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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. |
Which of the following statement are correct -A. friction acting on a cylinder without sliding one on a inclinded surface is always upward along the incline irrespective of any external force acting on it.B. friction acting on acylinder without sliding on an inclined surface is may be upward may be downwards depending on the external force acting on it.C. friction acting on a cylinder rolling without sliding may be zero depending on the external force acting in it.D. nothing can be said exactly about it as it depends on the friction coefficient on inclined plane. |
| Answer» Correct Answer - B::C | |
| 102. |
A plank with a uniform sphere placed on it. Rests on a smooth horizontal plane. Plank is pulled to right by a constant force `F`. It the sphere does not slip over the plank. .A. acceleration of centre of sphere is less than that of the plankB. acceleration of centre if sphere is grester than tha plank because friction acts rightward on the sphereC. acceleration of the centre of sphere may be towards left.D. acceleration of the centre of sphere relative to plank may be greater than that of the plank relative to floor. |
| Answer» Correct Answer - A::B::C | |
| 103. |
A disc of radius `10 cm` rotates in `XY` plane about an axis passing through its centre and perpendicular to the plane. At a particlular moment a point A on the disc has an acceleration `vec a_(A) = - 6 hat I m//s^(2)`. The angular acceleration of disc is :A. `48 rad//s^(2)` anticlockwiseB. `12 rad//s^(2)` anticlockwiseC. `48 rad//s^(2)` clockwiseD. `12 rad//s^(2)` clockwise. |
| Answer» Correct Answer - A | |
| 104. |
A uniform disc is rolling on a horizontal surface. At a certain instant `B` is the point of contact and `A` is at height `2 R` from ground, where `R` is radius of disc - .A. The magnitude of the angular momentum of the disc about `b` is thrice that about `A`B. The angular momentum of the disc about `A` anticlockwiseC. The angular momentum of the disc about `B` is clockwiseD. The angular momentum of the disc about about `A` is equal to that about `B`. |
| Answer» Correct Answer - A::B::C | |
| 105. |
A disc of radius `10 cm` rotates in `XY` plane about an axis passing through its centre and perpendicular to the plane. At a particlular moment a point A on the disc has an acceleration `vec a_(A) = - 6 hat I m//s^(2)`. The angular speed of disc is :A. `6 rad//s` clockwiseB. `3 rad//s` clockwiseC. `6 rad//s` anticlockwiseD. `3 rad//s` clockwise or anticlockwise |
| Answer» Correct Answer - C | |
| 106. |
A wheel of radius `r` rolling on a straight line, the velocity of its centre being `v`. At a certain instant the point of contact of the wheel the ground is `M` and `N` is the highest point on the wheel (diametricaly opposite to `M`). The incorrect statement is -A. The velocity of any point `P` of the wheel is proportional to `MP`B. Points of the wheel moving with velocity greater than `v` form a larger area of the wheel than points moving with velocity less than `v`.C. The point of contact `M` is instantaneously at restD. The velocities of any two parts of the wheel which are equildistant from centre are equal. |
| Answer» Correct Answer - D | |
| 107. |
The torque can be applied only about two points (a) centre of mass and (ii) point about which the body is rolling. The equation `a = r alpha` can always be applied in case of rolling.A. Statement-1 is True, Statement-2 is Ture , Statement -2 is a correct explanation for statement -1.B. Statement-1 is True, Statement-2 is True , Statement-2 is NOT a correct explanation fro Statement -1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - C (i) Conceptual. (ii) It can be applied only if body rolls and the surface has zero acceleration. |
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| 108. |
A ring of mass `M` and radius `R` sliding with a velocity `v_(0)` suddenly enters inthrough surface where the coefficient of friction is `mu`, as shown in figure. . Choose the correct statement (s).A. The ring starts its rolling motion when the centre of mass stationaryB. The ring starts rolling motion when the point of contact becomes stationaryC. The time after which the ring starts rolling is `v_(0)//2 mu g`.D. The rolling velocity is `v_(0)//2`. |
