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Physics NEET MCQ
Quiz 14
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Q.1
Two point masses m1 and m2 ( m >m2) attract each other with force which is inversely proportional to the square of the distance between them. The particles which are initially at rest, when released have centre of mass of the system moves
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a)moves towards m1
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b) moves towards m2
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c)move at right angles to the line joining m1 and m2
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d)does not move at all
Explanation
Since no external force acts on system centre of mass will not move Answer:(d)
Q.2
A heavy ball is thrown on rough horizontal surface in such a way that its slides with a speed Vo initially without rolling. It will roll without sliding when its speed falls to
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a) (2/7)Vo
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b) (3/7)Vo
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c) (5/7)Vo
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d) (7/5)Vo
Explanation
Retardation due to surface=a=F/m=µmg / m=µg ∴ Velocity after t sec, when sliding converted to rolling is V(t)=Vo - µgt --(i)Now, in case of rolling τ=µmg r ( here r is the radius of ball ) also τ=Iα Thus µmg r=Iα For sold sphere I=(2/5) mr2 Pure rolling starts when V(t)=rω(t) ∴ From equation (2) µgt=(2/5)V(t) Putting the value of µgt in equation (i) we get V(t)=Vo - (2/5)V(t) V(t) [ 1 + (2/5)]=Vo V(t)=(5/7)VoAnswer: (c)
Q.3
The centre of mass of a systems of two particles is
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a)on the line joining them and midway between them
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b) on the line joining them at a point whose distance from each particle is proportional to the square of the mass of that particle.
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c)on the line joining them at a point whose distance from each particle inversely proportional to the mass of that particle.
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d)On the line joining them at a point whose distance from each particle is proportional to the mass of that particle.
Explanation
Answer: (c)
Q.4
The ratio of the accelerations for a solid sphere(mass 'm' and radius 'R') rolling down an incline of angle 'θ'with out slipping and slipping down the incline without rolling is
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a) 2 : 5
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b) 7 : 5
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c) 5 : 7
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d) 2 : 3
Explanation
Acceleration of slipping down = gsinθ Acceleration without slipping = Answer:(c)
Q.5
A solid sphere is rotating about a diameter due to increase in room temperature, its volume increases by 0.5%, If no external torque acts. The angular speed of the sphere will.
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a)increase by nearly 1/3 %
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b)decrease by nearly 1/3 %
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c)increase by nearly (1/2 %
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d)decrease nearly by (1/2) %
Explanation
Volume V ∝ r3 --eq(1) Thus dV ∝3r2 --eq(2) dividing eq(2) by eq(1) we get dV/V=3 dr/r given dV/V=0.5% Thus dr/r=(1/3) × 0.5=1/6%No external torque so Iω=constantBut for sphere I=(2/5) mr2 (2/5) mr2ω=constant ∴ r2ω ∝ constant Thus r2 ∝ ω-1 --eq(3) taking derivative 2rdr ∝ (-1)ω-2 --eq(4) dividing eq(4) by eq(3) we get 2dr/r=- dω/ ω dω/ ω=- (2) × (1/6)=-1/3% Answer: (b)
Q.6
A circular disc of radius R is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through an angle of 60? and released. Its angularvelocity when it reaches the equilibrium position will be__
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a)
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b)
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c)
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d)
Explanation
