InterviewSolution
Saved Bookmarks
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. |
Starting from rest, the time taken by a body sliding down on a rough inclined plane at `45^(@)` with the horizontal is twice the time taken to travel on a smooth plane of same inclination and same distance. Then, the coefficient of kinetic friction isA. 0.25B. 0.33C. 0.5D. 0.75 |
|
Answer» Correct Answer - D `mu = tan theta[1-(1)/(n^(2))]` Here, `theta = 45^(@) " and " n=2` `therefore " "mu = tan 45^(@) [1-(1)/(2^(2))]=1-(1)/(4)=(3)/(4)=0.75` |
|
| 2. |
The vertical displacement of a block A in meter is given by `y=t^(2)//4` where t is in second. The downward acceleration `a_(B)` of a block B ( in `m//s^(2)`) is A. 4B. 8C. 10D. 5 |
|
Answer» Correct Answer - A `a_(A)=(d^(2)y)/(dt^(2))=1/2` `a_(B)=8a_(A)` by constrained relation |
|
| 3. |
Consider the following statement. When jumping from some height, you should bend your knees as you come to rest instead of keeping your legs stiff. Which of the following relations can be useful in explaining the statement?A. `Deltap_(1)=-Deltap_(2)`B. `DeltaE = -Delta(PE+KE)=0`C. `FDeltat=mDeltav`D. `Deltax prop DeltaF` |
|
Answer» Correct Answer - C `FDeltat = mDeltav` `implies " "F=(mDeltav)/(t)` By doing so time of change in momentum increases and impulsive force on knees decreases. |
|
| 4. |
Pseudo force isA. electromagnetic in natureB. a nuclear forceC. a gravitational forceD. None of the above |
|
Answer» Correct Answer - D Pseudo force is not a real force. |
|
| 5. |
A weight w is to be moved from the bottom to the top of an inclined plane of inclination `theta` to the horizontal. If a smaller force is to be applied to drag it along the plane in comparison to lift it vertically up, the coefficient of friction should be such that.A. `mugt tan((pi)/2-(theta)/2)`B. `mult tan((pi)/2-(theta)/2)`C. `mu lt tan theta`D. `mugt tan. (pi)/(4)` |
|
Answer» Correct Answer - B `F_(1)=w sin theta+mu w cos theta` `F_(2)=w` As, `F_(1) lt F_(2)` `:. mu lt tan((pi)/4-(theta)/2)` |
|
| 6. |
A man is standing on a sparing platform, Reading of spring balaance is `60kg` wt If man jumps outside the platform then the reading of the spring balance .A. Becomes `gt60 kg` wt initially and then decrease to zeroB. Becomes `llt60 kg` wt first and then decreases to zeroC. Becomes =60 kg wt fist and then decrease to zeroD. Remains decreasing till zero |
|
Answer» Correct Answer - A The man will first press the platform to get extra normal force required for the jump. Hence initially the reading will be greater than 60 kg wt and then decreases to zero. |
|
| 7. |
A smooth block is released at rest on a `45^(@)` incline and then slides a distance `d`. The time taken to slide is n times as much to slide on rough incline than on a smooth incline. The coefficient of friction isA. `mu_(s)=1-1/(n^(2))`B. `mu_(s)=sqrt(1-1/(n^(2)))`C. `mu_(k)=1-1/(n^(2))`D. `mu_(k)=sqrt(1-1/(n^(2)))` |
|
Answer» Correct Answer - C When fiction is absent `a_(1)=g sin theta` `:. S_(1)=1/2 a_(1)t_(1)^(2)`......(i) When friction is present `a_(2)= g sin theta-mu g cos theta` `:. S_(2)=1/2 a_(2)t_(2)^(2).....(ii)` From eq.(i) and (ii) `1/2 a_(1)t_(1)^(2)=1/2a_(2)t_(2)^(2)` or `a_(1)t_(1)^(2)=a_(2)(nt_(1))^(2) ( :. t_(2)=nt_(1))` or `a_(1)=n^(2)a_(2)` or `(a_(2))/(a_(1))=(g sin theta-mu g cos theta)/(g sin theta) =1/(n^(2))` or `(g sin 45^(@)-mu g cos 45^(@))/(g sin 45^(@))=1/(n^(2))` or `1-mu_(k)=1/(n^(2)) rArr ` or `mu_(k)=1-1/(n^(2))` |