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Answer» Correct Answer - B::C::D Velocity of `c.m` at any time before start rolling `v = v_(0) - mu gt` ….(1) and for angular velocity `tau = I alpha = (tau)/(I), omega = alpha t rArr omega = (tau)/(I) t` ….(2) Let at `t = t_(0)`, ring start pure rolling from (1),(2) & (3) `v_(0) - mu g t_(o) = R xx (tau)/(I) t_(0)` `v - mu g t_(0) = R xx (mu mg R)/(mR^(2)) t_(0)` `v_(0) - mu g t_(0) = mu g t_(o) rArr t_(o) = (V_(0))/(2 mu g)`. `(D)` `=v = v_(0) - mu g t_(0) rArr v = v_(0) -(v_(0))/(2)` `v = (v_(0))/(2)`. |
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| 109. |
A ring of mass `M` and radius `R` sliding with a velocity `v_(0)` suddenly enters inthrough surface where the coefficient of friction is `mu`, as shown in figure. . Choose the correct statement (s).A. The momentum of the ring is conservedB. The angular momentum of the ring is conserved about its centre of massC. The angular momentum of the ring conserved about any point on the horizontal surfaceD. The mechanical energy of the ring is conserved. |
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Answer» Correct Answer - C (A) Due friction `v_(0)` decreases so momentum will decrease. (B) Due to torque provided by friction `omega` increases so `J = I omega` will increase ( C) Angular moment about any point on horizontal surface `J = I omega + mv_(0) R` will remain conserved (D) Mechanical energy of ring will decrease. |
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| 110. |
A disc of circumference `s` is at rest at a point `A` on a horizontal surface when a constant horizontal force begins to act on its centre. Between `A` and `B` there is sufficient friction toprevent slipping, and the surface is smooth to the right of `B.AB = s`. The disc moves from `A` to `B` in time `T`. To the right of `B`, .A. the angular acceleration of the disc will disappear, linear acceleration will remain unchanged.B. linear acceleration of the disc will increaseC. the disc will make one rotation in time `T//2`D. the disc will cover a distance greater than `s` in further time `T`. |
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Answer» Correct Answer - B::C::D (B) To the right of `B`, friction vanishes, so linear acceleration of disc will increase. ( c) From `(A)` to `B` it complete one rotation in time `T` `theta = (1)/(2) alpha t^(2), 2 pi = (1)/(2) alpha T^(2), alpha = (4 pi)/(T^(2))` ....(1) & `omega = alpha T`...(2) from (1) & (2) `omega = (4 pi)/(T)` after `B, omega` becomes constant there fore time taken to complete one rotation after `B` `theta = omega t, 2 pi = (4 pi)/(T) t, t = (T)/(2)` (D) After `B` disc is linearly accelerated so it will cover greater distance than `s` in further time `T`. |
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| 111. |
A particle of mass `m` is moving in a plane along a circular path of radius `r`. Its angular momentum about the axis of rotation is `L`. The centripetal force acting on the particle is.A. `(L^2)/(mr)`B. `(L^2 m)/( r)`C. `(L^2)/(mr^3)`D. `(L^2)/(mr^2)` |
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Answer» Correct Answer - C ( c) `L = mvr rArr v = (L)/(mr)` `F_c = (mv^2)/(r) = (m)/(r) ((L)/(mr))^2 = (L^2)/(mr^3)`. |
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| 112. |
A block of mass `m` is held fixed against a wall by a applying a horizontal force `F`. Which of the following option is incorrect : .A. friction force = mgB. `F` will not produce torqueC. normal will not produce torqueD. normal reaction = F |
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Answer» Correct Answer - C This is the equilibrium of coplanar forces. Hence, `sum F_(x) = 0` :. `F = N` :, `sum F_(y) = 0` `f = mg` `sum tau_(C) = 0` :. `vec tau_(N) + vec tau_(f) = 0` :. Since `vec tau_(f) ne 0` :. `vec tau_(N) ne 0` :. Correct Answer is `(C)`. |
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| 113. |
