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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.
| 501. |
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 is case IIC. less than that in caseD. equal to that in case II |
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Answer» Correct Answer - C `F =ma, a =g ((m_(1)-m_(2))/(m_(1)+m_(2))),F -T =0` and `T =2mg` also `T -mg = ma^(1)`Finally `a lt a^(1)` . |
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| 502. |
A 500kg rocket is set for verticle firing. The exhaust speed is 800`ms^-2` . To give an initial upward acceleration of 20`ms^-2` , the amount of gas ejected per second to supply the needed thrust will be (g=10 `ms^-2` )A. `127.5"kg s"^(-1)`B. `187.5"kg s"^(-1)`C. `185.5 "kg s^(-1)`D. `137.5 "kg s"^(-1)` |
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Answer» Correct Answer - B |
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| 503. |
A cricket ball of mass 150 g is moving with a speed of `12ms^(-1)` and is hit by a bat so that the ball is turned back with velocity of `20ms^(-1)` if duration of contact between bat and ball is 0.01 s then: |
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Answer» According to given problem change in momentum of the ball `Deltap=p_(f)-p_(i)=m(v-u)=150xx10^(-3)(20-(-12))=4.8 N-s` So by impulse-momentum theorem, impulse I=change in moment `Deltap=4.8 N-s` And by time averaged definition of force in case of impulse `F_("av")=(I)/(Deltat)=(Deltap)/(Deltat)=(4.80)/(0.01)=480N` |
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| 504. |
A 5000 kg rocket is set for verticle firing. The relative speed of burnt gas is `800ms^(-1)`. To give an initial upwards acceleration of `20 ms^(-2)`, the amount of gas ejected per second to supply the needed thrust will beA. `127.5 kg s^(-1)`B. `137.5 kg s^(-1)`C. `187.5 kg s^(-1)`D. `185.5 kg s^(-1)` |
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Answer» Correct Answer - C `F=(dp)/(dt), m(g +a) = (dm)/(dt) vrArr(dm)/(dt)=(m(g+a))/(v)` . |
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| 505. |
A 500kg rocket is set for verticle firing. The exhaust speed is `800ms(-2)` . To give an initial upward acceleration of `20ms(-2)` , the amount of gas ejected per second to supply the needed thrust will be `(g=10ms(-2)`A. `127.5kgs^(-1)`B. `187.5kgs^(-1)`C. `185.5kgs^(-1)`D. `137.5kgs^(-1)` |
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Answer» Correct Answer - B `F(umd)/(dt)=m(g+a)` . `implies(dm)/(dt)=(m(g+a))/(u)=(5000xx(10+20))/(800)=187.5kg//s` . |
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| 506. |
A cricket ball of mass 250g collides with a bat with velocity `10m//s` and returns with the same velocity within `0.01` second. The force acted on bat isA. 25NB. 50NC. 250ND. 500N |
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Answer» Correct Answer - D Force `=((dv)/(dt))=(0.25xx[(10)-(-10)]]/(0.01)` . `=25xx20=500N` . |
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| 507. |
A shell is fired from a cannon with a velocity `v (m//sec.)` at an angle `theta` with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed (in `m//sec.`) of the other piece immediately after the explosion isA. `3vcos theta`B. `2vcos theta`C. `(3v)/(2)cos theta`D. `(sqrt(3v)cos theta)/(2)` |
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Answer» Correct Answer - A In case of projectile motion as the highest point `(v)_(vertical)=0` and `(v)_(horizontal)=vcostheta` . the initial linear momentum of the system will be `mvcostheta` . Now as force of blasting is internal and force of gravity is vertical, so linear momentum of the system along horizontal is conserved, i.e., `P_(1)+P_(2)=mvcostheta` or `m_(1)v_(1)+m_(2)v_(2)=mvcostheta` . But it is given that `m_(1)=m_(2)=m//2` and as one part retraces its path, `v_(1)=-vcostheta` . `(1)/(2)m(-vcostheta)+(1)/(2)mv_(2)=mvcostheta` . Solving, we get: `v_(2)=3_(v)costheta` . |
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| 508. |
Two masses `m_(1)` and `m_(2)` are connected by light inextensible string passing over a smooth pulley `P`. If the pulley moves vertically upwards with an acceleration equal to g then .A. Tension on the string is `(4m_(1)m_(2)g)/(m_(1)+m_(2))`B. Tension on the string is `(2m_(1)m_(2)g)/(m_(1)+m_(2))`C. The acceleration of mass `m_(1)` with respect to ground is `(3_(2)-m_(1))/(m_(1)+m_(2))g` .D. The acceleration of mass `m_(1)` with respect to ground is `(2(m_(2)-m))/(m_(1)+m_(2))g` |
