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Physics NEET MCQ
Quiz 16
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Q.1
A hoop of radius r and mass m rotating with an angular velocity ω0 is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip ? [ IIT mains 2013]
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a) (rω0)/4
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b) (rω0)/3
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c) (rω0)/2
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d) rω0
Explanation
From law of conservation of momentum Li = Lf Iω0 = Iω + mvr And v = ωr m r2 ω0 = mr2ω + mvr = 2mvr v=(rω0)/2 Answer:(c)
Q.2
A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is …[ IIT Advance 2014]
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a) always radially outwards.
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b) always radially inwards.
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c) radially outwards initially and radially inwards later.
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d) radially inwards initially and radially outwards later.
Explanation
As shown in figure force on the wire = Now as θ goes on increasing mgcosθ decreases and (mv2)/R increases thus Initially force on wire inward and then outward with increased v option “d” correct We can calculate the angle after which force become radially out ward Loss of potential energy = gain in kinetic energy When F = 0 inward and outward force is zero 3mgcosθ=2mg cosθ=2/3 cosθ ≥2/3 force out ward Answer:(d)
Q.3
Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies ω1 and ω2 and have total energies E1 and E2, respectively. Then variations of their momenta p with positions x are shown in the figures. If a/b=n2 and a/R=n then the correct equation(s) is(are) … [ IIT Advance 2015 ]
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a)
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b)
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c)
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d)
Explanation
At extreme position energy is only potential energy At equilibrium position energy is only kinetic energy Similarly mω2 =1 …(ii) From (i) and (ii) Answer:(d)
Q.4
A ring of mass M and radius R is rotating with angular speed ω about a fixed vertical axis passing through its centre O with two point masses each of mass M/8 at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is 8ω/9 and one of the masses is at a distance of 3R/5 from O. At this instant the distance of the other mass from O is …[IIT Advantage 2015]
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a) 2R/3
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b) R/3
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c) 3R/5
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d) 4R/5
Explanation
Let the other mass be at a distance x from the centre. Conserving angular momentum about the axis : on solving x =4R/5 Answer:(d)
Q.5
More than one correct option. The position vector of a particle of mass m is given by the following equation r(t) = αt3 i + βt2j. Where α = 10/3 m s-3, β = 5 m s-2 and m = 0.1 kg. At t = 1 s, which of the following statement(s) is(are) true about the particle?
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a) The velocity is given by v = (10i + 10j ) m/s
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b) The angular momentum L with respect to the origin is given by
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c) The force is given by F = (i + 2j)N
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d) The torque τ with respect to the origin is given by
Explanation
Derivative of r gives formula for v Option “a” correct Derivative of v gives acceleration Force = ma = 0.1(20i + 10j)=2i + j N Option “c” wrong Angular momentum L = m(r × v ) Angular momentum = (-5/3) k Option b correct Torque Option d correct Answer:(a, b, d)
Q.6
More than one correct option. Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length l=a√24 through their centres. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω. The angular momentum of the entire assembly about the point ‘O’ is L (see the figure). Which of the following statement(s) is(are) true? [IIT Advance 2016]
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a) The magnitude of angular momentum of the assembly about its centre of mass is
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b) The centre of mass of the assembly rotates about the z-axis with an angular speed of ω/5
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c) The magnitude of the z-component of L is 55 ma2ω
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d) The magnitude of angular momentum of centre of mass of the assembly about the point O is ma2ω
Explanation