Centre of mass of three particles of masses `1 kg, 2 kg and 3 kg` lies at the point `(1,2,3)` and centre of mass of another system of particles `3 kg and 3 kg` lies at the point `(-1, 3, -2)`. Where should we put a particle of mass `5 kg` so that the centre of mass of entire system lies at the centre of mass of first system ?A. `(0, 0,0)`B. `(1, 3,2)`C. `(-1, 2,3)`D. `(3, 1, 8)` |
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Answer» Correct Answer - D (d) According to the definition of centre of mass, we can imagine one particle of mass `(1 + 2 + 3) kg at (1,2,3)` , another particle of mass `(2 + 3) kg at (-1, 3, -2)` Let the third particle of mass `5 kg` put as `(x_3,y_3,z_3)` i.e., `m_1 = 6kg, (x_1,y_1,z_1) = (1,2,3)` `m_2 = 5 kg(x_2,y_2,z_2) = (-1, 3, - 2)` `m_1 = 5 kg(x_3,y_3, z_3) = (1,2,3)` Given, `(X_(CM), Y_(CM), Z_(CM) = (1,2,3)` Using `X_(CM) = (m_1 x_1 + m_2 x_2 + m_3 x_3)/(m_1 + m_2 + m_3)` `1 = (6 xx 1 + 5 xx(-1)+ 5x_3)/(6 + 5 + 5)` `5 x_3 = 16-1 = 15` or `x_3 = 3` Similarly, `y_3 = 1 and z_3 = 8`. |
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| 114. |
A small object of uniform density rolls up a curved surface with an initial velocity `v`. It reached upto maximum height of `3v^(2)//4g` with respect to the initial position. The object is - .A. ringB. solid sphereC. hollow sphereD. disc |
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Answer» Correct Answer - D `(1)/(2) mv^(2) + (1)/(2) I_(CM) ((v)/( r))^(2) = mg ((3v^(2))/(4g))` `I = (1)/(2) mr^(2)` The object is a disc. |
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| 115. |
The moment of inertia of a uniform disc about an axis passing through its centre and perpendicular to its plane is `1 kg-m^2`. It is rotating with an angular velocity `100 radia//sec`. Another indentical disc is gently placed on it so that their centres coincide. Now these two discs together continue to rotate about the same axis. Then the loss in kinetic energy in kilojoules is :A. `2.5`B. `3.0`C. `3.5`D. `4.0` |
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Answer» Correct Answer - A (a) `I_1 = 1 kg m^2, omega_1 = 100 rad//sec` Since the mass is doubled, `I_2 = (M_2 R^2)/(2) = (2 M.R^2)/(2) = 2 I_1 = 2 kgm^2` Conservation of angular momentum, `I_1 omega_1 = I_2 omega_2` or `1 xx 100 = 2 xx omega_2` or `omega_2 = 50 rad//sec` `E_1 = (1)/(2) I_1 omega_1^2 = (1)/(2) xx 1 xx (100)^2 = 5 xx 10^3 J` `E_2 = (1)/(2) I_2 omega_2^2 = (1)/(2) xx 2 xx (50)^2 = 2.5 xx 10^3 J` Loss in `KE = E_1 - E_2 = 5 xx 10^3 - 2.5 xx 10^3 = 2.5 kJ`. |
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| 116. |
The moment of inertia of a thin uniform rod of mass `M` and length `L` about an axis passing through its mid-point and perpendicular to its length is`I_0`. Its moment of inertia about an axis passing through one of its ends perpendicular to its length is.A. `I_0 + ML^2//4`B. `I_0 + 2 ML^2`C. `I_0 + ML^2`D. `I_0 + ML^2//2` |
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Answer» Correct Answer - A (a) The theorem of parallel axis for moment of inertia `I = i_(CM) + Mh^2` `I = I_0 + M((L)/(2))^2` =`I_0 + (ML^2)/(4)`. |
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| 117. |
Minimum moment of inertia of a uniform body is `I` about and axis. The axis must be passing through `COM` of body. Moment of inertia depends on distribution of mass about axis of rotation.A. Statement-1 is True, Statement-2 is Ture , Statement -2 is a correct explanation for statement -1.B. Statement-1 is True, Statement-2 is True , Statement-2 is NOT a correct explanation fro Statement -1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - B | |
| 118. |
One quarter sector is cut from a uniform circular disc of radius `R`. This sector has mass `M`. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. It moment of inertia about the axis of rotation is. .A. `(1)/(2) MR^(2)`B. `(1)/(4) MR^(2)`C. `(1)/(8) MR^(2)`D. `sqrt(2) MR^(2)` |
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Answer» Correct Answer - A Mass of the whole disc `= 4 M` Moment of inertia of the disc about the given axis. `(1)/(2) (4M) R^(2) = 2MR^(2)` :. Moment of inertia of quarter section of the disc. =`(1)/(4) (2MR^(2)) = (1)/(2) MR^(2)`. |
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| 119. |
