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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.
| 101. |
A block is gently placed on a conveyor belt moving horizontal with constant speed After `t = 4s` the velocity of the block becomes equal to velocity of the belt If the coefficient of friction between the block and the belt is `mu = 0.2`, then the velocity of the conveyor belt is .A. `8 ms^(-1)`B. `4 ms^(-1)`C. `6 ms^(-1)`D. `18 ms^(-1)` |
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Answer» Correct Answer - A (a) Due to friction (`a=mug)`, velocity of block will become equal to velocity of belt. Relative motion between two will stop. `therefore" "v=at=mug=0.2xx10xx4=8 ms^(-1)` |
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| 102. |
A block is gently placed on a conveyor belt moving horizontal with constant speed After `t = 4s` the velocity of the block becomes equal to velocity of the belt If the coefficient of friction between the block and the belt is `mu = 0.2`, then the velocity of the conveyor belt is .A. `8//s`B. `6m//s`C. `4m//s`D. `2m//s` |
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Answer» Correct Answer - a On account of friction `(a =mu g)` velocity of block will become equal to velocity of block will become equal to velocity of conveyor belt and the relative motion between the two will stop `upsilon = at = mu g t =0.2 xx 10 xx 4 = 8m//s` . |
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| 103. |
Two blocks of mass 4kg and 2kg are connected by a heavy string and placed on rough horizontal plane. The 2 kg block is pulled with a constant force F. the coefficent of friction between the blocks and the ground is 0.5. what is the value of F so that tension in the string is constant throughout the length? `(g=10m//s^(2))` A. 40NB. 30NC. 50ND. 60N |
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Answer» Correct Answer - b |
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| 104. |
In a laboratory experiment four students plotted graphs between force of limiting friction `(F)` and normal reaction `(R)` Which one is correct .A. .B. .C. .D. . |
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Answer» Correct Answer - d As `F prop R` therefore, choice (d) is the correct graph between `F` and `R` . |
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| 105. |
The mass of a lift is `2000kg` . When the tensioon in the supporting cable is `28000N` , then its acceleration is.A. `30ms^(-2)` downwardsB. `4ms^(-2)` upwardsC. `4ms^(-2)` downwardsD. `14 ms^(-2)` upwards |
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Answer» Correct Answer - B |
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| 106. |
For two bodies `A` and `B` of same material held on a horizontal plane force of limitting friction `F` varsus normal reaction `R` graphs are as shown in Which one has smoother surface in contact with the plane . .A. `A`B. `B`C. Both A and BD. Neither A nor B |
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Answer» Correct Answer - b As `mu = F//R =` slope of F -R curve which is greater for A therefore for body A force of frictionis smaller. `:.` Surface of body B in contact with the plane must be smoother |
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| 107. |
What are the conditions for maximum and minimum pull of a lift on a supporting cable ? |
| Answer» When the lift is in free fall , pull of the lift on supporting cable is zero (minimum) . When the lift is accelerating upward , pull ` = m (g + a )` = maximum. | |
| 108. |
In the above question the force of limitting friction isA. `3N`B. `5N`C. `4N`D. `zero` |
| Answer» Correct Answer - b | |
| 109. |
The minimum force required just to move a block on a rough horizontal surface is 10 N An applied force of 5 N fails to move the block. What are the values of static friction and dynamic friction ? |
| Answer» Static friction = 5 N Dynamic friction is slightly les than 10 N . | |
| 110. |
A minimum force of `7N` is required just to move a body on a rough surface What is the value of static friction when a force of `5N` is actually applied and there is no movement .A. `5N`B. `7N`C. `2N`D. `12N` |
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Answer» Correct Answer - a Static friction = applied force `=5N` . |
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| 111. |
The force of limiting friction between a body and the surface of contact is `5N` A force of `7N` is applied on the body and the actual motion starts The effective force of friction now is .A. zeroB. `5N`C. `7N`D. `lt5N` |
