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.
| 401. |
A cricket ball of mass `150g` has an initial velocity `(3 hati + 4hatj)ms^(-1)` and a final velocity `upsilon = -(3hati + 4hatj)ms^(-1)` after beigh hit The change in momentum (final momentum initial momentum) is (in kg `ms^(-1)`)A. zeroB. `-(0.45hati +0.6 hatj)`C. `-(0.9 hatj +1.2 hatj)`D. `-5(hati +hatj)hati` |
|
Answer» Correct Answer - C Given `u = (3 hati + 4hatj)m//s` and `v = - (3hati +4hatj) m//s` mass of the ball `=150 g = 0.15 kg` `DeltaP = mv -"mu"` `DeltaP =m(v -u) = - (0.15) [(3hati + 4hatj)-(3hati + 4hatj)]` `= (0.15) [-6hati -8hatj]` Hence `Deltap = - [0.9hati + 1.2hatj]` . |
|
| 402. |
The ratio of gravitational mass to inertial mass is equal to:A. `(1)/(2)`B. 2C. 1D. None of these |
|
Answer» Correct Answer - C `(M_("gravitatial")/(M_("intertial"))=1` |
|
| 403. |
Two bodies A (30 kg) and B (50 kg) tied with a light string are placed on a friction less table. A force F acting at B pulls this ystem with an acceleration of `2ms^(-2)`. The tension in the string is:A. 60N,60N,60NB. 100NC. 35ND. 140N |
|
Answer» Correct Answer - A `T=30xx2=60N` |
|
| 404. |
If action force acting on a body is gravitational in nature, then reaction force .A. will be a contact forceB. will be gravitational forceC. will be a gravitational or contact forceD. will be a force of any origin |
| Answer» Correct Answer - B | |
| 405. |
A block with mass `m_(1)` is placed on an inclined plane with slope angle `alpha` and is connected to a second hunging block with mass `m_(2)` by a cord passing over a small . Friction less pulley as shown in fig 7.247 . The coefficient of static friction is `mu_(2)` and the coefficient of kinetic friction is `mu_(s)` a. Find the mass `m_(2)` for which block `m_(1)` moves up plane at constant speed once it is set in motion b. Find the mass `m_(2)` for which block `m_(1)` moves down the plane at constant speed once it is set in motion c.For what range of `m_(2)` will the blocks remain at rest if they are released from rest? |
|
Answer» Correct Answer - `m_(1) sin theta +mu_(k)m_(1) cos theta, m_(1) sin theta - mu_(k)m_(1) cos theta` |
|
| 406. |
A `100g` iron ball having velocity `10m//s` collies with a wall at an angle `30^(@)` and rebounds with the same angle. If the period of contact between the ball and wall is `0.1` second, then the force experinced by the wall isA. 10 NB. 100 NC. `1.0`ND. `0.1`N |
|
Answer» Correct Answer - A |
|
| 407. |
USS 150) Two masses of 5 kg and 10 kg are connected to a pulley as shown. What will be the acceleration of the system (g= acceleration due to gravity) A. `g`B. `(g)/(2)`C. `(g)/(3)`D. `(g)/(4)` |
|
Answer» Correct Answer - C |
|
| 408. |
It is easier to pull a lawn roller than to push it because pulling .A. involves sliding frictionB. involves dry frictionC. increases the effective weightD. decreases normal reaction |
| Answer» Correct Answer - D | |
| 409. |
A ball falling with velocity `vecv_(i)=(-0.65hati-0.35hatj)ms^(-1)` is subjected to a net impulse `vecI=(0.6hati+0.18hatj)` Ns. If the ball has a mass of `0.275kg` calculate its velocity immediately following the impulse . |
|
Answer» `mvecv_(f)-mvecv_(i) = vecI, vecv_(f)=vecv_(i) +(vecI)/(m)` `vecv_(f) =-0.65hati-0.35hatj+(0.6hati+0.18hatj)/(0.275)` `vecv_(f) =-0.65hati-0.35hatj+2.18hati+0.655hatj` `vecv_(f) =(1.53hati +0.305hatj)ms^(-1)` . |
|
| 410. |
A body of mas `2kg` has an initial speed `5ms^(-1)` A force acts on it for some time in the direction of motion The force-time graph is shown. Find the final speed of the body . |
|
