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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.
| 1501. |
A particle is taken from point A to point B under the influence of a force field. Now it is taken back from B to A and it is observed that the work done in taking the particle from A to B is not equal to the work done in taking it from B to A. If `W_(nc)` and `W_c` are the work done by non-conservative and conservative forces present in the system, respectively, `DeltaU` is the change in potential energy and `Deltak` is the change in kinetic energy, thenA. (a) `W_(nc)-DeltaU=Deltak`B. (b) `W_c=-DeltaU`C. (c) `W_(nc)+W_(c)=Deltak`D. (d) `W_(nc)-DeltaU=-Deltak` |
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Answer» Correct Answer - A::B::C According to the work-energy theorem for a non-conservative system, we have `W_c+W_(nc)=DeltaK` Also, we know that the work done by a conservative force equals the decrease in potential energy, so `W_c=-DeltaU` `W_(nc)-DeltaU=DeltaK` |
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| 1502. |
Ram and Ali have been fast friends since childhood. Ali neglected studies and now has no means to earn money other than a channel whereas Ram has become an engineer. Now both are working in the same factory. Ali uses camel to transport the load within the factory. Due to low salary and degradation in health of camel, Ali becomes worried and meets his friend Ram and discusses his problems. Ram collected some data and with some assumptions concluded the following: i. The load used in each trip is `1000kg` and has friction coefficient `mu_k=0.1` and `mu_s=0.2`. ii. Mass of camel is `500kg`. iii. Load is accelerated for first `50m` with constant acceleration, then it is pulled at a constant speed of `5ms^-1` for `2km` and at last stopped with constant retardation in `50m`. iv. From biological data, the rate of consumption of energy of camel can be expressed as `P=18xx10^3v+10^4Js^-1` where P is the power and v is the velocity of the camel. After calculations on different issues, Ram suggested proper food, speed of camel, etc. to his friend. For the welfare of Ali, Ram wrote a letter to the management to increase his salary. (Assuming that the camel exerts a horizontal force on the load): The ratio of the energy consumed by the camel during uniform motion for the two cases when it moves with speed `5ms^-1` to the case when it moves with `10ms^-1`A. (a) `19/20`B. (b) `19/10`C. (c) `10/19`D. (d) `20/19` |
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Answer» Correct Answer - D `W=P Deltat` `P=18xx10^3v+10^4Js^-1` `P_5=18xx10^3xx5+10^4Js^-1=10^5Js^-1` and `Deltat_5=(2000m)/(5ms^-1)=400s` `P_10=18xx10^3xx10xx10^4Js^-1` `Deltat_10=(200m)/(10ms^-1)=200s` `W_5/W_10=(10^4(9+1)xx400)/(10^4(18+1)xx200)=20/19` |
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| 1503. |
A force `F=-K(yhati+xhatj)` (where K is a positive constant) acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point `(a, 0)`, and then parallel to the y-axis to the point `(a, a)`. The total work done by the force F on the particle isA. ZeroB. `2pibJ`C. 2 bJD. None of these |
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Answer» Correct Answer - B |
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| 1504. |
A body of mass `1 kg` begins to move under the action of a time dependent force `vec F = (2 t hat I + 3 t^(2) hat j) N`, where `hat i` and `hat j` are unit vectors along x-and y-axes. What power will be developed by the force at the time `t` ?A. `(2t^(2)+4t^(4))W`B. `(2t^(3)+3t^(4))W`C. `(2t^(3)+3t^(5))W`D. `(2t+3t^(3))W` |
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Answer» Correct Answer - c `(c )` According to question, a body of mass 1 kg begins to move under the action of time dependent force, `F=(2thati+3t^(2)hatj)N` Where` hati "and" hatj` are unit vectors along X and Y-axes. `thereforeF=ma` `implies a=F/m` `impliesa=((2thati+3^(2)thatj))/(1)" "(therefore m=1kg)` `impliesa=(2thati+3^(2)thatj)m//s^(2)` `therefore` accelaration, `a=(dv)/(dt)` `implies" "dv= a dt` Integrating both sides, we get `intdv=inta" "dt=int(2thati+3t hatj)dt` `v=t^(2)hati+t^(3)hatj` `therefore` Power developed by the force at the time t will be given as `P=F.v=(2thati+3thatj).(t^(2)hati+t^(3)hatj)` `=(2t.t^(2)+3t^(2).t^(3))` `P=(2t^(3)+3t^(5))W` |