|
| 8. |
A man is pulling a rope attached to a block on a smooth horizontal table. The tension in the rope will be the same at all pointsA. if and only if the rope is not acceleratedB. if and only if the rope is masslessC. if either the rope is not accelerated or is masslessD. always |
|
Answer» Correct Answer - C The tension in the rope will be same at all points if it is massless or if the string has mass, the tension will be the same in the string when it is not accelerating. |
|
| 9. |
In figure, block A is released from rest, when spring is at its natural unstretched length. For block B of mass M to leave contact with the ground at some stage, the minimum mass of A must be A. 3MB. `(M)/(2)`C. MD. cannot say |
|
Answer» Correct Answer - B The spring will exert maximum force when A is at lowest position, say at distance x from the present position. `therefore" "mgx=(1)/(2) kx^(2)` where, k is force constant of the spring, for the block B to leave contact, kx = Mg `therefore " "mgx=(1)/(2) Mgx` `therefore" "m=(M)/(2)` |
|
| 10. |
A block weighing w is held against a vertical wall be pressing horizontally with a force F. Then, F needed to hold the block is A. equal to wB. equal to `(w)/(mu)` onlyC. greater than `(w)/(mu)` onlyD. greater than or equal to `(w)/(mu)` |
|
Answer» Correct Answer - B `F-N+0` `implies " "F-N` Further, `" "f=muN=w` `implies mu F=w implies F=(w)/(mu)` |
|
| 11. |
A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. the weight of the block is `(g=10 m//s^(2))` A. 20NB. 50 NC. 100 ND. 2N |
|
Answer» Correct Answer - D Force `F=muR=mg` weight of block `=muR=0.2xx10=2N` |
|
| 12. |
A block of mass 0.18 kg is attached to a spring of force constant 2N/m. The coefficient of friction between the block and the floor is 0.1. Initially, the block is at rest and the spring is unstretched. An impulse is given to the block. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m/s is v=N/10. Then, N isA. 2B. 3C. 4D. 6 |
|
Answer» Correct Answer - D Here, `m=0.18 kg, k=2 N//m, mu=0.1, x=0.06m`. According to conservation of mechanical energy principle, we know Decrease in mechanical energy = Work done against friction `(1)/(2) mv^(2)-(1)/(2)kx^(2)=mumgx` `implies " "v=sqrt((2mumgx+kx^(2))/(m))` Substituting the values of m, `mu`, g, x, and k, we get `v = sqrt((2xx0.1 xx0.18xx9.8xx0.06+2xx0.06)/(0.18))` `v = (4 //10)m//s` So, N = 4 |
|
| 13. |
What is the minimum stopping distance for a vehicle of mass m moving with speed v along a level road. If the coefficient of friction between the tyres and the road is `mu`A. `(v^(2))/(2 mug)`B. `(2v^(2))/(mu g)`C. `(v^(2))/(mug)`D. none of these |
|
Answer» Correct Answer - A `0^(2)=V^(2)-2 mugs` `rArr s=(V^(2))/(2 mug). (A)` |
|
| 14. |
Water jet issues water from a nozzle of `2 cm^(2)` cross-section with velocity 30 cm/s and strikes a plane surface placed at right angles to the jet. The force exerted on the plane isA. 200 dyneB. 400 dyneC. 1800 dyneD. None of the above |
|
Answer» Correct Answer - C `thereforeF=(Deltap)/(Deltat)=v(Deltam)/(Deltat)=rhoAv(Deltax)/(Deltat)` `=rhoAv((vDeltat)/(Deltat))=1xx2xx(30)^(2)=1800` dyne |
|
| 15. |
A 0.1 kg body moves at a constant speed of 10 m/s. It is pushed by applying a constant force for 2 s. Due to this force, it starts moving exactly in the opposite direction with a speed of `4 m//s`. ThenA. the deceleration of the body is `7 m//s^(2)`B. the magnitude of change in momentum is 1.4 kg-m/sC. impulse of the force is 1.4 N-sD. All of the above |
