MCQ 511 Mark
If radius of the solid sphere is doubled by keeping its mass constant, the ratio of their moment of inertia about any of its diameter is
- A
$1: 8$
- B
$2: 5$
- C
$2: 3$
- ✓
$1: 4$
AnswerCorrect option: D. $1: 4$
(d) : Moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is,
$
I=\frac{2}{5} M R^2
$
When the radius of the solid sphere is doubled keeping the mass constant, the new moment of inertia is,
$
I^{\prime}=\frac{2}{5} M(2 R)^2=\frac{8}{5} M R^2 \quad \therefore \quad \frac{I}{I^{\prime}}=\frac{2 M R^2}{5} \times \frac{5}{8 M R^2}=\frac{1}{4}
$
View full question & answer→MCQ 521 Mark
The real force ' $F$ ' acting on a particle of mass ' $m$ ' performing circular motion acts along the radius of circle ' $r$ ' and is directed towards the centre of circle. The square root of magnitude of such force is ( $T=$ periodic time)
AnswerCorrect option: A. $\frac{2 \pi}{T} \sqrt{m r}$
(a) : In circular motion, the real force $F$ is provided by the centripetal force acting on the particle.
Now, $F=\frac{m v^2}{r}$ where $m$ is the mass of the particle and $r$ is the radius of the circular path.
$
\begin{aligned}
& \text { Now, } v=\frac{2 \pi r}{T} \quad \therefore F=\frac{m}{r} \times\left(\frac{2 \pi r}{T}\right)^2 \\
& \Rightarrow F=\frac{m r}{T^2} \times 4 \pi^2 \quad \therefore \sqrt{F}=\frac{2 \pi}{T} \sqrt{m r}
\end{aligned}
$
View full question & answer→MCQ 531 Mark
A stone of mass $1 kg$ is tied to a string $2 m$ long and is rotated at constant speed of $40 m s ^{-1}$ in a vertical circle. The ratio of the tension at the top and the bottom is [Take $g=10 m s ^{-2}$ ]
- A
$\frac{81}{79}$
- ✓
$\frac{79}{81}$
- C
$\frac{19}{12}$
- D
$\frac{12}{19}$
AnswerCorrect option: B. $\frac{79}{81}$
(b) : Given, $m=1 kg , L=2 m$ and $v_H=v_B=40 m s ^{-1}$
Tension at the top or the highest point,
$
T_H=\frac{m v_H^2}{L}-m g=\frac{1 \times(40)^2}{2}-1 \times 10=790 N
$
Tension at the bottom or the lowest point,
$
T_B=\frac{m v_B^2}{L}+m g=\frac{1 \times(40)^2}{2}+10=810 N
$
$\therefore \quad$ Ratio of the tension at the top and the bottom
$
\frac{T_H}{T_B}=\frac{790 N }{810 N }=\frac{79}{81}
$
View full question & answer→MCQ 541 Mark
The moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane is ' I '. It is rotating with angular velocity ' $\omega$ '. Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis, then loss in kinetic energy is
- A
$\frac{I \omega^2}{2}$
- ✓
$\frac{I \omega^2}{4}$
- C
$\frac{I \omega^2}{6}$
- D
$\frac{I \omega^2}{8}$
AnswerCorrect option: B. $\frac{I \omega^2}{4}$
(b) : From conservation of angular momentum
$
\begin{aligned}
& I_1 \omega_1=I_2 \omega_2 \quad\left(\because \omega_1=\omega, I_1=I \text { and } I_2=2 I\right) \\
& I \omega=2 I \omega_2 \quad \therefore \omega_2=\frac{\omega}{2} \\
& \text { New } K E=\frac{1}{2} 2 I\left(\frac{\omega}{2}\right)^2=\frac{I \omega^2}{4} \\
& \text { Change in } K E=\frac{1}{2} I \omega^2-\frac{I \omega^2}{4}=\frac{I \omega^2}{4}
\end{aligned}
$
View full question & answer→MCQ 551 Mark
A square frame $A B C D$ is formed by four identical rods each of mass ' $m$ ' and length ' $l$ '. This frame is in $X-Y$ plane such that side $A B$ coincides with $X$-axis and side $A D$ along $Y$-axis. The moment of inertia of the frame about $X$-axis is
- ✓
$\frac{5 m l^2}{3}$
- B
$\frac{2 m l^2}{3}$
- C
$\frac{4 m l^2}{3}$
- D
$\frac{m l^2}{12}$
AnswerCorrect option: A. $\frac{5 m l^2}{3}$
(a) : The moment of inertia of the square frame $A B C D$ about $x$-axis is
$
\begin{aligned}
& I=\frac{m l^2}{3}+\frac{m l^2}{3}+m l^2 \\
& \Rightarrow I=\frac{5}{3} m l^2
\end{aligned}
$
View full question & answer→MCQ 561 Mark
A disc has mass ' $M$ ' and radius ' $R$ '. How much tangential force should be applied to the rim of the disc so as to rotate with angular velocity ' $\omega$ ' in time ' $t$ '?
