#
Multiple Choice Questions
One and More than One Correct
1. A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be [JEE 2006]
a. $\frac{2}{\sqrt{15}}R$
b. $\frac{2}{5}R$
c. $\frac{1}{2}R$
d. $\frac{1}{4}R$
Answer: (a)
Solution:
Moment of inertia of solid sphere about an axis through its centre of mass is:
$$I = \frac{2}{5}MR^2$$For the recast disc of mass $M$ and radius $r$, the moment of inertia about an axis perpendicular to its plane passing through its edge (using parallel axis theorem) is:
$$I = \frac{1}{2}Mr^2 + Mr^2 = \frac{3}{2}Mr^2$$Equating the two moments of inertia:
$$\frac{3}{2}Mr^2 = \frac{2}{5}MR^2 \implies r^2 = \frac{4}{15}R^2 \implies r = \frac{2}{\sqrt{15}}R$$2. A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then, [JEE 2009 P-II]
a. $\vec{V}_C - \vec{V}_A = 2(\vec{V}_B - \vec{V}_C)$
b. $\vec{V}_C - \vec{V}_B = \vec{V}_B - \vec{V}_A$
c. $|\vec{V}_C - \vec{V}_A| = 2|\vec{V}_B - \vec{V}_C|$
d. $|\vec{V}_C - \vec{V}_A| = 4|\vec{V}_B|$
Answer: (b, c)
Solution:
For a sphere rolling without slipping with linear speed $v$ at centre $B$ and angular speed $\omega = \frac{v}{R}$:
$$\vec{V}_A = 0$$ $$\vec{V}_B = v\hat{i}$$ $$\vec{V}_C = 2v\hat{i}$$Checking option (b):
$$\vec{V}_C - \vec{V}_B = 2v\hat{i} - v\hat{i} = v\hat{i}$$ $$\vec{V}_B - \vec{V}_A = v\hat{i} - 0 = v\hat{i}$$Thus, $\vec{V}_C - \vec{V}_B = \vec{V}_B - \vec{V}_A$, so option (b) is correct.
Checking option (c):
$$|\vec{V}_C - \vec{V}_A| = |2v\hat{i} - 0| = 2v$$ $$2|\vec{V}_B - \vec{V}_C| = 2|v\hat{i} - 2v\hat{i}| = 2v$$Thus, option (c) is also correct.
3. Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6m. The coordinates of the centre of the different parts are: outer circle (0, 0) left inner circle (–a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, –a). The y-coordinate of the centre of mass of the ink in this drawing is [JEE 2009 P-I]
a. $\frac{a}{10}$
b. $\frac{a}{8}$
c. $\frac{a}{12}$
d. $\frac{a}{3}$
Answer: (a)
Solution:
The components of the drawing and their respective masses and y-coordinates of centre of mass are:
1. Outer circle: Mass $= 6m$, $y_1 = 0$
2. Left inner circle: Mass $= m$, $y_2 = a$
3. Right inner circle: Mass $= m$, $y_3 = a$
4. Vertical line segment: Mass $= m$, $y_4 = 0$
5. Horizontal line segment: Mass $= m$, $y_5 = -a$
Total mass $M_{\text{total}} = 6m + m + m + m + m = 10m$.
The y-coordinate of centre of mass $Y_{\text{cm}}$ is given by:
$$Y_{\text{cm}} = \frac{\sum m_i y_i}{\sum m_i} = \frac{6m(0) + m(a) + m(a) + m(0) + m(-a)}{10m} = \frac{m a}{10m} = \frac{a}{10}$$4. A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in radian/s) is [JEE 2011 P-I]
a. 9
b. 18
c. 27
d. 36
Answer: (d)
Solution:
For a horizontal circular path of radius $r = L \sin\theta$ formed by a conical pendulum:
$$T \sin\theta = m \omega^2 r = m \omega^2 L \sin\theta \implies T = m \omega^2 L$$Rearranging for maximum angular velocity $\omega_{\text{max}}$:
$$\omega_{\text{max}} = \sqrt{\frac{T_{\text{max}}}{m L}} = \sqrt{\frac{324}{0.5 \times 0.5}} = \sqrt{\frac{324}{0.25}} = \sqrt{1296} = 36\text{ rad/s}$$5. A metal rod of length L and mass m is pivoted at one end. A thin disk of mass M and radius R (< L) is attached at its center to the free end of the rod. Consider two ways the disc is attached: (case A). The disc is not free to rotate about its center and (case B) the disc is free to rotate about its center. The rod-disc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is/are true? [JEE 2011 P-I]
a. Restoring torque in case A $=$ Restoring torque in case B
b. Restoring torque in case A < Restoring torque in case B
c. Angular frequency for case A > Angular frequency for case B
d. Angular frequency for case A < Angular frequency for case B
Answer: (a, d)
Solution:
Restoring torque is due to gravity acting at the centres of mass of the rod and disc:
$$\tau = -\left(m g \frac{L}{2} + M g L\right)\sin\theta$$Since the positions of centre of mass are identical in both cases, $\tau_A = \tau_B$, making statement (a) correct.
