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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.
| 101. |
A sphere of radius `0.10 m` and mass `10 kg` rests in the corner formed by a `30^(@)` inclined plane and a smooth vertical wall. Choose the correct option A. `N_(1)=56.5 N`B. `N_(2)=113N`C. `f=0`D. `fne0` |
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Answer» Correct Answer - A::B::C If we take moments about an axis through the centre of the sphere. Only f can have a torque. `sum tau=0` , take `sum F_(y)=0` yields `N_(2)cos30^(@)=mg=(10kg)(9.8 m//s^(2))` `sum F_(x)=0` yeilds `N_(2) sin 30^(@)-N_(1)=0` or `N_(1)=56.5N,N_(2)=113 N` |
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| 102. |
An impulse `I` is applied at the end of a uniform rod if mass `m`. Then: A. `KE` of translation of the rod is `(I^(2))/(2m)`B. `KE` of rotation of the rod is `(I^(2))/(6m)`C. `KE` of rotation of the rod is `(3I^(2))/(2m)`D. `KE` of the rod is `(2I^(2))/m` |
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Answer» Correct Answer - A::C::D `KE_(T)=1/2mv^(2).....(1) & I=mv....(2)` so `KE_(T)=(I^(2))/(2m),` again `KE_(R)=1/2Iomega^(2)...(3)` we have `Ixxl/2=1/12 ml^(2)xxomegarArromega=(6I)/(ml)` Thus `KE_(R)=1/2xx1/12 ml^(2)xx(36I^(2))/(m^(2)l^(2))=(3I^(2))/(2m)` and `KE=KE_(T)+KE_(R)=(2I^(2))/m` |
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| 103. |
Two particles of masses `m_(1)` and `m_(2)` are connected with a rigid rod of length `l`. If a force `F` acts perpendicular to the rod then (`a_(1) & a_(2)` are instantaneous of `m_(1) & m_(2)`) A. `a_(2)=0`B. `a_(1)=F/(m_(1))`C. `a_(CM)=F/(m_(1)+m_(2))`D. `alpha=(F(m_(1)+m_(2)))/(m_(1)m_(2)l)` |
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Answer» Correct Answer - A::B::C `a_(cm)=F/(m_(1)+m_(2)),Fxxl/2=Ialpha` `a_(1)=a_(cm)+l/2alpha,a_(1)=a_(cm)-l/2 alpha` |
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| 104. |
A string is wrapped several time on a cylinder of mass `M` and radius `R`, the cylinder is pivoted about its axis of symmetery . A block mass of `m` tied to the string rests on support so that the string is black. The block is lifted upto a height `h` and the support is removed. (shown in figure) What is the angular velocity of cylinder just before the string becomes tautA. zeroB. `(sqrt(2gh))/R`C. `(sqrt(gh))/R`D. `(sqrt(gh))/R` |
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Answer» Correct Answer - A `Just before the string becomes taut, the block falls freely, so `v_(0)=sqrt(2gh)`. There is no tension in the string, so nothing causes the cylinder to spin, so `omega_(0)=0` |
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| 105. |
A body is freely rolling down on an inclined plane whose angle of inclination is `theta`. If `a` is acceleration of its centre of mass then following is correctA. `a= g sin theta`B. `altg sin theta`C. `a gt g sin theta`D. `a=0` |
| Answer» Correct Answer - B | |
| 106. |
When a ring is rolling and `V_4` are velocities of top most point, lowest point, end point of horizontal diameter, centre of ring respectively, the decreasing order of these velocities isA. `V_2, V_1, V_4,V_3`B. `V_2, V_1, V_3,V_4`C. `V_2, V_1, V_3,V_4`D. `V_1, V_3, V_4,V_2` |
| Answer» Correct Answer - D | |
| 107. |
A conical pendulum consists of a mass `M` suspended from a string of length `l`. The mass executes a circle of radius `R` in a horizontal plane with speed `v`. At time `t`, the mass is at position `Rhat i` and has velocity `v hat j`. At time `t`, the angular momentum vector of the mass `M` about the point from which the string suspended is A. `M upsilonRhatk`B. `m upsilonlhatk`C. `M upsilonl[(sqrt(l^(2)-R^(2))/lhati)+R/lhatk]`D. `-M upsilonl[(sqrt(l^(2)-R^(2))/lhati)+R/lhatk]` |
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Answer» Correct Answer - C `vec(r)=Rhati-sqrt(l^(2)-R^(2)hatk)` `vec(P)=Mupsilonhatj` `vec(L)=vec(r)xxvec(P)` |
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| 108. |
In a rectangle `ABCD (BC = 2AB)`. The moment of inertia is maximum along axis through A. `BC`B. `AB`C. `HF`D. `EG` |
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Answer» Correct Answer - B perpendicular distance is maximum when the axis of rotation passes through `AB`, hence `M.I.` about `AB` is maximum. |