- A
$\frac{M R \omega}{4 t}$
- ✓
$\frac{M R \omega}{2 t}$
- C
$\frac{M R \omega}{t}$
- D
$M R o t$
AnswerCorrect option: B. $\frac{M R \omega}{2 t}$
(b) : Torque of tangential force on a disc will rotate the disc
$
\tau=I \alpha \Rightarrow F \times R=\frac{M R^2}{2} \times \frac{\omega}{t} \Rightarrow F=\frac{M R \omega}{2 t}
$
View full question & answer→MCQ 571 Mark
A mass attached to one end of a string crosses topmost point on a vertical circle with critical speed. Its centripetal acceleration, when string becomes horizontal will be ( $g=$ gravitational acceleration)
Answer(b) ; For the motion in a vertical circle the velocity of position $B$ is
$
\begin{aligned}
& v=\sqrt{3 r g} \\
& \therefore \quad \text { Centripetal acceleration }= \\
& \frac{v^2}{r}=\frac{3 r g}{r}=3 g
\end{aligned}
$
View full question & answer→MCQ 581 Mark
A fly wheel at rest is to reach an angular velocity of $24 rad / s$ in 8 second with constant angular acceleration. The total angle turned through during this interval is
- A
$24 rad$
- B
$48 rad$
- C
$72 rad$
- ✓
$96 rad$
AnswerCorrect option: D. $96 rad$
(d) : Given : $\theta_0=0, \omega_0=0, \omega=24 rad / s , t=8 s$
$
\begin{aligned}
\omega & =\omega_0+\alpha t ; \omega-\omega_0=\alpha t \\
\Rightarrow \quad \alpha & =\frac{\omega-\omega_0}{t}=\frac{24}{8}=3 rad / s ^2 \\
\theta & =\theta_0+\omega_0 t+\frac{1}{2} \alpha t^2=\frac{1}{2} \times 3 \times 8^2=96 rad
\end{aligned}
$
View full question & answer→MCQ 591 Mark
A solid sphere of mass $2 kg$ is rolling on a friction horizontal surface with velocity $6 m / s$. It collides on the free end of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be (Force constant of the spring - $36 N / m$ ).
- A
$\sqrt{14} m$
- ✓
$\sqrt{2.8} m$
- C
$\sqrt{1.4} m$
- D
$\sqrt{0.7} m$
AnswerCorrect option: B. $\sqrt{2.8} m$
(b) : Kinetic energy of rolling solid sphere
$
\begin{aligned}
& =\frac{1}{2} m v^2+\frac{1}{2} I \omega^2=\frac{1}{2} m v^2+\frac{1}{2} \times \frac{2}{5} m r^2 \omega^2 \\
& =\frac{1}{2} m v^2+\frac{1}{5} m v^2=\frac{7}{10} m v^2 \\
& =\frac{7}{10} \times 2 \times(6)^2=\frac{7}{10} \times 2 \times 36=\frac{252}{5}
\end{aligned}
$
The potential energy of the spring of maximum compression $x=\frac{1}{2} k x^2$
or $\frac{1}{2} k x^2=\frac{252}{5}$ or $k x^2=\frac{252 \times 2}{5}$
or $\quad x^2=\frac{252 \times 2}{5 \times 36}=2.8 \quad$ or $\quad x=\sqrt{2.8} m$
View full question & answer→MCQ 601 Mark
A disc of moment of inertia $I_1$ is rotating in horizontal plane about an axis passing through a centre and perpendicular to its plane with constant angular speed $\omega_1$. Another disc of moment of inertia $I_2$ having zero angular speed is placed coaxially on a rotating disc. Now both the discs are rotating with constant angular speed $\omega_2$. The energy lost by the initial rotating disc is
- A
$\frac{1}{2}\left[\frac{I_1+I_2}{I_1 I_2}\right] \omega_1^2$
- B
$\frac{1}{2}\left[\frac{I_1 I_2}{I_1-I_2}\right] \omega_1^2$
- C
$\frac{1}{2}\left[\frac{I_1-I_2}{I_1 I_2}\right] \omega_1^2$
- ✓
$\frac{1}{2}\left[\frac{I_1 I_2}{I_1+I_2}\right] \omega_1^2$
AnswerCorrect option: D. $\frac{1}{2}\left[\frac{I_1 I_2}{I_1+I_2}\right] \omega_1^2$
(d) : As no external torque is applied to the system, the angular momentum of the system remains conserved.
