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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.
| 451. |
A body of mass 2kg is at rest on a horizontal table. The coefficient of friction between the body and the table is `0.3` . A force of 5N is applied on the body. The acceleration of the body is.A. `0ms^(-2)`B. `2.5ms^(-2)`C. `5ms^(-2)`D. `7.5ms^(-2)` |
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Answer» Correct Answer - A `f_(ms)=0.3xx2xx10=6N` applied force `ltf_(ms)` So, body would not move. |
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| 452. |
A uniform rope of length `(piR)/(2)` has been placed on fixed cylinder of radius R as shown in the fig. One end of the rope is at the top of the cylinder. The coefficient of friction between the rope and the cylinder is just enough to prevent the rope from sliding. Mass of the rope is M. (a) At what position, the tension in the rope is maximum? (b) Calculate the value of maximum tension in the rope. |
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Answer» Correct Answer - (a) `theta = 45^(@)` from vertical diameter . (b) `T_("max") = 2 ((sqrt2-1)/(pi))Mg` |
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| 453. |
A force of `150N` produces an acceleration of `2ms^(-2)` in a body and a force of `200N` produces an acceleration of `3ms^(-2)` The mass of the body and the coefficinent of kinetic friction are .A. `50kg,0.1`B. `25kg,0.1`C. `50kg,0.5`D. `50kg,0.2` |
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Answer» Correct Answer - A `F -f =ma, f =mu_(k) mg` |
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| 454. |
A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of `45^@` with the initial vertical direction isA. `Mg`B. `(Mg)/(sqrt2)`C. `Mg(sqrt2+1)`D. `sqrt2Mg` |
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Answer» Correct Answer - A `"Tan" theta = (F)/(mg)` |
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| 455. |
A bomb of mass `6kg` initially at rest explodes in to three identical fragments On of the fragments moves with a velocity of `10sqrt3 hatim//s` another frament moves with a velocity of `10 hatj m//s` then the thired fragment moves with velocity of magnitude .A. `30m//s`B. `20m//s`C. `15m//s`D. `5 m//s` |
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Answer» Correct Answer - B `P_(3) = - (P_(1)+P_(2)), P_(3) = sqrt(P_(1)^(2) +P_(2)^(2))` `m_(3)v_(3) =sqrt((m_(1)v_(1))^(2)+(m_(2)v_(2))^(2)` |
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| 456. |
A boy of mass M stands on a platform of mass m as shown in figure-2.210, supporting two stringsvia a massless support S. Strings are passing over the pulleys and other ends are connected to the platform as shown. Find the acceleration of the platform and the boy if he applies a constant force T to the string he isholding, which is connected to support S. |
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Answer» Correct Answer - `[(2T-(M+m)g)/(M+m)]` |
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| 457. |
A man and a cart move towards each other. The man weight `64kg` and the cart weighs `32kg`. The velocity of the man is `5.4km//hr` and that of the cart he jumps on to it The velocity of the cart carrying the man will be .A. `3km//hr`B. `30 km//hr`C. `1 .8km//hr`D. zero |
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Answer» Correct Answer - A `m_(1) u_(1) + m_(2)u_(2) = (m_(1)+m_(2))v` |
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| 458. |
A body of mass `2kg` moving on a horizontal surface with an initial velocity of `4 ms^(-1)` comes to rest after `2` second. If one wants to keep this body moving on the same surface with a velocity of `4 ms^(-1)` the force required isA. zeroB. 2 NC. 4 ND. 8 N |
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Answer» Correct Answer - C `0=4-axx2rArra=2m//s^(2)` `F=ma=2xx2=4N` |
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| 459. |
Velocity of particle of mass 2kg varies with time t accoridng to the equation `vecv=(2thati+4hatj)ms^(-1)`. Here t is in seconds. Find the impulse imparted to the particle in the time interval from t=0 to t=2s. |
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Answer» Impulse=Change in momentum`=m(vecv_(2)-vecv_(1))` `=2xx[(4hati+4hatj)-(4hatj)]=8hati kgms^(-1)` |
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| 460. |
A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is `0.5`, the maximum percentage of the length of the chain that can hang over one edge of the table isA. `20%`B. `25%`C. `35%`D. `15%` |
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Answer» Correct Answer - A `(l)/(L) xx 100 = (mu)/(mu+1) xx 100` |
