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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.
| 1. |
A body is allowed to slide down a frictionless track from rest position at its top under gravity. The track ends in a circular loop of diameter `D`. Then, the minimum height of the inclined track (in terms of `D`) so that it may complete successfully the loop isA. `7D//4`B. `9D//4`C. `5D//4`D. `3D//4` |
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Answer» Correct Answer - C Loss `P.E` = gain in `K.E , mgh = (1)/(2) m[sqrt(5 gr)]^(2)`. |
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| 2. |
A body of mass `2 kg` is projected with an initial velocity of `5 ms^(-1)` along a rough horizontal table. The work done on the body by the frictional forces before it is brought to rest isA. 250 JB. 25 JC. `-250 J`D. `-25 J` |
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Answer» Correct Answer - D `W_(f) = ((1)/(2) mv_(f)^(2) - (1)/(2) m v_(i)^(2))`. |
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| 3. |
A particle is projected along a horizontal field whose coefficient of friction varies as `mu=A//r^2`, where r is the distance from the origin in meters and A is a positive constant. The initial distance of the particle is `1m` from the origin and its velocity is radially outwards. The minimum initial velocity at this point so the particle never stops isA. `infty`B. `2 sqrt(gA)`C. `sqrt(2 gA)`D. `4 sqrt(gA)` |
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Answer» Correct Answer - C If the particle never stops than it may move till `x = infty` `(1)/(2) mv^(2) = underset(1) overset(infty) int mumgd.dx rArr (V^(2))/(2) = Ag underset(1) overset(infty) int (1)/(x^(2)) dx` so, `V = sqrt(2gA)`. |
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| 4. |
A particle of mass `m` is projected at an angle `alpha` to the horizontal with an initial velocity `u`. The work done by gravity during the time it reaches its highest point isA. `u^(2) sin^(2) alpha`B. `(m u^(2) cos^(2) alpha)/(2)`C. `(m u^(2) sin^(2) alpha)/(2)`D. ` - (m u^(2) sin^(2) alpha)/(2)` |
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Answer» Correct Answer - D Velocity at top `= u cos alpha` `W_(m) = DeltaKE = (1)/(2) m(u cos alpha)^(2) -(1)/(2) m u^(2) = -(m u^(2) sin^(2) alpha)/(2)`. |
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| 5. |
A car, moving with a speed of `50 km//hr`, can be stopped by brakes after at least `6 m`. If the same car is moving at a speed of `100 km//hr`, the minimum stopping distance isA. 6 mB. 2 mC. 18 mD. 24 m |
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Answer» Correct Answer - D Using work - energy theorem `W_("total") = Delta K` i.e.., work done by stopping forces should be equal to change in the kinetic energy of car. `-F_("stopping") xx "stopping distance" = (k_("final") - k_("initial"))` `rArr -F_("stopping") xx s=(0 -(1)/(2)mv_("initial")^(2))` `rArr F_("stopping") xx s = (1)/(2) mv_("inital")^(2)` As stopping force is same for both the cars, hence `s prop v_("initial")^(2)` `rArr (s_(1))/(s_(2)) = ((v_("initial")^(2))_(1))/((v_("Initial")^(2))_(2))` or `(6)/(s_(2))=((v_("initial")^(2))_(1))/((v_("initial")^(2))_(2)) = ((50)/(100))^(2) rArr s_(2) = 24 m`. |
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| 6. |
A particle is projected at `t = 0` from a point on the ground with certain velocity at an angle with the horizontal. The power gravitational force is plotted against time. Which of the following is the best representation ?A. B. C. D. |
| Answer» Correct Answer - C | |
| 7. |
A body is moved along a straight line by a machine delivering constant power . The distance moved by the body is time `t` is proptional toA. `t^(1//2)`B. `t^(3//4)`C. `t^(3//2)`D. `t^(1//4)` |
