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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.
| 201. |
A ballet dancer is rotating about his own vertical axis on smooth horizontal floor. `I, omega, L, E` are moment of inertia, angular velocity, angular momentum, rotational kinetic energy of ballet dancer respectively. If ballet dancer stretches himself away from his axis of rotation, thenA. `I` increases and `omega,E` decrease but `L` is constantB. `I` decreases, `omega` and `E` increases but `L` is constantC. `I` increase, `omega` decreases, `L` and `E` are constantD. `I` increases, `omega` increases but `L` and `E` are constant |
| Answer» Correct Answer - A | |
| 202. |
A circular wheel is rotating in horizontal plane without friction about its axis. If a body is gently attached to the rim of the wheel then following is false.A. Moment of inertia increases but angular momentum remains sameB. angular velocity decreases but angular momentum remains sameC. Rotational kinetic energy decreases but angular momentum remains sameD. Angular momentum increases but angular velocity remins same |
| Answer» Correct Answer - D | |
| 203. |
A circular disc is rotating about its own axis, the direction of its angular momentum isA. radialB. along axis of rotationC. along tangentD. perpendicular to the direction of angular velocity |
| Answer» Correct Answer - B | |
| 204. |
A uniform circular disc of radius `R` is rotating about its own axis with moment of inertia `I` at an angular velocity `omega` if a denser particle of mass `m` is gently attached to the rim of disc than its angular velocity isA. `omega`B. `Iomega(I+mR)`C. `(I+mR^(2))/(Iomega)`D. `(Iomega)/(I+mR^(2))` |
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Answer» Correct Answer - D `I_(1)omega_(1)=I_(2)omega_(2), I_(1)=I, I_(2)=I+mR^(2)` |
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| 205. |
A circular disc is rotating in horizontal plane about vertical axis passing through its centre without friction with a person standing on the disc at its edge. If the person gently walks to centre of disc then its angular velocityA. increasesB. decreasesC. does not changeD. becomes zero |
| Answer» Correct Answer - A | |
| 206. |
A circular disc is rotating about its own axis at a uniform angular velocity `omega`. The disc is subjected to uniform angular retardation by which its angular velocity is decreased to `omega/2` during `120` rotations. The number of rotations further made by it before coming to rest isA. `120`B. `60`C. `40`D. `20` |
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Answer» Correct Answer - C `alpha` is constant `alpha=(omega_(1)^(2)-omega_(2)^(2))/(2 theta) , theta_(2)=(theta_(1))/3` |
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| 207. |
A circular disc is rotating about its own axis at a uniform angular velocity `omega`. The disc is subjected to uniform angular retardation by which its angular velocity is decreased to `omega/2` during `120` rotations. The number of rotations further made by it before coming to rest isA. `120`B. `60`C. `40`D. `20S` |
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Answer» Correct Answer - C `alpha` is constant `alpha=(omega_(1)^(2)-omega_(2)^(2))/(2 theta)` `(omega-((omega)/2)^(2)-0)/(2 theta_(2)), ._(2)=(._(1))/3` |
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| 208. |
A circular disc is rotating without friction about its natural axis with an angular velocity `omega`. Another circular disc of same material and thickness but half the raduis is gently placed over it coaxially. The angular velocity of composite disc will beA. `(4omega)/3`B. `(8 omega)/9`C. `(7 omega)/8`D. `(16 omega)/17` |
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Answer» Correct Answer - D `I_(1)omega_(1)=(I_(1)+I_(2))omega_(2)` |
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| 209. |