Cenre of mass of circular disc will be at it centre. at a distance R from axis of rotation at rim Potential energy of the disc at 60°=mgR(1-cos60)=mgR/2 At equilibrium point kinetic energy=½ I ω2 By using parallel axis theorem I=½ MR2 + MR2 I=(3/2) MR2 Now P.E=kinetic energy Answer:(b)
Q.7
The moment of inertia of a hollow sphere of mass M and inner and outer radii R and 2R about the axis passing through its centre and perpendicular to its plane is
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a)
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b)
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c)
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d)
Explanation
We can obtain hollow sphere as it solid sphere of radius R is removed from a solid sphereof 2R mass of hollow sphere M=M1 – M2If ρ is the density then Moment of inertia of hollow sphere Answer: (d)
Q.8
Two strings of length l = 0.5 m each are connected to a block of a mass m = 2 kg at one end and their ends are attached to the point A and B, 0.5 m apart on a vertical pole which rotates with a constant angular velocity ω = 7 rad/sec. Find ratio T1 / T2 . Tension in the upper sting is (T1) and tension in lower string is (T2) [ use g = 9.8 m/s2]
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a) 1
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b) 3
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c) 9
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d) 6
Explanation
From resolution of vectors as shown in figure radius r = 0.5cosθ T1cosθ + T2cosθ = mω2( 0.5cosθ) T1 + T2 = 0.5 × m × ω2 --eq(1) And T1sinθ = T2sinθ + mg since θ = 30 from geometry of figure T1 = T2 + 2mg -- eq(2) From substituting value of T1 from eq(2) in eq(1) we get T2 + 2mg + T2 = 0.5 × m × ω2 2T2 = 0.5 × m × ω2 - 2mg 2T2 = 0.5 × 2 × 72 - 2 × 2 × 9.8 2T2 = 49 - 39.2 = 9.8 T2 = 4.9 From equation (2) T1 = 4.9 + 2 × 2 × 9.8 = 44.1 T1 / T2 = 44.1/4.9= 9 Answer: (c)
Q.9
From a uniform circular disc of radius R, a circular disc of radius R/6 and having centre at a distance +R/2 from the centre of the disc is removed. Determine the centre of mass of remaining
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a) -R/70
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b) +R/70
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c) -R/7
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d) +R/7
Explanation
Let us consider the disc is made up of two parts small and remaining part Then centre of mass of the disc is at (0,0,0) Thus (centre of mass of small disc) × ( mass of the small disc) = - (centre of remaining part of disc) × ( mass of remaining part) --eq(1) Let ρ be the mass per unit area then mass of small disc = m Mass of remaining part = M Let centre of mass of remaining part be at x Substituting values in equation (1) we get Answer: (a)
Q.10
A wheel is rotating at 900 rpm about its axis. When power is cut off it comes to rest in 1 minute, the angular retardation in rad / sec is ___
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a) π/2
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b) π/4
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c) π/6
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d) π/ 8
Explanation
900 rpm = 900/60 = 15 rotation per second = 30π rad/sec Form the formula ω = ωo - αt 0 = 30π - α × 60 α = π/2 Answer:(a)
Q.11
A thin uniform rod AB of mass M and length L is hinged at one end A to the horizontal floor. Initially it stands vertically and becomes horizontal. It is allowed to fall freely on the floor in the vertical plane. The angular velocity of the rod when its end B strikes the floor is ____
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a) √(g/l)
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b) √(2g/l)
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c) √(3g/l)
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d) 2√(g/l)
Explanation
Loss in P.E = gain in Rotational K.E. centre of Mass of rod is at L/2 Potential energy of rod = MgL/2 Kinetic energy = ½ I ω2 Moment of inertia of rod for axis of rotation at the end = ML2 / 3 thus we get Answer: (c)
Q.12
A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter42 cm. is removed from +ve x edge of the plate. Find the position of centre of mass of theremaining portion with respect to centre of mass of whole plate.