Consider an isolated system moving through empty space. The system consists of objects that interact with each other and can change location with respect to one another. Which of the following quatities can change in time ?A. The angular momentum of the systemB. The linear momentum of the systemC. Both the angular momentum and linear momentum of the systemD. Neither the angular momentum nor linear momentum of the system. |
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Answer» Correct Answer - D (d) As long as no net external force, or torque, acts on the system, the linear and angular momentum of the system are constant. |
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| 120. |
A solid cylinder rolls up an inclined plane of inclination `theta` with an initial velocity `v`. How far does the cylinder go up the plane ?A. `(3 v^2)/(2 g sin theta)`B. `(v^2)/(4 g sin theta)`C. `(3 v^2)/(g sin theta)`D. `(3 v^2)/(4 g sin theta)` |
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Answer» Correct Answer - D (d) Let the cylinder go up the plane upto a height `h`. Let `M` and `R` be the mass and radius of cylinder respectively. According to law of conservation of mechanical energy, we get. `(1)/(2) Mv^2 + (1)/(2) I omega^2 = Mgh` `(1)/(2) Mv^2 + (1)/(2) (MR^2)/(2) omega^2 = Mgh` (because For a solid cylinder, `I = (1)/(2) MR^2`) `(1)/(2) Mv^2 + (1)/(4) MR^2 omega^2 = Mgh` `(1)/(2) Mv^2 + (1)/(4) Mv^2 = Mgh` `(because v = R omega)` `(3)/(4) Mv^2 = Mgh` `h = (3 v^2)/(4 g)` ...(i) Let `s` be distance travelled by the cylinder up the plane. `sin theta (h)/(s)`or `s = (h)/(sin theta) = (3 v^2)/(4 g sin theta)` (Using (i)). |
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| 121. |
A body of mass `m` and radius `r` is released from rest along a smooth inclined plane of angle of inclination `theta`. The angular momentum of the body about the instantanoues point contact after a time `t` from the instant of release of equal is :A. `mgrt cos theta`B. `mgrt sin theta`C. `(3//2) mgrt sin theta`D. none of these |
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Answer» Correct Answer - B (b) Since the surface is frictionless, the body does not ratate about its centre of mass. Only it slides parallel to the surface. i.e., `omega = 0` and `v (g sin theta) t` The angular momentum after time `t` : `L = mvr = m(g sin theta t) r = mgrt sin theta`. |
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| 122. |
A projectile of mass `m` is launched with an initial velocity `vec v_i` making an angle `theta` with the horizontal as shown in figure. The projectile momentum of the particle about the origin when it is at the highest point of its trajectory is. .A. `(-m_i^3 sin theta^2 cos theta)/(2 g) hat k`B. `(m_i^3 sin theta^2 cos theta)/(2 g) hat k`C. `(-m_i^3 sin theta^2 cos theta)/(g) hat k`D. `(2 m_i^3 sin theta^2 cos theta)/(g) hat k` |
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Answer» Correct Answer - A (a) At the highest point of the trajectory, `x = (1)/(2) R = (v_i^2 sin 2 theta)/(2 g)` and `y = h_(max) = ((v_i sin theta)^2)/(2 g)` The angular momentum is then `vec L_1 = vec r_1 xx m vecv_1` =`[(v_i^2 sin 2 theta)/(2 g)hati +((v_i sin theta)^2)/(2g)hat j]xx mv_(x i) hat i` =`(-mv_i^2 sin theta^2 cos theta)/(2 g) hat k`. |
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| 123. |
Average torque on a projectile of mass `m` (initial speed `u` and angle of projection `theta`) between initial and final positions `P` and `Q` as shown in figure, about the point of projection is : .A. `1(m u^2 sin 2 theta)/(2)`B. `(m u^2 cos theta)/(2)`C. `m u^2 sin theta`D. `m u^2 cos theta` |
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Answer» Correct Answer - A (a) `vec tau. Delta t = Delta vec L`…(i) Here `Delta t = time of flight = (2 u sin theta)/(g)` Change in angular about point of projection (initially it is zero) `|vec(Delta L)| = |vec L_f = vec L_i|= (m u sin theta)` Range =`((m u sin theta)(u^2 sin 2 theta))/(g) = (m u^2 sin theta sin 2 theta)/(g)` Now `|vec tau_(av)| = |vec(Delta L)/(Delta t)| = (m u^3 sin theta sin 2 theta)/(g) xx (g)/(2u sin theta)` =`(m u^2 sin 2 theta)/(2)`. |
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| 124. |
