physical independence of force is a Consequcnce ofA. first law of motionB. second law of motionC. third law of motionD. all of these laws |
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Answer» Correct Answer - A Physical independence of force is a consequence of first law of motion. |
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| 215. |
A body of mass m moves along X-axis such that its position coordinate at any instant t is ` x = at^(4) - bt^(3) + 2 ct^(2) - (3d)t` where a,b,c,d are constants What is the force acting on the particle at instant ? |
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Answer» Correct Answer - `F = m (12at^(2) - 6bt +2c)` Here, `x = at^(4) - bt^(3) + 2 ct^(2) - 3 d (t)` velocity, `upsilon = (dx)/(dt) = 4 at^(3) - 3 bt^(2) + 2 ct - 3 d` Acceleration, `a = (d upsilon)/(dt) = 12 at^(2) - 6 bt + 2 c` Force `F = ma = m(12 at^(2) - 6 bt + 2 c)` . |
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| 216. |
A cricket ball of mass `150g` has an initial velocity `(3 hati + 4hatj)ms^(-1)` and a final velocity `upsilon = -(3hati + 4hatj)ms^(-1)` after beigh hit The change in momentum (final momentum initial momentum) is (in kg `ms^(-1)`)A. zeroB. `-(0.45hati + 0.6hatj)`C. ` -(0.9hati + 1.2hatj)`D. ` -5(hati + hatj)` |
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Answer» Correct Answer - c Here, `m = 150g = 0.15kg , vec(u) = (3hati + 4hatj)m//s, vec(upsilon) = - (3hati + 4hatj)m//s` Change in momentum `vec (p) = m vec(upsilon) - m vec(upsilon) = 0.15 [-(3hati + 4hatj) -(3hati + 4hatj)] = - 0.30 [3hati + 4hatj)]` `vec(p) = - (0.9hati + 1.2hatj) kgm//s` . |
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| 217. |
A cricket ball of mass 150 g has an intial velocity u = `(3 hati + 4 hatj)ms^(-1)` and a final velocity `v = -(3hati + 4 hatj) ms^(-1)` , after being hit. The change in momentum (final momentum - initial momentum ) is (in `Kg ms^(1)`)A. zeroB. `-(0.45 hati + 0.6 hatj)`C. `-(0.9 hatj + 1.2 hatj)`D. `-5(hati + hatj ) hati` |
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Answer» Correct Answer - C Given , `" " v = (3 hati + 4 hatj)` m/s and `" " v = - (3 hati + 4 hatj)` m/s Mass of the ball = 150 g = `0.15` kg `Delta`p = Change in momentum = Final momentum - initial momentum = mv - mu = `m(v - u) = (0.15) [-(3hati + 4 hatj ) - (3hati + 4 hatj)]` = `(0.15)[-6 hati - 8 hatj]` = `-[0.15 xx 6 hati + 0.15 xx 8 hatj]` = `-[0.9 hati + 1.20 hatj ]` Hence , `" " Delta p = -[0.9 hat i + 1.2 hatj]` |
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| 218. |
A cricket ball of mass 150 g has an initial velocity `bar(u)=(3hati-4hatj)ms^(-1)` and a final velocity `bar(v)=-(3hati-4hatj)ms^(-1)` after being hit. The change in momentum (final momentum -initial momentum) is `("in kg ms"^(-1))`A. zeroB. `-(0.45hati+0.6hatj)`C. `-(0.9hati+1.2hatj)`D. `-5(hati+hatj)` |
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Answer» Correct Answer - C Here, m=150g =0.15kg `bar(u)=(3 hati+4 hatj)ms^(-1)` `bar(v)=-(3 hati+4 hatj)ms^(-1)` Initial momentum, `bar(p)_(i)=mbar(u)` `bar(p)_(f)=(0.15kg) (3 hati+4 hatj)ms^(-1)=(0.45hati+0.6hatj)kg ms^(-1)` Final momentum, `bar(p)_(i)=mbar(u)` `bar(p)_(f)=(0.15kg)(-3hati-4hatj)ms^(-1)` `=-(-0.45hati-0.6 hatj)kg ms^(-1)` Change in momentum, `Deltabar(p)=bar(p)_(f)-bar(p)_(i)` `=(-0.45hati-0.6hatj)kg ms^(-1) -(0.45 hati+0.6 hatj) kg ms^(-1)` `=(-0.9hati+1.2hatj)kg ms^(-1) ` |
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| 219. |
In the pervious question (6), the magnitude of the momentum transferred during the hit isA. zeroB. `0.75 kg -ms^(-1)`C. `1.5 kg ms^(-1)`D. `14 kg ms^(-1)` |
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Answer» Correct Answer - B (b) By previous solution `Deltap=-(0.9hati+1.2hatj)` Magnitude =`|Deltap|=sqrt((0.9)^(2)+(1.2)^(2))` `=sqrt(0.81+1.44)=1.5 kg ms^(-1)` |
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| 220. |
What is the funcation of shokers in scoots ? |
| Answer» When a scooter moves on an uneven road , impulsive forces are exerted by the road The funcation of shockers is to increase the time of impact . This would reduce the jerk experienced by the rider of the vehicle . | |
| 221. |