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a)- 7 cm
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b)+ 9 cm
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c)- 9 cm
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d)+ 7 cm
Explanation
Let the centre of circular plate before cutting smaller plate be at (0,0) From figure it is clear that centre of mass of smaller circular plate is at 7 cm Let ρ be the surface mass density of the plate Mass of smaller plate=π(21)2ρ Mass of bigger pate=π[(28)2 - (21) 2]ρ Mass of bigger pate=π[(49)(7)]ρ From definition of cetre of mass Thus π(21)2ρ × 7=- π[(49)(7)]ρ × x 212=-72 × x x=-9cm Answer: (c)
Q.13
A particle performing uniform circular motion has angular momentum L., its angular frequency is doubled and its K.E. halved, then the new angular momentum is
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a) L/2
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b)L/4
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c)2L
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d)4L
Explanation
Kinetic energy E=½ Iω2 E=½ Iω × ω But Iω= L E=½ Lω L=2E/ω for given L'=2(E/2)/ 2ω L'=L/4Answer: (b)
Q.14
A solid sphere is rotating about a diameter at an angular velocity ω. if it cools so that its radius reduces to 1/n of its original value. Its angular velocity becomes_____
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a) ω/n
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c)nω
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d)n2ω
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b) ω/n²
Explanation
No external torque acts on it thus angular moment is conserved Use formula I1ω1=I2ω2 And I=(2/5) mr2 Answer:(d)
Q.15
Two blocks of masses 10 kg an 4 kg are connected by a spring of negligible mass and placedon a frictionless horizontal surface. An impulse gives velocity of 14 m/s to the heavier block inthe direction of the lighter block. The velocity of the centre of mass is
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a) 30 m/s
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b) 20 m/s
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c) 10 m/s
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d) 5 m/s
Explanation
From the formula for centre of mass Answer: (c)
Q.16
A circular disc of mass m and radius r is rolling on a smooth horizontal surface with a constant speed v. Its kinetic energy is _____
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a)(1/4) mv2
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b) (1/2) mv2
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c)(3/4) mv2
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d) mv2
Explanation
Since disc is rolling it will have two types of energy liner kinetic and rotational kinetic energy E=½ mv2 + ½ Iω2 since it is rolling without sliding ω=v/R and I=(1/2) MR2 on substituting values of I and ω we get option "c"Answer: (c)
Q.17
Two disc of same thickness but of different radii are made of two different materials such thattheir masses are same. The densities of the materials are in the ratio 1:The moment of inertiaof these disc about the respective axes passing through their centres and perpendicular to theirplanes will be in the ratio.
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a) 1 : 3
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b) 3 : 1
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c)1 : 9
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d)9 : 1
Explanation
Moment of inertia of disc I=½ MR2 Given that masses are same, and thickness is same , if ρ is mass per unit area Now mass=π R2 ρ Thus R2 ∝ (1/ρ) And I ∝ R2 thus I ∝ 1/ρ Thus correct option is "b" Answer: (b)
Q.18
A body of mass m is tied to one end of spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimeter. If the angularvelocity is doubted, the elongation in the spring is 5 cm. The original length of spring is...
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a)16 cm
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b)15 cm
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c)14 cm
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d)13 cm
Explanation
Let original length be L and spring constant be k Centrifugal force=force of spring mω2r=k × x here x is increase in length of spring case I: Elongation x=1 mω2 ( L + 1 )=k --eq(1) case II : elongation x=5 cm and angular speed in doubled 4mω2 ( L + 5 )=5k --eq(2) on solving equation (1) and (2) we get L=15 cm Answer: (b)
Q.19
A mass m rotating freely in horizontal circle of radius 1m on frictionless smooth table supports a stationary mass 2m, attached to the other end of the string passing through smooth hole O in the table, hanging vertically. Find the angular velocity of rotation
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a)√(2g)
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b)√(g)
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c)√(3g)
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d)√(4g)
Explanation
Tension in the string T=2mg Centrifugal force on the mass m=mω2 r Thus to continue rotational motion of mass m Tension=centrifugal force 2mg=mω2 r ω=√(2g/r) but r=1 m Thus ω=√(2g) Answer:(a)
Q.20
Three particles of the same mass lie in the (X, Y) plane, The (X, Y) coordinates of their positions are (1, 1), (2, 2) and (3, 3) respectively. The (X,Y) coordinates of the centre of mass are
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a) (1, 2)
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b) (2, 2)
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c) (1.5, 2)
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d) (2, 1.5)
Explanation
Answer: (b)
Q.21
A Pulley of radius 2 m is rotated about its axis by a force F = (20 t - 5t2 ) N where t is in sec applied tangentially. If the moment of inertia of the Pulley about its axis of rotation is 10 Kgm2 , the number of rotations made by the pulley before its direction of motion is reversed is
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a) more than 3 but less then 6
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b) more than 6 but less then 9
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c) more than 9
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d) Less then 3
Explanation
Here direction of Motion will be reversed when force F = 0 or 20 t – 5t2 = 0 or t = 4sec. If α is angular acceleration then torque τ = Iα = F.r OR 10 × α = (20 t – 5t2 ) × 2 OR α = 4t - t2 2π = 44/7 rad = 1 rotation Therefore 64/3 rad = 3.4 rotation option "a" is correct Answer: (a)
Q.22
A cubical block of side a is moving with velocity V on a horizontal smooth plane as shown in figure. It hits a ridge at point O. The angular speed of the block after it hits O is ...