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Answer» Correct Answer - d Effective force of friction = dynamic friction which is slightly less than force of limitting friction `(=5N)` . |
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| 112. |
A bullet of mass `7kg` is fired into a block of metal weighing `7kg` The block is free to move Caluculte initial velocity of the bullet if the velocity of the block with the bullet in is `0.7m//s` . |
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Answer» Correct Answer - `700.7m//s` . According to the principle of conseration of linear momentum, linear momentum lost by bullet = linear momentum gained by block with bullet inside `(7)/(1000) xxupsilon =(7 + (7)/(1000))xx 0.7` `upsilon = (7007)/(1000) xx (0.7 xx 1000)/(7) = 700.7 m//s` . |
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| 113. |
A car of mass `m` is moving on a level circular track of radius `R` if `mu_(s)` represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by.A. `sqrt(mu_(s)mRg)`B. `sqrt(Rg//mu_(s))`C. `sqrt(mRg//mu_(s))`D. `sqrt(mu_(s)Rg)` |
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Answer» Correct Answer - d For smooth driving of the car force of friction = centripetal force `mu_(s) R = mu_(s) (mg) = (m upsilon^(2))/(R)` `upsilon = sqrt(mu_(s) Rg` . |
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| 114. |
The breaking strength of the cable used to pull a body is 40 N. A body of mass 8 kg is resting on a table of coefficient of friction `mu=0.20`. The maximum acceleration which can be produced by the cable is ( take , `g=ms^(-2)` )A. `6 ms^(-2)`B. `3 ms^(-2)`C. `8 ms^(-2)`D. `8 ms^(-2)` |
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Answer» Correct Answer - B (b) `a_("max")=(T_("max")-mu mg)/(m)=()40-0.2xx8xx(10)/(8)=3 ms^(-2)` |
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| 115. |
A bullet of mass `7g` is fired into block weighing `7kg`, which is free to move. Calculate initial velocity of bullet if the velocity of the block with the bullet inside is `0.7m//s`. |
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Answer» Correct Answer - `700.7ms^(-1)` |
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| 116. |
The minimum force required just to move a block on a rough horizontal surface is `5N` The block fails to move when a force of `3N` is applied on it Static frition is .A. `5N`B. `3N`C. `4N`D. `zero` |
| Answer» Correct Answer - b | |
| 117. |
A bullet of mass 4 g is fired with a velocity of 600 `ms^(-1)` into a block of metal weighing `9.996kg`. The block is free to move. Calculate the common velocity of the bullet and block after the impact. |
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Answer» Correct Answer - 24 `cms^(-1)` |
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| 118. |
A block of mass m is placed over a rogh surface with coefficents of static and kinetic as `mu_(s) and mu _(k) .` A force F is applied to move block .Surface f is the force of fricty=tion between block and ound . Then chose the wrong opition (s). A. `fle mu_(s)mg "till the block not move"`B. `f=mu_(k)` mg when the block starts movingC. `f=F` when block Is not movingD. `fltF` when block is moving |
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Answer» Correct Answer - a,c,d |
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| 119. |
A cubical block rests on an inclined plance of `mu = (1)/sqrt(3)` . Determine the angle at which the block just slides down the incline . |
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Answer» Here , ` mu = (1)/sqrt(3) , alpha = ? ` When the block just slides down the incline , `theta = alpha` = angle of repose , when `tan alpha = mu = (1)/(sqrt (3)) :. alpha = 30^(@) ` . |
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| 120. |
The rear side of a truck is open and a box of mass `20kg` is placed on the truck `4m` away from the open end `mu = 0.15` and `g = 10m//s^(2)` The truck starts from rest with an acceleration of `2m//s^(2)` on a straight road The box will fall off the truck when it is at a distance from the starting point equal to .A. `14m`B. `8m`C. `16m`D. `4m` |
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Answer» Correct Answer - c Max acc. Of the block ` = mu g = 0.5 xx 10` `=1.5m//s^(2)` Acc. Of truck `=2m//s^(2)` `:.` Relative acc. Of box w.r.t truck `a_(r) = 1.5` `-2.0 = -0.5m//s^(2)` (backwards) It will fall off the truck in a time `t = sqrt((2l)/(a_(r))) =sqrt(((2xx4))/(0.5)) = 4s` Displacement of truck during this time `s = (1)/(2)at^(2) = (1)/(2)xx2xx4^(2) =16m` . |