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| 1505. |
A partical moves from a point `(-2hati + 5hatj)` to `(4hati + 3hatj)` when a force of (4hati + 3hatj) N` is applied . How much work has been done by the force?A. 8JB. 11JC. 5JD. 2J |
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Answer» Correct Answer - c (c ) Position vectors of the particles are `r_(1)=12hati+5hatj "and" r^(2)=4hatj +3hatk` `therefore`Displacement of the particles, `Deltas=r_(2)-r_(1)` `=4hatj +3hatk-(-2hati+5hatj)=2hati-hatj+3hatk` Force on the particle ,`F=4hati +3hatj` N `therefore` work done, `W=F.Deltas=(4hati+3hatj).(2hati-hatj+3hatk)` `=8-35J` |
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| 1506. |
A force `vecF=(4hati+5hatj)N` acts on a particle moving in XY plane. Starting from origin, the particle first goes along x-axis to the point `(5.0)` m and then parallel to the Y-axis to the point `(5,4` m. The total work done by the force on the particle is A. `-20J`B. `+20J`C. `-40J`D. `+40J` |
| Answer» Correct Answer - D | |
| 1507. |
A block of mass 4 kg is released from the top of an inclined smooth surface as shown in figure If spring constant of spring is `100M//m` and block comes to rest after compressiong spring by 1 m, find the distance travelled by the block before it comes t o rest. |
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Answer» Let d be the distance travelled by block along the plane Height it comes down in downward direction, `h=d sin 30^(@).` When block comes to rest, decreases in its PE will be stored in the spring. `mgh1/2ky^(2)` (y is the compressin in t he spring) Let d be the distance travelled by the block along the plane and height it comes in downward direction is `h=d sin30^(@)` When block comes to rest, the decreases in its PE will be stored in the spring. `thereforemgh=1/2ky^(2)` (y is compression in spring) `mgd sin30=1/2ky^(2)` `4xx10xxd/2=1/2xx100xx1^(2)` `impliesd=2.5m` |
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| 1508. |
A simple pendulum with bob of mass m and length x is held in position at an angle `theta_(1)` and then angle `theta_(2)` with the vertical. When released from these positions, speeds with which it passes the lowest postions are `v_(1)&v_(2)` respectively. Then ,`(v_(1))/(v_(2))` is.A. `(1-costheta_(1))/(1-costheta_(2))`B. `sqrt((1-costheta_(1))/(1-costheta_(2)))`C. `sqrt((2gx(1-costheta_(1)))/(1-costheta_(2)))`D. `sqrt((1-costheta_(1))/(2gx(1-costheta_(2))))` |
| Answer» Correct Answer - B | |
| 1509. |
Given in fig are examples of some potential energy functions in one dimension. The total enrgy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant. |
| Answer» We know, that total energy `E=K.E+P.E` or `K.E=E-P.E` and kinetic energy can never be negative. The object can not exists in the region, where its K.E. Would become negative. A) IN the region between x=0 and x=a, potential energy is zero. So, kinetic energy will be negative in this region. Thus, the particle cannot be present in the region xgta. The minimum total energy that the particle can have in this case is zero. | |
| 1510. |
An object of mass `m` is tied to a string of length `l` and a variable force F is applied on it which brings the string gradually at angle thit `theta` with the vertical. Find the work done by the force `F` . |
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Answer» The various forces involved in the problem are shown in figure. As the object is in equilibrium at every stage, therefore, `mg=Tcostheta and F=Tsin theta` `(F)/(mg)=(Tsintheta)/(Tcostheta)=tantheta` `F=mg tan theta` ….(i) Work done by the force, `W=int_(0)^(theta)vecF.dvecs=int_(0)^(theta)F ds cos theta` `=int_(0)^(theta)mg tantheta xx(ld theta). costheta` `=mgl int_(0)^(theta)sind theta=mgl[-costheta]_(0)^(theta)` `=-mgl(cos theta-cos 0^(@))` `W=mgl(1-costheta)` |
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| 1511. |
