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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 cart of mass `M` is tied to one end of a massless rope of length `10 m`. The other end of the rope is in the hands of a man of mass `M`. The entire system is on a smooth horizontal surface. The man is at `x=0` and the cart at `x=10 m`. If the man pulls the cart by the rope, the man and the cart will meet at the pointA. `x = 0`B. `x = 5 m`C. `x = 10 m`D. They will never meet |
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Answer» Correct Answer - B |
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| 202. |
A rope lies on atable such that a part of it hangs down the table, when the length of hanging part is 1/3 of entire length the rope just begins to slide. The coefficient of friction between the rope and the table is:-A. `2//3`B. `1//2`C. `1//3`D. `1//6` |
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Answer» Correct Answer - B `muxx(2m)/(3)g=(m)(3)g` `mu(1)/(2)` |
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| 203. |
In the given figure if `T_(1)=2T_(2)` = 50 N then find the value of T. |
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Answer» As given in figure [Fig.1] `T_(3)=2T_(1)=2(2T_(2))=4T_(2)` and `T_(4)=2T_(2)` `thereforeT=T_(3)+T_(4)=4T_(2)+2T_(2)=6T_(2)=6xx(50)/(2)=150N` |
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| 204. |
An object is thrown vertically upward with some velocity. If gravity is turned off at the instant the object reaches the maximum height what happens ? .A. The object continues to move in a straight lineB. The object will be at restC. The object falls back with uniform velocityD. The object falls back with unform acceleration |
| Answer» Correct Answer - B | |
| 205. |
In the arrangement shown in figure-2.124 pulley A and B are massless and the thread is inextensible. Mass of pulley C is equal to m. If friction in all the pulleys negligible, then : A. Tension in thread is equal to `1//2 mg`B. Acceleration of pulley C is equal to `g//2` (downward)C. Acceleration of pulley A is equal to `g//2` (upward)D. Acceleration of pulley A is equal to 2 g (upward) |
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Answer» Correct Answer - D |
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| 206. |
A ball is dropped vertically from `a` height `d` above the ground . It hits the ground and bounces up vertically to a height ` (d)//(2). Neglecting subsequent motion and air resistance , its velocity `v` varies with the height `h` above the ground asA. B. C. D. |
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Answer» Correct Answer - A |
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| 207. |
A bicycle moves on a horizontal road with some acceleration. The forces of friction between the road and the front and rear wheels are `F_(1)` and `F_(2)` respectively.A. Both `F_(1)` and `F_(2)` act in the forward directionB. Both `F_(1)` and `F_(2)` act in the reverse directionC. `F_(1)` acts in the forward direction, `F_(2)` act in the reverse directionD. `F_(2)` acts in the forward direction, `F_(1)` act in the reverse direction |
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Answer» Correct Answer - D |
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| 208. |
When a force F acs on a body of mass m the acceleration product in the body is a . If htree equal forces `F_(1)=F_(2)=F_(3)=F` act on the same body as shown in figure the accleration produced is A. `(sqrt2 -1)a`B. `(sqrt2 +1)a`C. `sqrt2a`D. a |
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Answer» Correct Answer - A `a = (F)/(m), F_(R) = (sqrt(F_(1)^(2) +F_(2)^(2))-F_(3)),a^(1)=(F_(R))/(m)` |
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| 209. |
A rider on horse back falls when horse starts running all of a sudden becauseA. Rider is taken backB. Rider is suddenly afraid of fallingC. Inertia of rest keeps the upper part of body at rest whereas lower part of the body moves forward with the horseD. None of the above |
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Answer» Correct Answer - C |
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| 210. |
Find the acceleration of the prism of mass M and that of the bar of mass m shown in figure. |
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Answer» Correct Answer - `(g)/((1+(M)/(m)cot^(2)theta)), (g)/((tan theta +(M)/(m)cot theta))` |
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| 211. |
Consider the situation shown in figure. Find the acceleration of the system and the tension in the strings. |
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Answer» Correct Answer - `(2m-mu_(1)(M+m)g)/(M+m[5+1(mu_(2)-mu_(1))])` |
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| 212. |
