MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
NEET
Physics NEET MCQ
Quiz 13
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
Angular momentum of the body is conserved [ MPPET 1995]
0%
a) Always
0%
b) never
0%
c) in presence of external torque
0%
d) in absence of external torque
Explanation
Answer: (d)
Q.2
You are given two circular disc which have equal weight and equal thickness. they are made up of different metals having densities d1 and dthere radii are R1 and R2 respectively. For disc that will have more moment of inertia about the central axis than the other, the condition is [ MPPET 1993]
0%
a) d1 > d2
0%
b) R1 > R2
0%
c)d1> d2 and R1 > R2
0%
d)d1 < d2 and R1 < R2
Explanation
given that mass of both the disc is same thus I1 / I2=R1 / R2 ∴ If I1 > I2 , R1 >R2 Answer: (b)
Q.3
If the moment of inertia of disc about the tangent in its plane is I, its moment of inertia about the tangent perpendicular to the plane will be [ Raj. PET 1996]
0%
a)6I /5
0%
b)3I/4
0%
c)3I/2
0%
d)5I/4
Explanation
moment of inertia of the disc about tangent in its plane I=(5/4)MR2 MR2=4I/5 Moment of inertia of the disc in perpendicular plane to the disc=(3/2)MR2 Substituting value of MR2 we get Moment of inertia of the disc in perpendicular plane to the disc=6I/5 Answer:(a)
Q.4
The rotational and translation kinetic energy of rolling body are same, the body is [ RajPET 1996]
0%
a) Disc
0%
b) Sphere
0%
c) Cylinder
0%
d) Ring
Explanation
given ½ mv2=½ I ω2 angular velocity ω=v/R substituting in above equation we get ½ mv2=½ I ( v/R)2 I=MR2 above formula for moment of inertia is for ring Answer: (d)
Q.5
A body having moment of inertia about its axis of rotation equal to 3 kg m2 i rotating with angular velocity equal to 3rad/s. Kinetic energy of this rotating body is the same as that of body of mass 27kg moving with a speed of [ SCRA 1994]
0%
a)1.0 ms2
0%
b) 0.5 ms2
0%
c)1.5 ms2
0%
d)2.0 ms2
Explanation
given :Rotational kinetic energy=Translation kinetic energy of 27kg mass ∴ ½ I ω2=½ Mv2 ½ 3 × 32=½ 27 v2 ∴ v=1 Answer: (a)
Q.6
A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and moment of inertia about it is I. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be [ MPPMT 1994]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Loss in potential energy of mass=Rotational kinetic energy of wheel + Translation energy of mass mgh=½I ω2 + ½ ½ mv2 we know that v=ω R 2mgh=Iω2 + mR2ω2 ∴ ω2 (I + mR2)=2mgh Answer: (b)
Q.7
A rigid body rotates about a fixed axis with variable angular velocity equal to ω=α - βt, at time t, α and β are constants. find the angle turned through by the body before coming to rest
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
When body comes to rest ω=0 0=α - βt or time taken to stop is t=α/βNow dθ / dt=ω=α - βt dθ=(α - βt)dt ∴ Angle turned in this time is Answer:(a)
Q.8
Before jumping in water from above a swimmer bends his body to [ MPPMT 1994]
0%
a) increase moment of inertia
0%
b) decrease moment of inertia
0%
c) decrease the angular momentum
0%
d) reduce the angular velocity
Explanation
Answer: (b)
Q.9
A particle of mass m=5 is moving with a uniform speed v=3√2 in the XY plane along the straight line Y=X +The magnitude of the angular momentum about origin is
0%
a)zero
0%
b) 30 units
0%
c)75 units
0%
d)40√2 units
Explanation
Equation Y=X + 4 represent line with intercept on the y axis at 4 units According to formula L=mvr [ Here r is the perpendicular distance between the origin and line of action of moment of inertia as shown in figure by red line ] Since slope = 1 therefor tanθ=1 or θ= 45. From geometry of figure r=4sin45=4/√2 L=5 × (3/√2) × (4/√2) L=30 unitsAnswer: (b)