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Answer» Correct Answer - A::D `T = (4m_(1)m_(2)g)/(m_(1)+m_(2))` `a=(2(m_(2)-m_(1))g)/(m_(1)+m_(2))rArra_(ground)=(3m_(2)-m_(1))/(m_(1)+m_(2))g` . |
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| 509. |
A force-time graph for the motion of a body is shown in the figure. The change in the momentum of the body between zero and 10sec is A. `zero`B. `4kgm//s`C. `5kgm//s`D. `3kgm//s` |
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Answer» Correct Answer - D As `F=(dp)/(dt)` `intdp=intFdt` `Deltap` =area under `F-t` graph `Deltap=(1)/(2)xx8xx1-2xx8.5=3kgm//s` |
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| 510. |
A 2kg toy car can move along an x axis. Graph shows force `F_(x)` , acting on the car which being at rest at time `t=0` . The velocity of the particle at `t=0s` is: A. `-im//s`B. `-1.5im//s`C. `6.5im//s`D. `13im//s` |
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Answer» Correct Answer - C `intdp=p_(f)-p_(i)=intFdt` =Area under the curve. `p_(i)=0` Net Area `=16-2-1=13N-s` `=V_(f)=(13)/(2)=6.5im//s` [As momentum is position, particle is moving along positive `x` axis.] |
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| 511. |
You are thrown outer side when your car suddenly takes a turn. Which law of Newtn is involved in this ? .A. Third lawB. Second lawC. First lawD. Law of gravitation |
| Answer» Correct Answer - C | |
| 512. |
Three masses of 1 kg, 6 kg and 3 kg are connected to each other with threads and are placed on a table as shown in figure. What is the accelration with which the system is moving? Take `g=10ms^(-2)`: A. ZeroB. `2 ms^(-2)`C. `4 ms^(-2)`D. `3 ms^(-2)` |
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Answer» Correct Answer - B `3g-T_(2)=3a` `T_(2)-T_(1)=6a` `T_(1)-g=a` `therefore 2g=10a` `a=2m//s^(2)` |
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| 513. |
If `A` and `B` moves with acceleration a as shown in diagram calculate accelration of `C` with respect to `B` .A. `2a`B. `a sqrt2`C. `3a`D. `4a` |
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Answer» Correct Answer - D `Ta_(B)+2Ta_(A) -Ta_(C) =0, Ta+2Ta -Ta_(C) =0` `a_(C) =3a` . |
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| 514. |
Figure-2.202 shows a two block constrained motion system. Block A has a mass M and B has a mass m. If block A is pulled toward right horizontally with an external force F, find the acceleration of the block B relative to ground. |
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Answer» Correct Answer - `[[a=(f)/(M+m)]]l` |
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| 515. |
Which of the following groups of forces could be in equibriumA. 3 N, 4 N, 5 NB. 4N, 5 N, 10 NC. 30N, 40 N, 80 ND. 1N, 3 N, 5 N |
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Answer» Correct Answer - A |
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| 516. |
Two partical tied to different strings are whirled in a horizontal circle as shown in figure. The ratio of lengths of the string so that they complete their circular path with equal time priod is: A. `sqrt((3)/(2))`B. `sqrt((2)/(3))`C. 1D. None of these |
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Answer» Correct Answer - B Since `T=2pisqrt((Lcostheta)/(g))` :. `T_(1)=T_(2)` implies `L_(1)costheta=L_(2)costheta_(2)` `:. (L_(1))/(L_(1))=(costheta_(2))/(costheta_(1))=(cos45^(@))/(cos30^(@))` `(L_(1))/(L_(2))=(sqrt(2))/(sqrt(3))` |
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| 517. |
A car is being driven on a tilted ground. The ground makes an angle `theta` with the horizontal. The driven drives on a circle of radius R. The coefficient of friction between the tires and the ground is `mu`. (a) What is the largest speed for which the car will not slip at point A? Assume that rate of change of speed is zero. (b) What is the largest constant speed with which the car can be driven on the circle without slipping? |
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Answer» Correct Answer - (a) `[g^(2) R^(2) (mu^(2) cos^(2) theta - sin^(2) theta)]^(1//4)` `sqrt(gR(mu"cos"theta - "sin"theta))` |
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| 518. |
