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Physics NEET MCQ
Centre Of Mass And Rotational Motion. Mcq
Quiz 4
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Q.1
A solid sphere is rotating in free space. if the radius of the sphere is increased keeping mass same which one of the following will not be affected? [ AIEEE 2004]
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a) Angular velocity
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b) Angular momentum
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c)Moment of inertia
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d)Rotational kinetic energy
Explanation
Angular momentum remains same since external torque is zeroAnswer: (b)
Q.2
A particle performing uniform circular motion has angular frequency if doubled and its kinetic energy halved, then the new angular momentum is [ AIEEE 2003]
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a) L/4
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b) 2L
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c)4L
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d)L/2
Explanation
Kinetic energy E=½ I ω2 but L=Iω ⇒ I=L/ω∴ E=½ ( L/ω) × ω2=½ Lω∴ E /E'=L × ω / L'× ω'Given E'=E/2 and ω'=2ω∴ L'=L /4 Answer:(a)
Q.3
The moment of inertia of a solid sphere and a spherical shell of equal masses about their diameters are equal. the ration of their radii are .. [ CBSE 1997]
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a) 5:3
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b) 3:5
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c) √5 : √3
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d) √3 : √5
Explanation
Moment of Inertia of solid sphere I1= (2/5)m r21 Moment of Inertia of Hollow sphere I2= (2/3) m r22 Given I1 = I2 ∴ (2/5) m r21 = (2/3) m r22 ∴ r1 / r2 = √5 / √ 3 Answer: (c)
Q.4
A body A of mass M while falling vertically downwards under gravity breaks into two parts, a body B of mass (1/3)M and a body C of masses (2/3)M. The centre of mass of body B and C taken together shifts compared to that of body A towards [ AIEEE 2005]
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a) does not shift
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b) depends on height of breaking
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c) body B
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d) body C
Explanation
Does not shift as no external force acts. The centre of mass of the system continues its original path. It is only the internal forces which comes into play while breaking Answer: (a)
Q.5
A 'T' shaped object with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C. [ AIEEE 2005]
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a)(3/2)l
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b) (2/3)l
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c)l
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d)(4/3)l
Explanation
centre of mass of 2l length of vertical side will be at (0,l) and centre of mass of horizontal side will be at (0,2l)let mass of horizontal be m, then mass of vertical will be 2m To have a linear motion, the force F has to be applied at the centre of massi.e. the point 'P' has to be at the centre of mass Answer: (d)
Q.6
A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is ω. Its centre of mass rises to a maximum height of [ AIEEE 2009]
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a)
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b)
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c)
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d)
Explanation
The moment of inertia of the rod about O is (1/3) ml2The maximum angular speed of the rod is when the rod is instantaneously vertical.The energy of the rod in this condition is ½ I ω2 where I is the moment of inertia of the rod about O. When the rod is in its extreme portion, its angular velocity is zero momentarily. In this case, the energy of the rod is mgh, where h is the maximum height to which the centre of mass ( C.M. rises)∴ mgh=½I ω2by substituting the value of I in above equation Answer: (c)
Q.7
A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincides. The centre of mass of the new disc is α/R from the centre of the bigger disc. The value of α is [ AIEEE 2007]
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a) 1/4
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b) 1/3
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c)1/2
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d)1/6
Explanation
Let the mass per unit area be σ Then the mass of the complete disc=Σ[π(2R)2]=4πσR2 The mass of the removed disc m2=σ( πR2)Let us consider the above situation to be a complete disc of radius 2R on which a disc of radius of NEGATIVE mass is superimposed. Let O be the origin. Then the above figure can be redrawn keeping in mind the concept of centre of mass as : Now according to formula for centre of mass Substituting values of masses and position vectors in above equations we get Answer:(b)
Q.8
For the given uniform square lamina ABCD, whose centre is O, [ AIEEE 2007]
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a) IAC=√2 IEF
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b) √2 IAC=IEF
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c) IAD=3 IEF
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d) I AC=IEF
Explanation
By the theorem of perpendicular axis Iz=Ix + Iy By symmetry of figure Ix=Iy we get Iz=2 Iy --eq(1)∴ IEF =Iz / 2 { As IEF = IY}Again by same theorem Iz=IAC + IBD=2 IAC (By symmetry of figure IAC=IBD) ∴ I AC=Iz / 2 --eq(2) Thus from eq(1) and eq(2) we get IEF=IAC Answer: (d)
Q.9
A circular disc X of radius R is made from iron plate of thickness t and another disc Y of radius 4R is made from an iron plate of thickness t/Then the relation between the moment of inertia Ix and Iy is [ AIEEE 2003]
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a)IY=32IX
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b) IY=16IX
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c)IY=IX
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d)IY=64IX
Explanation
We know that density ρ=mass(M)/ volume(V) Thus M=ρ × VM=ρ × (πR2 × t) The moment of inertia of a disc is given by I=½ MR2 I=½( ρ × πR2 × t)R2 I=½ ( π ρt R2) I ∝ t × R2 Now IX ∝ tX × R4X Given thickness of Y is one fourth of thickness of X and radius is four times the radius of xand IY ∝ (t/4) × (4R)2Answer: (d)