The time taken by a body to slide down a rough `45^(@)` inclined plane is twice that required to slide down a smooth `45^(@)` inclined plane. The coefficient of kinetic friction between the object and rough plane is given byA. `(1)/(3)`B. `(3)/(4)`C. `sqrt((3)/(4))`D. `sqrt((2)/(3))` |
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Answer» Correct Answer - B `t = sqrt((25)/(a)) prop (1)/sqrt(a)` `:. (t_(1))/(t_(2)) = sqrt((a_(2))/(a_(1)))` `(2)/(1) = sqrt((g sin theta )/(g sin theta - mu_(K) g cos theta)) = sqrt((1)/(1 - mu_(K)))` as `sin theta = cos theta `at `45^(@)` On solving the above equation, we get `mu_(K) = (3)/(4)` |
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| 222. |
What is the principle of working of a rocket ? |
| Answer» A rocket is based on the principle of conservtion of linear momentum . | |
| 223. |
Is large brake on a bicycle wheel more effective than a small one ? Explain |
| Answer» No the large brake on a bicycle wheel is not more effective than a small one because the force of friction is independent of the surface area of contact . However chances of failure of brake decrease when area of contact is larger Therefore , large brakes are perferred . | |
| 224. |
A body of mass `0.25` kg is projected with muzzle velocity `100ms^(-1)` from a tank of mass 100 kg. What is the recoil velocity of the tankA. `5 ms^(-2)`B. `25 ms^(-1)`C. `0.5 ms^(-1)`D. `0.25 ms^(-1)` |
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Answer» Correct Answer - D (d) Using law of consevation of momentum, we get `100xxv=0.25xx100` `implies v=0.25 ms^(-1)` |
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| 225. |
There blocks a,b and c of massses 10 kg ,10 kg and 20 kg respectivaly are arrenged as shown in figure all the surface are frictionless and string is inextensible A constant force F= 20N MN is applied on block a as shown .pulley and string are light . part of block a as shown . pulley and string are light . part of the string connecting both pulleys is vertical and part of hte stings connecting pulleys with masses a and b are horizontal . A. Acceleration of block b `0.5 m//s^(2)`B. Acceleration of block b is `1m//s^(2)`C. Tension in the string is 10ND. Acceleration of block c is `0.5m//s^(2)` |
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Answer» Correct Answer - b,c |
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| 226. |
A particle of mass 21 g attached to a string of `70cm` length Keeping the string always taut the ball describes a horizontal circle of radius `15cm` Calculate the angular speed of the ball . |
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Answer» Correct Answer - `14509 dyne` . Here, `m = 21 g r = 70 cm, T = 2 s, F =` ? `F = mromega^(2) = mr((2pi)/(T))^(2) = 4pi^(2) (mr)/(T_(2))` `=4 xx 9.87 xx(21xx70)/(2xx2) = 14509 "dyne"`. |
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| 227. |
An iron block of sides `50cm xx 8cm xx 15cm` has to be pushed along the floor. The force required will be minimum when the surface in contact with ground is A. `8"cm"xx15"cm surface"`B. `50"cm"xx15"cm surface"`C. `8"cm"xx50"cm surface"`D. force is same for all surface |
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Answer» Correct Answer - D Force will be same for all the surfaces. |
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| 228. |
A block `A` of mass `2 kg` rests on another block `B` of mass `8 kg` which rests on a horizontal floor. The coefficient of friction between `A` and `B` is `0.2` while that between `B` and floor is `0.5`.When a horizontal floor `F` of `25 N` is applied on the block `B` the force of friction between `A` and `B` isA. `3 N`B. `4 N`C. `2 N`D. zero |
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Answer» Correct Answer - D Maximum value of friction between `B` and ground `= mu N = mu (m_(A) + m_(B)) g` `= (0.5) (2 + 8) (10)= 50N` Since applied force `F = 25 N` is les than `50 N`. Therefore system wil not move and force of friction between `A` and `B` is zero |
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| 229. |
A block placed on a horizontal surface is being pushed by a force`F` making an angle `0` with the verticle If the friction coefficient is `mu` how much force is needed to get the block just started Discuss the situation when tan ` theta lt mu `. |