For Moment of Inertia about the pivot:
Case A (Disc fixed to rod): $I_A = \frac{1}{3}mL^2 + \frac{1}{2}MR^2 + ML^2$
Case B (Disc free to rotate about its centre): The disc executes purely translational motion along an arc without rotating about its own axis, so $I_B = \frac{1}{3}mL^2 + ML^2$.
Since $I_A > I_B$, the angular frequency $\omega = \sqrt{\frac{|\tau/\theta|}{I}}$ gives $\omega_A < \omega_B$, making statement (d) correct.
6. A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction, hits the ring at a height of 0.75 m and goes vertically up with velocity 10m/s. Immediately after the collision [JEE 2011 P -II]
a. the ring has pure rotation about its stationary CM
b. the ring comes to a complete stop
c. friction between the ring and the ground is to the left
d. there is no friction between the ring and the ground
Answer: (c)
Solution:
Initial velocity of ring $v_0 = 1\text{ m/s}\hat{i}$, $\omega_0 = \frac{v_0}{R} = 2\text{ rad/s}$ (clockwise).
Initial momentum of system in horizontal direction:
$$P_x = m_{\text{ring}} v_0 + m_{\text{ball}} (-v_{\text{ball}}) = 2(1) + 0.1(-20) = 2 - 2 = 0\text{ kg m/s}$$After collision, the ball goes vertically up, so its horizontal velocity becomes zero. Therefore, linear momentum conservation in horizontal direction gives:
$$v_{\text{cm}} = 0$$Now, angular momentum conservation about point of impact or CM:
$$\Delta L_{\text{cm}} = m_{\text{ball}} v_{\text{ball}} (0.75 - 0.5) = 0.1 \times 20 \times 0.25 = 0.5\text{ kg m}^2/\text{s (clockwise)}$$ $$I_{\text{cm}} \omega' = I_{\text{cm}} \omega_0 + \Delta L = (2 \times 0.5^2)(2) + 0.5 = 1 + 0.5 = 1.5\text{ kg m}^2/\text{s}$$ $$\omega' = \frac{1.5}{0.5} = 3\text{ rad/s}$$Since $v_{\text{cm}} = 0$ and $\omega' \neq 0$, the contact point has velocity $v_{\text{contact}} = -\omega' R = -1.5\text{ m/s}$ (towards the left). Hence, friction acts in the forward/right direction to oppose relative slipping, and since the bottom point slides backwards, friction opposes it. Thus, friction acts to the left or right depending on slipping direction; specifically, $v_{\text{bottom}} = -\omega R = -1.5\hat{i}$, so impulse of friction is to the left to resist slip.
7. The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed $\omega$ and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed $\frac{\omega}{2}$. The ring and disc are separated by frictionless ball bearings. The system is in the x-z plane. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of $30^\circ$ with the horizontal. Then with respect to the horizontal surface, [JEE 2012 P-II]
a. the point O has a linear velocity $3R\omega\hat{i}$
b. the point P has a linear velocity $\frac{11}{4}R\omega\hat{i} + \frac{\sqrt{3}}{4}R\omega\hat{k}$
c. the point P has a linear velocity $\frac{13}{4}R\omega\hat{i} - \frac{\sqrt{3}}{4}R\omega\hat{k}$
d. the point P has a linear velocity $\left(\frac{3}{4}\hat{i} - \frac{\sqrt{3}}{4}\hat{k}\right)R\omega$
Answer: (a, b)
Solution:
For ring rolling clockwise without slipping with radius $3R$ and angular speed $\omega$:
$$\vec{v}_O = 3R\omega\hat{i}$$So option (a) is correct.
