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.
| 501. |
If the elevator in shown figure is moving upwards with constant acceleration `1 ms^(-2)`, tension in the string connected to block A of mass 6 kg would be ( take, g`=10 ms^(-2)`) A. 60 NB. 66 NC. 54 ND. 42 N |
|
Answer» Correct Answer - B (b) The tension T in the string is T=6(g=a)=6(10+1)=66N |
|
| 502. |
An elevator and its load have a total mass of 800 kg. If the elevator, originally moving downwards at `10 ms^(-1)`, is brought to rest with constant deceleration in a distance of 25 m, the tension in the supporting canble will be (`g=10 ms^(-2)`)A. 8000 NB. 6400 NC. 11200 ND. 9600 N |
|
Answer» Correct Answer - D (d) Given, `u=10 ms^(-1),v=0`, So using, `v^(2)=u^(2)-2as` `0=(10)^(2)-2(a)(25)` `therefore a=2 ms^(-2)` (upwards) Now, `T-800g=800a` `therefore T=800(10+2)=9600 N` |
|
| 503. |
A person of mass 50 kg stands on a weighing scale on a lift . If the lift is descending with a downward acceleration of `9ms^(-2)` what would be the reading of the weighing scale? `(g =10ms^(-2)) ` . |
|
Answer» Reading of weighing scale = apparent wt. of person ` = R = m (g-a) = 50 (10 - 9) = 50 N = 5 kg ` . |
|
| 504. |
An elevator weighing 5000 kg is moving upward and tension in the supporting cable is 5000 N . Find upward acceleration . Starting from rest , how far does it rise in 10 secound ? |
|
Answer» Here , ` m = 5000 kg `, ` R= 50000 N , a = ? , u = 0 , s = ? , t = 10 s ` ` As m (g +a) = R ` ` :. 5000 (9.8+ a ) = 50000 , a = 0.2 m//s^(2) ` Again , `s = ut + (1)/(2)/= at^(2) = 0 + (1)/(2) xx 0.2 xx 10^(2) = 10 m` . |
|
| 505. |
An elevator and its load weigh a total of `800kg` Find the tension `T` in the supporting cable when the elevator originally moving downwards at `20m//s` is brought to rest with constant retardation in a distnce of `50m`. |
|
Answer» Correct Answer - `1.014xx10^(4)N` |
|
| 506. |
If block `A` of the pulley system is moving downward with a speed of `1 m//s` while block `C` moving up at `0.5m//s` determine the speed of block `B` |
|
Answer» Let `B` and `C` both move upwards (alongwith there pulleys) with speed `nu_(B)` and `nu_(c)` then we can see that, `A` will move downward with speed , `2nu_(B) + 2nu_(C)`. So `:. nu_(B) = (nu_(A))/(2) - nu_(C)` Substituting the value we have,`nu_(B) = 0` |
|
| 507. |
In the adjoining figure all surface are frictionless . What force `F` must by applied to `M_(1)` to keep `M_(3)` free from rising or falling? |
| Answer» `FBD` of `M_(2)` and `M_(3)` in accelerated frame of reference is shown in figure | |
| 508. |
Shown the position time graph of a particle of mass 4 kg What is the force on the particle for `t lt 0 , 0 lt t lt4 s` motion only . |
|
Answer» (a) (i) for `t lt 0` the position time graph is AO which means displacement of the particle must be zero (ii) for `0 lt t lt 4 s` the position time graph OB has a constant slope Therefore velocity of the particle is constant in this interval i.e particle has zero acceleration Hence force on the particle must be zero (iii) For `t gt 4 s` the position time graph BC is parallel to time axis Therefore the particle remains at a distance of 3 m from the origin i.e., it is at rest Hence force on the particle is zero (b) Impulse at `t = 0` We know impulse = change in linear momentum Before t = 0 particle is at rest i.e. u = 0 After t = 0 particle has a constant velocity `upsilon = (3)/(4) = 0.75 m//s ` `:.` Impulse ` = m (upsilon - u ) = 4 (0.75 - 0) = 3 kg ms^(-1)` Impulse at `t = 4 s ` Before t = 4 s particle has a constant velocity `u = 0 .75 m//s ` After t = 4 s particle is at rest i.e., `upsilon = 0` `:.` Impulse ` = m (upsilon - u ) = 4 (0 - 0 .7 5 ) = -3 kg ms^(-1)`. |
|
| 509. |
An elevator starts from rest with a constant downward acceleration and covers `2.5m` in first second If the lift weighs `200kg` what would be the tension in the ropes of the lift ? |
|
Answer» Correct Answer - `960N` . Here, `u =0, s = 2.5 m, t = 1 s, m t =1 s, m =200 kg` `R=`? `s = ut + (1)/(2) at^(2) = (1)/(2)at^(2)` `a = (2s)/(t^(2)) =(2xx2.5)/(1^(2)) =5 m//s^(2)` `R =m (g -a) = 200 (9.8 -5)` `=960N` . |