Connecting rod is mass less and do not rotate about its axis. It is given that set rolling without slipping on the surface . So sin component of angular velocity will be the angular velocity along Z axis Option a Axis of rotation is passing through centre of mass. Not considered axis perpendicular to rod. But axis of rotation of disc Moment of inertia of disc = mR2/2 Angular momentum L = I ω option a correct Angular velocity of CM = angular velocity of any of the disc. Since set rolling without slipping on the surface . So sin component of angular velocity will be the angular velocity along Z axis X2 = l2 + a2 X2 = 24a2 + a2 X=5a Sinθ = a/x = a/5a = 1/5 Angular velocity along z axis is ωsinθ =ω/5 Option b correct , c is wrong and d is wrong as component of ω along z axis is ωsinθ Answer:(a, b)
Q.7
PARAGRAPH [IIT Advance 2016] A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω is an example of a non-inertial frame of reference. The relationship between the force Frot experienced by a particle of mass m moving on the rotating disc and the force Fin experienced by the particle in an inertial frame of reference is Where Vrot is the velocity of the particle in the rotating frame of reference and is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed ω about its vertical axis through its centre. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis ω= ωk . A small block of mass m is gently placed in the slot at r =R/2 i at t = 0 and is constrained to move only along the slot. Q237A) The distance r of the block at time t is
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a)
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b)
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c)
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d)
Explanation
Terms in Frot is Fin + 2m(Vrot × ω) represents the normal reactions Fin=N1= mgk Another normal reaction Total reaction Answer:(d)
Q.8
More than one correct option. A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at x = 0, in a co-ordinate system fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is x and the velocity is v. At that instant, which of the following options is/are correct ? [ IIT Advance 2017]
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a) The x-component of dispalcement of the centre of mass of the block M is -mR/(m+M)
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b) The position of the point mass is
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c) The velocity of the point mass m is
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d) The velocity of the block M is
Explanation
Small object is slipping and not rolling thus object will leave the curved surface at end of the path Conservation of centre of mass, centre of mass of system will not change The small object be at a distance x from x=0 M is moved by -x final position of m =- x , initial position –R Change in possition of m = final position – initial position = (-x) –(-R) = R - x (option b is wrong) Since change in C.M. of system = 0 Mx = m(R-x) Option a is correct Option c From law of conservation of momentum Now loss of P.E. of m = gain in K.E. of m + gain in K.E. of M Option c is correct Velocity of M = Option d is wrong Answer:(a,c)
Q.9
Consider regular polygons with number of sides n = 3, 4, 5 ..... as shown in the figure. The center of mass of all the polygons is at height h from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is Δ. Then Δ depends on n and h as : …[ IIT Advance 2017]
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a)
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b)
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c)
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d)
Explanation
All sheps are symmetric, hence CM is at geometric centre When polygon rotate and polygone atten position shown by dotted line then centre of mass is at height bo, Now bo = h/cosθ As θ = π/n Answer:(c)
Q.10
More than one correct option A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vertical is θ. Which of the following statements about its motion is/are correct ? [ IIT Advance 2017]
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a) When the bar makes an angle θ with the vertical, the displacement of its midpoint from the initial position is proportional to (1 – cosθ)
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b) The midpoint of the bar will fall vertically downward
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c) Instantaneous torque about the point in contact with the floor is proportional to sinθ
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d) The trajectory of the point A is a parabola
Explanation
Initial position of central point L/2 Mid point will be located at position M, Distance OM Option a correct No force is acting along x at mid point and cm is located thus option b is correct Torque = Force ×perpendicular distance between the line of force and fix point Torque = mg OB = Mg (L/2) sinθ [ Option c wrong ] Centre of mass follows straight path Position M = (L/2) cosθ Position of point A y-axis position depends on position of B thus y= Lsin(π/2 –θ) = Lcosθ x-axis position depends on position of M as M is falling vetically down thus x= L/2 sinθ Now cos2θ + sin2θ = 1 Thus trajectory is ellips [ Option d is wrong] Answer:(a, b)
Q.11