Two masses `M_(1)` and `M_(2)` are connected by light string, which passes over the top of a smooth plane inclined at `30^(@)` to the horizontal, so that one mass rests on the plane and the other hangs vertically as shown in figure. It is found that `M_(1)`, hanging vertically can draw `M_(2)` up the full length of the plane in half the time in which `M_(2)` hanging vertically draws `M_(1)` up. Find `M_(1)//M_(2)`. Assume pulley to be smooth. Initially at time `t = 0` smooth masses `M_(1) = 15 kg` and `M_(2) = 10 kg` are held at rest and then they released. if after one second. the string snaps, find the further time taken for the 15 kg mass to return to its original position on the plane. Take `g = 10 m//s^(2)` |
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Answer» Correct Answer - `3//2, 0.69` sec |
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| 213. |
With increase of temperature the friction force acting between two surfaces .A. increasesB. decreasesC. remians sameD. may increase or decrease |
| Answer» Correct Answer - B | |
| 214. |
If man is walking direction of friction is .A. opposite to the direction of motionB. same as the direction of motionC. perpendicular to that of direction of motionD. `45^(@)` to the direction of motion |
| Answer» Correct Answer - B | |
| 215. |
A rope of length 5 m is kept on frictionless surface and a force of 5 N is applied to one of its end. Find the tension in the rope at 1 m from this endA. 1 NB. 3 NC. 4 ND. 5 N |
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Answer» Correct Answer - C |
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| 216. |
An aircraft is moving with a velocity of `300ms^(-1)` . If all the forces acting on it are balanced, thenA. It still moves with the same velocityB. It will be just floating at the same point in spaceC. It will fall down instantaneouslyD. It will lose its velocity gradually |
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Answer» Correct Answer - A |
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| 217. |
Figure shows a system of four pulleys with two masses `m_(A) =3kg` and `m_(B)=4 kg`. At an instant force acting on block A if block `B` is going up at an acceleration of `3 m//s^(2)` and pulley `Q` is going down at an acceleration of `1 m//s^(2)` is .A. `7N` acting upwardB. `7N` acting downwardC. `10.5 N` acting upwardD. `10.5 N` acting downward |
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Answer» Correct Answer - D Writing constraint for the string `2x_(A) +2x_(B) -x_(Q) +k =l` Differnting twice w.r.t time `2a_(A)+2a_(B) -a_(Q)=0` Taking downward direction to be positive `2a_(A)+2(-3) -1 =0, 2a_(A) =7, a_(A) =3.5 m//s^(2)` Positive sign indicates that block is acceleration at `3.5m//s^(2)` in downward direaction. So, force acting `F =3 xx 3.5 =10.5 N` (downward) . |
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| 218. |
Find the maximum value of (m//m) in the situation shown in figure so that the system remains at rest. Friction coefficient of both the contacts is `mu` , string is massless an pulley is friction less. A. `(costheta)/(sintheta-mucostheta)`B. `(sintheta)/(sintheta-mucostheta)`C. `(mucostheta)/(sintheta-mucostheta)`D. `(mu)/(sintheta-mucostheta)` |
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Answer» Correct Answer - D The system is at rest `(F_(net)=0)` For maximum `M//m` , Limiting friction will be acting on both blocks (at contact surface). `F=Net` pulling force on the whole system `mumg+muMgcostheta=Mgsintheta` `Mg(sintheta-mucostheta)=mumg` `(M)/(m)=(mu)/((sintheta-mucostheta))` |
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| 219. |
Two masses 2 kg and 3 kg are attached to the end of the string passed over a pulley fixed at the top. The tension and acceleration areA. `(7g)/(8),(g)/(8)`B. `(21g)/(8),(g)/(8)`C. `(21g)/(8),(g)/(5)`D. `(12g)/(5),(g)/(5)` |
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Answer» Correct Answer - D |
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| 220. |
Pulling force making an angle `theta` to the horizontal is applied on a block of weight `W` placed on a horizontal table. If the angle of friction is `alpha`, then the magnitude of force is `alpha` , then the magnitude of force required to move the body is equal toA. `(W"Cos"phi)/("Cos"(theta-phi))`B. `(W"sin"phi)/("Cos"(theta-phi))`C. `(W"Tan"phi)/("Sin"(theta-phi))`D. `(W"Sin"phi)/("Tan"(theta-phi))` |
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Answer» Correct Answer - B `N +P sin theta = mg, f = mu_(k) N` . |
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| 221. |
A block of mass `sqrt3 kg` is kept on a frictional surface with `mu =(1)/(2sqrt3)`. The minimum force to be applied as shown the move the block is .A. `5N`B. `20N`C. `10N`D. `20//3N` |
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Answer» Correct Answer - B `N =mg + F sin theta, F cos theta = f,f = mu N` |