A solid cylinder is rolling down the inclined plane without slipping. Which of the following is `//` are correctA. The friction force is dissipativeB. The displacement of the disc until rolling beginsC. the velocity when rolling beginsD. the work done by the force of friction |
| Answer» Correct Answer - C::D | |
| 210. |
A wheel of radius `0.2 m` rolls without slip ping with a speed `10 m//s` on ahorizontal road. When it is at a point `A` on the road, a small lump of mud separates from the wheel at its highest point `B` and drops at point `C` on the ground. The distance `AC` is A. `10/7`B. `20/7`C. `30/7`D. `40/7` |
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Answer» Correct Answer - D `R=2vxxT=2vsqrt((2h)/g)` and `h=2R` |
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| 211. |
A solid cylinder of mass `m` rolls without slipping down an inclined plane making an angle `theta` with the horizontal. The frictional force between the cylinder and the incline isA. `mg sintheta`B. `(mg sin theta)/3`C. `mg cos theta`D. `(2 mg sin theta)/3` |
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Answer» Correct Answer - B `f=mgsin theta((k^(2))/(k^(2)+R^(2)))` |
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| 212. |
A wheel of radius `R` rolls without slipping on the horizontal surface with speed `v_(0)`. When the contact point is `P` on the road, a small patch of mud separates from the wheel at its highest point strikes the road at point `Q`. Find distance `PQ`.A. `vsqrt(r/g)`B. `2vsqrt(r/g)`C. `4vsqrt(r/g)`D. `vsqrt((3r)/g)` |
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Answer» Correct Answer - C `R=2vxxT=2vsqrt((2h)/g)` |
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| 213. |
A disc of mass `m` is connected with an ideal spring of stiffness `k`. If it is released from rest, it rolls without sliding on an inclination plane. The maximum elongation of the spring is: A. `(mg sin theta)/k`B. `(2mg sin theta)/(3k)`C. `(3mg sin theta)/k`D. `(2 mg sin theta)/k` |
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Answer» Correct Answer - D We use work enregy principale `mgsin thetaxxx-1/2kx^(2)=0 rArrx=(2mg sin theta)/k` |
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| 214. |
A thin metal disc of radius `0.25m` and mass `2kg` starts from rest and rolls down an inclined plane. If its rotational kinetic energy is `4J` at the foot of the inclined plane, then its linear velocity at the same point isA. `1.2 ms^(-1)`B. `2.8ms^(-1)`C. `20 ms^(-1)`D. `2 ms^(-1)` |
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Answer» Correct Answer - B `KE_(rot)=1/2Iomega^(2)=1/2((MR^(2))/2)omega^(2)=(mv^(2))/4` |
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| 215. |
A thin metal disc of radius `0.25m` and mass `2kg` starts from rest and rolls down an inclined plane. If its rotational kinetic energy is `8J` at the foot of the inclined plane, then its linear velocity of centre of mass of disc isA. `2 m//s`B. `4m//s`C. `6m//s`D. `8m//s` |
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Answer» Correct Answer - B `KE_(rot)=1/2Iomega^(2)=1/2((MR^(2))/2)omega^(2)` |
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| 216. |
A ring and a disc of same mass roll without slipping along a horizontal surface with same velocity. If the `K.E.` of ring is `8J`, then that of disc isA. `2J`B. `4J`C. `6J`D. `16 J` |
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Answer» Correct Answer - C `KE=1/2mv^(2)(1+(k^(2))/(R^(2)))` |
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| 217. |
A massless thin hollow sphere is completely filled with water of mass `m`. If the hollow sphere rolls with a velocity `v`. The kinetic energy of the sphere of water is : (Assume water is non viscous) A. `1/2 mv^(2)`B. `1/3 mv^(2)`C. `7/10 mv^(2)`D. `5/6 mv^(2)` |
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Answer» Correct Answer - A Liquid gets only translatory motion in side any roll mosum. This `kE=1/2mv^(2)` |
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| 218. |