The arrangement shown in figure is in equilibrium with all strings vertical. The end A of the string is tied to a ring which can be slid slowly on the horizontal rod. Pulley `P_(1)` is rigidly fixed but `P_(2)` can move freely. A mass m is attached to the centre of pulley `P_(2)` through a thread. Pulleys and strings are mass less. (a) Which block will move up as A is moved slowly to the right? (b) Will the block of mass m have horizontal displacement? (c) Is it possible, for a particular position of A, that M has no acceleration but m does have an acceleration? If this happens when string from `P_(2)` to A makes an angle `theta` with vertical, find the acceleration of m at the instant.A. `B.C.D. |
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Answer» Correct Answer - (a) Block with mass M will move up. (b) yes (c) `g (1 - "cos"theta)` |
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| 519. |
A girl ridding a bicycle along a straight road with a speed of `5 ms^(-1)` throws a stone of mass 0.5 kg which has a speed of `15ms^(-1)` with respect to the ground along her direction of motion. The mass of the girl and bicycle is ` 5 kg` . Does the speed of the bicycle change after the stone is thrown ? What is the change in speed, if so ?A. `0.5 m//s`B. `0.1m//s`C. `0.3m//s`D. `0.8m//s` |
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Answer» Correct Answer - B Given total mass of girl bicycle and stone `m_(1) = (50 + 0.5) kg =50.5kg` Velocity of bicycle `u_(1) =5m//s`, Mass of girl and bicycle ` m = 50 kg` Yes the speed of the bicycle changes after the stone is thrown. Let after throwing the stone the speed of bicycle be m//s According to law of conservation of linear momentum, `m_(1) u_(1) =m_(2) u_(2) +mv` `50.5xx5 =0.5 xx 15 + 50 +v` `252.5 -7.5 = 50v` or `v = (245.0)/(50)` `v = 4.9m//s` . |
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| 520. |
A particle P is attached to two fixed points `O_(1)` and `O_(2)` in a horizontal line, by means of two ight inextensible strings of equal length l. It is projected with a velocity just sufficient to make it describe a circle, in a vertical plane, without the strings getting slack and with the angle`lt O_(2)O_(1)P = lt O_(1)O_(2)P = theta`. When the particle is at its lowest point, the string `O_(2)P` breaks and the subsequent path of the particle was found to be a circle of radius l `"cos" theta`. Find `theta`. |
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Answer» Correct Answer - `theta = tan^(-1) ((1)/(sqrt5))` |
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| 521. |
A boy sitting on the topmost berth in the compartment of a train which is just going to stop on a railway station, drop an apple aiming at the open hand of his brother sitting vertically below his hands at a distnace of about 2 meter. The apple willl fallA. Precisely on the hand of his brotherB. Slightly away from the hand of his brother in the direction of motion of the trainC. Slightly away from the hand of his brother in the direction opposite to the direction of motion of the trainD. None of the above |
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Answer» Correct Answer - B |
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| 522. |
Which of the following sets of concurrent force may be in equilibrium?A. `F_(1)=3` `N` , `F_(2)=5` `N` , `F_(3)=9` `N`B. `F_(1)=3` `N` , `F_(2)=5` `N` , `F_(3)=1` `N`C. `F_(1)=3` `N` , `F_(2)=5` `N` , `F_(3)=19` `N`D. `F_(1)=3` `N` , `F_(2)=5` `N` , `F_(3)=6` `N` |
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Answer» Correct Answer - D Range of resultant of `F_(1)` and `F_(2)` varies between `(3+4)=8N` and `(5-3)=2N` . It means for some value of angle `(theta)` , resultant `6can` be obtained. So the resultant of `3N` , `5N` and `6N` may be zero and the force may be in equilibrium. |
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| 523. |
A block of mass `m` is stationary on a horizontal surface. It is connected with a string which has no tension. The coefficient of friction between the block and surface is `m` . Then , the frictional force between the block and surface is: A. zeroB. `mumg`C. `(mumg)/(mu)`D. None of these |
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Answer» Correct Answer - A `f_(s)=mumg` , `Fltf_(s)` . If applied force is less than limiting friction force then frictional force is equal to the applied force. `f=F` . |
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| 524. |
A block `A` with mass 100kg is resting on another block `B` of mass 200kg. As shown in figure a horizontal rope tied to a wall hold it. The coefficient of friction between `A` and `B` is `0.2` while coefficient of friction between `B` and the ground is `0.3` . the minimum required force `F` to start moving `B` will be. |