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| 222. |
The coefficient of friction between a hemispherical bowl and an insect is `sqrt(0.44)` and the radius of the bowl is `0.6m`. The maximum height to which an insect can crawl in the bowl will be .A. `0.4m`B. `0.2m`C. `0.3m`D. `0.1m` |
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Answer» Correct Answer - D `h =r (1 -cos theta) =r [1 -(1)/(sqrt(mu_(s)^(2)+1))]` |
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| 223. |
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 force of 25 N is applied on the block B. the force of friction between A and B isA. ZeroB. 3.9 NC. 5.0 ND. 49 N |
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Answer» Correct Answer - A |
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| 224. |
Two blocks of equal mass have been placed on two faces of a fixed wedge as shown in figure. The blocks are released from position where centre of one block is at a height h above the centre of the other block. Find the time after which the centre of the two blocks will be at same horizontal level. There is no friction anywhere. |
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Answer» Correct Answer - `2sqrt((2h)/(g))` |
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| 225. |
A pendulum has a bob connected to a light wire. Bob ‘A’ is in equilibrium in the position shown. The string is horizontal and is connected to a block B resting on a rough surface. The block B is on verge of sliding when `theta = 60^(@)`. (a) Is equilibrium possible if `theta` were `70^(@)`? (b) With `theta = 60^(@)`, calculate the ratio of tension in the pendulum wire immediately after the string is cut to the tension in the wire before the string is cut. |
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Answer» Correct Answer - (a) No (b) `1:4` |
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| 226. |
A helicopter of mass M = 15000 kg is lifting a cubical box of mass m = 2000 kg. The helicopter is going up with an acceleration of `a = 1.2 m//s^(2)`. The four strings are tied at mid points of the sides of the square face PQRS of the box. The strings are identical and form a knot at K. Another string KH connects the knot to the helicopter. Neglect mass of all strings and take `g = 10 m//s^(2)`. Length of each string AK, BK, CK and DK is equal to side length of the cube. (a) Find tension T in string AK. (b) Find tension `T_(0)` in string KH. (c) Find the force (F) applied by the atmosphere on the helicopter. Assume that the atmosphere exerts a negligible force on the box. (d) If the four strings are tied at P,Q,R and S instead of A, B, C & D, how will the quantities T, `T_(0)` and F change? Will they increase or decrease? Assume that length of the four identical strings remains same. |
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Answer» Correct Answer - (a) 6467 N (b) 22400 N (c) 190400 N (d) `T_(0)` and F do not change . T will increase . |
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| 227. |
A parachutist of weight ‘ w ’ strikes the ground with his legs fixed and comes to rest with an upward acceleration of magnitude 3 g . Force exerted on him by ground during landing isA. wB. 2 wC. 3 wD. 4 w |
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Answer» Correct Answer - D |
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| 228. |
A block of mass 10 kg is suspended through two light spring balances as shown in figure A. Both the scales will read 10 kgB. Both the scales will read 5 kgC. The upper scale will read 10 kg and the lower zero.D. The reading may be anything but their sum will 10 kg. |
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Answer» Correct Answer - A |
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| 229. |
A block of mass 4 kg is suspended through two light spring balances A and B is series. Then A and B will read respectively.A. 4 kg and zero kgB. Zero kg and 4 kgC. 4 kg and 4 kgD. 2 kg and 2 kg |
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Answer» Correct Answer - C |
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| 230. |
A block of mass 10 kg is suspended through two light spring balances as shown in figure A. Both the scales will read `10kg`B. Both the scales will read `5kg`C. The upper scale will read `10kg` and the lower zeroD. The reading may be anything but their sum will be `10kg` |
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Answer» Correct Answer - A As the spring balances are light, so tension in both the spring will be same and equal to weight of block suisoended. |
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| 231. |