|
| 510. |
An elevator weighs `4000kg` When the upward tension in the supporting cable is `48000 N` what is the upward acceleration? Starting from rest, how far does it rise in `3` seconds ? |
|
Answer» Correct Answer - `2.2 ms^(-2) ; 9.9m` . Here, `m = 400kg, T = 48000 N a = ?` As `T =m (g +a)` `:. g + a = (T)/(m) = (48000)/(4000) = 12` `a = 12 - g = 12 -9.8 = 2.2 ms^(-2)` (ii) `u = 0, s = ? T = 3 sec` From `s = ut + (1)/(2)at^(2) = 0 + (1)/(2) xx2.2 xx3^(2) = 9.9m` . |
|
| 511. |
A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rupe is tied at the end B, and it is pulled horizontally with a force. If the rope AB makes an angle `theta` with the vertical in equilibrium then the tension in the string AB is A. F `sin theta`B. `F//sin theta`C. `F cos theta`D. `F// cos theta ` |
|
Answer» Correct Answer - b |
|
| 512. |
A chain of mass M and length L is held vertical by fixing its upper end to a rigid support. The tension in the chain at a distance y from the rigid support is:A. MgB. `((L-x)/(L))Mg`C. `((L)/(L-x))Mg`D. `(x)/(L)Mg` |
|
Answer» Correct Answer - B (b) `T_(x)=`mass of rope of length `(L-x)xxg=(M)/(L)(L-x)g` |
|
| 513. |
The co-efficient of friction (both static and kinetic ) between the inclined plane and the block A of mass 1 kg is `1//sqrt(3)` whereas there is no friction between block B (1 kg ) and the inclined plane. The system starts from rest. When the system covers a distance s=30 m on the inclined plane, the cord between A and B is burnt. The velocity of A when it covers a further distance of 20 m on the inclined plane is A. ZeroB. `sqrt(120) m//s`C. `sqrt(150) m//s`D. `sqrt(300) m//s` |
|
Answer» Correct Answer - C |
|
| 514. |
How do we save petrol when the types of the motor cycle are fully inflated ? |
| Answer» When the tyres are fully inflated , deformation of tyres will be smail As a result of it value of force of rolling friction will reduce ,. Due to it the motorcycle will cover more distance for the given petrol consumed in it Hence we save petrol when the tyres of the motorcycle are fully inflated . | |
| 515. |
A rocket is set for vertical firing If the exhaust speed is `1200ms^(-1)` , how much gas must be ejected per second to supply the thrust needed (i) to overcome the weight of rocket (ii) to give to the rocket an initial vertical upward acceleration of `29.6 m//s^(2)` Given mass of rocket = 6000 kg. |
|
Answer» Here, ` u=1200 m//s, (dm)/(dt) = ` ? ` m_(0) = 6000 kg ` (i) To overcome the weight of rocket, Thrust , ` F = u (dm)/(dt) = m_(0) g ` ` (dm)/(dt) = (m_(0) g)/(u) = (6000 xx 9.8)/(1200) = 49 kg s^(-1) ` ` a = 29 .6 m//s^(2) u ((dm)/(dt)) = m_(0) (a +g) ` ` ((dm)/(dt)) = (m_(0) (a +g))/(u) = (6000 (29.6 + 9.8))/(1200) ` ` = 197 kg // s ` . |
|
| 516. |
The distance travelled by a body is directly proportional to time Is any external force acting on it ? |
| Answer» When `s prop t` , acceleration = 0 Therefore , no exernal force is acting on the body . | |
| 517. |
A particle moves in the X-Y plane under the influence of a force such that its linear momentum is `oversetrarrp(t)=A[haticos(kt)-hatjsin(kt)]`, where A and k are constants. The angle between the force and the momentum isA. `0^@C`B. `30^@`C. `45^@`D. `90^@` |
|
Answer» Correct Answer - D `oversetrarrp(t)=A[haticos(kt)-hatjcos(kt)]` `oversetrarrF=(doversetrarrp)/(dt)=Ak[-hatisin(kt)-hatjcos(kt)]` Here, `oversetrarrF.oversetrarrP=0` But `oversetrarrF.oversetrarrp=Fpcostheta` `:.` `costheta=0impliestheta=90^@` |
|
| 518. |
A constant force acts on a body of mass 10 g moving with a velocity of `500 ms^(-1)` . The body is found to move with same velocity but in the opposite direction after 10 s . Calculate the magnitude of the force . |
|
Answer» Correct Answer - 1 N opposite to the motion |
|
| 519. |