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| 121. |
A block is resting on a horizontal plate in the xy plane and the coefficient of friction between block and plate is `mu`. The plate begins to move with velocity `v=bt^(2)` in x direction At what time will the block start sliding on the plate?A. `(mub)/(g)`B. `(mubg)/(2)`C. `(mug)/(b)`D. `(mug)/(2b)` |
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Answer» Correct Answer - d |
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| 122. |
A railway engine weighing 40 metric ton is travelling along a level track at a speed of `54 km H^(-1)` What additional power is required to maintain the same speed up an incline of 1 in 49 Take `g = 9.8 m//s^(2)` and `m u = 0.1`. |
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Answer» Here , `m = 40` metric ton ` = 40 xx 10^(3) kg` `upsilon 54 km//h = (54 xx 1000) /( 60 xx 60 ) m //s = 15 m//s`, `g = 9.8 m // s^(2)` `mu = 0. 1 , sin theta = 1//49 , (P_(2) - P_(2) ) = ? ` On a level track , `P_(1) = mu mg xx upsilon = 0.1 xx 4 xx 10^(4) xx 9.8 xx 15` `5.88 xx 10^(5) "watt"` Up the incline , ` P_(2) = mg (sin theta + mu cos theta) xx upsilon` ` 4 xx 10^(4) xx 9.8 ((1)/(49) + 0.1 xx 1 ) xx 15 ` ` 7 .08 xx 10^(5) W ` Additional power required ` = P_(2) - P_(1) = (7 .08 - 5 .88 ) 10^(5) W ` ` = 1.20 xx 10^(5) W = 120 k W ` . |
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| 123. |
What is the maximum value of force F such that the block shown in does not move ? |
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Answer» The body diagram of the block is shown in Clearly `R = mg + F sin 60^(@)` The block will not move , when `F cos 60^(@) le f` (force of friction) `(F)/(2) le mu R ` `(F)/(2) le mu (mg + F sin 60^(@))` `(F)/(2) le (1)/(2 sqrt(3))(sqrt3g + F sqrt3/2)` `(F)/(2) le ((g)/(2) + (F)/(4))` or `(F)/(2) - (F)/(4) le(g)/(2)` or `F//4 le (g)/(2)` `F le 2 g` Hence `F_(max) = 2 g = 2 xx 10 = 20 N` |
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| 124. |
The rear side of a truck is open and a box of 40 kg mass is placed 5 m from the open end The coefficient of friction between the box and the surface below it is 0.15 On a straight road the truck starts from rest and accelerates with `2 m//s^(2)` At what distance from the starting point does the box fall off the truck ? Ignore the size of the box . |
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Answer» Here , mass of box , ` m = 40 kg ` acceleration of truck ,`a = 2 m//s^(2)` distance of box from open end , s = 5 m coefficient of friction ` mu = 0.15` Force on the box due to accelerated motion of truck `F = ma = 40 xx 2 = 80 N` This force is in forward direction Reaction F of this force on box = 80 N in backward dirction This reaction is opposed by force of limiting friction `f = mu R = mu mg = 0.15 xx 40 xx 9.8 = 58.8` N in the forward direction `:.` Net force on the box in backward direction `p = F - f = 80 - 58 .8 = 21 .2` N backwards acceleration of block `a = (p)/(m)` ` a = (21.2)/(40) = 0 .53 m//s^(2)` If t is time taken by the box to travel s = 5 m and fall off the truck then from ` s = ut + (1)/(2) at^(2)` `5 = 0 + (1)/(2) xx 0.53 t^(2)` ` t = sqrt((2 xx 5 )/(0.53 ))= 4.34 s` If the truck travels a distance x during this time then from `s = ut + (1)/(2) at^(2)` `x = 0 + (1)/(2) xx 2 (4.34)^(2) = 18.84 m` . |
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| 125. |
A man of `50 kg` mass is standing in a gravity free space at a height of `10m` above the floor. He throws a stone of `0.5 kg` mass downwards with a speed `2m//s`. When the stone reaches the floor, the distance of the man above the floor will beA. `20m`B. `9.9m`C. `10.1m`D. `10m` |
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Answer» Correct Answer - c Here, `m_(1) = 50 kg, m_(2) = 0.5kg, upsilon_(2)m//s` `S = 10 m, upsilon_(1) =`? From the coservation of liner momentum `m_(1) upsilon_(1) = m_(2) upsilon_(2)` Therefore `50 xx upsilon_(1) = 0.5 xx 2` `upsilon_(1) = (1)/(50)m//s` Time taken by the stone to cover the distance of `10m` and reach the floor (in gravity free space) ` t = (S)/(upsilon_(2) = (10)/(2) =5s` distance covered by the man (in the opposite direction) in `5 sec = upsilon_(1) xx t = (1)/(5) xx 5 =0.1m` Distance of the man above the floor `= 10+ 0.1 = 10.1m` . |
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| 126. |
A ball moving translationally collides elastically with another, stationary, ball of the same mass. At the moment of impact the angle between the straight line passing through the centres of the balls and the direction of the initial motion of the striking ball is equal to `alpha=45^@`. Assuming the balls to be smooth, find the fraction `eta` of the kinetic energy of the striking ball that turned into potential energy at the moment of the maximum deformation. |
