Rotational Motion
261 Questions
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
The following bodies,
(1) a ring
(2) a disc
(3) a solid cylinder
(4) a solid sphere,
of same mass 'm' and radius 'R' are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ___________. [Mark the body as per their respective numbering given in the question]
(1) a ring
(2) a disc
(3) a solid cylinder
(4) a solid sphere,
of same mass 'm' and radius 'R' are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ___________. [Mark the body as per their respective numbering given in the question]
Correct Answer: 4
Explanation:
$a = {{g\sin \theta } \over {\left( {1 + {I \over {m{R^2}}}} \right)}}$
IR = mR2, aR = g sin$\theta$/2
ID = ${{m{R^2}} \over 2}$, aD = ${2 \over 3}$ g sin$\theta$
ISC = ${{m{R^2}} \over 2}$, aSC = ${2 \over 3}$ g sin$\theta$
ISC = ${2 \over 5}$mR2, aSS = ${5 \over 7}$ g sin$\theta$
S = ut + ${1 \over 2}$at2,
t = $\sqrt {{{2S} \over a}} $
$ \therefore $ t $\propto$ ${1 \over {\sqrt a }}$
solid sphere will take minimum time.
IR = mR2, aR = g sin$\theta$/2
ID = ${{m{R^2}} \over 2}$, aD = ${2 \over 3}$ g sin$\theta$
ISC = ${{m{R^2}} \over 2}$, aSC = ${2 \over 3}$ g sin$\theta$
ISC = ${2 \over 5}$mR2, aSS = ${5 \over 7}$ g sin$\theta$
S = ut + ${1 \over 2}$at2,
t = $\sqrt {{{2S} \over a}} $
$ \therefore $ t $\propto$ ${1 \over {\sqrt a }}$
solid sphere will take minimum time.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
A solid disc of radius 'a' and mass 'm' rolls down without slipping on an inclined plane making an angle $\theta$ with the horizontal. The acceleration of the disc will be ${2 \over b}$g sin$\theta$ where b is ____________. (Round off to the Nearest Integer) (g = acceleration due to gravity, $\theta$ = angle as shown in figure)
Correct Answer: 3
Explanation:
We know that, on an inclined plane
Acceleration, $a = {{g\sin \theta } \over {1 + {I \over {m{R^2}}}}}$
$ \Rightarrow a = {{g\sin \theta } \over {1 + {1 \over 2}}}$ [$\because$ For disc, $I = {{m{R^2}} \over 2}$]
$ \Rightarrow a = {2 \over 3}g\sin \theta $ ....... (i)
As per question, acceleration of the disc will be ${2 \over b}g\sin \theta $.
Comparing it with Eq. (i), we get
$b = 3$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
A force $\overrightarrow F $ = ${4\widehat i + 3\widehat j + 4\widehat k}$ is applied on an intersection point of x = 2 plane and x-axis. The magnitude of torque of this force about a point (2, 3, 4) is ___________. (Round off to the Nearest Integer)
Correct Answer: 20
Explanation:
$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F $
$ = \left[ {(2 - 2)\widehat i + (0 - 3)\widehat j + (0 - 4)\widehat k} \right] \times (4\widehat i + 3\widehat j + 4\widehat k)$
$ = ( - 3\widehat j - 4\widehat k) \times (4\widehat i + 3\widehat j + 4\widehat k)$
$ = - 16\widehat j + 12\widehat k$
$|\overrightarrow \tau |\, = 20$ units
$ = \left[ {(2 - 2)\widehat i + (0 - 3)\widehat j + (0 - 4)\widehat k} \right] \times (4\widehat i + 3\widehat j + 4\widehat k)$
$ = ( - 3\widehat j - 4\widehat k) \times (4\widehat i + 3\widehat j + 4\widehat k)$
$ = - 16\widehat j + 12\widehat k$
$|\overrightarrow \tau |\, = 20$ units
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its center and is at rest initially. The disk is acted upon by a constant force F = 20 N through a massless string wrapped around is periphery as shown in the figure.
Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s$-$1.
The value of n, to the nearest integer, is __________.
[Given : In one complete revolution, the disk rotates by 6.28 rad]
Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s$-$1.
The value of n, to the nearest integer, is __________.