A jet airplane is travelling at `200 ms^(-1)` . The engine of the plane takes in 250 `m^(3)` of air of mass 32 kg each second . The air is used to burn 3 kg of fuel each second . The energy is used to compress the products of combustion to eject them at the rear of the plane at `500 ms^(-1)` relative to the plane . Find the power of the jet plane in H.P.(1 H.P = 746 W). [Hint : Resultant thrust = forward thrust - backward thrust and power = force `xx` velocity ] |
|
Answer» Correct Answer - 2975 HP |
|
| 520. |
A scale is adjusted to zero . Particles fall from a height h = 250 m before colliding with the balance pan of the scale . The collisions are elastic . If each particle has a mass of m = 100 mg and collisions occur at the rate n = 1000 particles per second , what is the scale reading in kg ? |
|
Answer» Correct Answer - `2 mn sqrt(2gh) = 1.43` kg |
|
| 521. |
In a circus , the diameter of globe of death is 20 m . From what minimum height must a motor cyclist start in order to go around the globe successfully ? . |
|
Answer» Here , ` r = (20)/(2) m = 10 m , h = ? ` On rolling down the incline , Loss in PE . = Gain in K . E ` mgh = (1)/(2) m upsilon^(2) :. upsilon = sqrt(2 gh) ` For looping the loop , minimum velocity at the lowest point ` = sqrt (5 gr ) ` `:. sqrt( 5 gr )= sqrt(2 gh ) ` ` h = (5)/(2) r = (5)/(2) xx 10 = m` . |
|
| 522. |
A person sitting in a carriage at rest pushes it from within. Will the carriage move ? |
| Answer» No . This is because push is an internal force which cannot produce motion . | |
| 523. |
A stone , when thrown on a glass window smashes the window pane to pieces , but a bullet from the gun passes through making a clean hole. Way ? |
| Answer» This is because velocity of bullet from the gun is very large . It takes very little time to cross the windown pane . Particles of window pane near the hole have to little time to share the motion of the bullet . Reverse is the case when stone is thrown with hand . | |
| 524. |
Assertion: If a body is momentarily at rest, it means that force or acceleration are necessarily zero at that instant. Reason: Force on a body at a given time is determined by the direction of motin only.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason is the not correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false. |
|
Answer» Correct Answer - D If a body is momentarily at rest i.e. v = 0 at an instant, it does not mean that force or acceleration are necessarily zero at that instant. For example, when a ball thrown upward reaches its maximum height, v=0 but the force continues to be its weight mg and the acceleration is g. Force is not always in the direction of motion. |
|
| 525. |
If the net force acting on a body be zero , will it remain necessarily at rest ? |
| Answer» No the body may be moving uniformly along a staight line . | |
| 526. |
Three books (X,Y and Z) rest on a table. The weight of each book is indicated. The net force acting on book Y is A. 4N downB. S N upC. 9N downD. None of these |
|
Answer» Correct Answer - d |
|
| 527. |
Forces of `5sqrt2` and `6sqrt2N` are acting on a body of mass 1000 kg at an angle of `60^(@)` to each other. Find the acceleration, distance covered and the velocity of the mass after 10 s. |
|
Answer» Correct Answer - `0.03349ms^(-2),0.1349ms^(-1),0.6745m` |
|
| 528. |
Five equal forces each of `20N` are acting at a point in the same plane. If the angles between them are same, the resultant of these forces isA. `F=10N`B. `F=5N`C. `90^(@) lt alphalt180^(@)`D. `180^(@) lt alphalt270^(@)` |
|
Answer» Correct Answer - b,c |
|
| 529. |
In the arrangement shown, end A of light intextensible string is pulled with constant velocity v. the velocity of block B is A. `V//2`B. vC. `v//3`D. 3v |
|
Answer» Correct Answer - c |
|
| 530. |
A minimum force F is applied to a block of mass 102 kg to prevent it from sliding on a plane with an inclination angle `30^(@)` with the horizontal. If the coefficients of static and kinetic friction between the block and the plane are 0.4 and 0.3 respectively, then the force F isA. 157 NB. 224 NC. 315 ND. zero |
|
Answer» Correct Answer - A (a) `mg sin theta=(102)(10) sin 30^(@)=510N` `mu , mg cos theta =(0.4)(102)(10) cos 30^(@)` =353 N `F=510-353=157N` |