We have two spheres, one of which is hollow and the other solid. They have identical massses and moment of intertia about theur respective diameters. The ratio of their radius is given by.A. `5 : 7`B. `3 : 5`C. `sqrt(3) : sqrt(5)`D. `sqrt(3) : sqrt(7)` |
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Answer» Correct Answer - C ( c) `I_h = I_s rArr (2)/(3) MR_h^2 = (2)/(5) MR_s^2 rArr (R_h)/(R_s) = (sqrt(3))/(sqrt(5))`. |
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| 125. |
The moments of inertia of two rotating bodies `A` and are `I_A` and `I_B(I_A gt I_B)`. If their angular momenta are equal then.A. Kinetic energy of `A` = Kinetic energy of `B`B. Kinetic energy of `A` gt Kinetic energy of `B`C. Kinetic energy of `A` lt Kinetic energy of `B`D. Kinetic energy of the two bodies cannot be compared with given data. |
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Answer» Correct Answer - C ( c) `I_A omega_A = I_B omega_B` (Given) :. `(omega_A)/(omega_A) = (I_B)/(I_A)` …(i) Kinetic energy `= (1)/(2) I omega^2` :. `((K.E.)_A)/((K.E.)_B) = ((1)/(2) I_A omega_A^2)/((1)/(2) I_B omega_B^2)` =`(I_A)/(I_B)xx ((I_B)/(I_A))^2` (Given (i)) =`(I_B)/(I_A)` As `I_A gt I_B` (Given) :. `(K.E.)_A lt (K.E.)_B`. |
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| 126. |
Two bodies have their moments of inertia `I` and `2 I` respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio.A. `1 : 2`B. `sqrt(2) : 1`C. `2 : 1`D. `1 : sqrt(2)` |
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Answer» Correct Answer - D (d) As said, `(KE)_(rot)` remains same. i.e., `(1)/(2) I_1 omega_1^2 = (1)/(2) I_2 omega_2^2` `rArr (1)/(2 I_1)(I_1 omega_1)^2 = (I)/(2 I_2) (I_2 omega_2)^2` `rArr (L_1^2)/(I_1) = (L_2^2)/(I_2)` `rArr (L_1)/(L_2) = sqrt((I_1)/(I_2))` but `I_1 = I, I_2 = 2 I` :. `(L_1)/(L_2) = sqrt((I)/(2 I)) =(1)/(2)` or `L_1 : L_2 = 1 : sqrt(2)`. |
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| 127. |
A circular disc A of radius r is made from aniron plate of thickness t and nother circular disc B of rdius 4r is made fro an iron plate of thickness t/4. The relatiion between the moments of inertia `I_A and I_B` isA. `I_A gt I_B`B. `I_A = I_B`C. `I_A lt I_B`D. depends on the actual values of `t` and `r`. |
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Answer» Correct Answer - C ( c) `m_A = volume xx density = pi r^2 t rho` `I_A = (1)/(2) m_A r^2 = (1)/(2) pi r^4 rho t` Similarly `I_B = (1)/(2) pi (4 r)^4 rho((t)/(4)) = 64 I_A` Clearly `I_B gt I_A`. |
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| 128. |
A circular disc `X` of radius `R` is made from an iron of thickness `t`, and another disc `Y` of radius `4R` is made from an iron plate of thickness `t//4`. Then the relation between the moment of mertia `I_(x)` and `I_(Y)` is :A. `I_(Y) = 32 I_(x)`B. `I_(Y) = 16 I_(x)`C. `I_(Y) = I_(x)`D. `I_(Y) = 64 I_(x)`. |
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Answer» Correct Answer - D Mass of disc `(X), mx = pi R^(2) t rho` where `rho` = density of material of disc :. `1_(X) = (1)/(2) m_(X) R^(2) = (1)/(2) R^(2) t rho R^(2)` `1_(X) = (1)/(2) pi rho R^(4)` Again mass of disc `(Y)` `m_(y) = pi = (4 R)^(2) (t)/(4) rho = 4 pi R^(2) t rho` and `I_(Y) = (1)/(2) m_(Y) (4R^(2)) = (1)/(2) 4 pi R^(2) t rho. 16 R^(2)` `rArr I_(Y) = 32 pi t rho R^(4)`...(ii) :. `(l_(y))/(l_(x)) = (32 pi t rho R^(4))/((1)/(2) pi rho t R^(4))` `rArr = 64` :. `1_(Y) = 64 1_(X)`. |
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| 129. |
Three bodies, a ring, a soild cylinder and a soild sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity ?A. RingB. Solid cylinderC. Solid sphereD. All reach the ground with same velocity |
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Answer» Correct Answer - C ( c) According to conservation of mechanical energy or `mgh = (1)/(2) mv^2 (1 + (k^2)/(R^2))` `v^2 = ((2 gh)/(1 + (k^2)/(R^2)))` It is independent of the mass of the rolling body. For a ring, `k^2 = R^2` `v_(ring) = sqrt((2 gh)/(1 + 1)) = sqrt(gh)` For a solid cylinder, `k_2 = (R^2)/(2)` `v_(cylinder) = sqrt((2 gh)/(1 + (1)/(2))) = sqrt((4g h)/(3))` For a solid sphere, `k^2 = (2)/(5) R^2` `v_(sphere) = sqrt((2 gh)/(1 + (2)/(5))) = sqrt((10 gh)/(7))` Among the given three bodies the solid sphere has the greatest and the ring has the least velocity at the bottom of the inclined plane. |