A block of mass 10 kg is suspended through two light spring balances as shown in figure A. Both the scales will read 10 kg.B. Both the scales will read 5kgC. The upper scale will read 10 kg and the lower zero.D. The readings may be anything but their sum will be 10 kg. |
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Answer» Correct Answer - A Since springs are massless, their reading will be same |
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| 232. |
Block A of mass M is placed on an incline plane, connected to a string, passing over a pulley as shown in the fig. The other end of the string also carries a block B of mass M. The system is held in the position shown such that triangle APQ lies in a vertical plane with horizontal line AQ in the plane of the incline surface. Find the minimum coefficient of friction between the incline surface and block A such that the system remains at rest after it is released. Take `theta = alpha = 45^(@)` |
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Answer» Correct Answer - `(sqrt(5 - 2sqrt2))/(sqrt(2)-1)` |
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| 233. |
The mass of a body measured by a physical balance in a lift at rest is found to be m . If the lift is going up with an acceleration a , its mass will be measured asA. `m(1-(a)/(g))`B. `m(1+(a)/(g))`C. `m`D. Zero |
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Answer» Correct Answer - C |
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| 234. |
In the system shown in fig, mass of the block is `m_(1) = 4` kg and that of the hanging particle is `m_(2) = 1` kg. The incline is fixed and surface is smooth. Block is initially held at the top of the incline and the particle hangs a distance d = 2.0 m below it. [Assume that the block and the particle are on same vertical line in this position]. System is released from this position. After what time will the distance between the block and the particle be minimum ? Find this minimum distance. [ `g = 10 m//s^(2)`.] |
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Answer» Correct Answer - 1.0 s , 1.0 m |
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| 235. |
As shown in the figure, two equal masses each of `2kg` are suspended from a spring balance. The reading of the spring balance will be. A. ZeroB. 2 kgC. 4 kgD. Between zero and 2 kg |
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Answer» Correct Answer - B |
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| 236. |
A player kicks a football of mass `0.5` kg and the football begins to move with a velocity of 10 m/s . If the contact between the leg and the football lasts for `(1)/(50)`sec , then the force acted on the football should beA. 2500 NB. 1250 NC. 250 ND. 625 N |
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Answer» Correct Answer - C |
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| 237. |
When the speed of a moving body is doubledA. Its acceleration is doubledB. Its momentum is doubledC. Its kinetic energy is doubledD. Its potential energy is doubled |
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Answer» Correct Answer - B |
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| 238. |
In the above situation all surface are frictionless system is released from rest. Then which of the following statements is/are correct .A. acceleration of wedges are zeroB. wedges accelerate towards rightC. Normal force exerted by ground on A is more than normal force exerted by ground on `B` .D. Tension in connecting string is nonzero |
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Answer» Correct Answer - A::C::D Force required to stop acceleration of wedges are mg sin `30 cos 60` which will be provided by string `N_(A) = Mg + mg cos^(2) 30 , N_(B) = Mg + mg cos^(2) 60` . |
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| 239. |
In the system of pulleys shown what should be the value of `m_(1)` such that 100 gm remains at rest : (Take `g = 10 m//s^(2)`) A. 180 gmB. 160 gmC. 100 gmD. 200 gm |
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Answer» Correct Answer - B |
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| 240. |
Three particles A, B and C of equal mass move with equal speed V along the medians of an equilateral triangle as shown in hgure. They collide at the centroid G of the triangle. After the collision, A comes to test, B retraces its path with the speed V. What is the velocity of C ? A. v, direction `bar(OA)`B. 2v, direction `bar(OA)`C. 2v, direction `bar(OB)`D. 2v, direction `bar(BO)` |
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Answer» Correct Answer - D Initial momentum of system is zero so, final momentum should also be zero. Finally `A` comes to rest after collision, so to have final momentum zeero, `B` and `C` should move in opposite directions. |
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| 241. |