A sphere cannot roll onA. a smooth horizontal surfaceB. a smooth inclined surfaceC. a rough horizontal surfaceD. a rough inclined surface. |
| Answer» Correct Answer - B | |
| 219. |
A solid sphere and a spherical shell roll down an incline from rest from same height. The ratio of times taken by them isA. `sqrt(21/25)`B. `21/25`C. `sqrt(25/21)`D. `25/21` |
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Answer» Correct Answer - A `t=(sqrt(2l(1+(k^(2))/(R^(2)))))/(g sin theta)` |
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| 220. |
A thin hollow sphere of mass `m` is completely filled with non viscous liquid of mass `m`. When the sphere roll-on horizontal ground such that centre moves with velocity `v`, kinetic energy of the system is equal to |
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Answer» Total enregy `=KE`+rotational `KE` `=1/2(2m)v^(2)+1/2(2/3mr^(2))omega^(2)` `=1/2(2m)v^(2)+1/3mv^(2)=4/3mv^(2)` |
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| 221. |
Four particles each of mass `1kg` are at the four corners of square of side `1m`. The M.I. of the system about a normal axis through centre of square isA. `6 kgm^(2)`B. `2 kgm^(2)`C. `1.25 kgm^(2)`D. `2.5 kgm^(2)` |
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Answer» Correct Answer - B `I=summr^(2), r=l/(sqrt(2))` |
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| 222. |
Three particles each `1kg` mass are placed at the corners of a right angled triangle `AOB`, `O` being the origin of the co-ordinate system `OA` and `OB` along `+ve` x-direction and `+ve` y-direction. The position vector of the centre of mass is (`OA = OB=1m`) (in meters)A. `(i+j)/3`B. `(i-j)/3`C. `(2(i+j))/3`D. `(i-j)` |
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Answer» Correct Answer - A `bar(v)_(cm)=x_(cm)hati+y_(cm)hatj` |
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| 223. |
Four particles, each of mass `1kg` are placed at the corners of a square of side `1m` in the `XY` plane. If the point of intersecting of the diagonals of the square is taken as the origin, the co-ordinates fo the centre of mass areA. `(1,1)`B. `(-1,1)`C. `(1,-1)`D. `(0,0)` |
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Answer» Correct Answer - D `x_(cm)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2)), y_(cm)=(m_(1)y_(1)+m_(2)y_(2))/(m_(1)+m_(2))` |
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| 224. |
A small solid cylinder of radius `r` is released coaxially from point A inside the fixed large cylindrical bawl of radius R as shown in figure. If the friction between the small and the large cylinder is sufficient enough to prevent any slipping then find. (a). What fractions of the total energy are translational and rotational when the small cylinder reaches the bottom of the larger one? (b). The normal force exerted by the small cylinder on the larger one when it is at the bottom.A. `7/5`B. `2/7`C. `5/7`D. `7/10` |
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Answer» Correct Answer - B `(1/2Iomega^(2))/(1/2mv^(2)(1+(k^(2))/(R^(2))))` |
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| 225. |
A small block of mass `m` is released from rest from position `A` inside a smooth hemispherical bowl of radius `R` as shown in figure. Choose the wrong option (s) A. acceleration to block is constant throughoutB. acceleration of block is `g` at `A`C. acceleration of block is `3g` at `B`D. acceleration of block is `2g` at `B` |
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Answer» Correct Answer - A::C Acceleration of block is not constant throught. Acceleration of block at `B` is `V^(2)//R` where `V^(2)=2 gR`. |
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| 226. |
For a body rolling along a level surface, without slipping the translational and rotational kinetic energies are in the ratio `2:1`. The body isA. Hollow sphereB. solid cylinderC. ringD. solid sphere |
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Answer» Correct Answer - B `(1/2mV^(2))/(1/2Iomega^(2))=2/1` |