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Answer» Correct Answer - C Here, power `P = F.v = (m (dv)/(dt)).v` `rArr vdv = (P)/(m) dt` Noe integrating, `int_(0)^(v) vdv = (P)/(m) int_(0)^(t) dt` `rArr (v^(2))/(2) = (Pt)/(m)` or `v = ((2Pt)/(m))^((1)/(2))` `rArr (dx)/(dt) = ((2P)/(m))^((1)/(2)) t^((1)/(2))` `rArr dx =((2P)/(m))^((1)/(2))t^((1)/(2)) dt` Integrating, we get `underset(0) overset(x) int dx = ((2P)/(m))^((1)/(2)) underset(0) overset(t) int t^((1)/(2)) dt` or `x = ((2P)/(m))^((1)/(2)) (t^((3)/(2)))/((3)/(2))` or `x prop t^((3)/(2))`. |
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| 8. |
A ball is thrown vertically upwards from the ground. Work done by air resistance during its time of flight isA. positive during ascent and negative during descentB. positive during ascent and descentC. negative during ascent and positive during descentD. negative during ascent and descent |
| Answer» Correct Answer - D | |
| 9. |
A particle of mass `100 g` is thrown vertically upwards with a speed of `5 m//s`. The work done by the force of gravity during the time the particle goes up isA. `-0.5 J`B. `-1.25 J`C. `1.25 J`D. `0.5 J` |
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Answer» Correct Answer - B `h = (u^(2))/(2g)` , workdone by gravity `= FS cos theta` here `F = mg, S = h, theta = 180^@`. |
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| 10. |
In the pulley-block system shown in figure, strings are light. Pulleys are massless and smooth. System is released from rest. In `0.3` seconds. . (a) work done on `2 kg` block by gravity is `6 J` (b) work done one `2 kg` block by string is `-2 J` ( c) work done on `1 kg` block by gravity is `-1.5 J` (d) work done on `1 kg` block string is `2 J`.A. only `a,d` are correctB. only `b,d` are correctC. only `a,b,c` are correctD. All are correct |
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Answer» Correct Answer - D `a = ((2m_(1)-m_(2)))/((2m_(1)+(m_(2))/(2)))g, T = ((3m_(1)m_(2))/(4m_(1)+m_(2)))g` On `2 kg` block, `W_(1) = m_(1)gx = m_(1)g ((1)/(2) at^(2))` , `W_(T) = T xx x = T ((1)/(2) at^(2))` On `1 kg` block, `W_(2) = m_(2) g "(x)/(2)`. |
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| 11. |
A block of mass `m = 1 kg` is released from point `A` along a smooth track as shown. Part `AB` is circular with radius `r_(1) = 4 m` and circular at `C` with radius `r_(2)`. Height of point `A` is `h_(1) = 2m` and of `c` is `h_(2) = 1m (g = 10 m//s^(2))`. . The work done by gravitational force from `A` to `C` isA. 10 JB. 20 JC. 30 JD. 40 J |
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Answer» Correct Answer - A `W_(mg) =mg(h_(1)-h_(2)) = Delta U`. |
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| 12. |
A block of mass `m = 1 kg` is released from point `A` along a smooth track as shown. Part `AB` is circular with radius `r_(1) = 4 m` and circular at `C` with radius `r_(2)`. Height of point `A` is `h_(1) = 2m` and of `c` is `h_(2) = 1m (g = 10 m//s^(2))`. . The force exerted by block on the track at `B` isA. 10 NB. 20 NC. 30 ND. 40 N |
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Answer» Correct Answer - B `v_(2) = sqrt(2gh_(1)) , N = mg + (mv_(2)^(2))/(r_(1))`. |
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| 13. |
A block of mass `m = 1 kg` is released from point `A` along a smooth track as shown. Part `AB` is circular with radius `r_(1) = 4 m` and circular at `C` with radius `r_(2)`. Height of point `A` is `h_(1) = 2m` and of `c` is `h_(2) = 1m (g = 10 m//s^(2))`. . The minimum safe value of `r_(2)` so that the block does not fly off the track at `C` isA. 1 mB. 2 mC. `1.5 m`D. 3 m |
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Answer» Correct Answer - B At `C N ge 0 (mv_(c)^(2))/(r_(2)) =mg` `mgh_(1)=(1)/(2) mv_(c)^(2)+mgh_(2) , r_(2)=2(h_(1)-h_(2))`. |
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| 14. |
A section of fixed smooth circular track of radius `R` in vertical plane is shown in the figure. A block is released from position `A` and leaves the track at `B` The radius of curvature of its trajectory just after it leaves the track `B` is ? A. RB. `( R)/(4)`C. `( R)/(2)`D. none of these |