Q.10
A tap can be operated easily using two fingers because : [ UGET 1995]
0%
a) the force available for the operation will be more
0%
b) this helps application of angular force
0%
c)the rotational effect is caused by the couple formed
0%
d)the force formed by one finger overcomes friction and the other finger the force for the operation
Explanation
Answer: (c)
Q.11
Two solid spheres each of mass M and radius R/2 are attached to weightless rod of length 2R, the moment of inertia about the axis passing through the centre of one of the sphere and perpendicular to the length of the rod be [ raj PMT 1996]
0%
a)(21MR2) / 5
0%
b)(2MR2) / 5
0%
c)(5MR2) / 2
0%
d)(5MR2) / 21
Explanation
Formula for moment of inertia about axis of sphere=(2/5)Mr2given r=R/2 Thus Moment of inertia about YY"=(2/5) M (R/2)2 =MR2 / 10 Moment of inertia of the other sphere about the axis YY"=(1/10)MR2 + M(2R)2=(41/10) MR2 / 10 Thus total moment of inertia about YY'=(MR2) /10 + (41MR2) /10 total moment of inertia about YY'=(42MR2) / 10=(21MR2)/ 5 Answer:(a)
Q.12
Two masses M and m are attached to a vertical axis by weightless threads of combined length l.They are set in rotational motion in a horizontal plane about this axis with constant angular velocity ω. If the tension in the threads are same during motion, the distance of M from the axis is [ MPPMT 1995]
0%
a) Ml/ (M + m)
0%
b) ml / (M + m)
0%
c) (M + m)l / M
0%
d) (M + m)l / m
Explanation
As shown in figure M and m are the two masses attached to axis of rotation O by mass less string Since tension is same centrifugal force due to both the masses is same Mxω2=m(l - x)ω2 x=ml / (M + m) Answer: (b)
Q.13
Two rings of the same radius and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the ring is ( mass of ring=m, radius of ring=r) [ MPPMT 1994]
0%
a)½ mr2
0%
b) mr2
0%
c)(3/2) mr2
0%
d)2mr2
Explanation
The axis is passing through the centre and perpendicular to the plane of one ring is the diameter of other ring. Therefore M.I. of system I=Mr2 + ½ Mr2 I=3Mr2 / 2Answer: (c)
Q.14
A disc of radius 33cm is hanged by a point at circumference by horizontal nail. Period of oscillation is 1.42 seconds, value of g by this experiment will be [ Raj. PMT 1997]
0%
a) 9.25 m/s²
0%
b) 9.68 m/s²
0%
c)9.86 m/s²
0%
d)100 m/s²
Explanation
periodic time is given by formula ( refer oscillation) Inertial factor = I , and restoring factor = mgd as point of suspension is at circumference By theorem of parallel axis By substituting value of I in equation for periodic time we get here d=33cm=0.33m , T=1.49 seconds by substituting the values and on simplification we get g=9.68m/s2Answer: (b)
Q.15
A smooth wire is bent into a vertical circle of radius a. A bead P can slide smoothly on the wire. The circle is rotated about diameter AB as axis with speed ω as shown in figure . the bead P is at rest with respect to the circular ring in the position shown. Then ω2 is equal to
0%
a)2g/a
0%
b)2g/(a√3)
0%
c)g√3 /a
0%
d)2a/ (g√3)
Explanation
Let N b the normal reaction on the bead. resolving it into two components. we get Ncosθ=mg and Nsinθ=m(a/2)ω2 Thus tanθ=aω2 / 2g ω2=(2g/a ) tanθ From figure sinθ=a/2a=1/2 or θ=30° ∴ ω2=(2g/a) tan30 ω2=2g / (a√3) Answer:(b)
Q.16
The moment of inertia of rod ( length l, mass m) about an axis perpendicular to the length of the rod and passing through a point equidistant from its mid-point and one end is [ MPPMT 1999]