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| 227. |
Particle of masses `m, 2m,3m,…,nm` grams are placed on the same line at distance `l,2l,3l,…..,nl cm` from a fixed point. The distance of centre of mass of the particles from the fixed point in centimeters is :A. `((2n+1)l)/3`B. `l/(n+1)`C. `(n(n^(2)+l)l)/2`D. `(2l)/(n(n^(2)+l)l)` |
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Answer» Correct Answer - A `x_(cm)=(Summx)/(Summ)=((1^(2)+2^(2)+3^(2)+...+n^(2))l)/((1+2+3+...++n))` |
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| 228. |
A ceiling fan is rotating about its own axis with uniform angular velocity `omega`. The electric current is switched off then due to constant opposing torque is its angular velocity is reduced to `(2 omega)/3` as it completes `30` rotations. The number of rotations further it makes before coming to rest isA. `18`B. `12`C. `9`D. `24` |
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Answer» Correct Answer - D `alpha` is constant `theta=(omega_(1)^(2)-omega_(2)^(2))/(2alpha)` and `(theta_(1))/(theta_(2))=(omega^(2)-(2omega//3)^(2))/((2omega//3)^(2)-0)` |
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| 229. |
A uniform thin rod of length `1m` and mass `3kg` is attached to a uniform thin circular disc of radius `30cm` and mass `1kg` at its centre perpendicular to its plane. The centre of mass of the combination from the centre of disc isA. `0.375 m`B. `2.532m`C. `3.513m`D. `2.125 m` |
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Answer» Correct Answer - A `X_(cm)=(m_(2)d)/(m_(1)+m_(2))` |
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| 230. |
Particles of masses `1kg` and `3kg` are at `2i + 5j + 13k)m` and `(-6i + 4j - 2k)m` then instantaneous position of their centre of mass isA. `1/4(-16i+17j+7k)m`B. `1/4(-8i+17j+7k)m`C. `1/4(-6i+17j+7k)m`D. `1/4(-6i+17j+5k)m` |
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Answer» Correct Answer - A `vec(r)_(cm)=(m_(1)vec(r)_(1)+m_(2)vec(r)_(2))/(m_(1)+m_(2))` |
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| 231. |
The centre of mass of a non uniform rod of length L, whose mass per unit length varies as `rho=(k.x^2)/(L)` where k is a constant and x is the distance of any point from one end is (from the same end)A. `(3L)/4`B. `L/8`C. `K/L`D. `(3K)/L` |
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Answer» Correct Answer - A `dm=(kx^(2))/Ldx, x_(cm)=(int_(0)^(L)xdm)/(int_(0)^(L)(kx^(2))/Ldx)=(((L^(4))/(4))/((L^(3))/(3)))=(3L)/4` |
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| 232. |
A uniform bar of length `6a` and mass `8m` lies on a smooth horizontal table. Two point masses `m` and `2m` moving in the same horizontal plane with speeds `2v` and `v` respectively, strike the bar (as shown in figure) and stick to the bar after collision. Calculate (a) velocity of the centre of mass (b) angular velocity about centre of mass and (c) total kinetic energy, just after collision. |
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Answer» (a) As `F_(ext)=0` linear momentum of the system is conserved i.e., `-2mxxv+mxx2v+0=(2m+m+8m)xxV` or `V=0` i.e., velocity of centre of mass is zero. (b) As `tau_(ext)=0` angular momentum of the system is conserved. i.e, `m_(1)v_(1)r_(1)+m_(2)v_(2)r_(2)=(I_(1)+I_(2)+I_(3)) omega` `2mva+m(2v)(2a)=[2m(a)^(2)+8mxx(6a)^(2)//12]omega` ie. `6mva=30ma^(2) omegarArromega=(v/(5a))` (c) From (a) and (b) it is clear that, the system has no translatory motion but only rotatory motion. `E=1/2Iomega^(2)=1/2(30ma^(2))[v/(5a)]^(2)=3/5mv^(2)` |
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| 233. |
The centre of mass of the letter `F` which is cut from a uniform metal sheet from point `A` is A. `15//7,33//7`B. `15//7,23//7`C. `22//7,33//7`D. `33//7,22//7` |
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Answer» Correct Answer - A `m=sigmaA, X_(cm)=(Sigmam_(i)x_(i))/(Sigmam_(i)),Y_(cm)=(Sigmam_(i)y_(i))/(Sigmam_(i))` |
