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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.
| 1. |
A heavy box of mass 20 kg is pulled on a horizontal surface by applying a horizontal force. If the coefficient of kinetic friction between the box and the horizontal surface is 0.25, find the force of friction exerted by the horizontal surface on the box.A. 5 NB. 10 NC. 50 ND. 200 N |
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Answer» Correct Answer - C Since, F = Frictional force = `muN` |
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| 2. |
Block A weighs `4 N` and block weigh `8 N` The coefficient of kinetic friction is `0.25` for all surface find the force `F` to slide `B` at a constant speed when (a)`A` rest on `B` and moves with it (b) `A` is held at rest and (c )`A and B` are connected by a light cord passing over a smooth putting as shown in fig 7.31 (a - c) restively. |
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Answer» (a) When A moves with B the force opposing the motion is the only force of friction between B and S the horizontal and as velocity of system is constant , `F = f_(1) = muR_(1) = 0.25 ( 4 + 8) = 3N` (b) When A is held stationary , the friction opposing the motion is between A and B , B and S . So , `F = muR_(1) + muR_(2) = 3 + 0.25(4)` i.e., `" " F = 3 + 1 = 4N` (c) In this situation for dynamic equilibrium of B `F = muR_(1) + muR_(2) + T " "...........` (i) while for the uniform motion of A . ` T = muR_(2) " " ..........` (ii) Substituting T from Eqn. (ii) in (i) , we get `F = muR_(1) + 2muR_(2) = 3 + 2 xx 1 = 5 N` |
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| 3. |
A block slides down a rough inclined plane of slope angle `theta` with a constnat velocity. It is then projected up the same plane with an intial velocity v the distance travelled by the block up the plane coming to rest is . |
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Answer» Correct Answer - `v_(0)^(2)// ( 4 "g sin" theta)` `v^(2) = v_(0)^(2) - 2as ` `v_(0)^(2) = 2as ` `s = (v_(0)^(2))/(2 xx 2 "g sin " theta )` As a = g sin `theta` + g cos `theta xx mu` = g sin `theta + ("sin" theta)/("cos" theta) xx "g cos" theta ` a = 2 g sin `theta` |
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| 4. |
A block of mass M = 4 kg is kept on a smooth horizontal plane. A bar of mass m = 1 kg is kept on it. They are connected to a spring as shown & the spring is compressed. Then what is the maximum compression in the spring for which the bar will not slip on the block when released if coefficient of friction between them is 0.2 & spring constant = 1000 N/m : (Take `g=10m//s^(2)`) A. 1 cmB. 1 mC. 1.25 cmD. 10 cm |
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Answer» Correct Answer - A `kx = (M+m)a kx = 5a " " …….. (i)` as `mumg=ma " " ……. (ii)` `a=2m//sec^(2)` Putting value of a in eqn. (i), we get x. |
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| 5. |
A block takes a twice as much time to slide down a `45^(@)` rough inclined plane as it takes to slide down a similar smooth , plane . The coefficient of friction is :A. `(3)/(4)`B. `(sqrt3)/(2)`C. `(1)/(4)`D. `(1)/(3)` |
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Answer» Correct Answer - A l = `(1)/(2) a_(1) t_(1)^(2) = (1)/(2) a_(2)t_(2)^(2)` `(1)/(2) ( "g cos" theta) t^(2) = (1)/(2) g ( "sin" theta - mu "cos" theta ) 4t^(2)` `"cos" 45^(@) = 4 ( "sin" 45^(@) - mu "cos" 45^(@))` `(1)/(sqrt2) = 4 ((1)/(sqrt2) - (mu)/(sqrt2))` `1 = 4 (1 - mu)` `mu = (3)/(4)` |
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| 6. |
A ball of mass `m` is rotating in a circle of radius `r` with speed `v` inside a smooth cone as shown in figure. Let `N` be the normal reaction on the ball by the cone, then choose the correct option: A. `N cos theta = mg`B. `g sin theta = (v^(2))/(r) cos theta `C. `N sin theta - (mv^(2))/(r) = 0`D. none of these |
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Answer» Correct Answer - A::B::C `N = mg cos theta (mv^(2))/(r) = mg sin theta ` |
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| 7. |
A block lying on a long horizontal conveyer belt moving at a constant velocity receives a velocity `v_(0) = 5m//s` relative to the ground in the direction opposite to the direction of motion of the conveyer. After `t = 4s`, the velocity of the block becomes equal to the velocity of the belt. The coefficient of friction between the block and the belt is `mu = 0.2`. Determine the velocity v of the conveyer belt.A. 3 m/sB. 5 m/sC. 4 m/sD. 7 m/s |