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Answer» Correct Answer - `eta=(1)/(2) cos^(2) alpha` |
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| 127. |
Find an expression for, the acceleration and speed of a trolley A under the action of a constant force F if it contains sand that pours out through the floor at the rate of `mu` per second. The speed of the trolley at t=0 was v=0 and the initital mass of sand and trolley was `M_(0)` |
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Answer» Correct Answer - `v=(F)/(mu)" In"(M_(0))/(M_(0)-mut)` |
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| 128. |
Sand drops from a stationary hopper at the rate of 5 kg `s^(-1)` on to a conveyor belt moving with a constant speed of `2 m s ^(-1)` . What is the force required to keep the belt moving and what is the power delivered by the motor moving the belt ? [ Hint: Tangetial force exerted on the belt = `u_("rel") (dM)/(dt)` where `u_("rel")` is the tangential relative velocity of the sand .] |
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Answer» Correct Answer - 10 N , 20 W |
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| 129. |
A mam of mass 50 g stands on a frame of mass 30 g. He pulls on a light rope which passes over a pulley. The other end of the rope is attached to the frame. For the system to be in equilibrium what force man must exert on the rope? A. 40 gB. 80 gC. 30 gD. 50 g |
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Answer» Correct Answer - A (a) Force exerted by man on rope transfers to it in the form of tension. Net upwards force on the system is 2T or 2F. Net downward force is (50+30)g=80g. For equilibrium of system, 2F=80g or F=40g |
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| 130. |
A man pulls a block of mass equal to himself with a light string .The coefficient of friction between the man and the floor is greater than that between the block and the floorA. if the block does not move , then the man also does not moveB. the block can move even when the man is stationaryC. if both move then the acceleration of the block is greater than the acceleration of manD. if both move then the acceleration of man is greater than the acceleration of block |
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Answer» Correct Answer - A::B::C Since `mu_(1) gt mu_(2)` `:. (f_(1))_(max) gt (f_(2))_(max)` Further if both move, `a = (T -mu mg)/(m)` `mu` of block is less .Therefore its acceleration is move |
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| 131. |
The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle `theta` should be A. `0^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)` |
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Answer» Correct Answer - c If `T` is tension in each part of the string holding mass `sqrt2m` then in equilibrium `T cos theta + T cos theta = sqrt2 mg` `2 T cos theta = sqrt2 mg` But `T = mg , :. 2mg cos theta = sqrt 2mg` ` cos theta = (1)/sqrt2` `theta = 45^(@)` . |
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| 132. |
The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle `theta` should be : A. `0^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)` |
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Answer» Correct Answer - C ( c) Equilibrium of m: T=mg …..(i) Equilibrium of `sqrt(2) m: 2T cos theta = sqrt(2) mg` ……..(ii) Solving these two equations we get `theta=45^(@)` |
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| 133. |
Masses `M_1`, `M_2` and `M_3` are connected by strings of negligible mass which pass over massless and friction less pulleys `P_1` and `P_2` as shown in fig The masses move such that the portion of the string between `P_1` and `P_2` in parallel to the inclined plane and the portion of the string between `P_2` and `M_3` is horizontal. The masses `M_2` and `M_3` are 4.0kg each and the coefficient of kinetic friction between the masses and the surfaces 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 of the string `(g=9.8 m//sec^2, sin 37^@=3//5)` |
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Answer» Correct Answer - B::D (a) If `M_1`, `M_2` and `M_3` are considered as a system, then the force responsible to more them is `M_1g` and the retarding force is `(M_2gsintheta+muM_2gcostheta+muM_3g)`. These two should be equal as the system is moving with constant velocity. |