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| 461. |
A body of mass `2kg` moving on a horizontal surface with an initial velocity of `4 ms^(-1)` comes to rest after `2` second. If one wants to keep this body moving on the same surface with a velocity of `4 ms^(-1)` the force required isA. 8NB. 4NC. ZeroD. 2 N |
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Answer» Correct Answer - B `F=2xx(4)/(2)=4N` |
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| 462. |
A block of weight `200N` is pulled along a rough horizontal surface at contant speed by a force of `100N` acting at an angle `30^(@)` above the horizontal. The coefficient of kinetic friction between the block and the surface is .A. `0.43`B. `0.58`C. `0.75`D. `0.83` |
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Answer» Correct Answer - B `F cos theta = f_(k),f_(k) = mu_(k) N ,N = mg -F sin theta` |
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| 463. |
A block of mass `10kg` is placed on a rough horizontal surface having coefficient of friction `mu=0.5` . If a horizontal force of `100N` is acting on it, then acceleration of the will be.A. `0.5m//s^(2)`B. `5m//s^(2)`C. `10m//s^(2)`D. `15m//s^(2)` |
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Answer» Correct Answer - B `a=(Applied fo rce-Ki n etic f riction)/(Mass)` `=(100-0.5xx10xx10)/(10)=5m//s^(2)` |
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| 464. |
In the adjacent figure there is a cube having a smooth groove at an inclination of `30°` with horizontal in its vertical face. A cylinder A of mass 2kg can slide freely inside the grove. The cube is moving with constant horizontal acceleration `A_(0)` parallel to the shown face, so that the slider does not have acceleration along horizontal. A. The normal reaction acting on cube is zeroB. The value of `a_(0)` is `gsqrt(3)`C. The value of `a_(0)` is g.D. Acceleration of the particle in groimd frame is g |
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Answer» Correct Answer - A::B::D |
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| 465. |
Figure shown a fixed surface inclined at an angle `theta` to the horizontal. A smooth groove is cut on the incline along QR forming an angle `phi` with PR. A small block is released at point Q and it slides down to R in time t. Find t. |
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Answer» Correct Answer - `t = (1)/("sin"theta "sin"phi) sqrt((2d)/(g))` |
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| 466. |
In the arrangement shown in figure, a block A of mass m has been placed on a smooth wedge B of mass M. The wedge lies on a horizontal smooth surface. Another block C of mass `(M)/(4)` has been placed in contact with the wedge B as shown. The coefficient of friction between the block C and the vertical wedge wall is `mu= (3)/(4)`. Find the ratio `(m)/(M)` for which the block C will not slide with respect to the wedge after the system is released? |
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Answer» Correct Answer - `(m)/(M) = (20)/(3sqrt3 - 4) = 16.7` |
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| 467. |
Block B has mass m and is released from rest when it is on top of wedge A, which has a mass 3m. Determine the tension in cord CD needed to hold the wedge from moving while B is sliding down A. Neglect friction. A. `(mg)/(2) sin (2theta)`B. `(mg)/(2) sin (3theta)`C. `(mg)/(2) sin (3theta)`D. `(mg)/(2) sin (2theta)` |
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Answer» Correct Answer - A `T =mg sin theta cos theta, T =(mg)/(2) sin 2theta` . |
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| 468. |
Figure-2.162 shows a wedge of mass 2kg resting on a fritionless floor. A block of mass 1 kg is kept on the wedge and the wedge is given an acceleration of `5 m//s^(2)` towards right. Then : A. Block will remain stationary w.r.t. wedge.B. The block will have an acceleration of `1 m/sec^(2)` w.r.t. the wedge.C. Normal reaction on the blockis 11N.D. Net force acting on the wedge is 4 N. |
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Answer» Correct Answer - C |
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| 469. |
An inclined plane makes an angle `30^(@)` with the horizontal. A groove (OA) of length 5m cut in the plane makes an angle `30^(@)` with OX. A short smooth cylinder is free to slide down under the influence of gravity. The time taken by the cylinder to reach from A to O is `(g=10ms^(-2))`. A. `4s`B. `2s`C. `3s`D. `1s` |
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Answer» Correct Answer - B Acceleration of cylinder down the plane is `a=(gsin30^(@))(sin30^(@))=10((1)/(2))((1)/(2))=2.5m//s^(2)` time taken: `t=sqrt((2_(s)/(a)))=sqrt((2xx5)/(2.5))=2s` |