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Answer» Correct Answer - A `v_("Block") = v_(0) - at ` `a = mu g ` |
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| 8. |
Calculate `F` such that frictional force acting on all blocks zero. |
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Answer» Correct Answer - 4 In the given case F = 4 all block will have acceleration `2 m// sec^(2)` . So , no friction will act on blocks . |
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| 9. |
A ball weighing `10g` hits a hard surface vertically with a speed of `5m//s` and rebounds with the same speed The ball remains in contact with the surface speed The ball remains in contact with the surface for `0.01s` The average force exerted by the surface on the ball is .A. 100 NB. 10 NC. 1ND. `0.1` N |
| Answer» Correct Answer - B | |
| 10. |
(A) : Polishing a surface beyond a certain limit increases rather than decreases the frictional forces . (R) : When the surface is polished beyond a certain limit , the molecules exert strong attractive force . This is called surface adhesion , to overcome which additional force is required . Hence frictional force increases .A. If both A and R are true and R is the correct explanation of A .B. If both A and R are true but R is not correct explanation of A .C. If A is false but R is false .D. If A is false but R is true |
| Answer» Correct Answer - A | |
| 11. |
A uniform cylinder of height `h` and radius `r` is placed with its circular face on a rough inclined plane and the inclination of the plane to the horizontal is gradually increased. If `mu` is the coefficient of friction, then under what condition the cylinder will (a) slide before toppling (b) topple before sliding. |
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Answer» Consider the cylinder on inclined plane as shown in Fig . 7.73, then (A) it will slide if mg sin `theta gt mu R` i.e., mg sin `theta gt mu "mg cos" theta " " ["as" R = "mg cos" theta]` i.e., tan `theta gt mu " " …… (a)` (B) it will topple if ( mg sin `alpha) (h)/(2) gt "mg cos " alpha xx r " " ("replacing" theta by alpha)` or tan `alpha gt (2 r //h) " " ....... (b) ` So , from Eqns . (a) and (b) , it is clear that : (a) The cylinder will slide before toppling if `alpha lt theta ` i.e., tan `theta lt "tan" alpha` or `mu lt (2 r//h)` (b) The cylinder will topple before sliding if `alpha lt theta ` i.e., tan `theta gt "tan" alpha ` or `mu gt (2r //h)` |
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| 12. |
The figure shows th velocity and acceleration of a point like body at the initial moment of its motion. The acceleration vector of the body remain constant. The minimum radius of curvature of trajectory of the body is A. 2 metreB. 4 metreC. 8 metreD. 16 metre |
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Answer» Correct Answer - B Acceleration vector shall charge component of velocity v cos `theta` along acceleration vector `r = (v^(2))/(a_(n)) r ` is minimum when it is maximum this is possible when component of velocity parallel to acceleration vector is zero i.e., `v_(0)` cos `theta= 0 `. |
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| 13. |
A particle moves in a circle of radius 4.0 cm clockwise at constant speed of `2 ms^(-1)`. If `hat(x)` and `hat(y)` are unit acceleration vectors along X-axis and Y-axis respectively (in `cms^(-2)`), find the acceleration of the particle at the instant half way between P and Q. Refer to Fig.5.14 (a). A. `-4 (hatx + haty)`B. `4 (hatx + haty)`C. `-(hatx + haty) (1)/(sqrt2)`D. `(hatx + haty) 4` |
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Answer» Correct Answer - C Acceleration vector `veca = (v^(2))/(R) (-hatR)` `= - ((R "cos" 45 hatx + R "sin" 45 haty))/(R) = -(hatx + haty) (1)/(sqrt2)` |
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| 14. |
A `20 kg` block is originally at rest on a horizontal surface for which the coefficient of friction is `0.6`. A horizontal force `F` is applied such that it varies with time as shown in the figure (a) & (b). If the speed of the block after `10 s` is `8v` then find `v`. ( Take `g=10m//s^(2)`) |