A horizontal plane supports a plank with a block placed on it. A light elastic string is attached to the block, which is attached to a fixed point O. Initially, the cord is unstretched and vertical. The plank is slowly shifted to right until the block starts sliding over it. It occurs at the moment when the cord deviates from vertical by an angle `theta=0^@`. Work done by the force F equals A. (a) Energy lost against friction `F_1` plus strain energy in cordB. (b) Work done against total friction acting on the plank aloneC. (c) Work done against total friction acting on the plank plus strain energy in the cordD. (d) Work done against total friction acting on the plank plus strain energy in the cord minus work done by friction acting on the block |
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Answer» Correct Answer - A::B::D When the plank is shifted, the cord elongates and becomes inclined with vertical. Therefore, a tenstion is developed in the cord and that tension has two components, a vertical component and a horizontal component: Horizontal component tries to slide the block leftwards, relative to the plank. But friction `F_2` prevents that slipping. Hence, the block moves to the right with the plank. Since the block does not slip over the plank, no energy is lost against friction `F_2`. Hence, work done by the force F is used to overcome loss against friction `F_1` and to elongate the cord. Therefore, option (a) is correct. The cord elongates due to displacement of the block and the block gets displaced due to friction `F_2` acting on it. Hence, the friction `F_2` is responsible for the elongation of the cord. Therefore, strain energy stored in the cord is equal to work done by the friction on the block or work done against friction acting on the upper surface of the plank. Hence, the total work done by F is equal to energy lost against friction `F_1` plus work done against friction acting on upper surface of the plank. Therefore, option (b) is correct. Hence, it is obvious that option (c) is wrong. Since, strain energy stored in the cord is equal to work done by friction acting on the block, option (d) becomes same as option (b). Hence, it is also correct. |
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| 1512. |
What is pumped from a depth of 10 m and delivered through a pipe of cross section `10^(-2)m^(2)` upto a height of 10 m if it is needed to deliver a volume 0.2 `m^(3)` per second the power required will be :A. 19.6 kWB. 9.8 kWC. 39.2 kWD. 4.9 kW |
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Answer» Correct Answer - C |
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| 1513. |
Work done in time t on a body of mass m which is accelerated from rest to a speed v in time `t_(1)` as a function of time t is given bA. `(1)/(2)m(v^(2))/(t_(1)^(2))t^(2)`B. `(1)/(2)((mv)/(t_(1)))^(2)t^(2)`C. `m(v)/(t_(1))t^(2)`D. `(1)/(2)m(v)/(t)t^(2)` |
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Answer» Correct Answer - A (a) Work done `="Force"xx"displacement"` F=ma and `s=(1)/(2)at^(2)` Using,`" " s=ut+(1)/(2)at^(2)` `:. " " W=maxx(1)/(2)at^(2)=(1)/(2)ma^(2)t^(2)` `=(1)/(2)m((v)/(t_(1)))^(2)t^(2)=(1)/(2)m(v^(2))/(t_(1)^(2))t^(2) " " ("Using" a=(v)/(t_(1)))` |
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| 1514. |
A car of mass m has an engine which can deliver power P. The minimum time in which car can be accelerated from rest to a speed v is :-A. `(mv^(2))/(2P)`B. `Pmv^(2)`C. `2 Pmv^(2)`D. `(mv^(2))/(2)P` |
| Answer» Correct Answer - A | |
| 1515. |
Potential energy of a body is the energy possessed by the body by virtue of its position. P.E.`=mgh` where the symbols have their usual meaning. Kinetic energy of a body is the energy possessed by the body by virtual of its velocity. `K.E.=(1)/(2)m upsilon^(2)` Energy can neither be created nor be destroyed. However energy can be changed from one form to other, such that energy appearing in one form is equal to the energy disappearing in the other form. With the help of the passage given above, choose the most appropriate alternative for each of the following question `:` The body will attain this K.E. when it falls freely from a height ofA. `125 m`B. `250 m`C. `1250 m`D. `2500 m` |
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Answer» Correct Answer - A `m gh = K.E. = 1250 ` `h= (1250)/( mg)=(1250)/(1xx10)= 125m` |
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| 1516. |