|
| 531. |
A wire of mass `9.8 xx 10^(-3)` kg per metre passes over a frictionless pulley fixed on the top of an inclined frictionless plane which makes an angle of `30^(@)` with the horizontal Masses `M_(1)` and `M_(2)` are tied at the two ends of the wire The mass `M_(1)` rests on the plane and mass `M_(2)` hangs freely vertically downwards . The whole systemis in equilibrium Now a transverse wave propagates along the wire with a velocity of `100 m//s` If `g = 9.8 m//s^(2)` calculate the valuse of masses `M_(1)` and `M_(2)` . |
|
Answer» Here, ` m = 9.8 xx 10^(-3) kg//m` `theta = 30^(@) , g = 9.8 m//s^(2) ` ` upsilon = 100 m//s , M_(1) = ? M_(2) = ? ` The various forces acting on the system are shown in As the system of two masses is in equilibrium therefore , ` T = M_(1) g sin theta = M_(1) g sin 30^(@) = (M_(1) g )/(2) ` ` R = M_(1) g cos theta = M_(1) g cos 30^(@) = M_(1) g sqrt3/(2) ` Also ` T = M_(2) g` From (i) and (iii) , ` T = (M_(1) g )/(2) = M_(2) g ` ` M_(1) = 2 M_(2) ` Now the velocity of transverse waves is ` upsilon= sqrt((T)/(m) ` ` T = upsilon^(2) xx m = (100)^(2) xx 9.8 xx 10^(-3) = 9.8 N ` From (iii) `M_2 = (T)/(g) = (98)/(9.8) = 10 kg ` From `M_(1) = 2 M_(2) = 2 xx 10 kg = 20 kg ` |
|
| 532. |
A mass of 1kg is suspended by a thread. It is 1. lifted up with an accleration 4.9`m//s^(2)`. 2. lowered with an acceleration 4.9 `m//s^(2)`. The ratio of the tensions isA. `3 : 1`B. `1 : 3`C. `1 : 2`D. `2 : 1` |
|
Answer» Correct Answer - A |
|
| 533. |
A mass of `2kg` is suspended with thread `AB` (figure) Thread `CD` of the same type is attached to the other end of `2kg` mass. Lower thread is pulled gradually, harder and harder in the downward gradually, harder and harder in the downward direction so as to apply force on `AB`. which of the threads will break and why? |
|
Answer» The thread AB will break earlier than the thread CD . This is because force acting on thread CD = applied force and force acting on thread AB = (applied force + weight of 2 kg mass). Hence , force acting on thread AB is larger than the force acting on thread CD. |
|
| 534. |
Three identical blokcs are suspended on two identical springs one below the other as shown in figure. If thread is cut that supports block 1, then initially (choose one alternative only). A. the second ball falls with zero accelerationB. the first ball falls with maximum accelerationC. Both a and b are wrongD. Both a and b are correct |
|
Answer» Correct Answer - d |
|
| 535. |
In the figure shown, there is no relative motion between the two blokcs. Force of friction acting on 1 kg block is A. zeroB. 3NC. 6ND. 4N |
|
Answer» Correct Answer - c |
|
| 536. |
Assertion: Friction opposite relative motion and thereby dissiplates power in the from of heat. Reason: Friction is always an undesirable force.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason is the not correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false. |
|
Answer» Correct Answer - C We consider friction as an undesirable force sometimes, not always. In many situation, like in a machine with different moving parts, friction does have a negative role. It opposes relative motion and thereby dissipates power in the form of heat etc. |
|
| 537. |
In the arrangement shown in figure `[sin37^(@)=(3)/(5)]` A. dircetion o fforce of friction is up the planeB. the manitude of force o ffriction is zeroC. the tension in the string 40 ND. magnitude of force of friction is 56N |
|
Answer» Correct Answer - a,c |
|
| 538. |
A sphere of mass m, is held between two smooth inclined walls. For `sin37^(@)=3//5`, the normal reactions of the wall (2) is equal to A. mgB. `mg sin 74^(@)`C. `mg cos 74^(@)`D. None of these |
|
Answer» Correct Answer - a |
|
| 539. |
an elastic spring has a length `l_(1)` when tension in it is 4N. Its length is `l_(2)` when tension in it is 5N. What will be its length when tension in it is 9N?A. `5l_(1)-4l_(2)`B. `5l_(2)-4l_(1)`C. `4l_(1)+5l_(2)`D. `4l_(2)+5l_(1)` |