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| 470. |
A triangular wedge of mass M lies on a smooth horizontal table with half of its base projecting out of the edge of the table. A block of mass m is kept at the top of the smooth incline surface of the wedge and the system is let go. Find the maximum value of `(M)/(m)` for which the block will land on the table. Take `theta = 60^(@)`. |
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Answer» Correct Answer - 3 |
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| 471. |
A body is released from the top of a smooth inclined plane of inclination `theta`. It reaches the bottom with velocity `v`. If the angle of inclina-tion is doubled for the same length of the plane, what will be the velocity of the body on reach ing the ground .A. `v`B. `2v`C. `(2 cos theta)^(1/2) v`D. `(2 cos theta)^(1/2) v` |
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Answer» Correct Answer - C `v =sqrt(2glsin theta),v^(1)=sqrt(2glsin2theta)` `F_(UP) = mg (sin theta +mu_(k) cos theta)` . |
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| 472. |
Block A of mass m is placed over a wedge of same mass m. Both the block and wedge are placed on a fixed inclined plane. Assuming all surfaces to be smooth, the displacement of the block A in ground frame in 1s is `(gsin^(2)theta)/(x+sin^(2)theta)` then the value of x is: |
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Answer» Correct Answer - 1 Let acceleration of wedge in ground frame is a down the plane. The acceleration of block A will be a sin `theta` vertically downward `veca_(A//g)=veca_(A//B)+veca_(B//g).(1)` `[a_(A//g)]_(x)=[a_(A//B)]_(x)+[a_(B//g)]_(x)...(2)` From `FBD` of A it is clear that Block A cannot accelerate horizontally i.e in x-direaction becasue there is no force in x-direaction Block A can accelerate in y-direaction only. `[a_(A//g)]_(x)=0` There for `[a_(A//g)]=-[a_(B//g)]_(x)` That means for an observation on wedge block moves only `x gt 0`. For block `A`, mg -N =n (a sin theta) ....(3)` For block `B`, (N +mg) sin theta = ma...(4)` On solving eqns (3) and (4) we get `a=[(2gsintheta)/(1+sin^(2)theta)]` The acceleration of block A, `a_(A) a sin theta` `a=[(2gsintheta)/(1+sintheta)]sintheta=[(2gsin^(2)theta)/(1+sin^(2)theta)]` Displacement of block `A` in `1 s` is `S =0+(1)/(2)a_(A)t^(2)=(1)/(2)xx[(2gsintheta)/(1+sin^(2)theta]]xx(1)^(2)[(gsin^(2)theta)/(1+sin^(2)theta)]` . |
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| 473. |
A triangular wedge A is held fixed and a block B is released on its inclined surface, from the top. Block B reaches the horizontal ground in time t. In another experiment, the wedge A was free to slide on the horizontal surface and it took t’ time for the block B to reach the ground surface after it was released from the top. Neglect friction and assume that B remains in contact with A. (a) Which time is larger t or t´? Tell by simple observation. (b) When wedge A was free to move, it was observed that it moved leftward with an acceleration `(g)/(4)` and one of the two measured times (t & t´) was twice the other. Find the inclination `theta` of the inclined surface of the wedge. |
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Answer» Correct Answer - (a) `t gt t^(1)` (b) `theta = tan^(-1) ((1)/(12))` |
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| 474. |
A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is `5m//s^2`, the frictional force acting on the block is…………newtons.A. 2 NB. 3 NC. 5 ND. 6 N |
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Answer» Correct Answer - C |
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| 475. |
Coefficient of friction between two block shown in figure is `mu = 0.4`. The blocks are given velocities of `2 m//s` and `8 m//s` in the directions figure. Find (a) The time when relative motion between them will stop (b) the common velocities of blocks upto that instant. (c) Displacement of `1 kg` and `2 kg` block upto that instant `(g = 10 m//s^(2))`.A. `1 sec`B. `2 sec`C. `3 sec`D. `4 sec` |
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Answer» Correct Answer - A `a =mug, v =u +at,` |
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| 476. |
All surfaces are smooth The acceleration of mass m relative to the wedge is .A. `g sin theta`B. `g sin theta +a cos theta`C. `g sin theta -a cos theta`D. `a cos theta` |
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Answer» Correct Answer - B `F =ma, a = g sin theta + cos theta` . |
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| 477. |