Potential energy of a body is the energy possessed by the body by virtue of its position. P.E.`=mgh` where the symbols have their usual meaning. Kinetic energy of a body is the energy possessed by the body by virtual of its velocity. `K.E.=(1)/(2)m upsilon^(2)` Energy can neither be created nor be destroyed. However energy can be changed from one form to other, such that energy appearing in one form is equal to the energy disappearing in the other form. With the help of the passage given above, choose the most appropriate alternative for each of the following question `:` Kinetic energy of the body at the same time isA. `1250 J`B. `2500 J`C. `625 J`D. `25000 J` |
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Answer» Correct Answer - A `K.E. =(1)/(2) m upsilon^(2) =(1)/(2) xx1xx(50)^(2)= 1250 J` |
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| 1517. |
Potential energy of a body is the energy possessed by the body by virtue of its position. P.E.`=mgh` where the symbols have their usual meaning. Kinetic energy of a body is the energy possessed by the body by virtual of its velocity. `K.E.=(1)/(2)m upsilon^(2)` Energy can neither be created nor be destroyed. However energy can be changed from one form to other, such that energy appearing in one form is equal to the energy disappearing in the other form. With the help of the passage given above, choose the most appropriate alternative for each of the following question `:` A body of mass `1 kg` is allowed to fall freely under gravity. The momentum of the body 5 second after it starts falling isA. `100 kg ms^(-1)`B. `50 kg ms^(-1)`C. `150 kg ms^(-1)`D. `200 kg ms^(-1)` |
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Answer» Correct Answer - B Here,` m=1 kg u =0, t = 5 s , p=?` `upsilon=u+at` `upsilon=0+10xx5=50m//s` `p= m upsilon = 1xx50= 50 kg m s ^(-1)` |
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| 1518. |
Potential energy of a body is the energy possessed by the body by virtue of its position. P.E.`=mgh` where the symbols have their usual meaning. Kinetic energy of a body is the energy possessed by the body by virtual of its velocity. `K.E.=(1)/(2)m upsilon^(2)` Energy can neither be created nor be destroyed. However energy can be changed from one form to other, such that energy appearing in one form is equal to the energy disappearing in the other form. With the help of the passage given above, choose the most appropriate alternative for each of the following question `:` Velocity of the body on striking the ground will beA. `25 m//s`B. `12.5 m//s`C. `50 m//s`D. `100 m//s` |
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Answer» Correct Answer - C As `K.E.=(1)/(2) m upsilon^(2)` `:. upsilon = sqrt((2KE)/( m))= sqrt((2xx1250)/(1))= 50 m //s` |
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| 1519. |
Give example of a situation in which an applied force does not result in a change in kinetic energy. |
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Answer» In an atom, an electron revolve around the nucleus with a constant speed in a given orbit. The force applied is electrostatic force of attraction between the nucleus and electron. As speed `upsilon` is constant, `K.E. =(1)/(2)m upsilon^(2)=` constant |
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| 1520. |
If the force is always perpendicular to motion, then KE remains constantA. KE remains constantB. Work done is zeroC. velocity is constantD. speed is constant |
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Answer» Correct Answer - A::B::D Here, `theta=90^(@), W=Fs cos theta=Fs cos 90^(@)=0` So work done is zero. In uniform circulart motion, speed is always a constant but velocity changes as direction changes. If speed is constant, then `KE` is also constant. |
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| 1521. |
Consider the situation shown in figure. Mass of block A is m and that of blcok B is 2m. The force constant of string is k. Friction is absent everywhere. System is released from rest with the spring unstretched. Find (a) The maximum extension of the spring `x_(m)` (b) The speed of block A when the extension in the springt is `x=(x_(m))/(2)` (c ) The net acceleration of block B when extension in the spring is `x=(x_(m))/(4).` |