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Answer» Correct Answer - 3 `F - mu mg = m(dv)/(dt)` `"m" int_(0)^(v) dv = int_(3)^(10)dt - mu mg int_(3)^(10) dt ` `implies 20 v = 480 implies v = 24` |
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| 15. |
A composite inclined plane has three different inclined surfaces `AB, BC` and `CD` of heights `1 m` each and coefficients of friction `(1)/(sqrt(3)), (1)/(sqrt(8))` and `(1)/(sqrt(15))` respectively. A particle given an initial velocity at `A` along `AB` transverses the inclined surfaces with uniform speed, reaches `D` in `5s`. The initial speed given is `("in" m//s)` A. `1.6`B. 1.8C. 2.4D. 3 |
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Answer» Correct Answer - B We know that if the angle of inclination is `tan^(-1) mu` no resultant force acts on the body . `therefore " " AB = (1)/("sin" (tan^(-1) (1)/(sqrt3))) = 2m` `BC = (1)/("sin" (tan^(-1) (1)/(sqrt8))) = 3m` CD = `(1)/("sin" (tan^(-1) (1)/(sqrt15))) = 4m ` `therefore` ABCD = 9m , T = 5 s , v = 1.8 m/s |
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| 16. |
A body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficient of static and kinetic frictions `0.5` and `0.4` respectively . On applying the same force , what is the acceleration ?A. `0.98 m//s^(2)`B. `9.8 m//s^(2)`C. `0.54 m// s^(2)`D. `5.292 m//s^(2)` |
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Answer» Correct Answer - A ma = `f_(s) - f_(k) = (mu_(s) "mg" - mu_(k) "mg")` `(mu_(s) - mu_(k)) "mg"` `therefore " " a = (mu_(s) - mu_(k) )g = (0.5 - 0.4 ) xx 9.8 = 0.98 m//s^(2)` |
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| 17. |
A mass 1kg attached to the end of a flexible rope of diameter `d = 0.25 m`is raised vertically by winding the rope on a reel as shown. If the reel is turned uniformly at the rate of `2r.p.s.` What is the tension in rope. The inertia of rope may be neglected. A. 16.28 NB. 10 NC. 20 ND. 1 N |
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Answer» Correct Answer - A Tangential acceleration `(dv)/(dt) = omega (dr)/(dt)` `omega = 4 pi , (dr)/(dt) = 0.5 , a = 2 pi` T - mg = ma , T = mg + ma |
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| 18. |
A block of mass 10 kg is sliding on a surface inclined at `30^(@)` with horizontal . If coefficient of friction between the block and the surface is 0.5 , find acceleration produced in the block . Take ` g = 9.8 m //^(2)` .A. `0.68` mgB. `0.50` mgC. `0.74` mgD. `0.40` mg |
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Answer» Correct Answer - B f = Mg sin `theta` |
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| 19. |
In Fig.7.125, coefficient of friction between `m_(1)` and `m_(2)` is `mu` and that between `m_(1)` and the wall is zero. A force F is pressing the system against the wall. Minimum value of force required to hold the system in equilibrium is :A. `mu_(1)g`B. `mu_(1)g`C. `mu(m_(1)+m_(2))g`D. system cannot be held in equilibrium |
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Answer» Correct Answer - D No force in vertical direction on `m_(1)` as wall is smooth . |
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| 20. |
A block is moving on an inclined plane making an angle `45^(@)` with the horizontal and the coefficient of friction is `mu` . The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down . If we define `N = 10 mu` , then N is : |
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Answer» Correct Answer - 5 `F_(1) = 3F_(2)` mg (sin `45^(@) + mu` cos `45^(@)`) = 3 mg (sin `45^(@) - mu "cos" 45^(@))` `(1)/(sqrt2) ( 1+ mu) = (3)/(sqrt2) (1 - mu)` `mu = 0.5` `therefore " " N = 10 mu` = `10 xx 0.5` N = 5 |
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| 21. |
A block of mass m is on an inclined plane of angle `theta`. The coefficient of friction between the block and the plane is `mu` and `tanthetagtmu`. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from `P_1=mg(sintheta-mucostheta)` to `P_2=mg(sintheta+mucostheta)`, the frictional force f versus P graph will look likeA. B. C. D. |
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Answer» Correct Answer - A Given that tan `theta gt mu` it means the block will slide downward and force of friction will act up the plane , i.e., positive . The instant at which applied force P is equal to mg sin `theta` , tendency of the block to slide down the inclined plane will cease and frictional force vanishes . As P is further increased , the block will move up the inclined plane and hence friction will act down the plane , i.e., negative . |