$
\therefore \quad L_1=L_2
$
According to given problem,
$
I_1 \omega_1=\left(I_1+I_2\right) \omega_2
$
or $\omega_2=\frac{I_1 \omega_1}{\left(I_1+I_2\right)}$ .....(i)
Initial energy, $E_1=\frac{1}{2} I_1 \omega_1^2$ .....(ii)
Final energy, $E_2=\frac{1}{2}\left(I_1+I_2\right) \omega_2^2$ .....(iii)
Substituting the value of $\omega_2$ from equation (i) in equation (iii), we get
Final energy, $E_2=\frac{1}{2}\left(I_1+I_2\right)\left(\frac{I_1 \omega_1}{I_1+I_2}\right)^2$
$
=\frac{1}{2} \frac{I_1^2 \omega_1^2}{\left(I_1+I_2\right)}
$ .....(iv)
Loss of energy, $\Delta E=E_1-E_2$
$
\begin{aligned}
& =\frac{1}{2} I_1 \omega_1^2-\frac{1}{2} \frac{I_1^2 \omega_1^2}{\left(I_1+I_2\right)} \quad \text { (Using (ii) and (iv)) } \\
& =\frac{\omega_1^2}{2}\left(\frac{I_1^2+I_2 I_1-I_1^2}{\left(I_1+I_2\right)}\right)=\frac{1}{2} \frac{I_2 I_1}{\left(I_1+I_2\right)} \omega_1^2
\end{aligned}
$
View full question & answer→MCQ 611 Mark
A ceiling fan rotates about its own axis with some angular velocity. When the fan is switched off, the angular velocity becomes $(1 / 4)^{\text {th }}$ of the original in time $t$ and $n$ revolutions are made in that time. The number of revolutions made by the fan during the time interval between switch off and rest are (Angular retardation is uniform)
- A
$\frac{4 n}{15}$
- B
$\frac{8 n}{15}$
- ✓
$\frac{16 n}{15}$
- D
$\frac{32 n}{15}$
AnswerCorrect option: C. $\frac{16 n}{15}$
(c): The angular velocity is given as
$
\omega^2=\omega_0{ }^2+2 \alpha\left(\theta-\theta_0\right)
$
When the fan is switched off, $\theta=2 \pi n, \theta_0=0, \omega=\frac{\omega_0}{4}$
$
\begin{aligned}
\left(\frac{\omega_0}{4}\right)^2 & =\omega_0^2-2 \alpha(2 \pi n) \Rightarrow 2 \alpha(2 \pi n)=\frac{15}{16} \omega_0^2 \\
2 \pi n^{\prime} & =\left(\frac{\omega_0^2}{2 \alpha}\right)
\end{aligned}
$
When the fan come to rest
$
\begin{aligned}
& 0=\omega_0{ }^2-2 \alpha\left(2 \pi n^{\prime}\right) \\
& 2 \pi n^{\prime}=\left(\frac{\omega_0^2}{2 \alpha}\right) \quad \text { or } n^{\prime}=\frac{16}{15} n
\end{aligned}
$
View full question & answer→MCQ 621 Mark
A wheel of moment of inertia $2 kg m ^2$ is rotating about an axis passing through centre and perpendicular to its plane at a speed $60 rad / s$. Due to friction, it comes to rest in 5 minutes. The angular momentum of the wheel three minutes before it stops rotating is
- A
$24 kg m ^2 / s$
- B
- ✓
- D
$96 kg m ^2 / s$
Answer(c) : Given, $I=2 kg m ^2, \omega_0=60 rad / s$
$\omega=0 \quad$ (... it comes to rest)
$t=5$ minute $=5 \times 60=300$ second
$\omega=\omega_0+\alpha t$
or $\alpha=\frac{\omega-\omega_0}{t}=\frac{0-60}{300}=-\frac{1}{5} rad / s ^2$
For 2 minute (= 120 second)
$
\omega=\omega_0+\alpha t=60-\frac{1}{5} \times 120=60-24=36 rad / s
$
The angular momentum $L=I \omega=2 \times 36=72 kg m ^2 / s$
View full question & answer→MCQ 631 Mark
For a particle moving in vertical circle, the total energy at different positions along the path
Answer(a) : As no external force is applied on vertical circle
$\therefore \quad$The total energy at different positions along the path is conserved.