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| 134. |
Two blocks of masses M and m are connected with massless and frictionless pulleys as shown. Strings are also massless. If the magnitude of acceleration of M is `a_(0)` then magnitude of acceleration of m is A. `3 a_(0)`B. `7 a_(0)`C. `5 a_(0)`D. `a_(0)` |
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Answer» Correct Answer - B |
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| 135. |
A light string fixed at one end to a clamp on ground passes over a fixed pulley and hangs at the other side. It makes anangle of `30^(@)` with the ground. A monkey if makes an angle of `30^(@)` with the ground. A monkey of mass 5kg climbs up the rope. The clamp can tolerate is vertical force of 40N only. the maximum accelration in upward direction with which the monkey can climb safety is (Neglect friciton and take `g=10m//s^(2)`) A. `2m//s^(2)`B. `4m//s^(2)`C. `6m//s^(2)`D. `8m//s^(2)` |
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Answer» Correct Answer - c |
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| 136. |
A mass of 6 kg is suspended by a rope of length 2 m from the ceilling A force of 50 N in the horizontal direction is applied at the mid - point P of the rope as shown . What is the angle the rope makes with the vertical in equilibrium ? (Take = `10 ms^(-2)`) . Neglect mass of the rope . |
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Answer» Figure 5.8(b) and 5.8(c ) are known as free-body diagrams. Figure5.8(b) is the free-body diagram of W and Fig.5.8 (c ) is the free-body diagram of point P . Consider the equilibrium of the weight aaaaaaaaw. Clearly , `T_2` = `6 xx 10 `= 60 N. Consider the equilibrium of the point P under the action of three forces - the tensions `T_1` and `T_2`, and the horizontal force 50 N . The horizontal and vertical components of the resultant force must vanish seperately : `T_1` cos `theta` = `T_2` = 60 N `T_1` sin `theta` = 50 N which gives that tan `theta` = 5/6 or `theta ` = `tan^(-1)`(5/6) = `40^(@)` Note the answer does not depend on the length of the rope (assumed massles ) nor on the point at which the horizontal force is applied. |
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| 137. |
A uniform of rope length `L` and mass `M` is placed on a smooth fixed wedge as shown. Both ends of rope are at same horizontal level. The rope is initially released from rest, then the magnitude of initial acceleration of rope is. A. zeroB. `(cos alpha-cos beta)g`C. `(tan alpha-tan beta)g`D. `(sin alpha-sin beta)g` |
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Answer» Correct Answer - a |
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| 138. |
Fig show two bodies A and B of masses 2.5 kg and 2.8 kg respectively from a rigid support by two inextensible wires each of length 1.8 m . The upper wire is of negligible mass and lower wire is of mass `1.5 kg // m` . If the entire system moves upwards with an acceleration of `2 m//s^(2)`, find tension (i) at middle point p of upper wire (ii) at middle point Q of lower wire . Take `g =10 m//s^(2)` . |
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Answer» As the system moves upwards with an acceleration ` a = 2 m// s^(2) ` `:.` tension in the wire `T= R = m (g+a)` (i) For point P at point of upper wire `m= m_(A) + m_(B)` + mass of lower wire `= 2.5 + 2.8 + 1.8 xx 1.5 = 8.0 kg` `:. T = 8 (10+2) = 96 N` (ii) For point Q mid point of lower wire `m = m^(B)` + mass of half length of lower wire `= 2.8 + (1.8)/(2)xx 1.5 = 4.15 kg ` `:. T = m (g+a) = 4.15 (10+2) = 49.8 N` . |
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| 139. |
Two blocks of masses `2.9 kg` and `1.9 kg` are suspended from a rigid support `S` by two inextensible wires each of length `1 m` , as shown in figure.The upper wire has negligible mass and the lower wires and support have an uniformly distributed mass of `0.2 kg` .The whole system of blocks, wire and support have an upwards acceleration of `0.2 m//s^(2)`. Acceleration due to gravity is `9.8 m//s^(2)`. (a) Find the tension at the mid point of the lower wire. (b)Find the tension at the mid point of the upper wire. |
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Answer» As the system is accelerated upwards , therefore, tension at any point is given by ` T = m (g+a) ` for the point A , ` m = 2.9 + (0.2)/(2) = 3.0 ` ` :. T_(A) = m (g+a) = 3 (9.8+ 0.2 ) = 30 N ` For the point B ` m = 2.9 + 0.2 + 0.9 + 0 = 5 kg ` `:. T_(B) = m (g+ a ) = 5 (9.8+ 0.2) = 50 N` . |
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| 140. |