A wedge of mass m is kept on a smooth table and its inclined surface is also smooth. A small block of mass m is projected from the bottom along the incline surface with velocity u. Assume that the block remains on the incline and take `theta = 45^(@), g = 10 m//s^(2)`. (a) Find the acceleration of the wedge and the x and y components of acceleration of the block. (b) Draw the approximate path of the block as observed by an observer on the ground. At what angle does the block hit the table? (c) Calculate the radius of curvature of the path of the block when it is at the highest point. |
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Answer» Correct Answer - `veca_("wedge") = (g)/(3) hati` `a_("x block") = (-g)/(3)` `a_("y block") = (2g)/(3)` (b) The block hits the table normally . |
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| 478. |
A wedge is placed on the smooth surface of a fixed incline having inclination `theta` with the horizontal. The vertical wall of the wedge has height h and there is a small block A on the edge of the horizontal surface of the wedge. Mass of the wedge and the small block are M and m respectively. (a) Find the acceleration of the wedge if friction between block A and the wedge is large enough to prevent slipping between the two. (b) Find friction force between the block and the wedge in the above case. Also find the normal force between the two. (c) Assuming there is no friction between the block and the wedge, calculate the time in which the block will hit the incline. |
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Answer» Correct Answer - (a) `g "sin"theta` (b) `f = (1)/(2) mg "sin"2theta` `N = "mg"cos^(2)theta` (c) `t = sqrt((2h(M + m sin^(2)theta)/((M+m) g sin^(2)theta))` |
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| 479. |
A block is placed on an incline having inclination `theta` . There is a rigid L shaped frame fixed to the block. A plumb line (a ball connected to a thread) is attached to the end A of the frame. The system is released on the inline. Find the angle that the plumb line will make with vertical in its equilibrium position relative to the block when (a) the incline is smooth (b) there is friction and the acceleration of the block is half its value when the incline is smooth |
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Answer» Correct Answer - (a) `theta` (b) `tan^(-1) (("sin"theta. "cos"theta)/(2 - sin^(2) theta))` |
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| 480. |
A rod of mass M and length L lies on an incline having inclination of `theta = 37^(@)`. The coefficient of friction between the rod and the incline surface is `mu = 0.90`. Find the tension at the mid point of the rod. |
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Answer» Correct Answer - Zero |
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| 481. |
A 60 kg platform has been placed on a rough incline having inclination `theta = 37^(@)`. The coefficient of friction between the platform and the incline is `mu = 0.5`. A 40 kg man is running down on the platform so as to keep the platform stationary. What is the acceleration of the man? It is known that the man cannot manage to go beyond an acceleration of `7m//s^(2).["sin"37^(@) = (3)/(5)]` |
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Answer» Correct Answer - `5m//s^(2) le a le 7m//s^(2)` |
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| 482. |
Coefficient of friction between two block shown in figure is `mu = 0.4`. The blocks are given velocities of `2 m//s` and `8 m//s` in the directions figure. Find (a) The time when relative motion between them will stop (b) the common velocities of blocks upto that instant. (c) Displacement of `1 kg` and `2 kg` block upto that instant `(g = 10 m//s^(2))`. |
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Answer» (i) Frictional force between two blocks will oppose the relative motion. For 1 kg block friction support themotion `&` for 2 kg friction oppose the motion. Let common velocity be v then for `1kg v=2+a_(1)t` where `a_(1)=(mu(1g))/(2)=(0.4xx10)/(2)=2 ms^(-2)rArr2+4t=8-2trArr6t=5rArrt=1s` (ii) `v=2+4t=2+4xx1=6ms^(-1)` (iii) Displacement of 1kg block from initial point `s=ut+(1)/(2)at^(2)rArrs_(1)=2xx1+(1)/(2)xx4xx1^(2)=2+2=4m` Displacement of 2kg block from intial point `s=ut+(1)/(2)at^(2)rArrs_(2)=8xx1-(1)/(2)xx2xx1^(2)=8-1=7m` |
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| 483. |
All surfaces are smooth Find the horizontal displacements of the block and the wedge when the block slides down from top to bottom . |
| Answer» When the block slides down on the smooth wedge the wedge moves backwards In the horizontal direaction there is no external force , `vecF_(x) =0`. | |
| 484. |