[Given : In one complete revolution, the disk rotates by 6.28 rad]
Correct Answer: 20
Explanation:
$\alpha = {\tau \over I} = {{F.R.} \over {m{R^2}/2}} = {{2F} \over {mR}}$
$\alpha = {{2 \times 200} \over {20 \times (0.2)}} = 10$ rad/s2
${\omega ^2} = {\omega _0}^2 + 2\alpha \Delta \theta $
${(50)^2} = {0^2} + 2(10)\Delta \theta $
$ \Rightarrow \Delta \theta = {{2500} \over {20}}$
$\Delta \theta = 125$ rad
No. of revolution $ = {{125} \over {2\pi }} \approx 20$ revolution
$\alpha = {{2 \times 200} \over {20 \times (0.2)}} = 10$ rad/s2
${\omega ^2} = {\omega _0}^2 + 2\alpha \Delta \theta $
${(50)^2} = {0^2} + 2(10)\Delta \theta $
$ \Rightarrow \Delta \theta = {{2500} \over {20}}$
$\Delta \theta = 125$ rad
No. of revolution $ = {{125} \over {2\pi }} \approx 20$ revolution
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
Consider a frame that is made up of two thin massless rods AB and AC as shown in the figure. A vertical force $\overrightarrow P $ of magnitude 100 N is applied at point A of the frame.
Suppose the force is $\overrightarrow P $ resolved parallel to the arms AB and AC of the frame.
The magnitude of the resolved component along the arm AC is xN.
The value of x, to the nearest integer, is ___________.
[Given : sin(35$^\circ$) = 0.573, cos(35$^\circ$) = 0.819
sin(110$^\circ$) = 0.939, cos(110$^\circ$) = $-$ 0.342 J
Suppose the force is $\overrightarrow P $ resolved parallel to the arms AB and AC of the frame.
The magnitude of the resolved component along the arm AC is xN.
The value of x, to the nearest integer, is ___________.
[Given : sin(35$^\circ$) = 0.573, cos(35$^\circ$) = 0.819
sin(110$^\circ$) = 0.939, cos(110$^\circ$) = $-$ 0.342 J
Correct Answer: 82
Explanation:
Component along AC
= 100 cos 35$^\circ$ N
= 100 $\times$ 0.819 N
= 81.9 N
$ \approx $ 82 N
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is _______ $\times$ 10$-$1 kg m2.
Correct Answer: 8
Explanation:
MOI of AB about $P:{I_{ABp}} = {{{M \over 6}{{\left( {{l \over 6}} \right)}^2}} \over {12}}$
MOI of AB about O,
${I_{A{B_O}}} = \left[ {{{{M \over 6}{{\left( {{l \over 6}} \right)}^2}} \over {12}} + {M \over 6}{{\left( {{l \over 6}{{\sqrt 3 } \over 2}} \right)}^2}} \right]$
${I_{Hexago{n_0}}} = 6{I_{A{B_0}}} = M\left[ {{{{l^2}} \over {12 \times 36}} + {{{l^2}} \over {36}} \times {3 \over 4}} \right]$
$ = {6 \over {100}}\left[ {{{24 \times 24} \over {12 \times 36}} + {{24 \times 24} \over {36}} \times {3 \over 4}} \right]$
= 0.8 kgm2
= 8 $\times$ 10$-$1 kg-m2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
The linear mass density of a thin rod AB of length L varies from A to B as
$\lambda \left( x \right) = {\lambda _0}\left( {1 + {x \over L}} \right)$, where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is :
$\lambda \left( x \right) = {\lambda _0}\left( {1 + {x \over L}} \right)$, where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is :
A.
${2 \over 5}M{L^2}$
B.
${5 \over {12}}M{L^2}$
C.
${7 \over {18}}M{L^2}$
D.
${3 \over 7}M{L^2}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
Four point masses, each of mass m, are fixed at the corners of a square of side $l$. The square is
rotating with angular frequency $\omega $, about an axis passing through one of the corners of the square
and parallel to its diagonal, as shown in the figure. The angular momentum of the square about this
axis is :
A.
3m$l$2$\omega $
B.
4m$l$2$\omega $
C.
m$l$2$\omega $
D.
2m$l$2$\omega $
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is M, radius of its
top, R and height, H, then its moment of inertia about its axis is :
A.
${{M\left( {{R^2} + {H^2}} \right)} \over 3}$
B.
${{M{R^2}} \over 2}$
C.
${{M{R^2}} \over 3}$
D.