Answer» Area of OA `F = (1)/(2) xx 2 xx 4 =4` Area of ABG `F =2 xx 4 =8` Area of `BGHC =(1)/(2) (4 +2.5) xx 0.5 = 1.625` Area of `CDEH =2 xx 2.5 =5` Total area under `F-t` graph =Change in momentum `rArr m (v -u) =18.625` `rArr v = (18.625)/(2) +5 =14.25 ms^(-1)` . |
|
| 411. |
Assertion: A man who falls from aheight on a cement floor receive more injury than when he falls from the same height on a heap of sand. Reason: The impulse applied by a cement floor is more than the impulse by sand floor.A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false |
|
Answer» Correct Answer - C Impulse same as change in momentum is same. |
|
| 412. |
A particle of mass m is at rest at the origin at time `t=0` It is subjected to a force `F(t)=F_(0)e^(-bt)` in the X-direction. Its speed `V(t)` is depicted by which of the following curves |
|
Answer» As the force is exponentially decreasing its accelertion, rate of increase of velocity will decrease with time. Thus the graph of velocity will be an increasing curve with decreasing slope with time `a=(F)/(m) = (F_(0))/(m)e^(-bt)rArr(dv)/(dt) (F_(0))/(m)e^(-bt) rArrunderset(0)overset(v)(int)dv=underset(0)overset(t)int(F_(0))/(m)e^(-bt)dt` `rArrv=[[F_(0))/(m)((1)/(-b))e^(-bt)]_(0)^(t)=[[F_(0))/(m)((1)/(b))e^(-bt)]_(t)^(0)` `=(F_(0))/(mb)(e^(0)-e^(-bt))=(F_(0))/(mb) (1-e^(-bt))` So, velocity increases continuously and attains a maximum value `v_(max) = (F_(0))/(mb)` . |
|
| 413. |
A ball reaches a racket at `60m//s` along +X dirction and leaves the racket in the opposite direaction with the same speed. Assuming that the mass of the ball as `50gm` and the contact time is `0.02` second the force exerted by the racket on the ball is .A. `300N` along `+X` directionB. `300N` along `-X` directionC. `3,00,000 N` along `+X` directionD. `3,00,000 N` along `-X` direction |
|
Answer» Correct Answer - B `F_(avg) =ma,a = (v-u)/(t)` |
|
| 414. |
In the arrangement shown, the pulleys are fixed and ideal, the strings are light, `m_(1)gtm_(2)` and S is a spring balance which is itself massless. The readings of S (in units of mass) is A. `m_(1)-m_(2)`B. `(1)/(2)(m_(1)+m_(2))`C. `(m_(1)m_(2))/(m_(1)+m_(2))`D. `(2m_(1)m_(2))/(m_(1)+m_(2))` |
|
Answer» Correct Answer - D The tension in massless spring at every point is same |
|
| 415. |
In the system in the figure `m_(1)gtm_(2)` system is held at rest by thread `BC`. Just after the thread BC is burnt:A. acceleration of `m_(1)` will be equal to zeroB. acceleration of `m_(2)` will downwardsC. magnitude of acceleration of two blocks will be non-zero and unequal .D. magnitude of acceleration of both the blocks will be `((m_(1)-m_(2))/(m_(1)+m_(2)))g` . |
| Answer» Correct Answer - A | |
| 416. |
A body of mass `2kg` travels according to the law `x (t) = pt + qt^(2) + rt^(3)` where `p = 3ms^(-1),q = 4 ms^(-2)` and `r = 5ms^(-3)` . Find the force acting on the body at t=2 sec.A. `136N`B. `134N`C. `158N`D. `68 N` |
|
Answer» Correct Answer - A Given mass `=2kg` `x (t) =pt + qt^(2) + rt^(3)` `v =(dx)/(dt) = p + 2qt + 3rt^(2)` `a = (dv)/(dt) = 0 + 2q +6rt` ltbr gtat `t =2s, a =2q + 6 xx 2 xx r` `= 2q +12r` `= 2 xx 4 + 12 xx 5` `= 8 + 60 =68 m//s` Force ` =F =ma` `=2 xx 68 =136 N` . |
|
| 417. |
A bullet of mass 10 g is fired from a gun of mass 1 kg with recoil velocity of gun 5 m/s. The muzzle velocity will beA. `20m//s^(2)`B. `100m//s^(2)`C. `200m//s^(2)`D. `500m//s^(2)` |
|
Answer» Correct Answer - D |
|
| 418. |
Two blocks `A & B` attached to each other by a mass-less spring, are kept on a rough horizontal surface `mu=0.1`. A constant force `F=200 N` is applied on blcok `B` horizon tally as shown below. If a some instant the acceleration of `10 kg` mass is `12 m//s^(2)`, then the acceleration of `20 kg` mass is A. `2.5 m//s^(2)`B. `4.0 m//s^(2)`C. `3.6 m//s^(2)`D. `1.2 m//s^(2)` |
|