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Answer» In limitting equilibrium force of friction `f = mu R` In equilibrium along the horizontal `F sin theta = mu R` and along the verticle `F cos theta + mg = R = (F sin theta )/(mu` `:. Mg = F(( sin theta )/(mu) - cos theta)` or `F = (mu mg)/(sin theta - mu cos theta)` If tan `theta lt mu` `(sin theta - mu cos theta lt theta)` `:. F` is negative So for angles less angles less than `tan^(-1) mu` on cannot push the block ahed howsoever large the force may be . |
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| 230. |
A force `F=(6hati-8hatj+10hatk)`N produces acceleration of `sqrt(2) ms^(-2)` in a body. Calculate the mass of the body. |
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Answer» `therefore` Acceleration `a-=(abs(F))/(m)` `therefore=(|F|)/(a)=(sqrt(6^(2)+8^(2)+10^(2)))/(sqrt(2))=10ms^(-2)` |
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| 231. |
A rubber ball of mass `50g` falls from a height of `100cm` and rebounds to a height of `50cm` Find the impulse and average force between the ball and the ground, if time of contact is `0.1s` . |
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Answer» Correct Answer - `0.3778 Ns ; 3.78N` . Velocity on striking the ground `upsilon_(1) = sqrt(2 g h_(1)) = sqrt(2 xx 9.8 xx 1) = 4.427 m//s` velocity on rebouding `upsilon_(2) = sqrt(2 g h_(2))` `=sqrt(2 xx 9.8 xx (1)/(2)) = 3.13m//s` Impulse = change in linear momentum `= m upsilon _(1) - m (-upsilon_(2)) = m (upsilon_(1) + upsilon_(2))` `= 50 xx 10^(-3) (4.427 + 3.13) = 0.3778 Ns` `F = ("Impulse")/(t) = (0.3778)/(0.1) =3.78 N` . |
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| 232. |
A bullet of mass `50g` moving with a speed of `500 ms^(-1)` is brought to rest in `0.01s` Find the impulse and the average force . |
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Answer» Correct Answer - `-25 N_s, - 2500 N` . Here, `m = 50 g = 0.05 kg, u =500 m//s` `upsilon = 0, t = 0.01 s` Impulse = change in momentum `m (upsilon - m) = 0.05 (0-500)= - 25 Ns` As Impulse ` = Fxxt` `:. - 25 = F xx 0.01, F = -(25)/(0.01) = - 2500 N` . |
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| 233. |
In the figure two blocks are given velocities as shown. The lower block is very long. Find velocities of both the blocks when relative motion between them is ceased A. `2 m//s` (towards right)B. `1 m//s` (towards left)C. `2 m//s` (towards left)D. `1 m//s` (towards right) |
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Answer» Correct Answer - D |
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| 234. |
Block B moves to the right with a constant velocity `v_(0)`. The velocity of body A relative to B is A. `(v_(0))/(2)`, towards leftB. `(v_(0))/(2)`, towards rightC. `(3v_(0))/(2)`, towards leftD. `(3v_(0))/(2)`, towards right |
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Answer» Correct Answer - B (b) `T_(B)=3N " and " T_(A)=2N` `therefore" " v_(A)=(T_(B))/(T_(A)).v_(B)=(3)/(2)v_(0)`(towards right) `therefore" " v_(AB)=(3v_(0))/(2)-v_(0)=(v_(0))/(2),` towards right. In such cases, velocity and acceleration are in the ration of tensions. |
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| 235. |
The coefficient of friction between the tyres and road is 0.4. The minimum distance covered before attaining a speed of `8 ms^(-1)` starting from rest is nearly ( take `g=10 ms^(-2)`)A. 8mB. 4mC. 10mD. 16m |
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Answer» Correct Answer - A (d) Retardation, `a=(mu mg)/(m)=mug=4ms^(-2)` using, `v^(2)=u^(2)+2as` `(8)^(2)=(0)^(2)+2(4)(s)` `therefore s=8m` |
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| 236. |
Two blocks A and B each of mass m are placed on a smooth horizontal surface. Two horizontal force F and 2F are applied on blocks A and B, respectively, as shown in fig. Block A does not slide on block B. Then then the normal reaction acting between the two blocks is (assume no friction between the blocks) A. FB. `(F)/(2)`C. `(F)/(sqrt(3))`D. 3F |
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Answer» Correct Answer - D (d) `a=(2F-F)/(2m)=(F)/(2m)`( toward left) Writing equation of right hand side block, `2F-N sin 30^(@)=ma=(F)/(2)` `(N)/(2)=2F-(F)/(2)=(3F)/(2) implies N=3F` |
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| 237. |