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| 109. |
For the given uniform square lamina `ABCD`, where centre is `O`. Its moment of inertia about an axis `AD` is equal to how many times its moment of inertia about an axis `EF`? `1 sqrt(2) I_(AC)=I_(EF) " " 2. I_(AD)=3I_(EF)` `3.I_(AC)=4I_(EF). " " 4. I_(AC)=sqrt(2)I_(EF)` |
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Answer» `I_(EF)=I_(GH)` (due to symmetry) `I_(AC)=I_(BD)` (due to symmetry) `I_(AC)+I_(BD)=I_(0)` `rArr2I_(AC)=I_(0).....(1)` and `I_(EF)+I_(GH)=I_(0)` `rArr2I_(EF)=I_(0)...(2)` From eq(1) and (2), we get `I_(AC)=I_(EF)` `:. I_(AD)=I_(EF)+(md^(2))/4=(md^(2))/12+(md^(2))/4` `I_(AD)=(md^(2))/3=4I_(EF)` |
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| 110. |
A turn table is rotating in horizontal plane about its own axis at an angular velocity `90rpm` while a person is on the turn table at its edge. If he gently walks to the centre of table by which moment of inertia of system decreases by `25%`, then the time period of rotating of turn table isA. `0.5 sec`B. `1 sec`C. `1.5 sec`D. `2 sec` |
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Answer» Correct Answer - A `I_(1)omega_(1)=I_(2)omega_(2), I_(1)=100,I_(2)=100-25=75` |
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| 111. |
The linear velocity of a point on the surface of earth at a latitude of `60^@` isA. `800/3 m//sec`B. `(800 pi)/3 m//sec`C. `800xx5/18m//sec`D. `(2000 pi)/27 m//sec` |
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Answer» Correct Answer - D `v=romega,r=Rcos theta,omega=(2pi)/T` |
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| 112. |
An electron is moving with speed `2xx10^5 m//s` along the positive x-direction in the presence of magnetic induction `vecB = (hat i + 4 hat j - 3 hat k) T`. The magnitude of the force experienced by the electron in `N ( e = 1.6xx 10^(-19) C) ( vec F = q ( vec v xx vec B))`A. `18xx10^(13)`B. `28xx10^(-13)`C. `1.6xx10^(-13)`D. `73xx10^(-13)` |
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Answer» Correct Answer - C `vec(F)=e(vec(V)xxvec(B))` |
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| 113. |
Two particles of masses `p` and `q (p gt q)` are separated by a distance `d`. The shift in the centre of mass when the two particles are interchanged isA. `d(p+q)//(p-q)`B. `d(p-q)//(p+q)`C. `dp//(p-q)`D. `dq//(p-q)` |
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Answer» Correct Answer - B `x_(cm)=(m_(2)d)/(m_(1)+m_(2))=(pd)/(p+q)` `x_(cm)^(1)=(m_(1)d)/(m_(1)+m_(2))=(qd)/(p+q)`, shift `=x_(cm)^(1)~x_(cm)` |
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| 114. |
A rigid body consists of a `3kg` mass located at `vec r_1=(2hat i + 5 hat j)m` and a `2kg` mass located at `vec r_2 = (4hat i +2 hat j)m`. The position of centre of mass isA. `(14/5 hatj+19/5 hati)m`B. `(14/5 hati+19/5hatj) m`C. `(19/5 hati+14/5hatj) m`D. `0` |
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Answer» Correct Answer - B `vec(r)_(cm)=(m_(1)vec(r)_(1)+m_(2)vec(r)_(2))/(m_(1)+m_(2))` |
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| 115. |
A particle is moving with uniform speed `0.5 m//s` along a circle of radius `1m` then the angular velocity of particle is (in `rad s^(-1)`)A. `2`B. `1.5`C. `1`D. `0.5` |
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Answer» Correct Answer - D `omega=v/r` |
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| 116. |
The angular displacement of a particle is given by `theta =t^3 + t^2 + t +1` then,the angular acceleration of the particle at `t=2` sec is …….`rad s^(-2)A. `14`B. `16`C. `18`D. `24` |
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Answer» Correct Answer - A `alpha=(domega)/(dt)=6t+2=12+2=14 rad s^(-2)` |
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| 117. |
The angular displacement of a particle is given by `theta =t^3 + t^2 + t +1` then, its angular velocity at `t=2` sec is …… `rad s^(-1)`A. `27`B. `17`C. `15`D. `16` |
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Answer» Correct Answer - B `theta=t^(3)+t^(2)+t+1,omega=(dtheta)/(dt)=3t^(2)+2t+1` |
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| 118. |