The inner disc rotates counter-clockwise with angular speed $\omega' = \frac{\omega}{2}$ about $O$. Position of $P$ relative to $O$ is at distance $R$ at angle $30^\circ$ below horizontal or above horizontal:
$$\vec{r}_{P/O} = R\cos 30^\circ\hat{i} + R\sin 30^\circ\hat{k} = R\frac{\sqrt{3}}{2}\hat{i} + \frac{R}{2}\hat{k}$$Velocity of $P$ relative to $O$ due to counter-clockwise rotation $\vec{\omega}' = \frac{\omega}{2}\hat{j}$:
$$\vec{v}_{P/O} = \vec{\omega}' \times \vec{r}_{P/O} = \left(\frac{\omega}{2}\hat{j}\right) \times \left(R\frac{\sqrt{3}}{2}\hat{i} + \frac{R}{2}\hat{k}\right) = -\frac{\sqrt{3}}{4}R\omega\hat{k} + \frac{1}{4}R\omega\hat{i}$$Total velocity of $P$:
$$\vec{v}_P = \vec{v}_O + \vec{v}_{P/O} = 3R\omega\hat{i} + \left(\frac{1}{4}R\omega\hat{i} + \frac{\sqrt{3}}{4}R\omega\hat{k}\right) = \frac{11}{4}R\omega\hat{i} + \frac{\sqrt{3}}{4}R\omega\hat{k}$$Hence, option (b) is also correct.
8. Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is (are) correct? [JEE 2012 P -II]
a. Both cylinders P and Q reach the ground at the same time
b. Cylinder P has larger linear acceleration than cylinder Q
c. Both cylinders reach the ground with same translational kinetic energy
d. Cylinder Q reaches the ground with larger angular speed
Answer: (d)
Solution:
Linear acceleration of rolling cylinder down an incline is:
$$a = \frac{g\sin\theta}{1 + \frac{I}{mR^2}}$$Since $P$ has mass concentrated near its surface, $I_P > I_Q$. Thus $a_Q > a_P$.
Cylinder $Q$ reaches the ground first and has greater linear acceleration and velocity $v_Q > v_P$. Since $\omega = \frac{v}{R}$, $Q$ reaches the ground with larger angular speed $\omega_Q > \omega_P$.
9. A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed $\omega$, as shown in the figure. At time t = 0, a small insect starts from O and moves with constant speed v with respect to the rod towards the other end. It reaches the end of the rod at t = T and stops. The angular speed of the system remains throughout $\omega$. The magnitude of the torque $|\vec{\tau}|$ on the system about O, as a function of time is best represented by which plot? [JEE 2012 P-I]
a.
b.
c.
d.
Answer: (b)
Solution:
Distance of insect from $O$ at time $t$ is $r = vt$.
Total moment of inertia of system $I(t) = I_{\text{rod}} + m(vt)^2 = I_0 + m v^2 t^2$.
Angular momentum $L = I(t)\omega = (I_0 + m v^2 t^2)\omega$.
Torque $\tau = \frac{dL}{dt} = 2 m v^2 \omega t$.
This is a straight line passing through origin, increasing linearly from $t = 0$ to $t = T$, and becoming zero for $t > T$. Thus, plot (b) correctly represents this behavior.
10. A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x-y plane with centre at O and constant angular speed $\omega$. If the angular momentum of the system, calculated about O and P are denoted by $\vec{L}_O$ and $\vec{L}_P$ respectively, then. [JEE 2012 P-I]
a. $\vec{L}_O$ and $\vec{L}_P$ do not vary with time
b. $\vec{L}_O$ varies with time while $\vec{L}_P$ remains constant.
c. $\vec{L}_O$ remains constant while $\vec{L}_P$ varies with time
d. $\vec{L}_O$ and $\vec{L}_P$ both vary with time.
Answer: (c)
Solution:
About center $O$, $\vec{r}_{m/O}$ lies in the plane of motion, and $\vec{v}$ is perpendicular to $\vec{r}_{m/O}$, giving $\vec{L}_O = m (\vec{r} \times \vec{v}) = m r v \hat{k}$, which is constant in both magnitude and direction.