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Answer» When B is tries to move, by applying a force F, then the fricitional forces acting on the blokc B are `f_(1)` and `f_(2)` with limiting values, `f_(1)=(mu_(s))_(A)m_(A)g` and `l_(2)=(mu_(s))_(B)(m_(A)+m_(B))g` Then minimum value of F should be (for just tending to move), `F=f_(1)+f_(2)=0.2xx100g+0.3xx300g=110g=1100N` |
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| 525. |
By what acceleration the boy must go up so that `100kg` block remains stationary on the wedge. The wedge is fixed and is smooth `(g =10m//s^(2))` . |
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Answer» For the block to remain stationary `T =Mg sin theta = 100 xx 10 xx sin53` ` =100 xx 10 xx (4)/(5) = 800N` For man `, T -mg =ma` `T =ma (g +a) rArr 800 = 50 (10 +a) a = 6m//s^(2)` . |
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| 526. |
Five persons A, B, C, D, and E, are pulling a cart of mass 100kg on a smooth surface and cart is moving with acceleration `3m//s^(2)` in east direction. When person A stops pullling, it moves with acceleration `24m//s^(2)` in the north direction. The magnitude of acceleration of the cart when only A and B pull the cart keeping their direction, is:A. `26m//s^(2)`B. `3sqrt(71)m//s^(2)`C. `25m//s^(2)`D. `30m//s^(2)` |
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Answer» Correct Answer - C When all are pulling. `vec(F)_(net)=100xx3hati` . when `A` stop `vec(F)_(net)-vec(F)_(A)=100xx1(-hati)` . when `B` stop `vec(F)_(net)-vec(F)_(B)=100xx24hatj` . from these three get `vec(F)_(a)+vec(F)_(B)=(700hati-2400hatj)N` . Hence, acceleration of the cart, `vec(a)=(vec(F)_(A)+vec(F)_(B))/(m)=((700hati-2400hatj))/(100)m//s^(2)` . `vec(a)=(7hati-24hatj)m//s^(2)implies|vec(a)|=sqrt(7^(2)+24^(2))=25m//s^(2)` . |
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| 527. |
A particle of mass `4m` explodes into three pieces of masses m,m and `2m` The equal masses move along X-axis and Y-axis with velocities `4ms^(-1)` and `6ms^(-1)` respectively. The magnitude of the velocity of the heavier mass is |
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Answer» `M =4m, u =0, m_(1) =m, m_(2) =m, m_(3) =2m` `v_(1) = 4ms^(-1), v_(2) =6ms^(-1) ,v_(3) =` According to law of conservation of momentum `vecP_(1) + vecP_(2)+vecP_(3) =0` `vecP_(3) =- (vecP_(1) +vecP_(2)),|vecP_(3)|=|vecP_(1)+vecP_(2)|` `P_(3) =sqrt(P_(1)^(2)+P_(2)^(2)+2P_(1)P_(2)Costheta)` `P_(1)` and `P_(2)` are perpendicular to each other `P_(3) = sqrt(P_(1)^(2) +P_(2)^(2)), m_(3) v_(3) =sqrt((m_(1)v_(1))^(2)+(m_(2)v_(2))^(2))` `2mv_(3) = sqrt((mxx4)^(2)+(mxx6)^(6))` `2v_(3)=sqrt(16 +36) rArrv_(3) =sqrt13 ms^(-1)` . |
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| 528. |
A man is standing in a lift which goes up and comes down with the same constant acceleration. If the ratio of the apparent weights in the two cases is `2 : 1` , then the acceleration of the lift is : (Take `g = 10 m//s^(2)`)A. `3.3 ms^(-2)`B. `2.50 ms^(-2)`C. `2.00 ms^(-2)`D. `1.67 ms^(-2)` |
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Answer» Correct Answer - A |
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| 529. |
Two trolleys of mass `m` and `3 m` are attached by a spring. The spring was compressed and then released , they move off in opposite direction and comes to rest after covering distances `s_(1)` and `s_(2)` respectively. Assuming the coefficient of friction to be uniform , the ratio of distances `s_(1) : s_(2)` isA. `1:9`B. `1:3`C. `3:1`D. `9:1` |
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Answer» Correct Answer - B |
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| 530. |
A boy of 50 kg is in a lift moving down with an acceleration `9.8ms^(-2)`.The apparent weight of the body is `(g=9.8ms^(-2))`A. `50xx9.8 N`B. ZeroC. 50 ND. `(50)/(9.8)N` |
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Answer» Correct Answer - B |
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| 531. |
A boy of mass 40 kg is hanging from the horizontal branch of a tree. The tension in his arms is minimum when the angle between the arms is:-A. `0^(@)`B. `90^(@)`C. `120^(@)`D. `180^(@)` |
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Answer» Correct Answer - A Arms should be vertical for least tension. |
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| 532. |
Find the tension `T_(2)` in the system shown in fig. A. 1g NB. 2g NC. 5g ND. 6g N |
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Answer» Correct Answer - C `T_(3)=30N` `T_(2)=20+30=50N=5gN` |