${{M{H^2}} \over 3}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
A ring is hung on a nail. It can oscillate, without
slipping or sliding
(i) in its plane with a time period T1 and,
(ii) back and forth in a direction perpendicular to its plane,
with a period T2. The ratio ${{{T_1}} \over {{T_2}}}$ will be :
(i) in its plane with a time period T1 and,
(ii) back and forth in a direction perpendicular to its plane,
with a period T2. The ratio ${{{T_1}} \over {{T_2}}}$ will be :
A.
${{\sqrt 2 } \over 3}$
B.
${2 \over {\sqrt 3 }}$
C.
${2 \over 3}$
D.
${3 \over {\sqrt 2 }}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
A wheel is rotating freely with an angular speed
$\omega $ on a shaft. The moment of inertia of the
wheel is I and the moment of inertia of the
shaft is negligible. Another wheel of moment of
inertia 3I initially at rest is suddenly coupled to
the same shaft. The resultant fractional loss in
the kinetic energy of the system is :
A.
0
B.
${5 \over 6}$
C.
${1 \over 4}$
D.
${3 \over 4}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes
perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is :
A.
${1 \over 2}$
B.
${1 \over 4}$
C.
${1 \over 8}$
D.
${2 \over 3}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
Consider two uniform discs of the same thickness and different radii R1
= R and
R2 = $\alpha $R made of the same material. If the ratio of their moments of inertia I1 and I2 , respectively, about their axes is I1 : I2 = 1 : 16 then the value of $\alpha $ is :
R2 = $\alpha $R made of the same material. If the ratio of their moments of inertia I1 and I2 , respectively, about their axes is I1 : I2 = 1 : 16 then the value of $\alpha $ is :
A.
$\sqrt 2 $
B.
2
C.
$2\sqrt 2 $
D.
4
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
A uniform rod of length ‘$l$’ is pivoted at one of its ends on a vertical shaft of negligible radius.
When the shaft rotates at angular speed $\omega $ the rod makes an angle $\theta $ with it (see figure). To find $\theta $
equate the rate of change of angular momentum (direction going into the paper) ${{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta $
about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FH
and
FV
about the CM. The value of $\theta $ is then such that :
A.
$\cos \theta = {{2g} \over {3l{\omega ^2}}}$
B.
$\cos \theta = {{3g} \over {2l{\omega ^2}}}$
C.
$\cos \theta = {g \over {2l{\omega ^2}}}$
D.
$\cos \theta = {g \over {l{\omega ^2}}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
Moment of inertia of a cylinder of mass M,
length L and radius R about an axis passing
through its centre and perpendicular to the
axis of the cylinder is
I = $M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)$. If such a cylinder is to be made for a given mass of a material, the ratio ${L \over R}$ for it to have minimum possible I is
I = $M\left( {{{{R^2}} \over 4} + {{{L^2}} \over {12}}} \right)$. If such a cylinder is to be made for a given mass of a material, the ratio ${L \over R}$ for it to have minimum possible I is
A.
${3 \over 2}$
B.
$\sqrt {{3 \over 2}} $
C.
$\sqrt {{2 \over 3}} $
D.
${{2 \over 3}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
Two uniform circular discs are rotating
independently in the same direction around
their common axis passing through their
centres. The moment of inertia and angular
velocity of the first disc are 0.1 kg-m2 and 10
rad s–1 respectively while those for the second
one are 0.2 kg-m2 and 5 rad s–1 respectively. At
some instant they get stuck together and start
rotating as a single system about their common
axis with some angular speed. The kinetic
energy of the combined system is :
A.
${{20} \over 3}J$
B.
${{5} \over 3}J$
C.
${{10} \over 3}J$
D.
${{2} \over 3}J$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
A uniform cylinder of mass M and radius R is to
be pulled over a step of height a (a < R) by
applying a force F at its centre ‘O’
perpendicular to the plane through the axes of
the cylinder on the edge of the step (see
figure). The minimum value of F required is :
A.
$Mg\sqrt {1 - {{\left( {{{R - a} \over R}} \right)}^2}} $
B.
$Mg\sqrt {1 - {{{a^2}} \over {{R^2}}}} $
C.
$Mg{a \over R}$
D.
$Mg\sqrt {{{\left( {{R \over {R - a}}} \right)}^2} - 1} $
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
Shown in the figure is rigid and uniform one meter long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass ‘m’ and has another weight of mass 2 m hung at a distance of 75 cm from A. The tension in the string at A is :
A.
0.5 mg
B.
2 mg
C.
0.75 mg
D.