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| 130. |
A cylinder rolls up an inclined plane, reaches someheight and then rolls down (without slipping throught these motions). The direction of the frictional force acting on the cylinder are -A. Up the incline while ascending and down the incline whilse descendingB. Up the incline while ascending as well as descendingC. down the incline while ascending and up the incline while descendingD. down the incline while ascending as well as descending |
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Answer» Correct Answer - B During upward journey `v_(CM)` decreases due to `mg sin theta`. For pure rolling to maintain `v_(CM) = R omega`, torque of friction should be to decrease `omega`. During down ward journey `v_(CM)` increases due to `mg sin theta`. For pure rolling to maintain `v_(CM) = R omega`. torque of friction should be to increase `omega` so option (B) is correct choice. |
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| 131. |
A solid uniform disk of mass `m` rolls without slipping down a fixed inclined plane with an acceleration `a`. Find the frictional force on the disk due to surface of the plane. |
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Answer» Correct Answer - `1//2 ma` From previous question `f xx R = I alpha` `f = (mK^(2))/(R^(2)) a` for disc `= (K^(2))/(R^(2)) = (1)/(2)` `f = (ma)/(2)`. |
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| 132. |
Moment of inertia of a rectangular plate about an axis passing through `P` and perpendicular to the plate is `I`. Moment of `PQR` about an axis perpendicular to the plane of the plate : .A. about `P = I//2`B. about `R = I//2`C. about `P gt I//2`D. about `R gt I//2` |
| Answer» Correct Answer - C | |
| 133. |
A triangular plate of uniform thickness and density is made to rotate about an axis perpendicular to the plane of the paper and (i) passing through `A`, (ii) passing through `B`, by the application of some force `F` at `C` (mid - point AB) as shown in the figure. In which case angular acceleration is more ? .A. In case (i)B. In case (ii)C. Both (i) and (ii)D. None of these |
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Answer» Correct Answer - B (b) We can see that torque in both cases is `FL//2`. But in case (ii) moment of inertia is less. So angular acceleration is more in case (ii). |
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| 134. |
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+1)l)/(2)`D. `(2 l)/(n(n^2 + 1))` |
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Answer» Correct Answer - A (a) `x_(cm) =(m l + 2ml 2 l + 3m 3 l...nmnl)/(m + 2m _+...+nm)` =`(l[1 + 4 + 9 +...+ n^2])/(1 + 2 + ...+ n)` =`(l[n (n + 1)(2n + 1)//6])/(n(n + 1)//2) = ((2n + 1)l)/(3)`. |
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| 135. |
From a circular disc of radius R and mass 9 M , a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is A. `8 mR^(2)`B. `4 mR^(2)`C. `(40)/(9) mR^(2)`D. `(37)/(9) mR^(2)` |
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Answer» Correct Answer - B `I_("remaining") = I_("whole") - I_("removed")` `I =(1)/(2) (9m) (R^(2)) -[(1)/(2)m ((R)/(3))^(2) + m((2R)/(3))^(2)]` `I = 4mR^(2)`. |
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| 136. |
Four holes of radius `R` are cut from a thin square plate of side `4 R` and mass `M`. The moment of inertia of the remaining portion about z-axis is : .A. `(pi)/(12) MR^2`B. `((4)/(3) - (pi)/(4)) MR^2`C. `((8)/(3) - (10 pi)/(16))MR^2`D. `((4)/(3) -(pi)/(6)) MR^2`. |
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Answer» Correct Answer - C ( c) Area mass density : `sigma = (M)/((16 R^2))` Mass of each hole : `m_1 = sigma R^2 =(M)/((16 R^2)) pi R^2 = (pi M)/(16)` Distance between centre of plate and centre of hole `x = (4Rsqrt(2))/(4) = R sqrt(2)`. Moment of inertia of one hole about `z` axis : `I_1 = (1)/(2) m_1 R^2 + m_1 x^2`. Moment of inertia of whole plate about `z` axis : `I = (M(4 R)^2)/(6)` Required Moment of Inertia =`I - 4 I_1 =[(8)/(3) -(10 pi)/(16)] MR^2`. |