View full question & answer→MCQ 641 Mark
Let $M$ be the mass and $L$ be the length of a thin uniform rod. In first case, axis of rotation is passing through centre and perpendicular to the length of the rod. In second case axis of rotation is passing through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to second case is
- A
- ✓
$\frac{1}{2}$
- C
$\frac{1}{4}$
- D
$\frac{1}{8}$
AnswerCorrect option: B. $\frac{1}{2}$
(b) : In first case: moment of inertia $\left(I_1\right)=\frac{M L^2}{12}$
$\therefore$ Radius of gyration $\left(K_1\right)=\frac{L}{\sqrt{12}}$
In second case: $I_2=\frac{M L^2}{3}$
$\therefore \quad K_2=\frac{L}{\sqrt{3}}$ Hence, $\frac{K_1}{K_2}=\frac{L}{\sqrt{12}} \times \frac{\sqrt{3}}{L}=\frac{1}{\sqrt{4}}=\frac{1}{2}$
View full question & answer→MCQ 651 Mark
A ring and a disc roll on the horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is $4 J$ then total kinetic energy of the disc is
Answer(a) : Total kinetic energy of the ring when it tolls without slipping,
$
K_{\text {ring }}=K_T+K_R=\frac{1}{2} m v^2+\frac{1}{2} I_r \omega^2
$
$\begin{aligned} & \quad=\frac{1}{2} m v^2+\frac{1}{2} m r^2 \times \frac{v^2}{r^2} \quad\left(\because I_r=m r^2 \text { and } \omega=\frac{v}{r}\right) \\ & \quad=m v^2 \\ & \text { But } K_{\text {ring }}=4 J \text { (given) } \\ & \therefore \quad m v^2=4 J .....(i)\\ & \text { Similarly, } K_{\text {dise }}=\frac{1}{2} m v^2+\frac{1}{2} I_d \omega^2 \quad \text { } \\ & \quad=\frac{1}{2} m v^2+\frac{1}{2} \times \frac{m r^2}{2} \times \frac{v^2}{r^2} \quad\left(\because I_d=\frac{m r^2}{2}\right) \\ & \quad=\frac{3}{4} m v^2=\frac{3}{4} \times 4 J =3 J \quad \text { (Using (i)) }\end{aligned}$
View full question & answer→MCQ 661 Mark
In vertical circular motion, the ratio of kinetic energy of a particle at highest point to that at lowest point is
Answer(d) : In vertical circular motion,
kinetic energy of particle at highest point,
$
K_H=\frac{1}{2} m v_H^2=\frac{1}{2} m g r \quad\left(\because v_H=\sqrt{g r}\right)
$
and that at lowest point,
$
\begin{aligned}
& K_L=\frac{1}{2} m v_L^2=\frac{5}{2} m g r \\
& \therefore \quad \text { Required ratio }=\frac{K_H}{K_L}=\frac{1}{5}=0.2
\end{aligned} \quad\left(\because v_L=\sqrt{5 g r}\right)
$.