1 mg
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
A uniformly thick wheel with moment of inertia
I and radius R is free to rotate about its centre
of mass (see fig). A massless string is wrapped
over its rim and two blocks of masses m1 and
m2 (m1 $ > $ m2) are attached to the ends of the
string. The system is released from rest. The
angular speed of the wheel when m1 descents
by a distance h is :
A.
${\left[ {{{2\left( {{m_1} + {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}$
B.
${\left[ {{{{m_1} + {m_2}} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}gh$
C.
${\left[ {{{\left( {{m_1} - {m_2}} \right)} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}gh$
D.
${\left[ {{{2\left( {{m_1} - {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
Three solid spheres each of mass m and
diameter d are stuck together such that the lines
connecting the centres form an equilateral
triangle of side of length d. The ratio I0/IA of
moment of inertia I0 of the system about an axis
passing the centroid and about center of any
of the spheres IA and perpendicular to the plane
of the triangle is :
A.
${{13} \over {23}}$
B.
${{23} \over {13}}$
C.
${{15} \over {13}}$
D.
${{13} \over {15}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
A uniform sphere of mass 500 g rolls without
slipping on a plane horizontal surface with its
centre moving at a speed of 5.00 cm/s. Its
kinetic energy is :
A.
8.75 × 10–3 J
B.
1.13 × 10–3 J
C.
8.75 × 10–4 J
D.
6.25 × 10–4 J
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
Consider a uniform rod of mass M = 4m and
length $\ell $ pivoted about its centre. A mass m
moving with velocity v making angle $\theta = {\pi \over 4}$ to
the rod's long axis collides with one end of the
rod and sticks to it. The angular speed of the
rod-mass system just after the collision is :
A.
${{3\sqrt 2 } \over 7}{v \over \ell }$
B.
${3 \over 7}{v \over \ell }$
C.
${3 \over {7\sqrt 2 }}{v \over \ell }$
D.
${4 \over 7}{v \over \ell }$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
Mass per unit area of a circular disc of radius $a$ depends on the distance r from its centre as $\sigma \left( r \right)$ = A + Br
. The moment of inertia of the disc about the axis, perpendicular to the plane and
assing through its centre is:
A.
$2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)$
B.
$\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)$
C.
$2\pi {a^4}\left( {{{aA} \over 4} + {B \over 5}} \right)$
D.
$2\pi {a^4}\left( {{A \over 4} + {B \over 5}} \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
The radius of gyration of a uniform rod of length $l$, about an axis passing through a
point ${l \over 4}$ away from the centre of the rod,
and perpendicular to it, is :
A.
${1 \over 8}l$
B.
${1 \over 4}l$
C.
$\sqrt {{7 \over {48}}} l$
D.
$\sqrt {{3 \over 8}} l$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be:
A.
$r\sqrt {{3 \over {2gh}}} $
B.
$r\sqrt {{3 \over {4gh}}} $
C.
${1 \over r}\sqrt {{{4gh} \over 3}} $
D.
${1 \over r}\sqrt {{{2gh} \over 3}} $
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 5th September Evening Slot
A thin rod of mass 0.9 kg and length 1 m is
suspended, at rest, from one end so that it can
freely oscillate in the vertical plane. A particle
of move 0.1 kg moving in a straight line with
velocity 80 m/s hits the rod at its bottom most
point and sticks to it (see figure). The angular
speed (in rad/s) of the rod immediately after the
collision will be _________.
Correct Answer: 20
Explanation:
The given situation is shown in the following figure,
Applying law of conservation of angular momentum about pivotal point,
${L_i} = {L_f} \Rightarrow mvl = I\omega $
$ \Rightarrow mvl = \left( {{1 \over 3}M{L^2} + m{L^2}} \right)\omega $ ..... (i)
Given, m = 0.1 kg, v = 80 ms$-$1, M = 0.9 kg, l = 1 m
Substituting these values in eq. (i), we get
$0.1 \times 80 \times 1 = \left( {{{0.9 \times {1^2}} \over 3} + 0.1 \times {1^2}} \right)\omega $
$8 = \left( {{3 \over {10}} + {1 \over {10}}} \right)\omega \Rightarrow 8 = {4 \over {10}}\omega \Rightarrow \omega = 20$ rad/s
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 5th September Morning Slot
A force $\overrightarrow F = \left( {\widehat i + 2\widehat j + 3\widehat k} \right)$ N acts at a point
$\left( {4\widehat i + 3\widehat j - \widehat k} \right)$ m. Then the magnitude of torque
about the point $\left( {\widehat i + 2\widehat j + \widehat k} \right)$ m will be $\sqrt x $ N m.