|
Answer» Correct Answer - b |
|
| 540. |
A neutron of mass `1.67xx10^(27)` kg moving with a speed of `3xx10^(6)ms^(-1)` collides with a deutron of mass `3.34xx10^(-27)` kg which is at rest. After collision, the neutron sticks to the deutron and forms a triton. What is the speed of the triton? |
|
Answer» Correct Answer - `10^(6)ms^(-1)` |
|
| 541. |
A car is speeding on a horizontal road curving round with a radius 60 m The coefficient of friction between the wheels and the road in 0.5 The height of centre of gravity of the car from the road level is 0.3 m and the distance between the wheels is 0 .8 m . Calculate the maximum safe velocity for negotiating the curve . Will the car skid or topple if this velocity is exceeded ? |
|
Answer» Here , `r = 60 m,mu = 0.5 h = 0.3 m`, distance between the wheels , `2 xx = 0.8 m`, `x = 0.4 m` For no skidding, `tan theta = mu =(v^(2))/(r g)` `v = sqrt(mu rg)= sqrt(0.5 xx 60 xx9.8) =17.15 m//s` For toppling , `(m upsilon^(2))/(r) xx h = mg x` `upsilon = sqrt(grx)/(h) = sqrt(9.8 xx 60 xx 0.4)/(0.3)=28 m//s` Hence the maximum safe velocity for negotiating the curve is `17.15 m//s` Beyond this speed , skidding starts until the car topples at `v = 28 m//s` . |
|
| 542. |
A body under the action of a force `vec(F)=6hat(i)-8hat(j)N` acquires an acceleration of `5ms^(-2)` . The mass of the body isA. 2kgB. 5kgC. 4kgD. 6kg |
|
Answer» Correct Answer - A `"Given": overset(-)(F)=6 hati-hat8i N, a=5ms^(-2)` `therefore F=sqrt((6)^(2)+(-8))=10N` Mass of the body is, `m=(F)/(a)=(10N)/(5ms^(2))=2kg` |
|
| 543. |
An external force….. To keep a body…. According to….. . |
| Answer» is required , in unifrom motion , Aristotle . | |
| 544. |
An external force is required to keep a body in uniform motion This statement was given by.A. AristotleB. NewtonC. ArchimedesD. Einstein |
| Answer» Correct Answer - a | |
| 545. |
For a body of given mass graph between velocity of the body and its linear momentum isA. a straight line with slope =0B. a straight line with positive slopeC. a straight line with negative slopeD. a parabola |
| Answer» Correct Answer - b | |
| 546. |
Newton s first law definesA. force onlyB. inertia onlyC. both force and inertiaD. Neither force nor inertia |
| Answer» Correct Answer - c | |
| 547. |
Calculate the ration `m _(0)//m` for a rocket if it is to escape from the earth . Given escape velocity `= 11.2 km//s ` and exhaust speed of gases is `2 km// s` . |
|
Answer» Assuming that the rocket starts from rest, velocity acquired is given by ` upsilon = u log_(e) ((m_(0))/(m)) = 2.3026u log_(10) (m_(0)/(m)) ` Now , ` upsilon = 11.2 km//s ` and ` u=2 km//s ` ` :.` `11.2 = 2. 3026 xx 2 log_(10)(m_(0)/(m)) ` `log_(10) (m_(0)/(m)) = (11.2)/(2.23026 xx2) = 2.432 ` ` m_(0)/(m) = "antilog" 2.432 = 270.4 ` . |
|
| 548. |
A ball is travelling with uniform translatory motion. This means thatA. it is at rest .B. the path can be a straight line or circular and the ball travels with unifrom speed .C. all parts of the ball have the same velocity is constant .D. the center of the ball moves with constant velocity and the ball spins about its center uniformly . |
|
Answer» Correct Answer - c In uniform translatory motion all parts of the ball have the same velocity in magnitude and direction and this velocity is constant . |
|
| 549. |
A ball is travelling with uniform translatory motion . This means thatA. it is at restB. he path can be a straight line or circular and the ball travels with uniform speed.C. all parts of the ball have the same velocity (magnitude and direction) and the velocity is constantD. the centre of the ball moves with constant velocity and the ball spins about its centre uniformly. |
|
Answer» Correct Answer - C In uniform translatory motion, all parts of the ball have the same velocity in magnitude and direction, and this velocity is constant. |
|
| 550. |
A block is projected upwards on a rough inclined plane at 20 m/s. let the time in going up, then is equal to A. 4sB. 2sC. 8sD. 6s |
|
Answer» Correct Answer - b |
|