Q.10
A body falling vertically downwards under gravity breaks in two unequal parts of unequal masses. the centre of mass of the two parts taken together shifts horizontally towards
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a) lighter piece
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b) heavier piece
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c)depends on the vertical velocity at the time of breaking
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d)does 't shift horizontally
Explanation
Body is falling vertically down thus there is no initial momentum along the horizontal direction, the centre of mass cannot shift horizontally. Answer:(d)
Q.11
A sphere can not roll on
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a) a smooth inclined surface
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b) a smooth horizontal surface
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c) a rough inclined surface
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d) a rough horizontal surface
Explanation
A sphere cannot roll on a smooth inclined surface because of absence of the force of friction which produces torque for rolling Answer: (a)
Q.12
Two racing cars of mass m1 and m2 are moving in circles of radii r1 and r2 respectively. Their speeds are such that each makes a complete circle in the same length of time t. The ratio of the angular speed of the first to the second car is [ M.N.R. 1995]
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a)m1 : m2
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b) r1 : r2
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c)1:1
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d)m1r1 : m2r2
Explanation
Formula for angular velocity ω=2π/T Since periodic time for both the cars is same angular velocity is also same Answer: (c)
Q.13
A ring and disc have the same mass and radius. The ratio of their moment of inertia about an axis is [ punjab CET 1995]
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a) 1:1
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b) 2:1
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c)4:1
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d)1:2
Explanation
Answer: (b)
Q.14
A man is sitting with folded hands on a revolving table. Suddenly, he stretches his arms. Angular speed of the table would [ CPMT 1996]
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a) increase
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b) decrease
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c)remain the same
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d)nothing can be said
Explanation
On stretching arms, distance K increases I=MK2 increases. As Iω=constant, therefore, angular velocity ω decreases. Answer:(b)
Q.15
When a body rolls down an inclined plane, its potential energy is converted into [ CBSE 1996]
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a) translational K.E only
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b) rotational K.E only
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c) translational and rotational K.E
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d) none of above
Explanation
On rolling, the body has both translational as well as rotational motion. Therefore, potential energy of the body is converted into translational and rotational energy Answer: (c)
Q.16
A raw egg and a hard boiled egg are made to spin on a table with the same angular speed about the same axis. The ratio of the time taken by the two to stop is [ CPMT 1996]
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a) =1
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b) < 1
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c) > 1
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d) none of the above
Explanation
A raw egg has some fluid in it so on spinning the fluid is thrown outwards. Where as boiled egg is solid from inside. Therefore Ir > Ib As I× ω = constant ∴ ωr < ωb Thus ωr / ωb < 1 ∴ tr / tb < 1 Answer: (b)
Q.17
A couple produces a ... [ pre-medical dental, 1997]
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a) pure linear motion
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b) pure rotational motion
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c) both linear and rotational motion
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d) no motion
Explanation
A couple has turning effect and produce pure rotational motion Answer: (b)
Q.18
If sphere is rolling , the ratio of the rotational energy to the total kinetic energy is given by [ MEE BHU 1997]
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a) 7:10
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b) 2:5
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c) 10:7
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d) 2:7
Explanation
Total kinetic energy = KE. linear + K.E. Rotational Total K.E. E= ½ mv2 + ½ I ω2 For sphere I = (2/5) mr2 and v = ω r On taking the ratio of rotational kinetic energy to total kinetic energy we get Answer:(d)
Q.19
The least coefficient of friction for an inclined plane at an angle α with horizontal, in order that a solid cylinder will roll down it without slipping is [ CET 1997 punjab]
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a) 2/3( tan α)
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b) 2/7 (tan α)
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c)1/3 (tanα)
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d)4/3( tan α)
Explanation
Answer: (c)
Q.20
A mass m is moving with a constant velocity along a line parallel to the x-axis away from the origin. Its angular momentum with respect to the origin. Its angular momentum with respect to the origin is [ IIT 1997]
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a) zero
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b) constant
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c)goes on decreasing
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d)goes on increasing
Explanation