The value of x is _______.
$\left( {4\widehat i + 3\widehat j - \widehat k} \right)$ m. Then the magnitude of torque
about the point $\left( {\widehat i + 2\widehat j + \widehat k} \right)$ m will be $\sqrt x $ N m.
The value of x is _______.
Correct Answer: 195
Explanation:
$\overrightarrow \tau = \overrightarrow r \times F = (3\widehat i + \widehat j - 2\widehat k) \times (\widehat i + 2\widehat j + 3\widehat k)$
$ = \left| {\matrix{ i & j & k \cr 3 & 1 & { - 2} \cr 1 & 2 & 3 \cr } } \right|$
$ = \widehat i(3 + 4) - \widehat j(9 + 2) + \widehat k(6 - 1)$
$\overrightarrow \tau = 7\widehat j - 11\widehat j + 5\widehat k$
$\left| {\overrightarrow \tau } \right| = \sqrt {49 + 121 + 25} = \sqrt {195} $
$ \therefore $ $x = 195$
$ = \left| {\matrix{ i & j & k \cr 3 & 1 & { - 2} \cr 1 & 2 & 3 \cr } } \right|$
$ = \widehat i(3 + 4) - \widehat j(9 + 2) + \widehat k(6 - 1)$
$\overrightarrow \tau = 7\widehat j - 11\widehat j + 5\widehat k$
$\left| {\overrightarrow \tau } \right| = \sqrt {49 + 121 + 25} = \sqrt {195} $
$ \therefore $ $x = 195$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Morning Slot
A circular disc of mass M and radius R is rotating about its axis with angular speed ${\omega _1}$
. If another
stationary disc having radius ${R \over 2}$ and same mass M is droped co-axially on to the rotating disc.
Gradually both discs attain constant angular speed ${\omega _2}$
the energy lost in the process is p% of the
initial energy. Value of p is __________.
Correct Answer: 20
Explanation:
${I_f}{\omega _f} = {I_i}{\omega _i}$
${I_i} = {{M{R^2}} \over 2}$
${I_f} = {{M{R^2}} \over 2} + {{M{{(R/2)}^2}} \over 2}$
$ = {5 \over 4}.{{M{R^2}} \over 2}$
$\left[ {{{M{R^2}} \over 2} + {M \over 2}{{\left( {{R \over 2}} \right)}^2}} \right]\omega ' = \left( {{{M{R^2}} \over 2}} \right).\omega $
$ \Rightarrow $ $\left[ {{{M{R^2}} \over 2}.\left( {{5 \over 4}} \right)} \right]\omega ' = {{M{R^2}} \over 2}\omega $
$\omega = {4 \over 5}\omega $
loss of K.E. = ${{Loss} \over {{K_i}}} \times 100 $
= ${{{1 \over 2}I{\omega ^2} - {1 \over 2}\left( {{5 \over 4}I} \right){{\left( {{4 \over 5}\omega } \right)}^2}} \over {{1 \over 2}I{\omega ^2}}}$ $ \times $ 100
= ${{{\omega ^2} - {{16} \over {25}}{\omega ^2}\left( {{5 \over 4}} \right)} \over {{\omega ^2}}}$ $ \times $ 100 = $\left( {1 - {{80} \over {100}}} \right) \times 100$
= 20%
${I_i} = {{M{R^2}} \over 2}$
${I_f} = {{M{R^2}} \over 2} + {{M{{(R/2)}^2}} \over 2}$
$ = {5 \over 4}.{{M{R^2}} \over 2}$
$\left[ {{{M{R^2}} \over 2} + {M \over 2}{{\left( {{R \over 2}} \right)}^2}} \right]\omega ' = \left( {{{M{R^2}} \over 2}} \right).\omega $
$ \Rightarrow $ $\left[ {{{M{R^2}} \over 2}.\left( {{5 \over 4}} \right)} \right]\omega ' = {{M{R^2}} \over 2}\omega $
$\omega = {4 \over 5}\omega $
loss of K.E. = ${{Loss} \over {{K_i}}} \times 100 $
= ${{{1 \over 2}I{\omega ^2} - {1 \over 2}\left( {{5 \over 4}I} \right){{\left( {{4 \over 5}\omega } \right)}^2}} \over {{1 \over 2}I{\omega ^2}}}$ $ \times $ 100
= ${{{\omega ^2} - {{16} \over {25}}{\omega ^2}\left( {{5 \over 4}} \right)} \over {{\omega ^2}}}$ $ \times $ 100 = $\left( {1 - {{80} \over {100}}} \right) \times 100$
= 20%
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Evening Slot
An massless equilateral triangle EFG of side ‘a’ (As shown in figure) has three particles of mass m
situated at its vertices. The moment of inertia of the system about the line EX perpendicular to EG
in the plane of EFG is ${N \over {20}}$ ma2
where N is an integer. The value of N is _____.