In the above problem, if the lift moves up with a constant velocity of `2 m//sec` , the reading on the balance will beA. 2 kgB. 4 kgC. ZeroD. 1 kg |
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Answer» Correct Answer - A |
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| 242. |
A player catches a ball of 200 gm moving with a speed of 20 m/s. if the time taken to complete the cathc is 0.5s, the force exerted on the player, hand is :-A. 8 NB. 4NC. 2ND. 0N |
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Answer» Correct Answer - A `F=(0.2xx20)/(0.5)=8N` |
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| 243. |
The behaviour of a body under zero resultant force is given by .A. first law of motionB. second law of motionC. third law of motionD. law of gravitation |
| Answer» Correct Answer - A | |
| 244. |
Action and reaction:- (For a given system) (a) Act on the two different objects (b) Have opposite directions (c) zero resultantA. a,b,cB. b,c,dC. All of the aboveD. None of the above |
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Answer» Correct Answer - C All correct |
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| 245. |
A block is sliding along an inclined plane as shown in figure . If the acceleration of chamber is `a` as shown in figure. The time required to cover a distance `L` along inclined is A. `sqrt((2L)/(g sin theta-a cos theta))`B. `sqrt((2L)/(g sin theta+a sin theta))`C. `sqrt((2L)/(g sin theta+a cos theta))`D. `sqrt((2L)/(g sin theta))` |
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Answer» Correct Answer - C `f_("net")(downward) =mg sin theta + ma cos theta` ` =m (g sin theta + a cos theta)` `:.g_(eff) = g sin theta + a cos theta, T = sqrt((2L)/(g_(ef))` . |
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| 246. |
A horizontal force just sufficient to move a body of mass `4 kg ` lying on a rought horizontal surface is applied on it .The coefficient of static and kinetic friction the body and the surface are `0.8 and 0.6` respectively If the force contines to act even after the block has started moving the acceleration of the block in `ms^(-2)` is `(g = 10 ms^(-2))`A. `1//4`B. `1//2`C. 2D. 4 |
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Answer» Correct Answer - C The minimum force required to just move a body will be `f_(1)=mu_(s)mg` . After the motion is started, the friction will become kinetic. So the foprce which is responsible for the increase in velocity of the block `F=(mu_(s)-mu_(k))mg=(0.8-0.6)xx4xx10=8N` So `a=(F)/(m)=(8)/(4)=2m//sec^(2)` . |
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| 247. |
In the system shown in the figure AB and CD are identical elastic cords having force constant K. The string connected to the block of mass M is inextensible and massless. The pulley is also massless. Initially, the cords are just taut. The end D of the cord CD is gradually moved up. Find the vertical displacement of the end D by the time the block leaves the ground. |
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Answer» Correct Answer - `(5Mg)/(2K)` |
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| 248. |
Blocks A and B have dimensions as shown in the fig. and their masses are 8 kg and 1 kg respectively. A small block C of mass 0.5 kg is placed on the top left corner of block A. All surfaces are smooth. A horizontal force F = 18 N is applied to the block B at time t = 0. At what time will the block C hit the ground surface? Take `g = 10 m//s^(2)`. |
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Answer» Correct Answer - 2.9 s |
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| 249. |
A horizontal wooden block has a fixed rod OA standing on it. From top point A of the rod, two wires have been fixed to points B and C on the block. The plane of triangle OAB is perpendicular to the plane of the triangle OAC. There are two identical beads on the two wires. One of the wires is perfectly smooth while the other is rough. The wooden block is moved with a horizontal acceleration (a) that is perpendicular to the line OB and it is observed that both the beads do not slide on the wire. Find the minimum coefficient of friction between the rough wire and the bead. |
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Answer» Correct Answer - `mu_("min") = ("sin"theta)/(sqrt(cos^(2)theta + tan^(2)theta))` |
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| 250. |
Figure shown a wooden block at rest in equilibrium on a rought horizontal plane being acted upon by force `F_(1) = 10 N, F_(2) = 2 N`, as shown .IF`F_(1)` is removed , the resulting force acting on the block will beA. 2 N towards leftB. 2 N towards rightC. 0 ND. Cannot be determined |
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Answer» Correct Answer - C |
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