|
Answer» Correct Answer - D `a=(v-u)/(t)=(-4-10)/(2)=-7m//s^(2)` The magnitude of change in momentum `=|m(v-u)|=14kg-m//s`=Impulse Now, F = ma = 0.7 N |
|
| 16. |
A particle is on a smooth horizontal plane. A force F is applied whose F-t graph is given. Then, A. at `t_(1)`, acceleration is constantB. initially body must be in restC. at `t_(2)`, acceleration is constantD. both c and d are correct |
|
Answer» Correct Answer - D At `t_(2)`, force is constant So, acceleration is constant. At t = 0, force is zero. Hence, acceleration is zero. |
|
| 17. |
A bead of mass m is located on a parabolic wire with its axis vertical and vertex directed towards downwards as in figure and whose equation is `x^2=ay`. If the coefficient of friction is `mu`, the highest distance above the x-axis at which the particle will be in equilibrium isA. `mua`B. `mu^(2)a`C. `1/4 mu^(2)a`D. `1/2 mua` |
|
Answer» Correct Answer - C For the sliding not to occur when `tan theta le mu` `tan theta=(dy)/(dx) =(2x)/a=(2sqrt(ay))/a=2sqrt(y/a)` `:. 2sqrt(y/a) le mu` or `y le=(amu^(2))/4` |
|
| 18. |
Three weights w, 2w and 3w are connected to identical spring suspended from a rigid horizontal rod. The assembly of the rod and the weights fall freely. The positions of the weight from the rod are such thatA. 3w will be farthestB. w will be farthestC. all will be at the same distanceD. 2w will be farthest |
|
Answer» Correct Answer - C For w, 2w, 3w, apparent weight will be zero because the system is failing freely. So, the distance of the weights from the rod will be same. |
|
| 19. |
A wooden block and a metal coin are connected by a string as shown in figure. Both have same volume and both are placed in a beaker of water. Now the beaker is given acceleration towards right, then the string will tilt towards A. RightB. No whereC. leftD. Either right or left |
|
Answer» Correct Answer - A Metal coin has more inertia then the wooden block. Hence it will remain behind it. Thus string will tilt towards right. |
|
| 20. |
A force F is applied to the initially, stationary cart. The variation of force with time is shown in the figure. The speed of cart at t = 5 s is A. 10 m/sB. 8.33 m/sC. 2 m/sD. zero |
|
Answer» Correct Answer - B The equation of parabola is `x^(2)=4ay` Here, `t^(2)=4aF` When t = 5 s, F = 50 N (see graph) `therefore" "5^(2)=4axx50` `therefore " "a=(25)/(200)=(1)/(8)` `therefore t^(2) = 4 xx(1)/(8)F=(F)/(2)implies F=2t^(2)` Acceleration `=(F)/(m)=(2t^(2))/(m)=(2t^(2))/(10)=(t^(2))/(5)` `therefore(dv)/(dt)=(t^(2))/(5)" or "int_(0)^(v)dv=int_(0)^(5)(t^(2))/(5)dt` `therefore v = (1)/(5)[(t^(3))/(3)]_(0)^(5)=(125)/(15)=8.33 m//s` |
|
| 21. |
The mass m is placed on a body of mass M. There is no friction. The force F is applied on M on it moves with acceleration a. Then, the force on the top body is A. FB. maC. F - maD. None of these |
|
Answer» Correct Answer - D Since, no force of friction is present. So, no horizontal force is present on body of mass m. In vertical direction normal force balances weight of the body. Hence, net force of top body must be zero. Lesser contact force corresponds more comfortableness. |
|
| 22. |
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. `f_(a)=f_(b)`B. `f_(a)=f_(b)//2`C. `f_(a)=f_(b)//3`D. `f_(a)=2f_(b)` |
|
Answer» Correct Answer - C `f_(a)=((2m-m)/(2m+m))g=g/3` `f_(b)=(2mg-mg)/m=g` So, `f_(a)=f_(b)//3` |
|
| 23. |