Correct Answer: 25
Explanation:
I = m$a$2 + ${{m{a^2}} \over 4}$ = ${5 \over 4}m{a^2}$
Accoeding to the question,
${5 \over 4}m{a^2} = {N \over {20}}m{a^2}$
$ \Rightarrow $ N = 25
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Morning Slot
A person of 80 kg mass is standing on the rim
of a circular platform of mass 200 kg rotating
about its axis at 5 revolutions per minute (rpm).
The person now starts moving towards the
centre of the platform. What will be the
rotational speed (in rpm) of the platform when
the person reaches its centre _________.
Correct Answer: 9
Explanation:
${I_1}{\omega _1} = {I_2}{\omega _2}$
$ \Rightarrow $ $\left( {{{M{R^2}} \over 2} + m{R^2}} \right){\omega _1} = {{M{R^2}} \over 2}{\omega _2}$
$ \Rightarrow $ $\left( {1 + {{2m{R^2}} \over {M{R^2}}}} \right){\omega _1} = {\omega _2}$
$ \Rightarrow $ $\left( {1 + {{2 \times 80} \over {200}}} \right){\omega _1} = {\omega _2}$
$ \Rightarrow $ ${\omega _2} = \left( {1.8} \right){\omega _1}$
$ \Rightarrow $ 2$\pi $f2 = 2$\pi $f1 $ \times $ 1.8
$ \Rightarrow $ f2 = 5 $ \times $ 1.8 = 9
$ \Rightarrow $ $\left( {{{M{R^2}} \over 2} + m{R^2}} \right){\omega _1} = {{M{R^2}} \over 2}{\omega _2}$
$ \Rightarrow $ $\left( {1 + {{2m{R^2}} \over {M{R^2}}}} \right){\omega _1} = {\omega _2}$
$ \Rightarrow $ $\left( {1 + {{2 \times 80} \over {200}}} \right){\omega _1} = {\omega _2}$
$ \Rightarrow $ ${\omega _2} = \left( {1.8} \right){\omega _1}$
$ \Rightarrow $ 2$\pi $f2 = 2$\pi $f1 $ \times $ 1.8
$ \Rightarrow $ f2 = 5 $ \times $ 1.8 = 9
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Morning Slot
A body of mass m = 10 kg is attached to one
end of a wire of length 0.3 m. The maximum
angular speed (in rad s–1) with which it can be
rotated about its other end in space station is :
(Breaking stress of wire = 4.8 × 107 Nm–2 and
area of cross-section of the wire = 10–2 cm2) is:
(Breaking stress of wire = 4.8 × 107 Nm–2 and
area of cross-section of the wire = 10–2 cm2) is:
Correct Answer: 4
Explanation:
T = m${\omega ^2}l$
Breaking stress = $\sigma $ = ${{m{\omega ^2}l} \over A}$
$ \Rightarrow $ ${\omega ^2}$ = ${{4.8 \times {{10}^7} \times \left( {{{10}^{ - 2}} \times {{10}^{ - 4}}} \right)} \over {10 \times 0.3}}$ = 16
$ \Rightarrow $ $\omega $ = 4
Breaking stress = $\sigma $ = ${{m{\omega ^2}l} \over A}$
$ \Rightarrow $ ${\omega ^2}$ = ${{4.8 \times {{10}^7} \times \left( {{{10}^{ - 2}} \times {{10}^{ - 4}}} \right)} \over {10 \times 0.3}}$ = 16
$ \Rightarrow $ $\omega $ = 4
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Morning Slot
One end of a straight uniform 1m long bar is
pivoted on horizontal table. It is released from
rest when it makes an angle 30º from the
horizontal (see figure). Its angular speed when
it hits the table is given as $\sqrt n $ s-1, where n is
an integer. The value of n is _________.