Answer» Correct Answer - A |
|
| 419. |
Three block of masses `m_(1),m_(2)` and `m_(3)` kg are placed in contact with each other on a frictionless table. A force F is applied on the heaviest mass `m_(1)`, the acceleration of `m_(2)` will be A. `(F)/(m_(1))`B. `(F)/((m_(1)+m_(2))`C. `(F)/((m_(1)-m_(2))`D. `(F)/((m_(1)+m_(2)+m_(3))` |
| Answer» Correct Answer - D | |
| 420. |
A person is standing in an elevator. In which situation he finds his weight less ?A. The elevator moves upward with constant accleerationB. The elevator moves downward with constant acceleration.C. The elevator moves upward with uniform velocityD. The elevator moves downward with uniform velocity |
| Answer» Correct Answer - B | |
| 421. |
A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectivelyA. g,gB. g-a,g-aC. g-a,gD. a,g |
| Answer» Correct Answer - C | |
| 422. |
The force exerted by the floor of an elevator on the foot of a person standing there is more than the weight of the person if the elevator isA. (i), (ii)B. (ii), (iii)C. (iiI), (iv)D. (i), (iv) |
| Answer» Correct Answer - B | |
| 423. |
A person standing oin the floor of an elevator drops as coin. The coin reaches the floor of the elevator in a time `t_1` if the elevator is stationary and in the `t_2` if it is moving uniformly. ThenA. `t_(1) = t_(2)`B. `t_(1) lt t_(2)`C. `t_(1) gt t_(2)`D. `t_(1) lt t_(2)` or `t_(1) gt t_(2)` depending on whether the lift if going up or down. |
|
Answer» Correct Answer - A |
|
| 424. |
A person standing oin the floor of an elevator drops as coin. The coin reaches the floor of the elevator in a time `t_1` if the elevator is stationary and in the `t_2` if it is moving uniformly. ThenA. `t_(1)=t_(2)`B. `t_(1)ltt_(2)`C. `t_(1)gtt_(2)`D. `t_(1)ltt_(2)` or `t_(1)ltt_(2)` depending on whether the lift is going up or down |
|
Answer» Correct Answer - A In both cases, the acceleration of lift is zero. |
|
| 425. |
A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. (Consider `g=10m//s^2`).A. 16 NB. 20 NC. 22 ND. 4 N |
|
Answer» Correct Answer - B `0=u^(2)-2gh=u^(2)-2xx10xx2` `u^(2)=40` `u^(2)=0+2axx0.2rArr40=0.4arArra=100m//s^(2)` `F=ma=0.2xx100=20N` |
|
| 426. |
Two bodies of mass 3 kg and 4 kg are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is `(g=9.8m//s^(2))`A. `4.9m//s^(2)`B. `2.45m//s^(2)`C. `1.4m//s^(2)`D. `9.5m//s^(2)` |
|
Answer» Correct Answer - C |
|
| 427. |
In the figure all pulleys (P1, P2, P3 …….) are massless and all the blocks (1,2,3 …..) are identical, each having mass m. The system consist of infinite number of pulleys and blocks. Strings are light and inextensible and horizontal surfaces are smooth. Pulley P1 is moved to left with a constant acceleration of a0. Find the acceleration of block1. Assume the strings to remain horizontal. |
|
Answer» Correct Answer - `(3a_(0))/(2)` |
|
| 428. |
Three small discs are connected with two identical massless rods as shown in fig. The rods are pinned to the discs such that angle between them can change freely. The system is placed on a smooth horizontal surface with discs A and B touching a smooth wall and the angle ACB being `90^(@)`. A force F is applied to the disc C in a direction perpendicular to the wall. Find acceleration of disc B immediately after the force starts to act. Masses of discs are `m_(A) = m , m_(B) = 2 m , m_(C) = m `[ wall is perpendicular to the plane of the fig.] |
|
Answer» Correct Answer - `(F)/(5m)` |
|
| 429. |
When a body slides down an inclined plane with coefficient of friction as `mu_(k)`, then its acceleration is given by .A. `g(mu_(k) sin theta +cos theta)`B. `g(mu_(k) sin theta -cos theta)`C. `g(mu_(k) sin theta + mu_(k)cos theta)`D. `g (sin theta -mu_(k)cos theta)` |
|