| Answer» `x_(m)=(4mg)/(k),(b)v2gsqrt((m)/(3k)),(c)a=g/3`(downdards) | |
| 1522. |
Consider a situation as shown in the figure. The system is released from rest. When the block of mass `m` has falled a distance `L`, its speed becomes `(sqrt(gL))/(3)`. Find the friction coefficient `mu`. |
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Answer» When the block of mass `m` has descended a distance `L`, distance traveled by the block of mass `2m` is 2L. IF the speed of `m` is `v`, speed of `2m` will be `2v`. Work done against friction `=muxx2mgxx2L` `=4mumgL` Loss in `P.E=`gain in `K.E+`work donw against friction `mgL=(1)/(2)mv^2+(1)/(2)xx2m(2v)^2+4mumgL` `=(9)/(2)mv^2+4mumgL` `=(9)/(2)mxx(gL)/(9)+4mumgL` `(mgL)/(2)=4mumgLimpliesmu=(1)/(8)` |
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| 1523. |
As shown in figure a smooth rod is mounted just above a table top `10 kg` collar, which is able to slide on the rod with negligible friction is farstened to a spring whose other end is attached to a pivot at O. The spring has negligible mass, a relaxed length of `10 cm` and a spring constant of `500 N//m` the collar is released from rest at point A. (a) What is its velocity as it passes point B? . |
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Answer» Correct Answer - A::B::D (a) `K_(A) + U_(A) =K_(B) + U_(B)` `:. 0 + 1/2 xx 500 xx (0.5-0.1)^(2)` `1/2 xx 10 xx v^(2) + 1/2 xx 500 xx (0.3-0.1)^(2)` On solving we get `v=2.45 m//s` (b) `CO =sqrt((BO)^(2) + (BC)^(2))` `= sqrt((30)^(2) + (20)^(2))` `approx 36 cm` `=0.36 m` Again applying the equation, `K_(A) + U_(A) =K_(C ) + U_(C )` `0 + 1/2 xx 500 xx (0.5-0.1)^(2)` `=1/2 xx 10 xx v^(2) + 1/2 xx 500(0.36-0.1)^(2)` On solving, we get `v=2.15 m//s`. |
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| 1524. |
A machine delivers power to a body which is proportional to velocity of the body. If the body starts with a velocity which is almost negligible, then the distance covered by the body is proportional toA. `sqrt(v)`B. `3sqrt((v)/(2))`C. `v^(5//3)`D. `v^(2)` |
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Answer» Power `=Fv=m(dv)/(dt)(v)alphav` `rArr mv(dv)/(dt)=k_(0)v`, where `k_(0)=` constant `rArrm(dv)/(dt)=k_(0)rArrmv(dv)/(dx)=k_(0)` `rArr vdv=(k_(0))/(m)dx` `rArr` integrating both sides, `int_(0)^(v)vdv=(k_(0))/(m)int_(0)^(x)dx` `(v^(2))/(2)=(k_(0))/(m)x` Hence, displacement is proportional to square of instantaneous velocity. |
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| 1525. |
Two light vertical springs with equal natural length and spring constants `k_(1)` and `k_(2)` are separated by a distance `l`. Their upper end the ends `A` and `B` of a light horizontal rod `AB`. A vertical downwards force `F` is applied at point `C` on the rod. `AB` will remain horizontal in equilibrium if the distance `AC` is A. `(lk_(1))/(k_(2))`B. `(lk_(1))/(k_(2) + k_(1))`C. `(lk_(2))/(k_(1))`D. `(lk_(2))/(k_(1) + k_(2))` |
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Answer» Correct Answer - D `sum ("Moments about" C) =0` `:. (k_(1)x) AC=(k_(2)x) BC` `:. (AC)/(BC) =k_(2)/k_(1)`….(i) `AC + BC =l`….(ii) Soving these two equation we get, `AC=((k_(1))/(k_(1)+k_(2)))l` |
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| 1526. |
The potential energy of a two particle system separated by a distance `r` is given by `U(r ) =A/r` where A is a constant. Find to the radial force `F_(r)`, that each particle exerts on the other. |
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Answer» Correct Answer - A::B `F=-(dU)/(dr) =(A)/(r^(2))`. |
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| 1527. |
In the situation shown in figure all contact surfaces are smooth. The force constant of the spring is K. Two forces F are applied as shown. The maximum elongation produced in the spring is how many times of `F//K` (initially the spring is relaxed)? |
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Answer» Correct Answer - `(2)` Maximum elongation is given by `x_(max)=(2[F_1m_2+F_2m_1])/(K(m_1+m_2))` Here `F_1=F_2=F`, `m_1=m` and `m_2=M` Put the values and solve to get `x_(max)=2F//K` |
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| 1528. |