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Answer» Correct Answer - C Apply conservation of energy with `O` as reference level `mg (2r)/(5) +0 = (mgr)/(5)+(1)/(2)mv^(2)` `v = sqrt((2gr)/(5))` , Radius of curvature `= (v_(1)^(2))/(a_(n))` `(2gr//5)/(g cos37) = (r)/(2)`. |
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| 15. |
A block is freely sliding down from a vertical height `4 m` on smooth inclined plane. The block reaches bottom of inclined plane and then it decribes vertical circle of radius `1 m` along smooth track. The ratio of normal reactions on the block while it crossing lowest point and highest point of vertical circle isA. `6 : 1`B. `5 : 1`C. `3 : 1`D. `5 : 2` |
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Answer» Correct Answer - C Let `v_(2),v_(1)` be the velocities at lowest point, highest point of vertical circle. `v_(2)^(2) =2gh = 2g xx 4 = 8g` `v_(2)^(2) -v_(1)^(2) =4gr =4g xx 1=4g` `:. v_(1)^(2) = 4g` normal reaction at lowest point, `R_(2) = (mv_(2)^(2))/( r) + mg`, At highest point, `R_(1) =(mv_(1)^(2))/(r) -mg`. |
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| 16. |
The work done by a force `overlineF = 2 overline i- overline j-overline k` in moving an object from origin to a point whose position vector is `overline r = 3 overline i + 2 overline j - 5 overline k`.A. 1 unitB. 9 unitsC. 13 unitsD. 60 units |
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Answer» Correct Answer - B `W= vec F. vecS = vec F.(vecr_(2) - vec r_(1))`. |
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| 17. |
A pilot of mass `m` can bear a maximum apparent weight `7` times of `mg`. The aeroplane is moving in a vertical circle. If the velocity of aeroplane is `210 m//s` while diving up from the lowest point of vertical circle, then the minimum radius of vertical circle should beA. 375 mB. 420 mC. 750 mD. 840 m |
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Answer» Correct Answer - C At lowest point of vertical circle, `T_(max)=(mv^(2))/(r_(min)) + mg`. |
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| 18. |
A crane can lift up `10,000 kg` of coal in `1` hour form a mine of `180 m` depth. If the efficiency of the crane is `80 %`, its input power must be `(g = 10 ms^(-2))`.A. 5 kWB. 6.25 kWC. 50 kWD. 62.5 kW |
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Answer» Correct Answer - B `eta = (P_("out"))/(P_("in"))`, where `P_("out") = (W)/(t) = (mgh)/(t)`. |
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| 19. |
Calculate the power of a crane in watts, which lifts a mass of `100 kg` to a height of `10 m` in `20 s`.A. 590 wB. 480 wC. 490 wD. 600 w |
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Answer» Correct Answer - C `{:("Given",,"mass = m = 100",,),("kg",,,,),(,,"height = h = 10 m",,),("time duration t = 20 s",,,,),(,,"power =Rate of",,),("work done",,,,):}` =`("change of PE")/("time") = ("mgh")/(t)` `=(100 xx 9.8 xx 10)/(20)` `=5 xx 98 = 490 W`. |
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| 20. |
The potential energy `U` in joule of a particle of mass `1 kg` moving in `x-y` plane obeys the law`U = 3x + 4y`, where `(x,y)` are the co-ordinates of the particle in metre. If the particle is at rest at `(6,4)` at time `t = 0` then :A. the particle has constant accelerationB. the particle has zero accelerationC. the speed of particle when it crosses the y-axis is `10 m//s`.D. co-ordinates of particle at `t = 1 s` are `(4.5,2)`. |
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Answer» Correct Answer - A::C::D `vec F -(du)/(dx) hat i -(du)/(dy) hat j , = -3 hati - 4 hat j` `vec a = -a hat i - 4 hatj ,|vec a|=5` on `y` axis `x = 0`, after time `t` sec. the particle crosses `y` axis. `0-6=1//2 xx(-3_t)t^(2),t=2.v_(x)=(-3)xx2 =-6` `v_(y)=(-4)xx 2=-8.|vec v|= 10 m//s`. `t = 1 sec., x = 6 - 1//2 x 3 = 4.5` `y = 4- 1//2 x 4 = 2`. |
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| 21. |
The kinetic energy of a particle continuously icreses with timeA. the resultant force on the particle must be parallel to the velocity at all instants.B. the resultant force on the particle must be at an angle less than `90^(@)` with the velocity all the timeC. its height above the ground level must continuously decreasesD. the magnitude of its linear momentum is decreasing continuously |