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| 533. |
In the system shown in fig., block A and C are placed on smooth floors and both have mass equal to `m_(1)`. Blocks B and D are identical having mass `m_(2)` each. Coefficient of friction Between A and B and that between C and D are both equal to `mu`. String and pulleys are light. A horizontal force F is applied on block C and is gradually increased. (a) Find the maximum value of F (call it `F_(0)`) so that all the four blocks move with same acceleration. (b) Will the value of `F_(0)` increase or decrease if another block (E) of mass `m_(2)` is placed above block D and coefficient of friction between E and D is `mu`? |
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Answer» Correct Answer - `F_(0) = 2 mum_(2) g ((m_(2) + m_(1))/(2m_(2) + m_(1)))` ; increase |
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| 534. |
In the arrangement shows in fig. if the surface is smooth then find out acceleration of the block `m_(2)`. |
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Answer» Let accelration of `m_(2)` be a then acceleration of `m_(1)` will be `2a`. for `m_(1):T=m_(1)(2a)` and for `m_(2):m_(2)g-2T=m_(2)a` `rArrm_(2)g-4m_(1)a=m_(2)arArra=(m_(2)g)/(4m_(1)+m_(2))` |
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| 535. |
Compare the impulses exerted on a wall by the two objects, a golf ball and a lump of mud both having the same mass and the veloctiy .A. the golfball imparts greater impulseB. the lump of mud imparts the greater impulseC. both impart equal impulseD. nothing can be said |
| Answer» Correct Answer - A | |
| 536. |
A block of mass `m=2kg` is placed in equilibrium on a moving plank accelerating with `a=1m//s^(2)` . If coefficient of friction between plank and block `mu=0.2` . The friction force acting on theblock is:A. 2NB. 4NC. 3ND. None of these |
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Answer» Correct Answer - A In this case driving force=ma `=2xx1=2N` And resisting force `=f_(lim)=muN=4N` Hence the mass will not slide over the plank friction should be static nature `f=2N` . |
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| 537. |
When a person walks on a rough surface .A. the frictional force exerted by the surface keeps him moving .B. reaction of the force applied by the man on the surface keeps him movingC. the force applied by the man keep him movingD. weight of the man keeps him moving . |
| Answer» Correct Answer - A | |
| 538. |
The engine of a jet aircraft applies a thrust force of `10^(5)N` during take off and causes the plane to attain a velocity of 1 `km/sec` in 10 sec . The mass of the plane isA. `10^(2)kg`B. `10^(3)kg`C. `10^(4)kg`D. `10^(5)kg` |
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Answer» Correct Answer - B |
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| 539. |
A force of 50 dynes is acted on a body of mass 5 g which is at rest for an interval of 3 seconds, then impulse isA. `0.15xx10^(-3)`N-sB. `0.98xx10^(-3)`N-sC. `1.5xx10^(-3)` N-sD. `2.5xx10^(-3)` N-s |
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Answer» Correct Answer - C |
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| 540. |
A horizontal plane with the coefficient of friction `k` supports two bodies: a bar and an electric motor with a battery on a block. A thread attached to the bar is wound on the shaft of the electric motor. The distance between the bar and the electric motor is equal to `l`. When the motor is switched on, the bar, whose mass is twice as great as that of the other body, starts moving with a constant acceleration `w`. How soon will the bodies collide? |
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Answer» Correct Answer - `[t=sqrt((2L)/(3a+mug))]` |
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| 541. |
A stationary body mass `m` is slowly lowered onto a rough massive plat from moving at a constant velocity `v_(o)=4m//s`. On smooth surface. The distance the body will slide with respect to the plat from `(mu=0.2)` is: A. 4 mB. 6 mC. 12 mD. 8 m |
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Answer» Correct Answer - A |
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| 542. |
A body mass `2kg` has an initial velocity of 3 metre//sec along OE and it is subject to a force of `4N` in a direction perpendicular to OE. The distance of body from `O` after 4 sec will be: A. 12 metresB. 20 metresC. 8 metresD. 48 metres |