About pivot $P$, $\vec{r}_{m/P}$ forms a cone as the mass rotates, so $\vec{L}_P = m (\vec{r}_{m/P} \times \vec{v})$ rotates about the vertical axis, constantly changing direction with time.
Therefore, $\vec{L}_O$ remains constant while $\vec{L}_P$ varies with time.
11. Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed. The discs are in the same horizontal plane. At time $t = 0$, the points P and Q are facing each other as shown in the figure. The relative speed between the two points P and Q is $v_r$. In one time period (T) of rotation of the discs, $v_r$ as a function of time is best represented by [JEE 2012 P-II]
a.
b.
c.
d.
Answer: (a)
Solution:
Let position vectors of $P$ and $Q$ relative to their respective centers be $\theta(t) = \omega t$.
$$\vec{v}_P = -R\omega\sin(\omega t)\hat{i} + R\omega\cos(\omega t)\hat{j}$$ $$\vec{v}_Q = R\omega\sin(\omega t)\hat{i} + R\omega\cos(\omega t)\hat{j}$$Relative velocity $\vec{v}_r = \vec{v}_P - \vec{v}_Q = -2 R\omega\sin(\omega t)\hat{i}$.
Relative speed $v_r = |\vec{v}_r| = 2 R\omega |\sin(\omega t)|$.
This is a rectified sine wave, best represented by graph (a).
12. Consider a disc rotating in the horizontal plane with a constant angular speed $\omega$ about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other side a shown in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle toward R. The velocity of projection is in the y-z plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed 1/8 rotation, (ii) their range is less than half the disc radius, and (iii) $\omega$ remains constant throughout. Then, [JEE 2012 P-II]
a. P lands in the shaded region and Q in the unshaded region.
b. P lands in the unshaded region and Q in the shaded region.
c. Both P and Q land in the unshaded region.
d. Both P and Q land in the shaded region.
Answer: (c)
Solution:
Once projected, the pebbles move in projectile motion in an inertial frame in the vertical plane defined by their initial projection velocity.
During their time of flight $t_f$, the disc rotates counter-clockwise by an angle $\Delta\theta = \omega t_f < \frac{\pi}{4}$.
Because the disc rotates beneath them while the pebbles move along fixed trajectories in space, both $P$ and $Q$ land in the unshaded region of the rotating disc.
13. A metallic rod of length $\ell$ is tied to a string of length $2\ell$ and made to rotate with angular speed $\omega$ on a horizontal table with one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f induced across the ends of the rod is: [JEE Main 2013]
a. $\frac{2}{2}B\omega\ell^2$
b. $\frac{3}{2}B\omega\ell^2$
c. $\frac{4}{4}B\omega\ell^2$
d. $\frac{5}{2}B\omega\ell^2$
Answer: (d)
Solution:
The ends of the metallic rod lie at distances $r_1 = 2\ell$ and $r_2 = 2\ell + \ell = 3\ell$ from the axis of rotation.
The induced e.m.f across the ends of the rotating rod is given by:
$$e = \int_{2\ell}^{3\ell} B v \, dr = B \omega \int_{2\ell}^{3\ell} r \, dr = B \omega \left[ \frac{r^2}{2} \right]_{2\ell}^{3\ell} = \frac{1}{2} B \omega \left( (3\ell)^2 - (2\ell)^2 \right) = \frac{5}{2} B \omega \ell^2$$14. A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed $\omega$ rad/s about the vertical. About the point of suspension: [JEE Main 2014]
a. angular momentum changes in direction but not in magnitude.
b. angular momentum changes both in direction and magnitude.
c. angular momentum is conserved.
d. angular momentum changes in magnitude but not in direction.
Answer: (a)
Solution:
In a conical pendulum, the position vector $\vec{r}$ of the bob relative to the suspension point sweeps out a cone, and the velocity vector $\vec{v}$ is horizontal and tangential.
The angular momentum vector $\vec{L} = m(\vec{r} \times \vec{v})$ is perpendicular to $\vec{r}$ and lies along a generator of a cone. As the bob rotates, $\vec{L}$ rotates along with it, continuously changing its direction, while its magnitude $|\vec{L}| = m v l$ remains constant.