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Answer» Correct Answer - B `dW_(F)=vec(F).vec(ds)=dk gt 0rArr |vec(F)||vec(ds)|cos theta gt 0` `rArr 0 lt theta lt 90^(@)` `p=sqrt(2m(KE)), KE uarr ` so `p uarr`. |
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| 22. |
A body is falling under gravity . When it loses a gravitational potential energy `U`, its speed is`v`. The mass of the body shell beA. `2U//v^(2)`B. `2v//U^(2)`C. `2v//U`D. `U^(2)//2v` |
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Answer» Correct Answer - A `U_(i)+0=U_(f)+1/2 mv^(2)` `U_(i)-U_(f)=1/2 mv^(2)` `U=1/2 mv^(2)` `m=(2U)/(v^(2))` |
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| 23. |
A force `vec F = (5 vec i+ 3 vec j+ 2 vec k) N` is applied over a particle which displaces it from its origin to the point `vec r=(2 vec i- vec j) m`. The work done on the particle in joules is.A. `+ 10`B. `+ 7`C. `-7`D. `+ 13` |
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Answer» Correct Answer - B Here, `W = vec F. vec S` =`(5 hat i + 3 hat j + 3hat k).(2 hat i - hat j) = 10 -3 = 7 J`. |
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| 24. |
The work done by a force `vec F = 3 hat i -4hat j +5hat k` displaces the body from a point `(3,4,6)` to a point `(7,2,5)` isA. 15 unitsB. 25 unitsC. 20 unitsD. 10 units |
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Answer» Correct Answer - A `W = vec F. vec S = vec F.(vec r_(2) - vec r_(1))` |
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| 25. |
Two forces each of magnitude `10 N` act simultaneously on a body with their directions inclined to each other at an angle of `120^@`and displaces the body over `10 m` along the bisector of the angle between the two forces. Then the work done by force isA. 5 JB. 1 JC. 50 JD. 100 J |
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Answer» Correct Answer - C `W = FS = cos theta`. |
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| 26. |
A uniform chain of mass m & length L is kept on a smooth horizontal table such that `(L)/(n) `portion of the chaing hangs from the table. The work dione required to slowly bringsthe chain completely on the table isA. `mgL//16`B. `mgL//32`C. `3mgL//32`D. `mgL//8` |
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Answer» Correct Answer - B `W = (mgl)/(2n^(2))`. |
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| 27. |
An ideal spring with spring - constant `K` is bung from the colling and `a` block of mass `M` is attached to its lower end the mass is released with the spring initally unstetched . Then the maximum exlemsion in the spring isA. `(4 Mg)/(k)`B. `(2Mg)/(k)`C. `(Mg)/(k)`D. `(Mg)/(2k)` |
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Answer» Correct Answer - B Loss in `P.E` = Gain in `K.E + P.E` stored in spring `MgX_(max) =0+(1)/(2)kX_(max)^(2) , X_(max)=(2Mg)/(k)`. |
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| 28. |
A cyclist comes to a skidding stop in `10m`. During this process, the force on the cycle due to the road is `200N` and is directly opposite to the motion. a. How much work does the road do on the cycle? b. How much work does the cycle do on the road?A. `+ 200 J`B. ` - 200 J`C. zeroD. `- 20,000 J` |
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Answer» Correct Answer - C Here, work is done by the frictional force on the cycle and is equal to `200 xx 10 = - 2000 J` As the road is not moving, hence, work done by the cycle on the road = zero. |
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| 29. |
A particle, which is constrained to move along the x-axis, is subjected to a force from the origin as `F(x) = -kx + ax^(3)`. Here `k` and a are origin as `F(x) = -kx + ax^(3)`. Here `k` and a are positive constants. For `x=0`, the functional form of the potential energy `U(x)` of particle is.A. B. C. D. |