View full question & answer→MCQ 671 Mark
A hollow sphere of mass ' $M$ ' and radius ' $R$ ' is rotating with angular frequency ' $\omega$ '. It suddenly stops rotating and $75 \%$ of kinetic energy is converted to heat. If ' $S$ ' is the specific heat of the material in $J / kg K$ then rise in temperature of the sphere is (M.I. of hollow sphere $=\frac{2}{3} M R^2$ )
AnswerCorrect option: B. $\frac{R^2 \omega^2}{4 S}$
(b) : Kinetic energy of hollow sphere,
$
\begin{aligned}
& K=\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{2}{3} M R^2\right) \omega^2 \quad\left[\because \quad I=\frac{2}{3} M R^2\right] \\
& K=\frac{1}{3} M R^2 \omega^2
...(i)\end{aligned}
$
According to question,
Heat in the hollow sphere $=75 \%$ of $K$
$
\begin{aligned}
& M S \Delta T=\frac{75}{100} K \\
& M S \Delta T=\frac{3}{4} \times \frac{1}{3} M R^2 \omega^2 \\
& \therefore \quad \Delta T=\frac{R^2 \omega^2}{4 S}
\end{aligned}
$
View full question & answer→MCQ 681 Mark
A cord is wound around the circumference of wheel of radius ' $r$ '. The axis of the wheel is horizontal and moment of inertia about it is ' $r$. The weight ' $m g$ ' 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
- A
$[m g h]^{1 / 2}$
- B
$\left[\frac{2 m g h}{I+2 m r^2}\right]^{1 / 2}$
- ✓
$\left[\frac{2 m g h}{I+m r^2}\right]^{1 / 2}$
- D
$\left[\frac{m g h}{I+m r^2}\right]^{1 / 2}$
AnswerCorrect option: C. $\left[\frac{2 m g h}{I+m r^2}\right]^{1 / 2}$
(c) : Let speed of the block be $v$ when it is falling through a distance $h$. So it is also the speed of string and a particle at the rim of the wheel.
Suppose $\omega$ is the angular velocity of the wheel.
So, $\omega=\frac{v}{r}$ and
Kinetic energy of wheel $=\frac{1}{2} I \omega^2$
Using energy conservation principle,
Gravitational potential energy lost by the block
$=$ Gain in kinetic energy of block and wheel
$m g h=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2 ; m g h=\frac{1}{2} m(\omega r)^2+\frac{1}{2} I \omega^2$
$\omega^2\left(m r^2+I\right)=2 m g h \quad \therefore \omega=\left[\frac{2 m g h}{I+m r^2}\right]^{1 / 2}$ View full question & answer→MCQ 691 Mark
A solid cylinder has mass ' $M$ ', radius ' $R$ ' and length ' $I$ '. Its moment of inertia about an axis passing through its centre and perpendicular to its own axis is
- A
$\frac{2 M R^2}{3}+\frac{M l^2}{12}$
- B
$\frac{M R^2}{3}+\frac{M l^2}{12}$
- C
$\frac{3 M R^2}{4}+\frac{M l^2}{12}$
- ✓
$\frac{M R^2}{4}+\frac{M l^2}{12}$
AnswerCorrect option: D. $\frac{M R^2}{4}+\frac{M l^2}{12}$
(d) : Mass of cylinder $=M$, radius $=R$, length $=l$
Moment of inertia about an axis passing through centre of long solid cylinder and perpendicular to its own axis, $I=$ Moment of inertia of cylinder about its diameter +
Moment of inertia of thin rod about an axis passing through its centre of mass
$
I=\frac{M R^2}{4}+\frac{M l^2}{12}
$
View full question & answer→MCQ 701 Mark
Three identical spheres each of mass $1 kg$ are placed touching one another with their centres in a straight line. Their centres are marked as $A, B, C$ respectively. The distance of centre of mass of the system from $A$ is
- A
$\frac{A B+A C}{2}$
- B
$\frac{A B+B C}{2}$
- C
$\frac{A C-A B}{3}$
- D
$\frac{A B+A C}{3}$
Answer
As it is clear from the symmetry of the figure, that centre of mass of the system is at $B$.
$\therefore$ Its distance from $A$ is
$
x=\frac{m_A \times 0+m_B \times A B+m_C \times A C}{m_A+m_B+m_C}
$
Here, $m_A=m_B=m_C=1 kg$
$
\therefore x=\frac{A B+A C}{3}
$ View full question & answer→MCQ 711 Mark
The moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through one end is $I$. The same rod is bent into a ring and its moment of inertia about the diameter is $I_1$. The ratio $\frac{I}{I_1}$ is
- A
$\frac{4 \pi}{3}$
- ✓
$\frac{8 \pi^2}{3}$
- C
$\frac{5 \pi}{3}$
- D
$\frac{8 \pi^2}{5}$
AnswerCorrect option: B. $\frac{8 \pi^2}{3}$
(b) : Let $M$ be mass and $L$ be length of the rod.