Answer» Correct Answer - D `F = mg sin theta - mu_(k) mg cos theta` `a =g (sin theta - mu_(k) cos theta)` |
|
| 430. |
Which of the four arrangements in the figure correctly shows the vector addition of two forces `vecF_(1)` and `vecF_(2)` to yield the third force `vecF_(3)`A. B. C. D. |
|
Answer» Correct Answer - C |
|
| 431. |
A body is under the action of three forces `vecF_(1),vecF_(2)` and `vecF_(3)`. In which case the body cannot undergo angular acceleration ? .A. `vecF_(1),vecF_(2)` and `vecF_(3)` are concurrent, point of concurrence beign centre of mass .B. `vecF_(1)+vecF_(2)+vecF_(3) =0`C. `vecF_(1),vecF_(2)` is parallel to `vecF_(3)` but the three forces are not concurrent .D. `vecF_(1)` and `vecF_(2)` act at the same point but `vecF_(3)` acts at different point . |
| Answer» Correct Answer - A | |
| 432. |
Two forces `vecF_(1)=(3N)hati-(4N)hatj` and `vecF_(2)=-(1N)hati-(2N)hatj` act on a point object. In the given figure which of the six vectors represents `vecF(1)` and `vecF_(2)` and what is the magnitude of the net forces A. 4 and 8, sqrt(19)N`B. 3 and 6,5N`C. 5 and 7, `sqrt(40)N`D. 2 and 6,7 N |
|
Answer» Correct Answer - C `vecF_(1)=(3N)hati-(4N)hatj` This means (3N) along +x axis and (4N) along negative side of y axis. Hence vector 5 represents it. `vecF_(2)=-(1N)hati-(2N)hatj` This means (1N) along negative x-axis. (2N) along negative y-axis. hence force `|F|^(2)=(2)^(2)+(-6)^(2)=40` `therefore |F|=sqrt(40)N` |
|
| 433. |
A block of weight `W` is kept on a rough horizontal surface (friction coefficient `mu`). Two forces `W//2` each are applied as shown in the choose the correct statement .A. For `mugt(sqrt3)/(5)` block will move .B. For `mult(sqrt3)/(5)`, work done by frictional force is zero in ground frame .C. For `mu gt (sqrt3)/(2)` frictional force will do positive work (in ground frame) .D. For `mu le (sqrt3)/(2)` block will move . |
|
Answer» Correct Answer - D `N +(w)/(2) sin theta =w +(w)/(2), f_(max) =muN` . |
|
| 434. |
In the figure-2.205 shown in blocks A, B and C has masses `m_(A)= 5 kg, m_(B) = 5 kg` and `m_(C )= 10kg` respectively, find the acceleration of the three blocks. Assume all pulleys and strings are ideal. (Take `g = 10m//s^(2)`) |
|
Answer» Correct Answer - `[a_(A)=1.8m//s^(2)uarr,a_(B)=1m//s^(2)darr,a_(C )=4.2m//s^(2)larr]` |
|
| 435. |
If masses of the blocks A and B shown in figure-2.204 (a) and 2.204(b) are 10kg and 5kg respectively, find the acceleration of the two masses. Assume all pulleys and strings are ideal. (Take `g=9.8m/s^(2)`) |
|
Answer» Correct Answer - `[((20)/(3)),((10)/(3)).m//s^(2),(10)/(11)m//s^(2),(30)/(11)m//s^(2)]` |
|
| 436. |
A bar of mass m is pulled by means of a thread up an inclined plane forming an angle `theta` with the horizontal as shown in figure-2.211. The coefficient of friction is `mu`. Prove that (a)`alpha=tan^(-1)mu`, where `alpha` is the angle which the thread must from with the inclined plane for the tension of the thread to be minimum. |
|
Answer» Correct Answer - `T_("min")=(mgsintheta+mumgcostheta)/(sqrt(1+mu^(2))` |
|
| 437. |
Civil engineers bank a road to help a car negotiate a curve. While designing a road they usually ignore friction. However, a young engineer decided to include friction in his calculation while designing a road. The radius of curvature of the road is R and the coefficient of friction between the tire and the road is `mu`. (a) What should be the banking angle `(theta_(0))` so that car travelling up to a maximum speed `V_(0)` can negotiate the curve. (b) At what speed `(V_(1))` shall a car travel on a road banked at `theta_(0)` so that there is no tendency to skid. (No tendency to skid means there is no static friction force action on the car). (c) The driver of a car travelling at speed `(V_(1))` starts retarding (by applying brakes). What angle (acute, obtuse or right angle) does the resultant friction force on the car make with the direction of motion? |