Angular momentum of a body about a given axis is the product of linear momentum and perpendicular distance of line of action of linear momentum vector from the axis of rotation.in given problem object is moving parallel to x-axis away from origin thus perpendicular distance will be along y-axis and will not change thus r is constant Given that the object moves with constant velocity thus momentum is also constant Angular momentum=momentum × perpendicular distance between line of action of momentum Since momentum and perpendicular distance is constant Angular momentum is constant Answer:(b)
Q.21
Three identical metal balls each of radius r are placed touching each other on a horizontal surface such that an equilateral triangle is formed when such centres of three balls are joined. The centre of mass of system is located at [CBSEPMT 1995]
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a) horizontal surface
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b) centre of one of the balls
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c) line joining centre of any two balls
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d) point of intersection of medians
Explanation
The centre of mass of three identical balls will be at the point of intersection of medians of the triangle. Answer: (d)
Q.22
The moment of inertia of disc of mass M and radius R about an axis, which is tangential to circumference of disc and parallel to its diameter [ CBSE PMT 1999]
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a)(3/2) MR2
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b) (2/5) MR2
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c)(5/4) MR2
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d)(4/4) MR2
Explanation
Moment of inertia of disc about its diameter=(1/4)MR2Using theorem of parallel axes I=(1/4)MR2 + MR2 I=(5/4) MR2Answer: (c)
Q.23
The radius of gyration of a uniform rod of length L about on axis passing through its centre of mass [AMU 1999]
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a) L/ √12
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b) L2 / 12
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c)L/√3
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d)L/√2
Explanation
moment of inertia for uniform rod of length L=(1/12)M L2 Monet of inertia in terms of radius of gyration I=MK2 On comparing above equations we get Radius of gyration K=L / √12 Answer: (a)
Q.24
A wheel is rotating at 900 r.p.m about its axis. When power is cut off it comes to rest in 1 minute. The angular retardation is rad/s2 [ MP CEE 1999]
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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
Initial angular velocity 900 r.p.m One rotation=2π Thus 900 rotation per minute=900 × 2πInitial angular velocity per second ω=900 × 2π / 60=30π Final angular velocity=0Angular retardation=( Final angular velocity - Initial angular velocity) / 60 Angular retardation=π/2 Answer:(a)
Q.25
A solid cylinder rolls down a smooth inclined plane 4.8m high without slipping what is itslinear speed at the bottom of the plane, if it starts rolling from the top of the plane?(take g=10 m/s2)
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a) 4 m/s
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b) 2 m/s
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c) 10 m/s
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d) 8 m/s
Explanation
Linear velocity for solid cylinder K=R/√2 Answer: (d)
Q.26
A cylindrical solid of mass M has radius R and length L. Its moment of inertia about a generator is [ MP PMT 2009]
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a)
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b)
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c)½ MR2
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d)(3/2) MR2
Explanation
Generator is axis touching surface of cylinder and parallel to axis of cylinder. Using theorem of parallel axes I=I0 + MR2 I=½ MR2 + MR2I=(3/2) MR2Answer: (d)
Q.27
An earth satellite is moving around the earth in a circular orbit. In such a case, what is conserved? [ DEC 2001]
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a) velocity
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b) linear momentum
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c)angular momentum
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d)none of the above
Explanation
Since no external torque is acting in the motion of satellites around earth, angular momentum is conservedAnswer: (c)
Q.28
The acceleration of solid cylinder rolling down an inclined plane of inclination 30° is [ DPMT 2002]
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a) g/3
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b) g/2
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c)g
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d)g/4
Explanation
Acceleration of rolling cylinder without slipping along the plane is given by a=(2/3)g sinθ Given θ=30° a=(2/3) g sin 30=g/3 Answer:(a)
Q.29
When a torque acting upon a system is zero, which of the following will be constant? [ J and K CET 2002]
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a) force
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b) linear momentum
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c) linear impulse
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d) none of these
Explanation
since torque is zero Thus angular momentum is constant, but it is not given in option so option 'd' is correct. Answer: (d)
Q.30
An athlete throws a discus from rest to a final angular velocity of 15rad/s in 0.27s before releasing it. During acceleration, discuss moves in a circular arc of radius 0.81m. Acceleration of discuss before it is released is .... ms2 [ kerala PET 2004]
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a)45
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b) 182
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c)187
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d)192
Explanation
Here Initial angular velocity ω1=0 and final angular velocity ω2=15rad/sec time t=0.27 s and radius r=0.81m Angular acceleration=Change in angular velocity / time α=(ω2 - ω1 ) /t α =( 15 - 0) / 0.27=55.55 formula for tangential acceleration=r × angular accelerationa=0.81 × 55.55=45 m/s2Answer: (a)
0 h : 0 m : 1 s
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