Free Objective Test 01 Practice Test - 11th and 12th
Question 1
The angular velocity of the second's hand of a watch is
SOLUTION
Solution : B
We know that second's hand completes its revolution (2π) in 60 sec ∴ω=θt=2π60=π30 rad/sec
Question 2
The wheel of a car is rotating at the rate of 1200 revolutions per minute. On pressing the accelerator for 10 seconds, it starts rotating at 4500 revolutions per minute. The angular acceleration of the wheel is
SOLUTION
Solution : D
Angular acceleration (a) = rate of change of angular speed
=2π(n2−n1)t=2π(4500−120060)10=2π33006010×3602πdegreesec2=1980degree/sec2.
Question 3
Angular displacement θ of a flywheel varies with time as θ=at+bt2+ct3, then angular acceleration of the flywheel is given by
SOLUTION
Solution : D
Angular acceleration α=d2θdt2=d2dt2(at+bt2+ct3)=2b+6ct
Question 4
A wheel completes 2000 rotations in covering a distance of 9.5 km. The diameter of the wheel is
SOLUTION
Solution : A
Distance covered by wheel in 1 rotation =2πr=πD
(Where, D = 2r = diameter of wheel)
∴ Distance covered in 2000 rotation =2000πD=9.5×103m (given)
∴ D = 1.5 meter
Question 5
A wheel is at rest. Its angular velocity increases uniformly and becomes 60 rad/sec after 5 sec. The total angular displacement is
SOLUTION
Solution : D
Angular acceleration α=ω2−ω1t=60−05=12 rad/sec2
Now from θ=ω1t+12αt2=0+12(12)(5)2=150rad.
Question 6
A wheel initially at rest, is now rotated with a uniform angular acceleration. The wheel rotates through an angle θ1 in first one second and through an additional angle θ2 in the next one second. The ratio θ2θ1 is
SOLUTION
Solution : C
Angular displacement in first one second θ1=12α(1)2=α2...(i) [Fromθ=ω1t+12αt2]
Now again we will consider motion from the rest and angular displacement in total two seconds
θ1+θ2=12α(2)2=2α...(ii)
Solving (i) and (ii) we get θ1=12 and θ2=3α2 ∴θ2θ1=3.
Question 7
Let 'l' be the moment of inertia of an uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle θ with AB. The moment of inertia of the plate about the axis CD is then equal to
SOLUTION
Solution : A
Let Iz is the moment of inertia of square plate about the axis which is passing through the centre and perpendicular to the plane.
IZ=IAB+IA′B′=ICD+IC′D′ [By the theorem of perpendicular axis]
IZ=2IAB=2IA′B′=2ICD=2IC′D′
[As AB, A' B' and CD, C' D' are symmetric axis]
Hence ICD=IAB=I
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Question 8
Three rods each of length L and mass M are placed along X, Y and Z-axes in such a way that one end of each of the rod is at the origin. The moment of inertia of this system about Z axis is
SOLUTION
Solution : A
Moment of inertia of the system about z-axis can be find out by calculating the moment of inertia of individual rod about z-axis
I1=I2=ML23 because z-axis is the edge of rod 1 and 2
and I3=0 because rod in lying on z-axis
∴ISystem=I1+I2+I3=ML23+ML20+0=2ML23.
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Question 9
Three point masses each of mass m are placed at the corners of an equilateral triangle of side a. The moment of inertia of system about the side AB is
SOLUTION
Solution : C
The moment of inertia of system about AB side of triangle
I=IA+IB+IC
=0+0+mx2
=m(a√32)2=34ma2
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Question 10
Two identical rods each of mass 'M' and length 'l' are joined in crossed position as shown in figure. The moment of inertia of this system about a bisector would be
SOLUTION
Solution : B
Moment of inertia of system about an axes which is perpendicular to plane of rods and passing through the common centre of rods Iz=Ml212+Ml212=Ml26
Again from perpendicular axes theorem I2=IB1+IB2=2IB1=2IB2=Ml26 [As IB1=IB2]
∴IB1=IB2=Ml212.
Question 11
The moment of inertia of a rod of length 'l' about an axis passing through its centre of mass and perpendicular to rod is 'I'. The moment of inertia of hexagonal shape formed by six such rods, about an axis passing through its centre of mass and perpendicular to its plane will be
SOLUTION
Solution : C
Moment of inertia of rod AB about its centre and perpendicular to the length =ml212=l∴ml2=12l
Now moment of inertia of the rod about the axis which is passing through O and perpendicular to the plane of hexagon lrod=ml212+mx2 [From the theorem of parallel axes]
=ml212+m(√32l)2=5ml26
Now the moment of inertia of system lsystem=6×Irod=6×5ml26=5ml2
lsystem=5(12l)=60l [As ml2=12l]
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Question 12
The moment of inertia of HCl molecule about an axis passing through its centre of mass and perpendicular to the line joining the H+ and Cl− ions will be, if the interatomic distance is 1 oA
SOLUTION
Solution : B
If r1 and r2 are the respective distances of particles m1 and m2 from the centre of mass then
m1r1=m2r2⇒1×x=35.5×(L−x)⇒x=35.5(1−x)
⇒x=0.973oA and L−x=0.027oA
Moment of inertia of the system about centre of mass I=m2x+m2(L−x)2
I=1amu×(0.973oA)2+35.5amu×(0.027oA)2
Substituting 1 a.m.u. =1.67×10−27kg and 1oA=10−10m
I=1.62×10−47kgm2
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Question 13
Four masses are joined to a light circular frame as shown in the figure. The radius of gyration of this system about an axis passing through the centre of the circular frame and perpendicular to its plane would be
SOLUTION
Solution : C
Since the circular frame is massless so we will consider moment of inertia of four masses only.
I=ma2+2ma2+3ma2+2ma2=8ma2 .....(i)
Now from the definition of radius of gyration I=8mk2 .....(ii)
comparing (i) and (ii) radius of gyration k = a.
Question 14
Four spheres, each of mass 'M' and radius 'r' are situated at the four corners of square of side 'R'. The moment of inertia of the system about an axis perpendicular to the plane of square and passing through its centre will be
SOLUTION
Solution : B
M. I. of sphere A about its diameter IO′=25Mr2
Now M.I. of sphere A about an axis perpendicular to the plane of square and passing through its centre will be
IO=IO′+M(R√2)2=25Mr2+MR22
Moment of inertia of system (i.e. four sphere) =4IO=4[25Mr2+MR22]=25M[4r2+5R2]
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Question 15
The moment of inertia of a solid sphere of density ρ and radius R about its diameter is
SOLUTION
Solution : C
Moment of inertia of sphere about it diameter I=25MR2=25(43πR3ρ)R2 [AsM=Vρ=43πR3ρ]
l=8π15R5ρ=8×2215×7R5ρ=176105R5ρ