|
Answer» Correct Answer - (a) `theta_(0) = tan^(-1)((V_(0)^(2)/(Rg) - mu)/(1 + (muV_(0)^(2))/(Rg)))` (b) `V_(1) = sqrt(rg"tan"theta_(0))` (c) Obtuse |
|
| 438. |
A boy of mass `50kg` is standing on a weihing machine placed on the floor of a lift. The machine reads his weight in newtons. The reading of the machine if the lift is moving upwards with uniform speed of `10ms^(-1)` .A. `510N`B. `480N`C. `490N`D. `500N` |
|
Answer» Correct Answer - C When lift moves up or down with uniform speed The apparent weight = real weight `W =mg` . |
|
| 439. |
Two smooth block are placed at a smooth corner as shown in fig. Both the bloks are having mass m. We apply a force F on the block m. Block A presses block B in the normal direction, due to which pressing force on vertical wall will increase, and pressing force on the horizontal wall decreases, as we increases `F(theta = 37^(@)` with horizontal). As soon as the pressing force on the horizontal wall by block B become zero, it will lose contact with ground. If the value of F further increases, block B will accelerate in the upward direction and simulaneously block A will towards right. If both the blocks are stationary, the force exerted by ground of block A isA. `mg +(3F)/(4)`B. `mg -(3F)/(4)`C. `mg +(4F)/(3)`D. `mg -(4F)/(3)` |
|
Answer» Correct Answer - C `F =N sin 37^(@) , N_(A) = mg + N cos 37^(@) = mg + (4)/(3) F` . |
|
| 440. |
A block of mass `3kg` which is on a smooth inclined plane making angie of `30^(@)` to the horizontal is connected by cord passing over light frictionless pulley to second block of mass `2kg` hanging vertically. What is the acceleration of each block and waht is the tension of the cord ? .A. `0.98 m//s^(2), 17.6 N`B. `1.98 m//s^(2), 19.6 N`C. `0.49 m//s^(2), 9.8 N`D. `1.47 m//s^(2), 4.9 N` |
|
Answer» Correct Answer - A `m_(1) g -T =m_(1) a,T -m_(2) g sin theta = m_(2)a` . |
|
| 441. |
A bus moving on a level road with a velocity v can be stopped at a distance of x by the application of a retarding force `F` The load on the bus is increased by `25%` by boarding the passengers. Now if the bus is moving with the same speed and if the same retarding force is applied the distance travelled by the bus before it stops is . |
|
Answer» By using equations of motion `v^(2) -u^(2) =2as` `v^(2)-u^(2) = - 2 ((F)/(m))s -u^(2) = -2 ((F)/(m))s` `u^(2)=2 (Fs)/(m) rArrm = (2Fs)/(u^(2)) rArrm props rArr(m_(1))/(m_(2)) = (s_(1))/(s_(2))` Given `s_(1) =x, m_(1) = m,` and `m_(2) = m+ (25)/(100) (m) = m+ (m)/(4) = (5m)/(4)` `rArr (m)/(5m//4) = (x)/(s^(2)) rArr s_(2) =(5x)/(4) = (1.25x)m` . |
|
| 442. |
A block of mass `m` lying on a horizontal plane , is acted upon by a horizontal force p and another force `Q` inclined at an angle `theta` to the vertical .The block will remain in equilibrium if the coefficient of friction between it and the surface is (assume `p gt Q)` A. `(P+Qsintheta)//(mg+Qcostheta)`B. `(Pcostheta+Q)//(mg-Qsintheta)`C. `(P+Qcostheta)//(mg+Qsintheta)`D. `(P sintheta-Q)-(mg-Qcostheta)` |
|
Answer» Correct Answer - A Friction force `=mmuR=mu(mg+Qcostheta)` And horizontal push `=P+Qsintheta` For equilibrium, we have `mu(mg+Qcostheta)=P+Qsintheta` `mu=(P+Qsintheta)/(mg+Qcostheta)` |
|
| 443. |
Two unequal masses are connected on two sides of a light and smooth pulley as shown in figure. The system is released from rest. The larger mass is stopped `1.0` second after the system is set into motion and then released immediately. The time elapsed before the string is tight again is: Take `g=10m//s^(2)` A. `1//4s`B. `1//2s`C. `2//3s`D. `1//3s` |