Correct Answer: 15
Explanation:
By energy conservation
Ui + Ki = Uf + Kf
$ \Rightarrow $ $mg{l \over 2}\sin 30^\circ $ + 0 = 0 + ${1 \over 2}I{\omega ^2}$
$ \Rightarrow $ $mg{1 \over 2} \times {1 \over 2}$ = ${1 \over 2} \times {{m{{\left( 1 \right)}^2}} \over 3}{\omega ^2}$
$ \Rightarrow $ ${\omega ^2} = {{3g} \over 2}$
$ \Rightarrow $ $\omega = \sqrt {15} $
$ \therefore $ n = 15
Ui + Ki = Uf + Kf
$ \Rightarrow $ $mg{l \over 2}\sin 30^\circ $ + 0 = 0 + ${1 \over 2}I{\omega ^2}$
$ \Rightarrow $ $mg{1 \over 2} \times {1 \over 2}$ = ${1 \over 2} \times {{m{{\left( 1 \right)}^2}} \over 3}{\omega ^2}$
$ \Rightarrow $ ${\omega ^2} = {{3g} \over 2}$
$ \Rightarrow $ $\omega = \sqrt {15} $
$ \therefore $ n = 15
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Evening Slot
Consider a uniform cubical box of side a on a rough floor that is to be moved by applying
minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of
friction is $\mu $ = 0.4, the maximum possible value of 100 × ${b \over a}$
for box not to topple before moving
is .......
Correct Answer: 50
Explanation:
Torque about Center Of Mass
Fb + f${a \over 2}$ = N${a \over 2}$ ...(1)
For not sliding
F = f = $\mu $N ....(2)
N = Mg ...(3)
$ \therefore $ $\mu $N$\left( {b + {a \over 2}} \right)$ = Mg${{a \over 2}}$
$ \Rightarrow $ 4b = 3$a$
$ \Rightarrow $ b = 0.75$a$
which is not possible as as b can maximum be equal to 0.5$a$
$ \therefore $ ${b \over a}$ = 0.5
So 100 × ${b \over a}$ = 50
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
A uniform rod of length $\ell $ is being rotated in a horizontal plane with a constant angular speed about an axis
passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from
the axis, then which of the following graphs depicts it most closely?
A.
B.
C.
D.
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc
varies as $\left( {{{{\sigma _0}} \over r}} \right)$
, then the radius of gyration of the disc about its axis passing through the centre is:
A.
$\sqrt {{{{a^2} + {b^2} + ab} \over 2}} $
B.
$\sqrt {{{a + b} \over 3}} $
C.
$\sqrt {{{{a^2} + {b^2} + ab} \over 3}} $
D.
$\sqrt {{{a + b} \over 2}} $
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
A person of mass M is, sitting on a swing of length L and swinging with an angular amplitude $\theta $0. If the
person stands up when the swing passes through its lowest point, the work done by him, assuming that his
center of mass moves by a distance $\ell $($\ell $ << L), is close to;
A.
mg$\ell $(1 + $\theta $02)
B.
mg$\ell $
C.
mg$\ell $(1 + ${{\theta _0^2} \over 2}$)
D.
mg$\ell $(1 - $\theta $02)
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
The time dependence of the position of a particle of mass m = 2 is given by $\overrightarrow r \left( t \right) = 2t\widehat i - 3{t^2}\widehat j$
. Its angular
momentum, with respect to the origin, at time t = 2 is
A.
36 $\widehat k$
B.
- 48 $\widehat k$
C.
$ - 34\left( {\widehat k - \widehat i} \right)$
D.
$48\left( {\widehat i + \widehat j} \right)$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the
figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25
rotations per second in 5s, is close to :
A.
7.9 × 10–6 Nm
B.
4.0 × 10–6 Nm
C.
2.0 × 10–5 Nm
D.
1.6 × 10–5 Nm
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of ${{7M} \over 8}$
and is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere.
Let I1 be the moment of inertia of the disc about its axis and I2 be the moment of inertia of the new sphere
about its axis. The ratio I1/I2 is given by :
A.
65
B.
140
C.
185
D.
285
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
Two coaxial discs, having moments of inertia
I1 and I1/2, are rotating with respective angular
velocities $\omega $1 and
$\omega $1/2
, about their common axis.
They are brought in contact with each other and
thereafter they rotate with a common angular
velocity. If Ef and Ei are the final and initial total
energies, then (Ef - Ei) is:
A.
${{{I_1}\omega _1^2} \over {24}}$
B.
${{{I_1}\omega _1^2} \over {12}}$
C.
${3 \over 8}{I_1}\omega _1^2$
D.