0%
a) ml2 / 12
0%
b) 7ml2 / 48
0%
c) 13ml2 / 48
0%
d) 19ml2 / 48
Explanation
Momentum of inertia along the axis passing through the centre=Ml2 / 12 Axis is passing through the point equidistant from mid-point thus distance between axis passing through the centre and given axis is l/4 By applying parallel axis theorem Moment of Inertia=Ml2 / 12 + M(l/4)2 M.I=(7/48)Ml2 Answer: (b)
Q.17
railway tracks are banked on curves so that
0%
a)no frictional force may be produced between the track and wheels
0%
b) the train may not fall down inward
0%
c)necessary centripetal force may be obtained from the horizontal component of normal reaction due to the track
0%
d)none of the above
Explanation
Answer: (c)
Q.18
A motor cyclist rides on cylindrical wall of velodrome ( hollow metal cylinder kept with the axis vertical) of radius r rotating about its axis at the rate f rev/sec. the least coefficient of friction necessary so that he does not step down is
0%
a) g / (4π2f2r)
0%
b) (4π2f2r) / r
0%
c)(4π2f2) / gr
0%
d)gr/(4π2r2)
Explanation
From figure it is clear that N=mrω2 µN=mgµ mrω2=mg µ=g/(rω2) but ω=2πf ∴ µ=g / (4π2f2r)Answer: (a)
Q.19
If a particle of mass 'm' is moving in a horizontal circle of radius 'r' with a centripetal force ( -1/r2) the total energy is [ JIPMER 1998]
0%
a)-1/2r
0%
b) -1/r
0%
c)-2/r
0%
d)-4/r
Explanation
Given mv2 / r=+1/r2 mv2=1/r Kinetic energy ½ mv2=1/2r Potential energy=- ∫ F dr P.E=∫ F dr=∫ (1/r2)dr=-1/r Total energy=Kinetic energy + potential energy Total energy=1/2r - 1/r=-1/2r Answer:(a)
Q.20
A particle of mass m is executing uniform circular motion on a path of radius r. If p is the magnitude of its linear momentum, the radial force acting on the particle is : [ MPPMT 1994]
0%
a) pmr
0%
b) rm/p
0%
c) mp2 /r
0%
d) p² / rm
Explanation
F=mv2 /r F=m2 v2 / rm F=p2 / mr Answer: (d)
Q.21
A body of mass m hangs at one end of string of length l, the other end of which is fixed. It is given a horizontal velocity so that the string would just become slack, when it makes an angle of 60° with the upward drawn vertical. The tension in the string at this position is [ ISM Dhanbad 1994]
0%
a) 4.5 mg
0%
b) mg
0%
c)3m
0%
d)√3 mg
Explanation
The string just becomes slack at B ∴ TB=0 from figure By Energy conservation at A and B Therefore tension at A=mg + 3.5mg=4.5mgAnswer: (a)
Q.22
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a is varying with time t, a=k2rt2 where k is a constant. The power delivered to the particle by the forces acting on it is [ IIT 1994]
0%
a) 2πmk2r2t
0%
b) mk2r2t
0%
c)(mk4r2t5) / 3
0%
d)zero
Explanation
Given a=k2rt2=v2 / r∴ v2=k2r2t2Kinetic energy K=½ m v2 By substituting value of v2 from above we get K=½ mk2r2t2According ro work energy theorem W=ΔK W=½ mk2r2t2 - 0W=½ mk2r2t2Power P=dW/dt, taking derivative of W with respect to time P=½ mk2r2 (2t)P=mk2r2 tAnswer: (b)
Q.23
The tension in the string revolving in a vertical circle with a mass m at the end which is .. [ EAMCET 1995]
0%
a) mv2 / r
0%
b) (mv2 / r) - mg
0%
c)(mv2 / r) + mg
0%
d)mg
Explanation
Centrifugal and gravitational force both are down ward direction at lower position Answer:(c)
Q.24
A pendulum is suspended from the ceiling of a car rotating in ac circular path with an acceleration of 0.49m/sThe angle which the string makes with the vertical is
0%
a) tan⁻¹ ( 1/√3)
0%
b) tan⁻¹√3
0%
c) tan-1 0.05
0%
d) tan-1 0.005
Explanation
from figure tanθ=a/g θ=tan-1 0.05 Answer: (c)