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| 234. |
A uniform sphere has radius `R`. A sphere of diameter `R` is cut from its edge as shown. Then the distance of centre of mass of remaining portion from the centre of mass of the original sphere is A. `R//7`B. `R//14`C. `2R//7`D. `R//18` |
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Answer» Correct Answer - B shift`=(r^(3)d)/(R^(3)-r^(3))` |
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| 235. |
A thin uniform rod of length `L` is bent at its mid point as shown in the figure. The distance of the centre of mass from the point `O` is A. `L/2 sin (theta/2)`B. `L/2 cos(theta/2)`C. `L/4 sin (theta/2)`D. `L/4 cos (theta/2)` |
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Answer» Correct Answer - D `x_(cm)=(-m/2(L//4)-m/2(cos theta)L//4)/m` `y_(cm)=(m/2sin theta(L//4))/m,r_(cm)=sqrt(x_(cm)^(2)+y_(cm)^(2))` |
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| 236. |
A thin rod of mass `6 m` and length `6L` is bent into regular hexagon. The `M.I.` of the hexagon about a normal axis to plane and through centre of system isA. `ml^(2)`B. `3mL^(2)`C. `5mL^(2)`D. `11 mL^(2)` |
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Answer» Correct Answer - C `I_(rod)=(ml^(2))/12+m((sqrt(3))/2l)^(2), I_("system")=6I_(rod)` |
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| 237. |
A uniform rod o length one meter is bent at its midpoint to make `90^@`. The distance of centre of mass from the centre of rod is (in `cm`)A. `20.2`B. `13.4`C. `15`D. `25.36` |
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Answer» Correct Answer - D `d=L/4 sin (theta/2)` |
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| 238. |
Four bodies of masses `1,2,3,4 kg` respectively are placed at the comers of a square of side `a`. Coordinates of centre of mass are (take `1kg` at the origin,`2kg` on X-axis and `4kg` on Y-axis)A. `((7a)/10,a/2)`B. `(a/2,(7a)/10)`C. `(a/2,(3a)/10)`D. `((7a)/10,(3a)/2)` |
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Answer» Correct Answer - B `x_(cm)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2)) , y_(cm)=(m_(1)y_(1)+m_(2)y_(2))/(m_(1)+m_(2))` |
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| 239. |
A shell in flight explodes into `n` equal fragments `k` of the fragments reach the ground earlier than the other fragments. The acceleration of their centre of mass subsequently will beA. `g`B. `(n-k)g`C. `((n-k)g)/k`D. `((n-k))/ng` |
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Answer» Correct Answer - D `a_(cm)=(summ_(i)a_(i))/(summ_(i))` |
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| 240. |
A body of mass `m` is dropped and another body of mass `M` is projected vertically up with speed `u` simultaneously from the top of a tower of height `H`. If the body reaches the highest piont before the dropped body reaches the ground, then maximum height raised by the centre of mass of the system from ground isA. `H+(u^(2))/(2g)`B. `(u^(2))/(2g)`C. `H+1/(2g)((Mu)/(m+M))^(2)`D. `H+1/(2g)((m u)/(m+M))^(2)` |
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Answer» Correct Answer - C `h_(max)=H+((u_(cm))^(2))/(2g)` |
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| 241. |
An energy of `484 J` is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. Calculate moment of inertia of flywheel.A. `1.6 gkm^(2)`B. `0.3 kgm^(2)`C. `0.7 kgm^(2)`D. `1.2 kgm^(2)` |
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Answer» Correct Answer - C `W=1/2I2pi(n_(2)^(2)-n_(1)^(2))` |
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| 242. |
A boat of mass `40kg` is at rest. A dog of mass `4kg` moves in the boat with a velocity of `10m//s`. What is the velocity of boat (nearly)?A. `4m//s`B. `2m//s`C. `8m//s`D. `1 m//s` |
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Answer» Correct Answer - D `v_(b)=(mxxv)/(m+M)` |
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| 243. |