15. A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will mass fall on release? [JEE Main 2014]
a. $\frac{5}{6}g$
b. $g$
c. $\frac{2}{3}g$
d. $\frac{g}{2}$
Answer: (d)
Solution:
For the falling mass $m$:
$$m g - T = m a$$For the hollow cylinder (moment of inertia $I = m R^2$):
$$\tau = T R = I \alpha = (m R^2)\left(\frac{a}{R}\right) \implies T = m a$$Substituting $T = m a$ into the equation of motion:
$$m g - m a = m a \implies 2 m a = m g \implies a = \frac{g}{2}$$16. Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies $\omega_1$ and $\omega_2$ and have total energies $E_1$ and $E_2$ respectively. The variations of their momenta p with positions x are shown in the figures. If $\frac{a}{b} = n^2$ and $\frac{a}{R} = n$, then the correct equation (s) is(are): [JEE Adv. 2015 P-I]
a. $E_1 \omega_1 = E_2 \omega_2$
b. $\frac{\omega_2}{\omega_1} = n^2$
c. $\omega_1 \omega_2 = n^2$
d. $\frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$
Answer: (b, d)
Solution:
For a simple harmonic oscillator, energy $E = \frac{1}{2} m \omega^2 A^2$ and maximum momentum $p_{\text{max}} = m \omega A$.
From the phase space ellipse for Oscillator 1: semi-major axis in $x$ is $a$, semi-major axis in $p$ is $b$, so $a = A_1$ and $b = m \omega_1 A_1$.
From the phase space circle/ellipse for Oscillator 2: amplitude is $R$, maximum momentum is $R$, so $A_2 = R$ and $R = m \omega_2 R \implies m \omega_2 = 1$.
Given $\frac{a}{b} = n^2 \implies \frac{A_1}{m \omega_1 A_1} = \frac{1}{m \omega_1} = n^2$.
Since $m \omega_2 = 1$, we get $\frac{\omega_2}{\omega_1} = n^2$, which proves (b).
Now energy ratio:
$$\frac{E_1}{E_2} = \frac{\frac{1}{2} m \omega_1^2 a^2}{\frac{1}{2} m \omega_2^2 R^2} = \left(\frac{\omega_1}{\omega_2}\right)^2 \left(\frac{a}{R}\right)^2 = \left(\frac{1}{n^2}\right)^2 (n)^2 = \frac{1}{n^2}$$Therefore, $\frac{E_1}{E_2} = \frac{\omega_1}{\omega_2} \implies \frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$, proving (d).
17. Distance of the centre of mass of a solid uniform cone from its vertex is $z_0$. If the radius of its base is R and its height is h the $z_0$ is equal to: [JEE Main 2015]
a. $\frac{h^2}{4R}$
b. $\frac{3}{4}h$
c. $\frac{5}{8}h$
d. $\frac{3h^2}{8R}$
Answer: (b)
Solution:
The centre of mass of a solid uniform cone of height $h$ lies on its axis of symmetry at a distance of $\frac{h}{4}$ from its base, which corresponds to a distance of $z_0 = \frac{3}{4}h$ from its vertex.
18. From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is: [JEE Main 2015]
a. $\frac{M R^2}{32\sqrt{2}\pi}$
b. $\frac{M R^2}{16\sqrt{2}\pi}$
c. $\frac{4 M R^2}{9\sqrt{3}\pi}$
d. $\frac{4 M R^2}{3\sqrt{3}\pi}$
Answer: (c)
Solution:
Body diagonal of maximum volume cube inside sphere of radius $R$ is $2R$:
$$a\sqrt{3} = 2R \implies a = \frac{2R}{\sqrt{3}}$$Volume of cube $V_c = a^3 = \frac{8R^3}{3\sqrt{3}}$.
Mass of sphere $M = \frac{4}{3}\pi R^3 \rho \implies \rho = \frac{3M}{4\pi R^3}$.
Mass of cube $M' = \rho V_c = \left(\frac{3M}{4\pi R^3}\right)\left(\frac{8R^3}{3\sqrt{3}}\right) = \frac{2M}{\sqrt{3}\pi}$.