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| 22. |
If, in Q.19, coefficient of friction between `m_(1)` and the wall is `mu`,and `f_(1)` and `f_(2)` are respectively, the force of friction on `m_(1)` and `m_(2)`, then :A. `f_(1)` and `f_(2)` both are downward.B. `f_(1)` and `f_(2)` both are downward.C. `f_(1)` is upward and `f_(2)` upward.D. `f_(1)` is downward and `f_(2)` upward. |
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Answer» Correct Answer - B In upward direction . |
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| 23. |
Two thin roads are moving perpendicularly as shown in the figure. If the friction acting between them is `F_(R )` then the unit vector in the direction of friction force acting on the rod lying along x-axis is A. `((-hati - 2hatj))/(sqrt5)`B. `((hati + 2hatj))/(sqrt5)`C. `((3 hati + 2 hatj))/(sqrt13)`D. none of these |
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Answer» Correct Answer - B Relative velocity of rod on y-axis with respect to x-axis is `v hati + 2 v hatj`. The direction of friction on rod along x - axis is along relative velocity `hatf = ( hati + 2 hatj)/(sqrt5)` |
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| 24. |
A block of mass 2 kg is placed on the floor . The coefficient of static friction is `0.4` . If a force of `2.8` N is applied on the block parallel to floor , the force of friction between the block and floor is : (Taking g = `10 m//s^(2)`)A. 2.8 NB. 8 NC. 2 ND. zero |
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Answer» Correct Answer - A Force of friction , f = `mu "mg" = 0.4 xx 2 xx 10 = 8 N` As F `lt` f , the body will be in static equilibrium . `therefore " " f = 2.8 N` |
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| 25. |
A lineman of mass 60 kg is holding a vertical pole . The coefficient of the static friction between his hands and the pole is `0.5` . If he is able to climb up the pole , what is the minimum force with which he should press the pole with his hands ? ( g = `10 m//s^(2)`)A. 1200 NB. 600 NC. 300 ND. 150 N |
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Answer» Correct Answer - A `mu N = Mg` `N = ("Mg")/(mu)` |
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| 26. |
A disc of radius `R` has a light pole fixed perpendicular to the disc at its periphery which in turn has a pendulum of length `R` attached to its other end as shown in figure. The disc is rotated with a constant angular velocity `omega` The string is making an angle `45^(@)` with the rod. Then the angular velocity `omega` of disc is A. `((sqrt3g)/(R))^(1//2)`B. `((sqrt3g)/(2R))^(1//2)`C. `((g)/(sqrt3R))^(1//2)`D. `((sqrt2 g)/((sqrt2 + 1)R))^(1//2)` |
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Answer» Correct Answer - D The bob of the pendulum moves in a circle of radius `(R + R "sin" 45^(@)) = ((sqrt2 + 1)/(sqrt2))R` T sin `45^(@) m ((sqrt2 + 1)/(sqrt2)) R omega^(2)` T cos `45^(@) = "mg" " " omega = sqrt((sqrt2 g)/((sqrt2 + 1) R))` |
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| 27. |
A string of length 1m is fixed at one end and carries a mass of 1 kg at the other end. The string makes `2//pi` revolutions per second around a vertical axis passing through the fixed end. Calculate (i) angle of inclination of the string with the vertical, (ii) the tension in the strong and (iii) the linear velocity of the mass. |
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Answer» As shown in Fig. 7.67, for vertical equilibrium of mass m `T "cos" theta = "mg" …… (i) ` While for circular motion `T "sin" theta = (mv^(2)//r) = mr omega^(2) " " …….. (ii) ` (a) So , from Eqn . (ii) , `T = [ mr omega^(2)//"sin" theta ] = m L omega^(2)` [as r = L sin `theta`] `therefore " " T = 10^(-1) xx 1 xx 4^(2) = 1.6 N` [ as `omega = 2 pi f = 2 pi xx ( 2//pi) = 4]` (b) From Eqn. (i) , cos `theta = ("mg")/(T) = ( 10^(-1) xx 10)/(1.6) = (5)/(8)`, i.e., `" " theta = cos^(-1) (5 //8)` (c) `" " v = r omega = L "sin" theta xx omega ` [ as r = L sin `theta`] or `v = 1 xx 0.78 xx 4 = 3.12` m/s [ as sin `theta = 0.78`] |
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