An elevatore is moving up with a constant velocity. Two different masses are attached to two different springs of same spring constant. Say you do two operations expanding the spring or compressing the spring by the same distance. Which operation consumes more energy: expanding the spring or compressing it? |
| Answer» Both consume the same amount of energy assuming that the spring constants for both compression and expansion are same. This is because both the masses are at rest initially with respect to the car as the car moves with constant velocity. | |
| 1529. |
In the above question, if equal forces are applied on two springs, thenA. (a) More work is done on QB. (b) More work is done on PC. (c) Heir force constants will become equalD. (d) Equal work is done on both the springs |
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Answer» Correct Answer - A Spring Q has lesser force constant that P. So Q will develop less restoring force than P. As a result, Q will suffer more extension. Since force is same in both the cases, more work will be done on Q. |
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| 1530. |
In the following questions, each question contains Statement I(assertion) and StatementII(reason). Each questions has four choices a, b, c, and d out of which only one is correct. Statement I: A block of mass `m` starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now titled to an angle of `30^@` with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in mechanical energy in the second situation is smaller than that in the first situation. Statement II: The coefficient of friction between the block and the surface decreases with the increase in the angel of inclination.A. (a) Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I.B. (b) Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I.C. (c) Statement I is true, Statement II is false.D. (d) Statement I is false, Statement II is true. |
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Answer» Correct Answer - C Decrease in mechanical energy in case I will be `DeltaU_1=1/2mv^2` But decrease in mechanical energy in case II will be `DeltaU_2=1/2mv^2-mgh` `:. DeltaU_2ltDeltaU_1` Hence, Statement 1 is correct. More ones, statement 2 is wiring. Therefore, option (c) is correct. |
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| 1531. |
Mark the correct statement(s).A. (a) Total work done by internal forces of a system on the system is always zero.B. (b) Total work done by internal forces of a system on the system is sometimes zero.C. (c) Total work done by internal forces acting between the particles of a rigid body is always zero.D. (d) Total work done by internal forces acting between the particles of a rigid body is sometimes zero. |
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Answer» Correct Answer - B::C Consider a system of two point particles which are free to move on a smooth horizontal surface, then the work done by electrical forces acting on the two parties is non-zero. But it is not necessarily true always. For a rigid body, there is no relative displacement of particles, hence, internal forces do not do any work. |
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| 1532. |
Mark the correct statement(s).A. (a) The work-energy theorem is valid only for particlesB. (b) The work-energy theorem is an invariant law of physics.C. (c) The work-energy theorem is valid only in inertial frames of reference.D. (d) The work-energy theorem can be applied in non-inertial frames of reference too. |
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Answer» Correct Answer - B::D The statement given in option (b) is a very standard concept related to the work-energy theorem, Work-energy theorem can be applied to any stream. Work-eneryg theorem can be applied to non-inertial frames of reference also, provided the work done by inertial forces is also considered on right-hand side of the work-energy theorem. |
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| 1533. |
Assertion: Mass and energy are not conserved separately, but are conserved as a single entity called mass-energy. Reason: Mass and energy conservation can be obtained by Einstein equation for energy.A. If both assertion and reason are true and reason is a true explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion is and reason are false. |