When a man jumps down from a height of several storeys onto a stretched trapaulin , he receives no injury . Why ? |
| Answer» When a man jumps onto stretched trapaulin , the trapaulin gets depressed at the place of impact , increasing the time t of impact . Force experienced by the man F = Impulse/time decreases and the man is not hurt . | |
| 141. |
Three equal weight `A,B` and `C` of mass `2kg` each are hanging on a string passing over a fixed frictionless pulley as shown in the figure. The tension in the string connecting weights `B` and `C` is approximately A. zeroB. 13 NC. 3.3 ND. 19.6 N |
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Answer» Correct Answer - B (b) Acceleration of system `a=(" Net pulling force")/("Total mass")=(4g-2g)/(6)=(g)/(3)` Equation of motion of block C is `m_(C)g-T_("BC")=m_(C)a` `therefore T_("BC")=m_(C)(g-a)` `=2(9.8-(9.8)/(3))=13 N` |
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| 142. |
A balloon with mass `m` is descending down with an acceleration a `(where altg)` . How much mass should be removed from it so that it starts moving up with an acceleration a? |
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Answer» Correct Answer - `(2Ma)/(g +a)` . Let`F` be the upthrust of air in either case In(i) case, `Mg - F = Ma` …(i) In (ii) case `F - (M -m) g = (M - m) a` …(ii) using (i) `(Mg - Mg) - Mg + mg = Ma - ma` or `m (g +a) = 2 Ma` `m = (2Ma)/(g + a)` . |
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| 143. |
A uniform rope of length L , resting on frictionless horizontal table is pulled at one end by a force F . What is the tension in the rope at a distance x from the end where the force is applied ? [ Hint : Consider the motion of the entire rope and the motion of x length of rope using P = ma formula and third law of motion ] |
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Answer» Correct Answer - `T = F (1 - x//L)` |
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| 144. |
One end of string of length `l` is connected to a particle on mass `m` and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed `v` the net force on the particle (directed toward centre) will be (`T` reprents the tension in the string):A. TB. `T-(mv)^(2)/(l)`C. `T+(mv^(2))/(l)`D. 0 |
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Answer» Correct Answer - A The net force on the particle directed towards the centre is T. |
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| 145. |
A simple pendalum is suspended from a peg on a verticle wall . The pendulum is pulled away from the well is a horizental position (see fig) and released . The bell his the well the coefficient of resitution being `(2)/(sqrt(5)` what is the miximum number of colision after which the amplitube of secillections between less that `60` digree ? |
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Answer» Correct Answer - 4 |
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| 146. |
One end of string of length `l` is connected to a particle on mass `m` and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed `v` the net force on the particle (directed toward centre) will be (`T` reprents the tension in the string): |
| Answer» The net force on the particle directed towards the centre is T this provides the necessary centripetal force to the particle moving in the circle . | |
| 147. |
A constant retarding force of 50 N is applied to a body of mass 20 kg moving intially with a speed of `15ms^(-1)` How long does the body take to stop ? |
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Answer» Here, ` F = - 50 N, m = 20 kg , u = 15 ms^(-1) , upsilon = 0 , t = ? ` From ` F = ma , a = (F)/(m) = (-50)/(20) = - 2.5 ms^(-2) ` From ` upsilon = u + at ` ` 0 = 15 - 2.5 t ` ` t = (15)/(2.5) = 6 s`. |
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| 148. |
The linear momentum of a body changes at the rate of `10kgms^(-1)` per second Force acting on the body isA. 1NB. 10NC. 1kg fD. 10 kg f |
| Answer» Correct Answer - b | |
| 149. |
Linear momentum of a body is defined as….. . |
| Answer» product of mass of body and its velocity. | |
| 150. |
A force of `50N` is inclined to the vertical at an angle of `30^(@)` Find the acceleration it produes in a body of mass `2kg` which moves in the horizontal direction . |
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Answer» Correct Answer - `12.5m//s^(2)` Here, `F = 50N` angle with verticle `0 = 30^(@)` In the horizontal direction, `a = ("horizontal force")/("mass") = (F sin0)/(m) = (50sin30^(@))/(2)` `=12.5 m//s^(2)` . |
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