A block `A`is just placed inside a smooth box `B` as shown in figure. Now the box is given an acceleration `a = (2 hat(i) - 2 hat(j)) m//s ^(-2)`. Under the acceleration block `A` cannot remain in the position shown. Block will require `ma` force for moving with acceleration `a`.A. If both Accertion and Reason are true and the reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - B For `(-2hat(i))m//s^(2)` component of acceleration , a horizontal force in `- ve x -` direction is required which can be provided only by the right vertical wall of the box. |
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| 238. |
When a weel is rolling on a level , what is the directionv of fricctional force between the wheel and the road ? |
| Answer» As the wheel is moving forward the portion of the wheel in contact with the road moves backwards Hence the force of friction must be acting in the forward direction along the tangent acting in the forward direction , along tangent to the surface of the road and the wheel in contact . | |
| 239. |
Send is spread on tracks covered with snow. Why ? |
| Answer» Covering of snow laid tracks with sand increases the force of friction between the road and the wheels Therefore , chances of slipping & skidding reduce . | |
| 240. |
The distance traveled by a body is directly proportional to time Is any external force acting on it ? |
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Answer» Here , s prop t . ` :. s = kt` , Where k is constant .` upsilon = (ds)/(dt) = k ` ` :. a = (dupsilon)/(dt) = 0 ` External force ,` F = m xx a = zero` . |
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| 241. |
What type of friction is involved when anaxle rotates in a sleeve ? |
| Answer» It is sliding friction . | |
| 242. |
Assuming the gravity to be in negative z-direction, a force `F=vxxA` is exerted on a particle in addition to the force of gravity where v is the velocity of the particle and A is a constant vector in positive x-direction. With what minimum speed a particle of mass m be projected so that it continues to move undeflected with constant velocity ?A. `-(A)/(mg) hat j`B. `(A)/(mg) hat j`C. `(mg)/(A) hat j`D. `-(mg)/(A) hat j` |
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Answer» Correct Answer - D |
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| 243. |
A block of mass 15 kg is placed on a long trolly . The coefficient of friction between the block and trolly is 0.18 The trolly accelerates from rest at `0.5 m//s^(2)` for 20 seconds and then moves with a unifrom velocity Discuss the motion of the block as viewed by (a) a stationary observer on the ground (b) an observer moving with the trolly. |
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Answer» Here , ` m = 15 kg , mu = 0.18 , a = 0.5 ms^(-2) , t = 20 s ` Force on the block due to motion of the trooly ` F = ma = 15 xx 0.5 = 7.5 N ` Is is obviously in the forward driction (of motion of the trolly) Force of limitting friction on the block ` = F = mu R = mu mg = 0.18 xx 15 xx 9.8 = 26 . 46 N ` This opposes the motion of the block The block shall not move The force of stice friction F will adjust itsself equal and opposite to F the applied force Hence a stationary observer on the ground the block will appear to be at rest relative to the trolly When trolly moves with unifrom velocity forward force is zero force is zero friction alone is acting on the block (b) An observer moving with the trolly has accelerated motion The observer is theregfore non - inertical The law of inertia is no longer valid . |
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| 244. |
A rocket can move in air free space, but a jet plane cannot. Why? |
| Answer» A rocket is self contained , as both the fuel and the oxygen needed to burn the fuel are available inside the rocket . Therefore , a rocket can move in air free space . A jet plane has fuel only It needs to take oxygen from the atmosphere to burn the fuel . That is way a jet plance cannot operate in air free space . | |
| 245. |
Can a rocket operate in free space ? |
| Answer» Yes , principle of conservtion of linear momentum holds equally well in free space . | |
| 246. |