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Answer» Correct Answer - B The acceleration of the body perpendicular to `OE` is `a=(F)/(m)=(4)/(2)=2m//s^(2)` Displacement along `OE` , `s_(1)=vt=3xx4=12m` Displacement perpendicular to `OE` `s_(2)=(1)/(2)at^(2)=(1)/(2)xx2xx(4)^(2)=16m` the resultant displacement `s=sqrt(s_(1)^(2)+s_(2)^(2))=sqrt(144+256)=sqrt(400)=20m` |
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| 543. |
A light string of `70cm` has its two ends tied at the same level `50cm` apart. A force of `100N` is applied at a distance of `30cm` from `P` . The tension in part PR is. A. `18N`B. `8N`C. `0N`D. `80N` |
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Answer» Correct Answer - D Here `T_(1)cos(90-theta)`+`T_(2)costheta=100` And `T_(1)costheta=T_(2)cos(90-theta)` Solving, `T_(1)=80N` |
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| 544. |
A force of 100N is applied on a block of mass 3 kg as shown in the figure-2.164. The coefficient of friction between the surface and the block is 0.25.The frictional force acting on the block is: (Take `g=10m/s^(2)`) A. 15 N downwardsB. 25 N upwardsC. 20 N downwardsD. 30 N upwards |
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Answer» Correct Answer - C |
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| 545. |
A body of 5kg is moving with a velocity of 20m/s. If a force of 100N is applied on it for 10 s in the same direction as its velocity, what will now be the velocity of the body?A. 2000m/sB. 220m/sC. 240m/sD. 260m/s |
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Answer» Correct Answer - B `F=marArr100=5arArra=20m//s^(2)` `v=u+at=20+20xx10=220m//s` |
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| 546. |
A force of `100N` need to be applied parallel to a smooth inclined plane just to hold a body on it. The angle of inclination of the inclined plane is plane is `30^(@)` . How much horizontal force need to be applied to do the same?A. `50N`B. `87N`C. `100N`D. `115N` |
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Answer» Correct Answer - D Given that `mgsintheta=100N` . If horizontal force required be `F` , then component of `F` will balance the weight i.e., `Fcostheta=100` or `F=(100)/(costheta)=(100)/(cos30^(@)=115N` |
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| 547. |
A body of weight `w_1` is suspended from the ceiling of a room through as chain of weight `w_2`. The ceiling pulls the chain by a forceA. `w_(1)`B. `w_(2)`C. `w_(1)+w_(2)`D. `(w_(1)+w_(2))/(2)` |
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Answer» Correct Answer - C |
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| 548. |
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 ture and reason is the correct explanation of assertion.B. If both assertion and reason are ture but reason is not the correct explanation of assertion.C. If assertion is ture but reason is false.D. If both assertion and reason are false. |
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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 necesserily zero at that instant. For example, when a ball thrown upward reaches its maximum height, `v=0 ` but the force continue to be its weight `mg` and the acceleration is `g` . Force is not always inn the direction of motion. |
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| 549. |
Three forces `20 sqrt(2)N,20 sqrt(2) N` and `40N` are acting along `X,Y` and `Z-`axes respectively on a `5sqrt2 kg` mass at rest at the origin. The magnitude of its displacement after `5s` is .A. `50m`B. `25m`C. `60m`D. `100m` |
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Answer» Correct Answer - D `a =(F_(R))/(m),S=ut+(1)/(2)at^(2)` |
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| 550. |
A horizontal force `F` of variable magnitude and constant direction acts on a body of mass `m` which is initially at rest at a point `O` on a smooth horizontal surface. The magnitude of `F` is given by `F=beta+alphat` where `t` is the time for which the force has been acting on the distance of the body from `O` at time `t` , then `s` is equal to.A. `(1)/(2m)(betat+alphat^(2))t`B. `(1)/(2m)(beta+alphat^(2))`C. `((beta+alphat)t^(2))/(2m)`D. `t^(2)/(6m)(3beta+alphat)` |
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Answer» Correct Answer - D `a=(dv)/(dt)=(F)/(m)=(beta+alphat)/(m)` `int_(0)^(v)dv=int_(0)^(t)((beta+alphat)/(m))dt` implies `v=(betat)/(m)+(alphat2)/(2m)` Now `intds=intvdt` implies `int_(0)^(s)ds=int_(0)^(t)((betat)/(m)+(alphat^(2))/(2m))dt` implies `s=(t^(2))/(6m)(3beta+alphat)` |
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