A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque τ about an axis normal to the plane of the paper passing through the point Q. Which of the following options is/are correct ? [ IIT Advcance 2017]
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a) If the force is applied normal to the circumference at point X then τ is constant
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b) If the force is applied tangentially at point S then τ =0 but the wheel never climbs the step
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c) If the force is applied normal to the circumference at point P then τ is zero
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d) If the force is applied at point P tangentially then τ decreases continuously as the wheel climbs
Explanation
If force is applied at point X normal line of action passes through centre of sphere torque produced in zero. Thus torque is constant [Option a correct] τ ≠ 0 if force is tangential at S [ Option b wrong] Force applied normally at point P, torque due to force is zero zero [ Option c correct] If force is applied tangentially the torque is constant as F is constant [ Option d correct] Answer:(a, c)
Q.12
PARAGRAPH -2 One twirls a circular ring (of mass M and radius R) near the tip of one's finger as shown in FigureIn the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity ωThe rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is μ and the acceleration due to gravity is g. Q242A) The total kinetic energy of the ring is
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a)
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b)
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c)
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d)
Explanation
Normal force is produced due to centripital force. Produced because of circular path followed by C.M of ring around small circle of radius R. with velocity v Now Mg = f= µN Answer:(b)
Q.13
A solid cylinder of 500g and radius 10cm has moment of inertia about its natural axis ...[ AFMC 1997]
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a) 3.5 kg m2
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b) 5 × 10⁻³ kg m²
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c)2 × 10⁻³ kg m²
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d)2.5 × 10⁻³ kg m²
Explanation
M. I of Cylinder=½ MR2 On substituting the values we getM.I of Cylinder=2.5 × 10⁻³ kg m2 Answer:(d)
Q.14
A spherical solid ball of 1kg mass and radius 3cm is rotating about an axis passing through its centre with an angular velocity of 50 radian/sec. The kinetic energy of rotation is [ CPMT 1989]
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a) 4500J
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b) 90J
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d) (9/20) J
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c) 9×10⁻³ J
Explanation
Rotational kinetic energy E=½ I ω2 Here I=(2/5)mr2 m=1 kg r=3cm=3 × 10⁻²Thus I=(2/5) × 9 × 10⁻⁴ E=½ (2/5)× 9 × 10⁻⁴ × 502E=45 × 10⁻²=45/100E=(9/20) JAnswer: (d)
Q.15
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10⁻⁴ J by the end of the second revolution after the beginning of the motion?
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a) 0.1 m/s²
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b) 0.15 m/s²
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c) 0.18 m/s²
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d) 0.2 m/s²
Explanation
v2 = u2+ 2ats (s = 4πR as two revolution) 0.4 = 0 +2 at (4π×6.4×10⁻²) Answer:(a)
Q.16
A solid cylinder of 500g and radius 10cm has moment of inertia about its natural axis ...[ AFMC 1997]
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a) 3.5 kg m2
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b) 5 × 10⁻³ kg m²
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c)2 × 10⁻³ kg m²
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d)2.5 × 10⁻³ kg m²
Explanation
M. I of Cylinder=½ MR2 On substituting the values we getM.I of Cylinder=2.5 × 10⁻³ kg m2 Answer:(d)
Q.17
A spherical solid ball of 1kg mass and radius 3cm is rotating about an axis passing through its centre with an angular velocity of 50 radian/sec. The kinetic energy of rotation is [ CPMT 1989]
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a) 4500J
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b) 90J
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c) 9×10⁻³ J
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d) (9/20) J
Explanation
Rotational kinetic energy E=½ I ω2 Here I=(2/5)mr2 m=1 kg r=3cm=3 × 10⁻²Thus I=(2/5) × 9 × 10⁻⁴ E=½ (2/5)× 9 × 10⁻⁴ × 502E=45 × 10⁻²=45/100E=(9/20) JAnswer: (d)
Q.18
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10⁻⁴ J by the end of the second revolution after the beginning of the motion?
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d) 0.2 m/s²
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a) 0.1 m/s²
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b) 0.15 m/s²
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c) 0.18 m/s²
Explanation
v2 = u2+ 2ats (s = 4πR as two revolution) 0.4 = 0 +2 at (4π×6.4×10⁻²) Answer:(a)
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