Q.25
The string of pendulum of length l is displaced through 90° from the vertical and released. then the minimum strength of the string in order to withstand the tension, as the pendulum passes through the mean position is [ MP 1986]
0%
a)mg
0%
b) 3mg
0%
c)5mg
0%
d)6mg
Explanation
To reach the point of suspension ( 90° from vertical) From law of conservation of energy ½ mv2=mgh or v2=2gl tension at the lowest point T=mg + mv2 / l T=mg + 2mg=3mgAnswer: (b)
Q.26
A stone of mass 1kg tied to a light inextensible string of length L=(10/3) m is whirling in a circular path of radius L in a vertical plane If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if g is taken to be 10 m/s2 the speed of the stone at the highest point of the circle is [ CPMT 1998]
0%
a) 20 m/s
0%
b) 10√3 m/s
0%
c)5√2 m/s
0%
d)10 m/s
Explanation
maximum tension will be at lower point A Minimum tension will be at highest point B By taking the ratio of max. tension to min. tension we get According to law of conservation of energy energy at A = energy at B we get V2A = 4gL- V2A On solving both equations we get V2B=3gL Velocity at highest point=[3 × 10 × (10/3)]1/2=10 m/sAnswer: (d)
Q.27
A heavy small sized sphere is suspended by a string of length l. the sphere rotates uniformly in horizontal circle with the string making angle θ with the vertical. Then the time period of conical pendulum is
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
T cosθ=mg T sinθ=mrω2Taking ratio we get from figure r=lsinθ Answer:(c)
Q.28
A piece of stone is projected at angle θ with velocity u. If it executes nearly a circular motion at its maximum point for short time, the radius of the circle will be [CPMT]
0%
b) (u2cos2θ) / g
0%
c) (u2sin2θ) / g
0%
d) (u2cos2θ) / 2g
0%
a) u² / g
Explanation
The velocity at maximum point be ucosθ ( only horizontal component ), the centripetal acceleration is given by a=v2 / r ∴ g=(u2cos2θ) / rr=(u2cos2θ) / g Answer: (b)
Q.29
A block of mass M is tied to one end of mass less rope. The other end of the rope is in the hands of a man of mass 2M as shown in figure. The block and man are resting on a rough wedge of mass M as shown in figure. The whole system is resting on smooth horizontal surface. The man pulls the rope. Pulley is mass less and frictionless. What is the displacement of the wedge when the block meets the pulley ( man does not leave his position during the pull)
0%
a)0.5 m
0%
b) 1 m
0%
c)zero
0%
d)23 m
Explanation
Since there no external force acting on the system and initially the centre of mass of the system was at rest, hence the centre of mass will not move Let d be displacement of wedgecentre of mass at initial position from diagram centre of mass at final position When block touches the pulley and displacement of wedge from diagram Since centre of mass should not move as no external force is acting (Xcm)i=(Xcm)f∴ d=(1/2) m=0.5 mAnswer: (a)
Q.30
Two meteorites in free space are 8 m apart. Their masses are 7kg and 9kg. Then their centre of mass which lies on the line joining the centre of two bodies is
0%
a) At 4 meters from each meteorites
0%
b) At 3.5 meter from meteorites with a mass of 9kg
0%
c)At 3.5 meter from the meteorites with mass 7 kg
0%
d)At 8 meters from one end and 16 meter from the other end
Explanation
let x be the distance of centre of mass from 7kg meteorite then 7x=9(8 - x) 7x=72 - 9x16x=72 x=72/16=4.5 from 7 kg or 8 - 4.5=3.5 m from 9 kgAnswer: (b)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)