A triangular wedge W having mass M is placed on an incline plane with its face AB horizontal. Inclination of the incline is `theta`. On the flat horizontal surface of the wedge there lies an infinite tower of rectangular blocks. Blocks 1, 2, 3, 4 ………. have masses `M , (M)/(2) , (M)/(4) , (M)/(8)` ......... respectively . All surfaces are smooth. Find the contact force between the block 1 and 2 after the system is released from rest. Also find the acceleration of the wedge. |
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Answer» Correct Answer - `N_(12) = (Mgcos^(2)theta)/(1 + 2 sin^(2)theta)` `a = (3g "sin"theta)/(1 + 2 sin^(2)theta)` |
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| 485. |
Three point masses each of mass m are joined together using a string to form an equilateral triangle of side a. The system is placed on a smooth horizontal surface and rotated with a constant angular velcoity `omega` about a vertical axis passing through the centroid Then the tension in each string is .A. `ma omega^(2)`B. `3ma omega^(2)`C. `(maomega^(2))/(3)`D. `(maomega^(2))/sqrt(3)` |
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Answer» Correct Answer - C `F =mromega^(2),F =sqrt3T` |
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| 486. |
In the system shown in the figure, the string is light and coefficient of friction between the 10 kg block and the incline surface is `mu = 0.5`. Mass of the hanger, H is 0.5 kg. A boy places a block of mass m on the hanger and finds that the system does not move. What could be values of mass m? `"tan"37^(@) = (3)/(4) "and" g = 10m//s^(2)` |
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Answer» Correct Answer - `1.5 kg le m le 9.5 kg` |
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| 487. |
Figure shows two blocks in contact placed on an incline of angle `theta = 30^(@)`. The coefficient of friction between the block of mass 4 kg and the incline is m1, and that between 2 kg block and incline is `mu_(2)`. Find the acceleration of the blocks and the contact force between them if – (a) `mu_(1) = 0.5 , mu_(2) = 0.8` (b) `mu_(1) = 0.8 , mu_(2) = 0.5 ` (c) `mu_(1) = 0.6 , mu_(2) = 0.1 " "` [ Take `g = 10m//s^(2)`] |
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Answer» Correct Answer - (a) constant force = 0 , acceleration of 4 kg block is `0.7 m//s^(2)` and that of other block is zero (b) constant force 1.4 N , acceleration of both =0 (c) Constant force = 5.74 N , acceleration of both = `1.27 m//s^(2)` |
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| 488. |
If the coefficient of static friction between a table and a uniform massive rope is `mu`, what fraction of the rope can hang over the edge of a table without the rope sliding. |
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Answer» Correct Answer - `[(mu)/(1+mu)]` |
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| 489. |
Consider the shown arrangement. The coefficient of friction between the two blocks is 0.5 . There is no friction between the 4 kg block and of 12 N is applied on the 2 kg block as shown in the figure, the acceleration of the 4 kg block would be A. `2.5 m//s^(2)`B. `2 m//s^(2)`C. `5 m//s^(2)`D. None of these |
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Answer» Correct Answer - B |
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| 490. |
In the arrangement shown in figure, pulley and string are light. Friction coefficient between the two blocks is `mu` whereas the incline is smooth. Block A has mass m and difference in mass of the two blocks is`deltam`. Find minimum value of m for which the system will not accelerate when released from rest. |
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Answer» Correct Answer - `mu_("min") = (deltam)/(2m) "tan"theta` |
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| 491. |
A block on table shown in is just on the edge of slipping Find the coefficient of static friction between the blocks and table . |
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Answer» `f_(1) =T sin theta` `mu mg =T sin theta….(1)` `80 = T cos theta…..(2)` `(T sin 30^(@))/(T cos30^(@)) = (mumg)/(80),` `Tan 30^(@) = (mu40)/(80) , (1)/(sqrt3) = (mu)/(2) rArr mu =(2)/(sqrt3) = 1.15` . |
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| 492. |
A 40kg slab rests on a frictionless floor as shown in the figure. A 10kg block rests on the top of the slab. The static coefficient of friction between the block and slab is `0.60` while the kinetic friction is `0.04` . The 10kg block is acted upon by a horizontal force 100N. if `g=9.8m//s^(2)` , the resulting acceleration of the slab will be. A. `1m//s^(2)`B. `1.5m//s^(2)`C. `2m//s^(2)`D. `6m//s^(2)` |