|
Answer» Correct Answer - D Acceleration of the system: `a=(m_(2)-m_(1))/(m_(2)+m_(1))g=(10)/(3)m//s^(2)` Velocity of both the block at `t=1s` will be `v_(0)=at=(10)/(3)xx1=(10)/(3)m//s` Now at this moment, velocity of `2kg` block becomes zero, while that of `1kg` block is `(10)/(3)m//s` upwards. Hence, string becomes tight again when Displacement of `1kg` block=displacement of `2kg` block or `v_(0)t-(1)/(2)gt^(2)=(1)/(2)gt^(2)` . |
|
| 444. |
Assertion : The acceleration produced by a force in the motion of a body depends only upon its mass. Reason : Larger is the mass of the body, lesser will be the acceleration produced.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false. |
|
Answer» Correct Answer - B |
|
| 445. |
Two unequal masses are connected on two sides of a light and smooth pulley as shown in figure. The system is released from rest. The larger mass is stopped `1.0` second after the system is set into motion and then released immediately. The time elapsed before the string is tight again is: Take `g=10m//s^(2)` |
|
Answer» Net pulling force `=2g - 1g =10N` Mass being pulled `=2+1 =3kg` Acceleration of the system is `a =(10)/(30)m//s^(2)` Velocity of both the blocks at `t =1` s will be `v_(0) = at = ((10)/(3)) (1) = (10)/(3) m//s`. Now at this momnet velocity of `2kg` block becomes zero, while that of `1kg` block is `(10)/(3)m//s` upwards Hence, string becomes tight again when displacement of `1kg` block =displacement of `1kg` block =displacement of `2kg` block `v_(0) t- (1)/(2) g t^(2)= (1)/(2)g t^(2) rArrg t^(2)=v_(0)t` `t =(v_(0))/(g) = ((10//3))/(10) = (1)/(3) s` . |
|
| 446. |
A spring fo spring constant k is broken in the length ratio 1:3. The spring constant of larger part will beA. 4k/3B. 2k/3C. k/3D. 5k/3 |
|
Answer» Correct Answer - A For a spring kl = constant `k_(1)(l)/(4)=k_(2).(3l)/(4)=kl` `k_(2)=(4K)/(3)` |
|
| 447. |
Block A of mass `m_(A) = 200` g is placed on an incline plane and a constant force F = 2.2 N is applied on it parallel to the incline. Taking the initial position of the block as origin and up along the incline as x direction, the position (x) time (t) graph of the block is recorded (see figure (b)). The same experiment is repeated with another block B of mass `m_(B) = 500` g. Same force F is applied to it up along the incline and its position – time graph is recorded (see figure (b)). Now the two blocks are connected by a light string and released on the same incline as shown in figure (c). Find the tension in the string. `["tan"theta = (3)/(4), g = 10m//s^(2)]` |
|
Answer» Correct Answer - T = 0.49 N |
|
| 448. |
A uniform chain of mass M = 4.8 kg hangs in vertical plane as shown in the fig. (a) Show that horizontal component of tension is same throughout the chain. (b) Find tension in the chain at point P where the chain makes an angle `theta = 15^(@)` with horizontal. (c) Find mass of segment AP of the chain. [Take `g = 10 m//s^(2), "cos" 15^(@) = 0.96 , "sin"15^(@) = 0.25`] |
|
Answer» Correct Answer - (b) `T_(P) = 21. 65 N` (c) 3.05 kg |
|
| 449. |
A 20 kg monkey slides down a vertical rope with a constant acceleration of `7 ms^(-2)`. If `g = 10 m//s^(-2)`, what is the tension in the rope ?A. 140 NB. 100 NC. 60 ND. 30 N |
|
Answer» Correct Answer - C |
|
| 450. |
For a particle rotating in a vertical circle with uniform speed, the maximum and minimum tension in the string are in the ratio 5 : 3. If the radius of vertical circle is 2 m, the speed of revolving body is (`g=10m//s^(2)`)A. `sqrt(5) m//s`B. `4sqrt(5) m//s`C. `5m//s`D. `10m//s` |
|
Answer» Correct Answer - B |
|