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Answer» Correct Answer - D `F(x)=-kx +ax^(3) ,F_((x)) =0,` for `x = 0 x = sqrt((k)/(a)),` So slope is zero at `x = 0 x = sqrt((k)/(a))`. |
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| 30. |
A particle of mass in is moving in a circular with of constant radius `r` such that its contripetal accelenation `a_(c) ` is varying with time `t` as `a_(c) = K^(2) rt^(2)` where K` is a constant . The power delivered to the particles by the force action on it isA. zeroB. `mk^(2)r^(2) t^(2)`C. `mk^(2) r^(2) t`D. `mk^(2) rt` |
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Answer» Correct Answer - C `a_(c) = (v^(2))/(r) = k^(2) rt^(2)` or `v = krt` tangential acceleration is , `a_(t) = (dv)/(dt) = kr` hence `:. P = F_(t) v cos theta = ma_(t) v cos theta`. |
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| 31. |
Power supplied to a mass `2 kg` varies with time as `P = (3t^(2))/(2)` watt. Here `t` is in second . If velocity of particle at `t = 0 is v = 0`, the velocity of particle at time `t = 2s` will be:A. 1 m/sB. 4m/sC. 2m/sD. `sqrt(2) m//s` |
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Answer» Correct Answer - C Applying work energy theorem `K_(f)=K_(i)`+work done `1/2 mv^(2)=0+int_(0)^(2) Pdt=int_(0)^(2)(3t^(2))/2 dt` Solving we get v=2m/s. |
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| 32. |
Power supplied to a mass `2 kg` varies with time as `P = (3t^(2))/(2)` watt. Here `t` is in second . If velocity of particle at `t = 0 is v = 0`, the velocity of particle at time `t = 2s` will be:A. `1 m//s`B. `4 m//s`C. `2 m//s`D. `2 sqrt(2) m//s` |
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Answer» Correct Answer - C `Delta KE = W_("net")` or `K_(f) -K_(i) = int Pdt`. |
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| 33. |
Select the correct alternatives :A. Work done by static friction is always zeroB. Work done by kinetic friction can be positive alsoC. Kinetic enery of a system can not be increased without applying any external force on the systemD. Work energy theoram is valid in non-inerial frames also. |
| Answer» Correct Answer - B::D | |
| 34. |
A simple pendulum is oscillating with an angular amplitude `60^@`. If mass of bob is `50` gram, then the tension in the string at mean position is `(g = 10 ms^(-2))`A. 0.5 NB. 1 NC. 1.5 ND. 2 N |
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Answer» Correct Answer - B `T =mg+(mv^(2))/(r) =mg+(m)/(l)[2gl(1-cos theta)]`. |
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| 35. |
In the system shown in the figure the mass `m` moves in a circular arc of angular amplitude `60^@`. Mass `4 m` is stationary. Then : .A. the minimum value of coefficient of friction between the same of mass `4 m` and the surface of the table is `0.50`B. the work done by gravitational force in the block `m` is positive when it moves from `A` to `B`C. the power delovered by the tension when `m` moves from `A` to `B` is zeroD. The kinetic energy of `m` in position `B` equals the work done by gravitational force on the block when its from position `A` to `B`. |
| Answer» Correct Answer - A::B::C::D | |
| 36. |
A man carries a load of `50 kg` through a height of `40 m` in `25` seconds. If the power of the man is `1568 W`, his mass isA. 5 kgB. 1000 kgC. 200 kgD. 50 kg |
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Answer» Correct Answer - D `P =(W)/(t) = ((m+M)gh)/(t)`. |
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| 37. |
During inelastic collision between two bodies, which of the following quantities always remain conserved ?A. Total kinetic energy.B. Total mechanical energy.C. Total linear momentum.D. Speed of each body. |
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Answer» Correct Answer - C When we are considering the two bodies as system external force on the system will be zero. Hence, total linear momentum of the system remain conserved. |
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| 38. |