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Answer» Correct Answer - A Limiting friction between block and slab `=mu_(s)m_(A)g` `=0.6xx10xx10=60N` But applied force on block `A` is `100N` . So the block will slip over a slab. Now kinetic friction works between block and slab`Fk=mu_(k)m_(A)g=0.4xx10xx10=40N` This kinetic friction helps to move the slab :. Acceleration of slab `=(40)/(m_(B))=(40)/(40)=1m//s^(2)` . |
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| 493. |
A 40kg slab rests on a frictionless floor as shown in the figure. A 10kg block rests on the top of the slab. The static coefficient of friction between the block and slab is `0.60` while the kinetic friction is `0.40` . The 10kg block is acted upon by a horizontal force 100N. if `g=9.8m//s^(2)` , the resulting acceleration of the slab will be. A. `1.5m//s^(-2)`B. `2.0m//s^(-2)`C. `10m//s^(-2)`D. `1.0m//s^(-2)` |
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Answer» Correct Answer - D `f_(ms)=0.6xx10xx10N=60N` Since the applied force is dreater than `f_(ms)` therefore the block will be in motion . So, we should consider `f_(k)` `f_(k)=0.4xx10xx10N=40N` This would cause acceleration of `40kg` block Acceleration `=(4xx10)/(40kg)=1.0ms^(-2)` . |
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| 494. |
In the arrangement shown in the figure the coefficient of friction between the blocks C and D is `mu_(1) = 0.7` and that between block D and the horizontal table is `mu_(2) = 0.2`. The system is released from rest. [ Take `g = 10 ms^(–2)`] Pulleys and threads are massless. (a) Find the acceleration of the block C. (b) Block B is replaced with a new block. What shall be the minimum mass of this new block so that block C and D accelerate in opposite direction? |
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Answer» Correct Answer - (a) `2ms^(-1)` (b) 2.1 kg |
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| 495. |
The system shown is just on the verge of slipping. The coefficient of static friction between the block and the table top is : A. 0.5B. 0.95C. 0.15D. 0.35 |
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Answer» Correct Answer - D |
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| 496. |
The coefficient of static friction, `mu_(s)` between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless.`(g=10m//s^(2))` A. 4.0kg.B. 0.2kg.C. 0.4kg.D. 2.0kg. |
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Answer» Correct Answer - C `T=M_(B)g=mu_(s)M_(A)g` `therefore M_(B)=mu_(s)M_(A)=0.2xx2=0.4kg` |
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| 497. |
A cyclist on the ground goes round a ciruclar path of circumference 34.3 m in `sqrt(22)` second. The angle made by him, with the vertical, will be:-A. `45^(@)`B. `40^(@)`C. `42^(@)`D. `48^(@)` |
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Answer» Correct Answer - A `tan theta =(v^(2))/(rg)=(((34.3)/sqrt(22))^(2))/(((34.3)/(2pi))xx10)=1` `therefore theta=45^(@)` |
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| 498. |
A cork is submerged in water by a spring attached to the bottom of a bowl. When the bowl is kept in an elevator moving with acceleration downwards, the length of spring.A. IncreasesB. DecreasesC. Remains unchangedD. Data insufficient |
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Answer» Correct Answer - D |
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| 499. |
Figure shown two pulley arrangments for lifting a mass `m` . In case-1, the mass is lifting by attaching a mass 2m while in case-2 the mass is lifted by pulling the other end with a downward force `F=2mg` . If `a_(a)` and `a_(b)` are the accelerations of the two masses then (Assumme string is massless and pulley is ideal). A. `a_(a)=a_(b)`B. `a_(a)=a_(b)/(2)`C. `a_(a)=a_(b)/(3)`D. `a_(a)=2a_(b)` |
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Answer» Correct Answer - C In case-1, the acceleration of the system `a_(a)=((2m-m)/(2m+m))g=(g)/(3)` In case-2, the acceleration of the system `a_(a)=((2m-m)/(2m+m))g=(g)/(3)=g` So, `a_(a)=a_(b)//3` . |
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| 500. |
The pulley arrangements shown in figure are identical, the mass of the rope being negligible. In case I the mass m is lifted by attaching a mass `2m` to the other end of rope with a constant downward force `F =2mg`, where g is acceleration due to gravity The acceleration of mass m in case I is .A. ZeroB. More than that in case (b)C. Less than that in case (b)D. Equal to that in case (b) |
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Answer» Correct Answer - C In case (b) entire tension =2mg, hence acceleration is more. |
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