The angle between the vectors `(hat i + hat j + hat k)` and `( hat i - hat j -hat k)` isA. `sin^(-1)((sqrt(8))/3)`B. `sin^(-1)(1/3)+(pi)/3`C. `cos^(-1)((sqrt(8))/3)`D. `cos^(-1)sqrt(8/3)` |
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Answer» Correct Answer - A `sin theta=(|vec(A)xxvec(B)|)/(AB)` |
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| 119. |
Two identical thin uniform rods of length `L` each are joined to form `T` shape as shown in the figure. The distance of centre of mass from `D` is A. `0`B. `L//4`C. `3L//4`D. `L` |
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Answer» Correct Answer - C `X_(cm)=(m(L/2)+m(L))/(2m)` |
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| 120. |
A circular hole of radius `r` is made in a disk of radius `R` and of uniform thickness at a distance `a` from the centre of the disk. The distance of the new centre of mass from the original centre of mass is A. `(aR^(2))/(R^(2)-r^(2))`B. `(ar^(2))/(R^(2)-r^(2))`C. `(a(R^(2)-r^(2)))/(r^(2))`D. `(a(R^(2)-r^(2)))/(R^(2))` |
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Answer» Correct Answer - B `x=(r^(2)a)/(R^(2)-r^(2))` |
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| 121. |
A ball is attached to a string that is attached to a pole. When the ball is hit, the string wraps around the pole and the ball spirals inwards sliding on the frictionless surface. Neglecting air resistance, what happends as the ball swings around the pole? A. the mechanical energy and angular momentum are conservedB. the angular momentum of the ball is conserved and the mechanical energy of the ball increasedC. the angular momentum of the ball is conserved and the mechanical energy of the ball decreasedD. the mechanical energy the ball is conserved and angular momentum of ball decreases |
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Answer» Correct Answer - D Torque of tension about `O` is inward (clockwise sense) therefore angular momentum decrease. |
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| 122. |
A ball is attached to a string that is attached to a pole. When the ball is hit, the string wraps around the pole and the ball spirals inwards sliding on the frictionless surface. Neglecting air resistance, what happends as the ball swings around the pole? A. the mechanical energy and angular momentum are conservedB. the angular momentum of the ball is conserved and the mechanical energy of the ball increasedC. the angular momentum of the ball is conserved and the mechanical energy of the ball decreasedD. the mechanical energy the ball is conserved and angular momentum of ball decreases |
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Answer» Correct Answer - D The force is conservative. So `ME` is conserved (friction less) & Angular momentum `(L-Iomega)` decreases due to reverse swing compare to that of initial swing before strike. |
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| 123. |
A particle moves in a circular path with decreasing speed . Choose the correct statement.A. Angular momentum remains constantB. acceleration `(vec(a))` is towards the centreC. particle moves in a spiral path with decreasing radiusD. the direction of angular momentum remains constant |
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Answer» Correct Answer - D `L=mVR`, hear `V` decrease so `L` never be constant, also `vec(a)_(r) & vec(a)_(t)` both are acting & third point contradicts the given question. But direction `vec(L)` is always constants. |
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| 124. |
A wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle of `15^@` in time `t` sec. Then how much it rotates in the next `2t` sec |
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Answer» If angular acceleration is constant, we have `theta=omega_(0)t+1/2alphat^(2)rArr15^(@)=0+1/2 alphat^(2)rArr15^(@)=1/2 alphat^(2)....(1)` For the second condition (time `3 t sec`) `theta_(1)=1/2alpha(3t)^(2)......(2)` So, `Deltatheta=theta_(1)-theta_(2)=120^(@)` |
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| 125. |
When a disc rotates with uniform angular velocity, which of the following is not true ?A. The sense of rotation remains same.B. the orientation of the axis of rotation remains same.C. the speed of rotation is non-zero and remain same.D. the angular acceleration is zero. |