| Answer» From enstein equation `E = mc^(2)` it can be oserved that if mass is conservative then only energy is conservative and verse Thus, both connot be treated separately. | |
| 1534. |
A `15 gm` ball is shot from a spring whose spring has a force constant of `600 N//m`. The spring is compressed by `5 cm`. The greater possible horizontal range of the ball for this compression isA. `6.0 m`B. `12.0 m`C. `10.0 m`D. `8.0 m` |
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Answer» Correct Answer - B `R_(max) = (u^(2))/(g) at theta = 45^(@)` `R_(max) = ((m u^(2))/(2)) ((2)/(mg)) = ((1)/(2) kx^(2)) ((2)/(mg))` As `((1)/(2) m u^(2) = (1)/(2)kx^(2)) ((2)/(mg))` As `((1)/(2) m u^(2) = (1)/(2) kx^(2)) implies R_(max) = (kx^(2))/(mg)` Substituting the values `R_(max) = ((600) (5 xx 10^(-2))^(2))/((15 xx 10^(-3)) (10)) = 10.0m` |
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| 1535. |
A toy gun a spring of force constant `k`. When changed before being triggered in the upward direction, the spring is compressed by a distance `x`. If the mass of the shot is `m`, on the being triggered it will go up to a height ofA. `(Kx^(2))/(8g)`B. `(x^(2))/(8g)`C. `(Kx^(2))/(2 mg)`D. `(K^(2)x^(2))/(mg)` |
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Answer» Correct Answer - C According to law of conservation of energy , elastic potential energy strod in spring = gravational potential energy acquired by shot `(1)/(2) kx^(2) = mgh or h = (kx^(2))/(2 mg)` |
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| 1536. |
The kinetic energy of partical moving along a circule of radius `R` depends upon the distance covered `S` and given by `K = aS` where `a` is a constant. The the force acting on the partical isA. constantB. proportional to `v`C. proportional to `v^(2)`D. inversely proportional to `v` |
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Answer» Correct Answer - D ` KE prop x rArr KE =Cx rArr (1)/(2) mv^(2) =Cx` `(1)/(2) m2v(dv)/(dt) =C(dx)/(dt) rArr mv(dv)/(dt) =Cv` ` m(dv)/(dt) = rArr ma =C` `rArr F=C rarr `constant. |
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| 1537. |
Which of the following graph is correct between kinetic energy `E`, potential energy `(U)` and height `(h)` from the ground of the particalA. B. C. D. |
| Answer» Piotential energy increases and kinetic energy dicrease when the height of the partical increase it is clear from the graph (a). | |
| 1538. |
when the bob of a simple pendulum swings, the work done by tension in the string is :A. `gt 0`B. `lt 0`C. zeroD. maximum |
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Answer» Correct Answer - C Tension in the string is along the radius of circular path adopted by the body , while displacement of the bob is along the circumuforce of the path hence angle b,. Woon `vec F` and `vec s` in always `90^(@)` and to `W = 0` |
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| 1539. |
A particle falls from rest under gravity. Its potential energy with respect to the ground (PE) and its kinetic energy (KE) are plotted against time (t). Choose the correct graph.A. B. C. D. |
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Answer» Correct Answer - B |
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| 1540. |
A body is displaced from prigin to (1m,1m) by force `F=(2yhati + 3x^(2)hatj)` along two paths (a) `x=y` (b) `y=x^(2)` Find the work done along both paths. |
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Answer» `F=(2yhati + 3x^(2)hatj)` `dr=(dxhati + dyhatj)` `F.dr=(2ydx + 3x^(2)dy)` We cannot integrate F.dr or `(2ydx + 3x^(2)dy)` as such to find the work done. But along the given paths we can change this expression. (a) Along the path `x = y` `(2y dx + 3x^(2)dy) = (2x dx + 3y^(2)dy)` `W_(1) = int_(0,0)^(1m,1m,)F.dr =int_(0,0)^(1m,1m)(2xdx + 3y^(2)dy)` `=[x^(2) + y^(3)] _(0,0)^(1m,1m)` `=(1)^(2) + (1)^(3)=2J` (b) Along the path `y=x^(2)` `(2ydx + 3x^(2)dy) =(2x^(2)dx + 3ydy)` `:. W=(2)=int_(0,0)^(1m,1m)F.df =int(2x^(2)dx + 3y dy)` `=[2/3x^3 + 3/2y^(2)]_(0,0)^(1m,1m)` `=(13)/6 J` |
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| 1541. |