A block of mass `1 kg` moves towards a spring of force constant `10 N//m`. The spring is massless and unstretched. The coeffcient of friction between block and surface is `0.30`. After compressing the spring, block does not return back : `(g = 10 m//s)`.A. the maximum value of speed of block for which it is possible is `3.8 m//s`B. the maximum value of speed of block of which it is possible is `4.2 m//s`C. if `E_(i)` and `E_(f)` are initial and final mechanical energy, which is sum of kinetic energy and potential energy, than work done by friction on a system is `(E_(i) - E_(f))`.D. statement in option `( C)` is wrong. |
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Answer» Correct Answer - A::C The spring is compressed by `x` Block will not return if `mu mg ge Kx` So, `x_(max) = (mu mg)/(K) =((0.3)(1)(10))/(10) =0.30 m` work done against friction `= E_(i) - E_(f)` `mu mg(x+2)=(1)/(2) mv_(0)^(2) -(1)/(2) Kx^(2)` `(0.3)(1)(10)(0.3+2)=((1)/(2))(1) v_(0)^(2) -((1)/(2))(10)(0.3)^(2)` on solving, `v_(0) = 3.8 m//sec`. |
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| 39. |
A raindrop falling from a height `h` above ground, attains a near terminal velocity when it has fallen through a height `(3//4)h`. Which of the diagrams shown in figure correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground ?A. .B. .C. .D. . |
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Answer» Correct Answer - B When drop falls first velocity increases, hence, first `KE` also increases. After sometime speed (velocity) is constant this is called terminal velocity, hence, `KE` also become constant. `PE` decreases continuosly as the drop is falling continuosly. The variation in `PE` and `KE` is best represented by `(b)`. |
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| 40. |
A mass of `5kg` is moving along a circular path or radius `1m`. If the mass moves with 300 revolutions per minute, its kinetic energy would beA. `250 p^(2)`B. `100 p^(2)`C. `5 p^(2)`D. 0 |
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Answer» Correct Answer - A Given, mass `= m = 5kg` Radius `= 1 m = R` Revolution per minute `omega =300 rev//min` =`(300 xx 2pi)rad//min` =`(300 xx 2xx3.14) rad//60s` =`(300 xx 2 xx 3.14)/(60) rad//s = 10 pi rad//s` `rArr "Linear speed" = v = omega R` `=((300 xx 2 pi)/(60)) (1)` `=10 pi m//s` `KE = (1)/(2) mv^(2)` `=(1)/(2) xx 5 xx(10 pi)^(2)` `=100 pi^(2) xx 5 xx (1)/(2)` =`250 pi^(2) J`. |
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| 41. |
A body of mass `5 kg` is moved up over `10 m` along the line of greatest slope of a smooth inclined plane of inclination `30^@` with the horizontal. If `g = 10 m//s^(2)`, the work done will beA. 500 JB. 2500 JC. 250 JD. 25 J |
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Answer» Correct Answer - C `W = vec F. vec S = FS cos theta` , here `F = mg sin alpha` and `theta = 0^@, alpha = 30^@`. |
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| 42. |
A particle of mass m moving along a straight line experiences force F which varies with the distance x travelled as shown in the figure. If the velocity of the particle at `x_(0)sqrt((2F_(0)x_(0))/m)`, then velocity at `4x_(0)` is A. `2sqrt((2F_(0)x_(0))/m)`B. `2sqrt((F_(0)x_(0))/m)`C. `sqrt((F_(0)x_(0))/m)`D. none of these |
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Answer» Correct Answer - D Increases in KE =work done `1/2 mv_(2)^(2)-1/2 mx((2F_(0)x_(0))/m)` `=1/2(2F_(0)+F_(0))3x_(0)` `rArr v_(2)=sqrt((11F_(0)x_(0))/m)` |
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| 43. |
A particle `P` is initially at rest on the top `pf`a smooth hemispherical surface which is fixed on a horizontal plane. The particle is given a velocity `u` horizontally. Radius of spherical surface is `a`. .A. If the particle leaves the sphere, when it has fallen vertically by a distance of `(a)/(4)m u = (sqrt(ga))/(2)`.B. If the particle leaves the sphere at angle `theta` (fig) where `cos theta = (sqrt(3))/(2)`, then `u = (sqrt(ag))/(3)`C. If `u=0` and the particle just slides down the hemispherical surface, it will leave the surface when `cos theta = (2)/(3)`.D. The minimum value of `u`, for the object to leave the sphere without sliding over the surface is `sqrt(ag)`. |