${{{I_1}\omega _1^2} \over {6}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
A particle of mass m is moving along a
trajectory given by
x = x0 + a cos$\omega $1t
y = y0 + b sin$\omega $2t
The torque, acting on the particle about the origin, at t = 0 is :
x = x0 + a cos$\omega $1t
y = y0 + b sin$\omega $2t
The torque, acting on the particle about the origin, at t = 0 is :
A.
Zero
B.
+my0a $\omega _1^2$$\widehat k$
C.
$ - m\left( {{x_0}b\omega _2^2 - {y_0}a\omega _1^2} \right)\widehat k$
D.
m (–x0b + y0a) $\omega _1^2$$\widehat k$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
A thin disc of mass M and radius R has mass
per unit area $\sigma $(r) = kr2 where r is the distance
from its centre. Its moment of inertia about an
axis going through its centre of mass and
perpendicular to its plane is :
A.
${{M{R^2}} \over 3}$
B.
${{M{R^2}} \over 6}$
C.
${{2M{R^2}} \over 3}$
D.
${{M{R^2}} \over 2}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
A thin smooth rod of length L and mass M is
rotating freely with angular speed $\omega $0 about an
axis perpendicular to the rod and passing
through its center. Two beads of mass m and
negligible size are at the center of the rod
initially. The beads are free to slide along the
rod. The angular speed of the system , when
the beads reach the opposite ends of the rod,
will be :-
A.
${{M{\omega _0}} \over {M + 3m}}$
B.
${{M{\omega _0}} \over {M + m}}$
C.
${{M{\omega _0}} \over {M + 6m}}$
D.
${{M{\omega _0}} \over {M + 2m}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
Moment of inertia of a body about a given axis
is 1.5 kg m2. Initially the body is at rest. In order
to produce a rotational kinetic energy of
1200 J, the angular accleration of 20 rad/s2
must be applied about the axis for a
duration of :-
A.
2.5 s
B.
3 s
C.
5s
D.
2 s
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
The following bodies are made to roll up
(without slipping) the same inclined plane from
a horizontal plane. : (i) a ring of radius R, (ii)
a solid cylinder of radius
R/2 and (iii) a solid
sphere of radius
R/4 . If in each case, the speed
of the centre of mass at the bottom of the incline
is same, the ratio of the maximum heights they
climb is :
A.
20 : 15 : 14
B.
4 : 3 : 2
C.
2 : 3 : 4
D.
10 : 15 : 7
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
A stationary horizontal disc is free to rotate
about its axis. When a torque is applied on it,
its kinetic energy as a function of $\theta $, where $\theta $
is the angle by which it has rotated, is given as
k$\theta $2. If its moment of inertia is I then the
angular acceleration of the disc is :
A.
${k \over {4I}}\theta $
B.
${k \over {I}}\theta $
C.
${k \over {2I}}\theta $
D.
${2k \over {I}}\theta $
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
A rectangular solid box of length 0.3 m is held
horizontally, with one of its sides on the edge
of a platform of height 5m. When released, it
slips off the table in a very short time t = 0.01s,
remaining essentially horizontal. The angle by
which it would rotate when it hits the ground
will be (in radians) close to :-
A.
0.28
B.
0.02
C.
0.3
D.
0.5
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
A solid sphere and solid cylinder of identical
radii approach an incline with the same linear
velocity (see figure). Both roll without slipping
all throughout. The two climb maximum
heights hsph and hcyl on the incline. The ratio
hsph/hcyl is given by :-
A.
1
B.
14/15
C.
4/5
D.
2/$\sqrt5$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
An electric dipole is formed by two equal and
opposite charges q with separation d. The
charges have same mass m. It is kept in a
uniform electric field E. If it is slightly rotated
from its equilibrium orientation, then its angular
frequency $\omega$ is :-
A.
$\sqrt {{{qE} \over {md}}} $
B.
$\sqrt {{{qE} \over {2md}}} $
C.
$\sqrt {{{qE} \over {-2md}}} $
D.
$\sqrt {{{2qE} \over {md}}} $
Moment of inertia
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
A thin circular plate of mass M and radius R
has its density varying as $\rho $(r) = $\rho $0r with $\rho $0 as
constant and r is the distance from its centre.
The moment of Inertia of the circular plate about
an axis perpendicular to the plate and passing
through its edge is I = aMR2. The value of the
coefficient a is :
A.
${1 \over 2}$
B.
${3 \over 2}$
C.
${8 \over 5}$
D.
${3 \over 5}$