A block is given velocity 10 m/s along the fixed inclined as shown in figure from the bottom of the inclined plane. Find the total distance travelled along the inclined plane. If the coefficient of friction is equal to 0.5 A. `5m`B. `10m`C. `(25)/(3) m`D. `(50)/(3) m` |
|
Answer» Correct Answer - B Distance travelled along plane till it stops `s=((10)^(2))/(2(g sin 37^(@)+mu cos 37^(@)))=5m` But friction force is not sufficient to stope there, so it will come back to the bottom of the inclined plane. Total distance travelled =5+5=10 m |
|
| 24. |
Figure shows a 5 kg ladder hanging from a string that is connected with a ceiling and is having a spring balance connected in between. A boy of mass 25 kg is climbing up the ladder at acceleration `1(m)/(s^2)`. Assuming the spring balance and the string to be massless and the spring to show a constant reading, the reading of thespring balance is: (Take`g=10(m)/(s^2)`)A. `30 kg`B. `32.5 kg`C. `35 kg`D. `37.5 kg` |
|
Answer» Correct Answer - B If reading of spring balance is T, then applying NLM on (man+ ladder) system `T-(25+5) g =25a` `T-30 g=25a rArr T-300=25(1) rArr T=325 N=32.5 kg`. |
|
| 25. |
Two bodies have same mass and speed, thenA. their momentum are sameB. the ratio of momentums is not determinedC. the ratio of their magnitudes of momentum is oneD. Both a and b are correct |
|
Answer» Correct Answer - D (a) The magnitude of momentum are same to each other. (b) Momentum is a vector quantity and vector does not obey division law. So, the ratio of momentum is not determined. |
|
| 26. |
A heavy block of mass m is supported by a cord C attached to the ceiling, and another cord D is attached to the bottom of the block. If a sudden jerk is given to D, then A. cord C breaksB. cord D breaksC. cord C and D both BreakD. None of the cords breaks |
|
Answer» Correct Answer - B Cord D will breaks due to inertia offered by the block of mass m. |
|
| 27. |
The linear momentum p of a body moving in one dimension varies with time t according to the equation `p = a + bt^(2)`, where a and p are positive constant. The net force acting on the body isA. a constantB. proportional to `t^(2)`C. inversely proportional to tD. proportional to t |
|
Answer» Correct Answer - D Given, `p = a + bt^(2)` `(dp)/(dt) = 2bt` `because" "F = (dp)/(dt)` `therefore" "F=2bt " or "F prop t` |
|
| 28. |
A system consists of two cubes of mass `m_(1)`, and `m_(2)` respectively connected by a spring of force constant k. force (F) that should be applied to the upper cube for which the lower one just lifts after the force is removed, is A. mgB. `m_(2)g`C. `(m_(1)+m_(2))g`D. `(m_(1)m_(2))/(m_(1)+m_(2))g` |
|
Answer» Correct Answer - C Let `x_(0)` be the maximum compression, then `m_(1)g+F=kx_(0)" "...(i)` `"and "(1)/(2)kx_(0)^(2)=m_(1)gx_(0)" "...(ii)` If x is the maximum elongation, then `kx = m_(2)g" "...(iii)` `"and "-(1)/(2)kx^(2)=m_(1)g(x+x_(0))" "...(iv)` `"or "k(x_(0)-x)=2m_(1)g` From Eqs. (i) and (iii), we get `m_(1) g+ F- m_(2)g = k (x_(0) -x) =2m_(1)g` `"or "F = 2m_(1)g + m_(2) g-m_(1)g=(m_(1)+m_(2))g` |
|
| 29. |
A block of mass 4 kg is kept on ground. The co-efficient of friction between the block and the ground is 0.80. an external force of magnitude 30 N is applied parallel to the ground. The resultant force exerted by the ground on the block is `(g=10m//x^(2))`A. 40NB. 30NC. 0ND. 50N |
|
Answer» Correct Answer - D N=mg=40 `(f_(s))_(max)=muN=(0.8) (40)=32` `f_(S)=ext` force =30 `R^(2)=N^(2)+f_(s)^(2)=(50)^(2)` `:. R=50 Nt`. |
|