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Answer» Correct Answer - A::C::D Applying conservation of total energy `(1)/(2) m u^(2) + mga (1 -c os theta) = (1)/(2) mv^(2)` `mg cos theta -N =(mv^(2))/(a)` for particle to lose contact `N = 0` `v^(2) =ag cos theta , u^(2)+ga (2-3 cos theta) = 0`. |
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| 44. |
A hemispherical vessel of radius `R` moving with a constant velocity `v_(0)` and containing a ball, is suddenly haulted. Find the height by which ball will rise in the vessel, provided the surface is smooth :A. `(v_(0)^(2))/(2g)`B. `(2 v_(0)^(2))/(g)`C. `(v_(0)^(2))/(g)`D. None of these |
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Answer» Correct Answer - A `(1)/(2)mv_(0)^(2) =mgR(1-cos theta)` `(v_(0)^(2))/(2g) =R-R cos theta = "required height"`. |
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| 45. |
A chain of length `l = pi R//4` is placed. On a smooth hemispherical surface of radius `R` with one of its ends fixed at the top of the sphere. Mass of chain is `sqrt(pi kg)` and `R = 1m.(g = 10 m//s^(2))`. The tangential acceleration of the chain when its starts sliding down.A. `(40)/(pi)(1-(1)/(sqrt(2)))`B. `(20)/(pi)(1-(1)/(sqrt(2)))`C. `10(1-(1)/(sqrt(2)))`D. zero |
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Answer» Correct Answer - A Tangential force on differential element `dF_(t) = dmg sin theta` `F_(t)=int_(0)^(pi//4) ((m)/(l) R d theta)g sin theta = (gR)/(l)[1 -cos 45^@)`. |
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| 46. |
The potential energy of a partical veries with distance `x` as shown in the graph. The force acting on the partical is zero atA. CB. BC. B and CD. A and D |
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Answer» Correct Answer - C `F=-(dU)/(dx)=0` at B and C |
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| 47. |
The potential energy of a force filed `vec(F)` is given by `U(x,y)=sin(x+y)`. The force acting on the particle of mass m at `(0,(pi)/4)` isA. 1B. `sqrt(2)`C. `1/(sqrt(2))`D. 0 |
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Answer» Correct Answer - A `(delU)/(dely)=cos(x+y).` `(delU)/(dely)=cos(x+y)` `vec(F)=-cos(x+y)hati-cos(x+y)hatj` `=-cos(0+(pi)/4)hati-cos(0+(pi)/4)hatj` `|vec(F)|=1` |
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| 48. |
A steel ball of radius `2 cm` is initially at rest. It is struck head on by another stell ball of radius `4 cm` travelling with a velocity of `81 cm//s`. If the collision is elastic their respective final velocities areA. `63 cm//s,144 cm//s`B. `144 cm//s, 63 cm//s`C. `19 cm//s, 100 cm//s`D. `100 cm//s, 19 cm//s` |
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Answer» Correct Answer - B As mass, `m = rho(volume) = rho (4)/(3) pi r^(3)` `m alpha r^(3)` so, `v_(1)=((r_(1)^(3) -r_(2)^(3))/(r_(1)^(3)+r_(2)^(3))) u_(1)` and `v_(2) = (2 r_(1)^(3) u_(1))/(r_(1)^(3) + r_(2)^(3))`. |
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| 49. |
A body starts from rest and is acted on by a constant force. The ratio of kinetic energy gained by it in the first five seconds to that gained in the next five seconds isA. `2 : 1`B. `1 : 1`C. `3 : 1`D. `1 : 3` |
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Answer» Correct Answer - D `KE = (1)/(2) mv^(2) = (1)/(2) m(g t)^(2) (because v = g t)` `(K_(1))/(K_(o)) = (t_(1)^(2))/(t_(2)^(2) -t_(1)^(2))` where `t_(1) = 5 sec` and `t_(2) = 10 sec`. |
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| 50. |
Two stones of masses `m` and `2 m` are projected vertically upwards so as to reach the same height. The ratio of the kinetic energies of their projection isA. `2 : 1`B. `1 : 2`C. `4 : 1`D. `1 : 4` |
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Answer» Correct Answer - B `KE = (1)/(2) mv^(2)`, when two bodies reach the same height, `v_(1) = v_(2) , (KE_(1))/(KE_(2)) = (m_(1))/(m_(2)) (because v = sqrt(2 gh))`. |
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