Moment of inertia of the rod about the perpendicular axis passing through one end is
$
l=\frac{1}{3} M L^2
$
When it is bent into a ring of radius $R$, then
$
R=\frac{L}{2 \pi}
$. . . . . (i)
$
2 \pi R=L
$
Moment of inertia of the ring about the diameter is
$
\begin{aligned}
I_1=\frac{1}{2} M R^2 & =\frac{1}{2} M\left(\frac{L}{2 \pi}\right)^2 \\
& =\frac{1}{8 \pi^2} M L^2
\end{aligned}
$(Using (i))
The required ratio is $\frac{I}{I_1}=\frac{\frac{1}{3} M L^2}{\frac{1}{8 \pi^2} M L^2}=\frac{8 \pi^2}{3}$
View full question & answer→MCQ 721 Mark
An object of radius $R$ and mass $M$ is rolling horizontally without slipping with speed $v$. It then rolls up the hill to a maximum height $h=\frac{3 v^2}{4 g}$. The moment of inertia of the object is ( $g=$ acceleration due to gravity)
- A
$\frac{2}{5} M R^2$
- ✓
$\frac{M R^2}{2}$
- C
$M R^2$
- D
$\frac{3}{2} M R^2$
AnswerCorrect option: B. $\frac{M R^2}{2}$
(b) : The kinetic energy of the rolling object is converted into potential energy at height $h\left(=\frac{3 v^2}{4 g}\right)$
According to the law of conservation of mechanical energy, we get
$
\begin{aligned}
& \frac{1}{2} M v^2+\frac{1}{2} I \omega^2=M g h \\
& \frac{1}{2} M v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2=M g\left(\frac{3 v^2}{4 g}\right) \quad\left(\because \omega=\frac{v}{R}\right)
\end{aligned}
$
$\begin{aligned} & \frac{1}{2} M v^2+\frac{1}{2} I \frac{v^2}{R^2}=\frac{3}{4} M v^2 \\ & \frac{1}{2} I \frac{v^2}{R^2}=\frac{3}{4} M v^2-\frac{1}{2} M v^2=\frac{1}{4} M v^2 ; I=\frac{1}{2} M R^2\end{aligned}$ View full question & answer→MCQ 731 Mark
A thin uniform rod of mass M and length L has a small block of mass M attached at one end. The moment of inertia of the system about an axis through its CM and perpendicular to the length of the rod is
- A
$\frac{13}{12} ML ^2$
- B
$\frac{1}{3} ML ^2$
- ✓
$\frac{5}{24} ML ^2$
- D
$\frac{7}{48} ML ^2$
AnswerCorrect option: C. $\frac{5}{24} ML ^2$
$\frac{5}{24} ML ^2$
View full question & answer→MCQ 741 Mark
The radius of gyration k for a rigid body about a given rotation axis is given by
- A
$k=\frac{1}{M} \int r d m$
- ✓
$k^2=\frac{1}{M} \int r^2 d m$
- C
$k^2=\frac{1}{M} \int r d m$
- D
$k=\frac{1}{M} \int r^2 d m$.
AnswerCorrect option: B. $k^2=\frac{1}{M} \int r^2 d m$
$k^2=\frac{1}{M} \int r^2 d m$
View full question & answer→MCQ 751 Mark
A small object, tied at the end of a string of length r, is launched into a vertical circle with a speed$2 \sqrt{g r}$ at the lowest point. Its speed when the string is horizontal is
- A
$>3 \sqrt{g r}$
- B
$=3 \sqrt{g r}$
- ✓
$=2 \sqrt{g r}$
- D
AnswerCorrect option: C. $=2 \sqrt{g r}$
$=2 \sqrt{g r}$
View full question & answer→MCQ 761 Mark
A small bob of mass m is tied to a string and revolved in a vertical circle of radius r. If its speed at the highest point is$\sqrt{3 r g}$, the tension in the string at the lowest point is
View full question & answer→MCQ 771 Mark
A conical pendulum of string length L and bob of mass m performs UCM along a circular path of radius r. The tension in the string is
- ✓
$\frac{m g L}{\sqrt{L^2-r^2}}$
- B
$\frac{m g L}{\sqrt{L^2+r^2}}$
- C
$\frac{m g L}{\sqrt{2} r}$
- D
$\frac{m r g \tan \theta}{L}$.
AnswerCorrect option: A. $\frac{m g L}{\sqrt{L^2-r^2}}$
$\frac{m g L}{\sqrt{L^2-r^2}}$
View full question & answer→MCQ 781 Mark
Two particles with their masses in the ratio 2 : 3 perform uniform circular motion with orbital radii in the ratio 3 : 2. If the centripetal force acting on them is the same, the ratio of their speeds is
- A
$4 : 9$
- B
$1: 1$
- ✓
$3: 2$
- D
$9: 4$
AnswerCorrect option: C. $3: 2$
$3: 2$
View full question & answer→