A bucket of water is rotated in a vertical circle so that surface of water is at a distance r from the axix of rotation What is the minimum angular velocity so that the water does not spill out ? |
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Answer» For water not to spill weight of water must be balanced by centrifugal force (`m r omega^(2)` ) i.e `m r omega^(2) = mg , omega= sqrt(g//r)` . |
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| 247. |
Newtone established that a force never occurs singly in natrue Forces always occur in pairs as a result of mutual interaction between to bodies According to Newton s thired law to every action there is always an equal and opposite reaction The force of action and reaction may appear due to actual physical contact of the two bodies or even from a distance The law is applicable whether the bodies are at rest or they are in motion. Read the above passage and answer the following questions (i) It is difficult to walk on sand or ice Why? (ii) What does the law imply in day to day life ? |
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Answer» (i) On pushing sandy ground th sand slips Therefore force of reaction from sandy ground reduces Similarly friction between our feet and ice is small Thereforc force of reaction from ice is small That is why it is difficult to walk on sand or ice (ii) In day to day life this law implies that love begets love if you give love to anydody near or far you get love from hin/her in the equal measure/proportion The same is true for ill feelings They also return to you in the same proportion . |
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| 248. |
Though friction opposes relative motion yet in certain cases friction is also the cause of motion In fact without friction , motion cannot be started , stopped or transferred from one body to the other . Read the above passage and answer the following questions (i) Give one example where friction cause motion (ii) Give the direction of friction on fornt wheel and rear wheel of a bicycle when it is pedalled and when peddalling is stopped (iii) frictions is a necessary evail What does this imply in day to day life ? |
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Answer» (i) When a person pushes the ground backwards rough surface of ground reacts and exerts a forward force due to friction which causes motion (ii) When a bicycle is pedalled the rear wheel moves by the force communicated to it by pedalling while front wheel moves by itself Therefore force of friction on rear wheel is in forward direction and force of friction on front wheel is in the bacward direction However when pedalling is stopped both the wheels move by themselves Therefore force of friction on both the wheels is in the opposote direction (ii) Friction is a necessary evil because inspite of many disadvantages of frictions we cannot do without friction the same is true in day to day life Mad rush to surge ahead of our peers gives rise to cut throat competiton which is an evil like frictions but is necessary for progress . |
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| 249. |
In which of the following cases the net force is zero .A. a drop of rain falling down with terminal velocityB. a cork of mass `20g` floating in waterC. a car moving with constant speed of `60km//h` on a rough roadD. in a tug of war game if one team applies more force than other |
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Answer» Correct Answer - a,b,c If a rain drop falls with terminal velocity weight = upthrust + viscous drog `:. F _("net") = 0` For a cork floating in water weight = upthrust `:. F_("net") =0` In case of constant speed of a car, `a =0` In a tug of war game if one team applies more force than other then net force will be the resultant of two forces Which will not be zero `F _("net") ne 0` . |
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| 250. |
A body is in translatory equilibrium when .A. resultant force on it is zeroB. it at restC. it is in unifrom motionD. it is in an accelerated motion |
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Answer» Correct Answer - a,b,c, A body is in translatory equilibrium when resultant force on it is zero or it is at rest or if it is in unifrom motion In all the three cases net equilibrium. |
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