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Answer» Correct Answer - D We know that angular acceleration `alpha=(domega)/(dt)` given `omega`=const. Where `omega` is angular velocity of the disc `rArralpha=(domega)/(dt)=0/(dt)=0` Hence, angular acceleration is zero |
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| 126. |
A motor car is moving in a circular path with uniform speed `v`. Suddenly the car rotates through an angle `theta`. Then, the magnitude of change in its velocity isA. `2 v cos (theta/2)`B. `2 v sin (theta/2)`C. `2 v tan(theta/2)`D. `2 v sec (theta/2)` |
| Answer» Correct Answer - B | |
| 127. |
If the mass of earth and radius suddenly become `2` times and `1//4th` of the present value, the length of the day becomesA. `24 h`B. `6h`C. `3//2 h`D. `3h` |
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Answer» Correct Answer - D `I_(1)omega_(1)=I_(2)omega_(2)` |
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| 128. |
The arrangement shown in figure consists of two identical, uniform, solid cylinders, each of mass `m`, on which two light theads are wound symmetrically. Find the tensions of each thread in the process of motion. The friction in the axle of the upper cylinder is assumed to be absent. A. `4.9 N`B. `9.8 N`C. `8.8 N`D. `5.8 N` |
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Answer» Correct Answer - A `T=(mg)/10` |
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| 129. |
Moment of inertia of a body about an axis is `4kg-m^2`. The body is initially at rest and a torque of `8Nm` starts acting on it along the same axis. Workdone by the torque in `20s`, in joules is |
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Answer» `tau=IalpharArralpha=(tau)/I=2, theta=1/2 alphat^(2)=400`, `omega=tau theta=3200 J` |
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| 130. |
(i) Explain why friction is necessary to make the disc to roll in the direction indicated. (ii) Give the direction of frictional force at `B`, and the sense of frictional torque, before perfect rolling begins. (iii) What is the force of friction after perfect rolling begins? |
| Answer» (i) To roll a disc, a torque is required which in turn requires a tangential force to on it. As the force of friction is the only tangential force acting on the disc, so it is necessarily required for the rolling of a disc. (ii) Frictional force at `B` opposes the velocity of `B`. therefore, frictional force is in the same direction as the arrow. the sence of frictional torque is such as to oppose the angular motion. by right hand rule, `vec(omega)_(0)` act into the plane of paper and `vec(tau)` out of the paper. (iii) frictional force decreases the velocity of the point of constant `B`. perfect rolling begins when this velocity is zero at which the force of friction is zero. | |
| 131. |
The moment of inertia of a then circular disc about an axis passing through its centre and perpendicular to its plane is `I`. Then, the moment of inertia of the disc about an axis parallel to its diameter and touching the edge of the rim is |
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Answer» `I=(MR^(2))/2rArrMR^(2)=2I` `M.I.` of the disc about tangent in a plane `=5/4MR^(2)=5/2 I` |
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| 132. |
The moment of inertia of a disc of mass `M` and radius `R` about an axis. Which is tangential to sircumference of disc and parallel to its diameter is. |
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Answer» About the tangent parallel to the diameter `I=I_(g)+MR^(2)=(MR^(2))/4+MR^(2)=5/4 MR^(2)` |
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| 133. |
The point `P` of a string is pulled up with an acceleration `g`. Then the acceleration of the hanging disc (w.r.t ground) over which the string is wrapped, is A. `(2 g)/3 darr`B. `g/3 uarr`C. `(4g)/3 darr`D. `g/3 darr` |
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Answer» Correct Answer - D `2mg-T=ma, TR=Ialpha` `alpha=a/R,`solving `alpha=g/3` |
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| 134. |