A body `X` with a momentum `p` collides with another identical stationary body `Y` one dimensionally. During the collision, `Y` gives an impulse `J` to body `X`. Then coefficient of restitution isA. `(2J)/(P) - 1`B. `(J)/(P) + 1`C. `(J)/(P) - 1`D. `(J)/(2P) - 1` |
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Answer» Correct Answer - B `u_(1) = p//m, u_(2) = 0, v_(1) = (p - J)//m, v_(2) = J//m` Now apply `e = (v_(2) - v_(1))/(u_(1) - u_(2))` |
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| 1542. |
A smooth rubber cord of length `l` whose coefficient of elasticity is k is suspended by one end from the point O (figure). The other end is fittetd with a catch B. A small sleeve A of mass m starts falling from the point O. Neglecting the masses of the thread and the catch, find the maximum elongation of the cord. |
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Answer» Correct Answer - A::B |
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| 1543. |
A cord is used to lower vertically a block of mass `M`, a distance `d` at a constant downward acceleration of `(g)/(4)`, then the work done by the cord on the block isA. `mg d//4`B. `2 Mg d//4`C. `-3 Mg d//4`D. Mg d |
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Answer» Correct Answer - C |
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| 1544. |
A ball is attached to a horizontal cord of length `l` whose other end is fixed. (a) If the ball is released, what will be its speed at the lowest point of its path? (b) A peg is located a distance `h` directly below the point of attachment of the cord. If `h=0.75l`, what will be the speed of the ball when it reaches the top of its circular path about the peg? |
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Answer» Correct Answer - `[sqrt(2gL),sqrt(gL)]` |
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| 1545. |
The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere? |
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Answer» Rocket Total Energy (T.E.)=Potential energy (P.E)+Kinetic energy (K.E) =mgh+1/2mv2 The reduction in the rocket’s mass causes a drop in the total energy. Therefore, the heat energy required for the burning is obtained from the rocket. |
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| 1546. |
Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why? |
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Answer» Gravitational force is a conservative force. Since the work done by a conservative force over a closed path is zero, the work done by the gravitational force over every complete orbit of a comet is zero. |
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| 1547. |
An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth? |
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Answer» When an artificial satellite, orbiting around earth, moves closer to earth, its potential energy decreases because of the reduction in the height. Since the total energy of the system remains constant, the reduction in P.E. results in an increase in K.E. Hence, the velocity of the satellite increases. However, due to atmospheric friction, the total energy of the satellite decreases by a small amount. |
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| 1548. |
A pump is used to pump a liquid of density `rho` continuously through a pipe of cross, sectional area A. If liquid is flowing with speed v, then average power of pump isA. `1/3rhoAV^(2)`B. `1/2rhoAV^(2)`C. `2rhoAV^(2)`D. `1/2rhoAV^(3)` |
| Answer» Correct Answer - D | |
| 1549. |
A particle is placed at the origin and a force F=Kx is acting on it (where k is a positive constant). If `U_((0))=0`, the graph of `U (x)` verses x will be (where U is the potential energy function.)A. B. C. D. |
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Answer» Correct Answer - A `F=kx` `U=-int_(0)^(x)Fdx=-int_(0)^(x)kxdx=-(1)/(2)kx^2` U v/s x graph will be a parabola, open downward |
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| 1550. |
A particle is placed at the origin and a force F=Kx is acting on it (where k is a positive constant). If `U_((0))=0`, the graph of `U (x)` verses x will be (where U is the potential energy function.)A. B. C. D. |
| Answer» Correct Answer - B | |