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| 137. |
Assertion: A quick collision between two bodies is more violent that show collision , even when initial and final velocity are identical. Reason: The rate of change of momentum determine that force is small or large.A. If both assertion and reason are true and reason is the true explanation of assertion.B. If both assertion and reason are reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - A (a) Momentum `p = mv` or `p prop v`. i.e., momentum is directly proportional to its velocity, so the momentum is greater in a quicker collision between two bodies than in slower one. Hence, due to greater momentum quicker collision between two between two bodies will be more violent even initial and final velocities are identical. |
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| 138. |
A ring mass `m` and radius `R` has three particle attached to the ring as shown in the figure. The centre of the centre `v_(0)`. Find the kinetic energy of the system. (Slipping is absent). . |
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Answer» Correct Answer - `6mv_(0)^(2)` About instantaneous axis of rotation rolling system can be considered as pure rotation `KE = (1)/(2) I omega^(2)` …(1) here `I` = moment of inertia about instantaneous axis of rotation `I = 2m (sqrt(2) R)^(2) + m(2R)^(2) + m(sqrt(2) R)^(2) + I_(ring)` `I = 2m (sqrt(2) R)^(2) + m(2R)^(2) + m(sqrt(2) R)^(2) + mR^(2)` `I = 12 mR^(2)` putting the value of `I` in equation (1) `KE = 6mR^(2) omega^(2)` `KE = 6m(R omega)^(2)` `KE = 6mv_(o)^(2)`. |
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| 139. |
A wodden log mass `M` and length `L` is hinged by a frictionless nail at `O.A` bullet of mass `m` strikes with velocity `v` and strikes with velocity `v` and sticks to it . Find angular velocity of the system immediately after after the collision about `O`. . |
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Answer» Correct Answer - `omega = (3mv)/((3m + M) L)` Angular momentum of the system about point `O` will remain conserved. `L_(i) = L_(f)` `mvL = I omega rArr mvL = [mL^(2) + (ML^(2))/(3)) omega` `omega = (3mv)/(L(3 m + M))`. |
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| 140. |
In `Q.137`, the tension in the cord in the above problem is :A. `(mg)/(M + m)`B. `(Mmg)/(M + 2m)`C. `(M + 2m)/(Mmg)`D. none of these |
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Answer» Correct Answer - B (b) According to problem : `T = (Ma)/(2) = (Mmg)/(M + 2 m)`. |
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| 141. |
An automobile engine develops `100` kilo`-`watt, when rotating at a speed of `1800 rev//min`. Find the torque developed by it. |
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Answer» The power delivered by the torque `tau` exerted on rotating body is given by `P = tau omega` or `tau = (P)/(omega)` Here `P = 100 KW = 100, 000` Watt `omega = ((1800)/(60)) xx 2 pi` =`60 pi rad//sec`, `tau = (10^(5))/(60 xx 3.14) = 531 N-m`. |
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| 142. |
In a rectangle `ABCD, AB = 21` and `BC = 1`. Axes `xx` and `yy` pass through centre of the rectangle. The moment of inertia is least about : .A. DBB. BCC. xxD. yy |
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Answer» Correct Answer - A (a) Because distribution of mass is closest around the axis `BD` in this case. |
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| 143. |
Four point masses, each of value `m`, are placed at the corners of square `ABCD` of side `l`. The moment of inertia of this system about an axis passing through `A` and parallel to `BD` is -A. `2ml^(2)`B. `sqrt(3) ml^(2)`C. `3 ml^(2)`D. `ml^(2)` |
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Answer» Correct Answer - C `I = 2m (l//sqrt(2))^(2) + m(sqrt(2) l)^(2) = 3 ml^(2)`. |
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| 144. |