Two blocks of msses 10 kg and 30 kg are placed along a vertical line. The first block is raised through a height of 7 cm. By what distance should the second mass be moved to raise the centre of mass by 1 cm?A. `1 cm up`B. `1 cm down`C. `2 cm down`D. `2 cm up` |
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Answer» Correct Answer - B `Deltay_(cm)=(m_(1)Deltay_(1)+m_(2)Deltay_(2))/(m_(1)+m_(2))` |
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| 135. |
`I` is moment of inertia of a thin square plate about an axis passing through opposite corners of plate. The moment of inertia of same plate about an axis perpendicular to the plane of plate and passing through its centre isA. `I//2`B. `I//sqrt(2)`C. `sqrt(2)I`D. `2I` |
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Answer» Correct Answer - D `I=(ML^(2))/12,I_(z)=I_(x), :. I^(1)=(ML^(2))/6=2I` |
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| 136. |
The moment of inertia of a thin square plate of mass `1.2 kg ` is `0.2 kg m^(2)` when its is made to rotate about an axis perpendicular to plane of plate and through a corner of plate. The side length of plate isA. `0.2 m`B. `0.4m`C. `0.5 m`D. `0.8m` |
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Answer» Correct Answer - C `I=(MR^(2))/6+M(L/(sqrt(2)))^(2)=0.2` |
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| 137. |
Solid sphere, hollow sphere, solid cylinder and hollow cylinder of same mass and same radii are simultaneously start rolling down from the top of an inclined plane. The body that takes longest time to reach the bottom isA. solid sphereB. hollow sphereC. solid cyliderD. hollow cylinder |
| Answer» Correct Answer - D | |
| 138. |
A uniform solid sphere of radius `r` is rolling on a smooth horizontal surface with velocity `v` and angular velocity `omega = (v=omega r)`. The sphere collides with a sharp edge on the wall as shown in figure. The coefficient of friction between the sphere and the edge `mu=1//5`. Just after the collision the angular velocity of the sphere becomes equal to zero. The linear velocity of the surface just after the collision is equal to A. `v`B. `v/5`C. `(3v)/5`D. `v/6` |
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Answer» Correct Answer - A Impulse provided by the edge in the horizontal direction `-intNdt =-mV^(1)-(MV)....(1)` Friction impulse in the vertical direction `muRintNdt=2/5 mR^(2)(V/R)....(2)` from eq(1) and (2) we get `int Ndt=2mV` and `V^(1)=V` |
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| 139. |
A rolling body is connected with a trolley car by a spring of stiffness `k`. It does not slide and remains in equilibrium relative to the acceleration trolley car. If the trolly car is stopped after a time `t=t_(0)`. (the rolling body touches the trolly) A. `tau_(C)ne0` for `ctltt_(0)`B. `f=0` for `tltt_(0)`C. `x=(ma)/k`, where `x`=deformation of the springD. `(KE)_(max)=1/2ma^(2)t_(0)^(2)`, where`(KE)_(max)` is the maximum `KE` of the rolling body |
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Answer» Correct Answer - B `f=0` for `tgtt_(0)` until it can stop no friction acts because it neither slides nor rotates due to action of the rolling |
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| 140. |
A linear impulse `int Fdt` acts at a point `C` of the smooth rod `AB`. The value of `x` is so that the end `A` remains stationary just after the impact is A. `l/4`B. `l/3`C. `l/6`D. `l/5` |
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Answer» Correct Answer - C Let `J` be the impulse acting on the rod `J=mmv_(cm), Jx=1/12 ml^(2) omega`. Since the end `A` is stationary `V_(A)=V_(cm)-1/2 omega=J/m-((12 Jx)/(ml^(2)))l/2=0` |
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| 141. |
Masses `1 kg, 1.5 kg,2 kg` and `Mkg` are situated at `(2,1,1),(1,2,1),(2,-2,1)` and `(-1,4,3)`. If their centre of mass is situated at `(1,1,3//2)`, the value of `M` isA. `1kg`B. `2kg`C. `1.5 kg`D. `3kg` |
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Answer» Correct Answer - C `x_(cm)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3)+m_(4)x_(4))/(m_(1)+m_(2)+m_(3)+m_(4))` |
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| 142. |