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a) 3V/4a
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b) 3V/2a
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c) √(3V)/ √(2a)
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d) zero
Explanation
r = √2 ( a/2) or r2 = a2/2 Li = MV × (a/2) Final momentum = ( momentum of inertia about CM. + Mr2)ω For cube Moment of inertia about C.M = Ma2 /6 Net torque about O is zero Initial angular momentum = momentum × perpendicular distance of velocity vector from O ∴ angular momentum (L) about O will be conserved or Li = Lf MV (a/2) = ( Icm + Mr2)ω Answer: (a)
Q.23
Two uniform rod of equal length but different masses are rigidly joined to form L shaped body which is then pivoted as shown. If in equilibrium the body is in the shown configuration ratio M/m will be ..
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a) 2
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b) 3
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c) √2
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d) √3
Explanation
Gravitational force mg and Mg is acting done ward direction Cenetre of mass of each arm is at (l/2). Thus perpendicular distance between fix point O and line of action of force is (l/2)sin60 and (l/2)sin30. Net torque about O should be zero Thus Answer:(d)
Q.24
A light rod carries three equal masses A, B and C as shown in the figure the velocity of B in vertical position of rod if it is released from horizontal position as shown in the figure is ...
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a)
0%
b)
0%
c)
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d)
Explanation
Loss in PE = gain in angular kinetic energy Let I be the moment of inertia at fixed point Loss in PE = E Answer: (d)
Q.25
Two identical hollow spheres of mass M and radius R are joined together and the combination is rotated about an axis tangential to one sphere and perpendicular to the line connecting their centres. The moment of inertia of the combination is ________.
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a)10 MR2
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b) (4/3) MR2
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c)(32/3) MR2
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d)(34/3) MR2
Explanation
Answer: (d)
Q.26
Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of …..
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a) 0.4 m from mass of 0.3 kg
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b) 0.98 m from mass of 0.3 kg
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c)0.7 m from mass of 0.7 kg
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d)0.98 m from mass of 0.7 kg
Explanation
let x be the distance from 0.3 kg mass I=0.3x2 + 0.7 (1.4 – x)2For minimum work moment of inertia of the system should be minimum is dI/dt=0dI/dt=0.3 × (2x) – (0.7) × 2 × (1.4 – x) 0.3 ×(2x) – (0.7) × 2 × (1.4 – x)=0 0.6x - 1.96 + 1.4x=0 2x=1.96x=0.98 m from mass 0.3kg Answer: (b)
Q.27
A smooth sphere A is moving on a friction-less horizontal plane with angular speed ωand centre of mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA and ωB respectively, Then
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a) ωA < ωB
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b) ωA=ωB
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c) ωA=ω
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d) ω=ωB
Explanation
As it is head-on elastic collision between two identical balls there fore they will exchange their linear velocity is A comes to rest and B starts moving with linear velocity V.As there is no friction any where, torque on both the spheres about their centre of mass is zero and their angular velocities remains unchanged.Therefore ωA=ω and ωB=0 Answer:(c)
Q.28
Consider a body as shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J=MV is imparted to the body at one of its ends, what would be its angular velocity. What is V ?
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a) V / L
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b) 2V / L
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c) V / 3L
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d) V / 4L
Explanation
Given system of two particles will rotate about its centre of mass. Initial angular momentum=momentum × Perpendicular distance between direction of momentum and centre of mass=mV(L/2) Final angular momentum=2Iω=2M(L/2)2 ω By law of conservation of angular momentum Answer: (a)
Q.29
A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be.
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a)
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b)
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c)
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d)
Explanation
From law of conservation of energy mgh=½Mv2 + ½ I ω2 Use following formula to find ω V=ω × R and For cylinder I=½ MR2Answer: (c)
Q.30
Identify the correct statement for the rotational motion of a rigid body
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a) Individual particles of the body do not undergo accelerated motion
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b) The centre of mass of the body remains unchanged.
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c)The centre of mass of the body moves uniformly in a circular path
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d) Individual particle and centre of mass of the body undergo an accelerated motion
Explanation
Answer:(b)
0 h : 0 m : 1 s
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