A flywheel of mass `25 kg` has a radius of `0.2 m`. It is making 240 rpm. What is the torque necessary to bring it to rest in `20 s` ? If the torque is due to a force applies tangentially on the rim of the wheel, what is the magnitude of the force ? Assume that mass of flywheel is concentrated at its rim. |
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Answer» `alpha=(2pin)/t=(2pixx4)/20` Torque `tau=Ialpha=(MR^(2))/2 alpha=0.2 pi Nm` |
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| 244. |
If a cylinder is rolling down the incline with sliding -A. after some time it may start pure rollingB. after sometime it must start pure rollingC. it may be possible that it will never start pure rollingD. cannot conclude anything |
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Answer» Correct Answer - A::C If friction is enough to support pure rolling then its starts pure rolling. Other wise does not do this |
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| 245. |
A shaft rotating at 3000 rpm is transmitting a power of `3.14 KW`. The magnitude of the driving torque isA. `6Nm`B. `10 Nm`C. `15 Nm`D. `22Nm` |
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Answer» Correct Answer - B `p=tauomega` |
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| 246. |
An automobile engine develops `100` kilo`-`watt, when rotating at a speed of `1800 rev//min`. Find the torque developed by it.A. `350`B. `440`C. `531`D. `628` |
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Answer» Correct Answer - C `p=tauomega` |
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| 247. |
If the centre of mass of three particles of masses of `1kg, 2kg, 3kg` is at `(2,2,2)`, then where should a fourth particle of mass `4kg` be placed so that the combined centre of mass may be at `(0,0,0).` |
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Answer» Let `(x_(1),y_(1),z_(1)),(x_(2),y_(2),z_(2))` and `(x_(3),y_(3),z_(3))` be the positions of masses `1kg,2kg,3 kg` and let the co-ordinates of centres of mass of the three particle system is `(x_(cm),y_(cm),z_(cm))` respectively. `x_(CM)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))implies2=(1xxx_(1)+2xxx_(2)+3xxx_(3))/(1+2+3)` Or `x_(1)+2x_(2)+3x_(3)=12....(1)` Suppose the fourth particle of mass `4 kg` is placed at `(x_(4),y_(4),z_(4))` so that centre of mass of new system shifts to `(0,0,0)`. for `x` coordinates of new centre of mass we have. `0=(1xxx_(1)+2xxx_(2)+3xxx_(3)+4xxx_(4))/(1+2+3+4)` `rArrx_(1)+2x_(2)+3x_(3)+4x_(4)=0...(2)` From equations `(1)` and `(2)` `12+4x_(4)=0impliesx_(4)=-3` similarly, `y_(4)=-3` and `z_(4)=-3` Therefore `4 kg` should be placed at `(-3,-3,-3)`. |
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| 248. |
Mass of thin long metal rod is `2kg` and its moment of inertia about an axis perpendicular to the length of rod and passing through its one end is `0.5 kg m^2`. Its radius of gyration isA. `20 cm`B. `40 cm`C. `50 cm`D. `1m` |
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Answer» Correct Answer - C `I=(mL^(2))/2, :. K=L/(sqrt(3))` |
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| 249. |
The moment of inertia of a uniform thin rod of length `L` and mass `M` about an axis passing through a point at a distance of `L//3` from one of its ends and perpendicular to the rod isA. `(7ML^(2))/48`B. `(ML^(2))/6`C. `(ML^(2))/9`D. `(ML^(2))/3` |
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Answer» Correct Answer - C `I=(ML^(2))/12+M[L/6]^(2)` |
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| 250. |
The moment of inertia of a uniform rod of length `2l` and mass `m` about an axis `xy` passing through its centre and inclined at an angle `alpha` is A. `(ml^(2))/3 sin^(2) alpha`B. `(ml^(2))/12 sin^(2) alpha`C. `(ml^(2))/2 cos^(2)alpha`D. `(ml^(2))/2 cos^(2) alpha` |
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Answer» Correct Answer - A Take small element and use integration |
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