The densitis of two solid spheres A and B of the same radii R very with radial distance `r as p_A(r ) = k ((r )/(R )) and p_B(r ) = k((r )/(R )^5,` respectively, where k is a constant . The moments of inertia of the inividual spheres about axes passing throgh their centres are `I_A and I_B` respectively. if `(I_B)/(I_A) = (n)/(10),` the value of n is |
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Answer» Correct Answer - 6 `I=int2/3rho4pir^(2)r^(2)dr` `I_(A)propint(r)(r^(2))(r^(2))dr` `I_(B)propint(r^(5))(r^(2))(r^(2))dr` `:. (I_(B))/(I_(A))=6/10` |
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| 143. |
Find position of centre of mass of four particle system, which are at the vertices of parallelogram, as shown in figure. |
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Answer» From figure `DC=bsin theta,OD=bcos theta` `x_(CM)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3)+m_(4)x_(4))/(m_(1)+m_(2)+m_(3)+m_(4))` `x_(CM)=([a+bcos theta)]/2`, similarly `y_(CM)=(bsin theta)/2` `:. [x_(CM),y_(CM)]=[(a+bcostheta)/2,(bsintheta)/2]` |
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| 144. |
When a constant torque is applied on a rigid body, thenA. the body moves with linear accelerationB. the body rotates with constant angular velocityC. the body rotates with constant angular accelerationD. the body undergoes equal angular displacement in equal intervals of time |
| Answer» Correct Answer - C | |
| 145. |
If `vec A= 3i + j +2k` and `vec B = 2i - 2j +4k` and `theta` is the angle between the two vectors, then `sin theta` is equal toA. `2/3`B. `2/(sqrt(3))`C. `2/(sqrt(7))`D. `2/(sqrt(13))` |
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Answer» Correct Answer - C `sin theta=(|vec(A)xxvec(B)|)/(AB)` |
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| 146. |
A uniform solid of mass `m` is placed on a sheet of paper on a horizontal surface. The coefficient of friction between paper and sphere is `mu`. If the paper is pulled horizontally with an acceleration A. the tension in the string is equal to `mg sin theta`B. force acting on the cylinder is `(mg sin theta)/2`C. tension in the string is equal to `(mg sin theta)/2`D. frictional force acting on the cylinder is zero |
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Answer» Correct Answer - B::D `ma-f=ma_(0)` `fR=Ialpha, I=(2mR^(2))/5,a_(0)=Ralpha` |
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| 147. |
Identical blocks each of mass `M` and length `L` are placed one above the other such that each extends out by a maximum length as shown in figure. Find the maximum extension of the `n^(th)` block from the top, so that the blocks will not fall. A. `L//5`B. `L//4`C. `L//3`D. `L` |
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Answer» Correct Answer - B `CM` of bricks, above each brick must to be beyond its edge `x_(cm)=(Sigmam_(i)x_(i))/(Sigmam_(i)), x_(cm)=L` `x_(1)=a+L/2,x_(2)=2a+L/2,x_(3)=3a+L/2` (or) `a=L/n` |
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| 148. |
Figure shows an arrangements of masses hanging from a ceilling. In equilibrium each rod is horizontal, has negligible mass and extends three times as far to the right of the wire supporting is as to the left. If mass `m_(4)` is `48 kg,` then `(m_(3))/(m_(2))` is |
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Answer» Correct Answer - 4 `m_(2)gxx1=m_(1)gxx3`i.e., `m_(2)=3m_(1)` `4m_(1)gxx3=m_(3)g i.e., m_(3)=12 m_(1)` `16m_(1)gxx3=m_(4)g` `m_(1)=1 kg` |
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| 149. |
The angular momentum of rotating body is increased by `20%`. What will be the increase in its rotational kinetic energy? |
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Answer» Kinetic energy `KE=(L^(2))/(2I)rArrE propL^(2)` `(DeltaE)/E=(120/100)^(2) (or) (DeltaE)/D=0.44` `(DeltaE)/Exx100=44%` |
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| 150. |
Consider a uniform square plate of of side and mass `m`. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is -A. `5/6 ma^(2)`B. `1/12 ma^(2)`C. `7/12 ma^(2)`D. `2/3 ma^(2)` |
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Answer» Correct Answer - C Using parallel axes theoram, `I=I_(G)+Mr^(2)=(MI^(2))/12+(MI^(2))/2=(7Ml^(2))/12` |
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