A girl sits on a rolling chair, when she stretch her arms horizontally, her speed is reduced. Principle of conservation of angular momentum is applicable in this situation.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - A (a) If friction in the rotational mechanism is neglected, there is no extermal torque about the axis of rotation of the chair and hence `I omega` is constant. Where `I` is constant, where `I` is moment of inertoa and `omega` is angular velocity. Stretching the arms increases `I` about the rotation , results in decreasing the angular speed `omega`. Bringing the arms closer, body has the opposite effect. |
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| 145. |
A student sits on a freely rotating stool holding dumbbells, each of mass `5.0 kg` (Fig). When his arms are extended horizontally (Fig a), the dumbbells are `1.0 m` from the axis of rotation and the student rotate with an angular speed of `1.0 rad//s`. The moment of inertia of the student plus stool is `5.0 kg.m^2` and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position `0.50 m` from the rotationa are (Fig.) The new angular speed of the student is. .A. `1.5 rad//s`B. `2.5 rad//s`C. `2.0 rad//s`D. `1.25 rad//s` |
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Answer» Correct Answer - C ( c) The total angular momentum of the system of the student, the stool, and the weight about the axis of rotation is given by `I_(total) = I_(weights) + I_(student) = 2(mr^2) + 5.0 kg.m^2` Before : `r = 1.0 m` Thus, `I_i = 2(5.0 kg)(1.0 m)^2 + 5.0 kg.m^2 = 15 kg.m^2` After : `r = 0.50 m` Thus, `I_i = 2(5.0 kg)(0.50 m)^2 + 5.0 kg.m^2 = 7.50 kg.m^2` We now use conservation of angular momentum `I_f omega_f = I_i omega_i` or `omega_f = ((I_i)/(I_f)) omega_i = ((15.0)/(7.5))(1.0 rad//s) = 2 rad//s`. |
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| 146. |
An experimentor stands on a stool capable of rotating about a vertical axis and holds a rotating wheel of moment of intertia `I_(1)` about its axis and angualr velocity `epsilon_(1)` with its axis coinciding with the vertical axis of the stool. The moment of inertia of the stool and the experiment about the vertical axis is `I_(2)`. Calculate the change in kinetic energy of the system if the experiment rotates the wheel through i) `180^(@)` and ii) `90^(@)` |
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Answer» Correct Answer - `(2I_(1)^(2)omega_(1)^(2)//I_(2)), (I_(1)^(2)omega_(1)^(2))/(2(I_(2) + I_(1)/I_(2)))` |
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| 147. |
A particle at angular position `theta_(0)` inside a fixed smooth hemispherical bowl of radius r is projected horizontally with velocity `v_(0)`. Calculate its value so that the particle may rise to the top. |
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Answer» Correct Answer - `v_(0)=sqrt(2gr//costheta_(0))` |
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| 148. |
A student sits on a stool that is free to rotate about a vertical axis. He holds out his arms horizontally, with a 4-kg weight in each hand. The stool is set in rotation with angular speed of 0.5 revolution per second. Calcualate the angular speed of the student is 90 cm and his rotational inertia is 7kg `m^(2)`. [Hint: Use conservation principle] |
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Answer» Correct Answer - 0.96 rev/sec. |
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| 149. |
The motion of the centre of mass of a system of two particles is unaffected by their internal forces.A. irrespective of the actual direction of the internal forces.B. only if they are along the line joining the particlesC. only if they are at right angles to the joining the particlesD. only if they are obliquely inclined to the line joining the particle. |
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Answer» Correct Answer - A (b) Only in this case, the total effect of internal forces can be cancelled. |
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| 150. |
A long light thread is wond partly around cylinder of radius r and mass m partly on a simialr cylinder but free to turn about a fixed axis. The movable one is allowed to fall from rest. Find the velocity of the cylinder as a functiono of the height h through which it has fallen. Does the cylinder fall without slipping? |
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Answer» Correct Answer - `V=sqrt8gh//5`;No it falls with slipping |
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