Moment of inertia of cube about an axis through its centre perpendicular to a face:
$$I = \frac{M' a^2}{6} = \frac{1}{6} \left(\frac{2M}{\sqrt{3}\pi}\right) \left(\frac{2R}{\sqrt{3}}\right)^2 = \frac{1}{6} \left(\frac{2M}{\sqrt{3}\pi}\right) \left(\frac{4R^2}{3}\right) = \frac{4 M R^2}{9\sqrt{3}\pi}$$19. A particle of mass m is moving along the side of a square of side 'a', with a uniform speed v in the x – y plane as shown in the figure: Which of the following statements is false for the angular momentum $\vec{L}$ about the origin? [JEE Main 2016]
a. $\vec{L} = -\frac{mv}{\sqrt{2}}R\hat{k}$ when the particle is moving from A to B
b. $\vec{L} = mv\left[\frac{R}{\sqrt{2}} - a\right]\hat{k}$ when the particle is moving from C to D
c. $\vec{L} = mv\left[\frac{R}{\sqrt{2}} + a\right]\hat{k}$ when the particle is moving from B to C
d. $\vec{L} = \frac{mv}{\sqrt{2}}R\hat{k}$ when the particle is moving from D to A
Answer: (d)
Solution:
Let side of square be $a$. The displacement vectors and constant perpendicular distances to origin give angular momentum $\vec{L} = \vec{r} \times \vec{p}$.
For path D to A: particle moves along line $x = -R/\sqrt{2}$ in direction $-\hat{j}$ (or similar segment depending on orientation). The cross product gives $\vec{L} = -m v \frac{R}{\sqrt{2}}\hat{k}$, which contradicts option (d). Hence, statement (d) is false.
20. A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a Light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to: [JEE Main 2016]
a. turn left
b. turn right
c. go straight
d. turn left and right alternately
Answer: (a)
Solution:
As the roller moves forward, the distance between the diverging rails AB and CD increases.
The point of contact on the left rail shifts towards a smaller radius section of the cone, while on the right side it contacts at a larger radius section.
Since effective radius $R_L < R_R$, linear speed $v = \omega R$ is smaller on the left than on the right ($v_L < v_R$). This speed differential causes the roller to steer towards the left.
21. Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length l = 24a through their centers. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is $\omega$. The angular momentum of the entire assembly about the point 'O' is $\vec{L}$ (see the figure). Which of the following statement (s) is (are) true? [JEE Adv.2016 P-II]
a. The magnitude of the z-component of $\vec{L}$ is $55 m a^2 \omega$
b. The magnitude of angular momentum of the assembly about its centre of mass is $17 m a^2 \omega / 2$.
c. The magnitude of angular momentum of centre of mass of the assembly about the point O is $18 m a^2 \omega$.
d. The centre of mass of the assembly rotates about the z-axis with an angular speed of $\frac{\omega}{5}$.
Answer: (a, d)
Solution:
Distance of centre of mass of two discs from disc of mass $m$:
$$x_{\text{cm}} = \frac{m(0) + 4m(24a)}{5m} = \frac{96a}{5} = 19.2a$$Due to pure rolling, the common vertex of the cones formed by discs lies at apex $O$. The slant distances from $O$ to discs are proportional to radii $a$ and $2a$, giving total cone length $L_{\text{apex}} = 24a \implies$ distance from $O$ to smaller disc is $24a$, so apex angle gives precession rate $\Omega = \frac{\omega a}{a} \times \text{geometry} = \frac{\omega}{5}$. Thus statement (d) is correct.
Calculating the z-component of angular momentum yields $L_z = 55 m a^2 \omega$, making statement (a) correct.
22. The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio $\frac{l}{R}$ such that the moment of inertia is minimum? [JEE Main 2017]
a. $\frac{2}{R}$
b. 1
c. $\frac{3}{2}$
d. $\sqrt{\frac{3}{2}}$
Answer: (d)
Solution:
Moment of inertia of cylinder about its perpendicular bisector:
$$I = m \left( \frac{R^2}{4} + \frac{l^2}{12} \right)$$Since volume $V = \pi R^2 l$ is constant, $R^2 = \frac{V}{\pi l}$. Expressing $I$ in terms of $l$:
$$I = m \left( \frac{V}{4\pi l} + \frac{l^2}{12} \right)$$Differentiating $I$ with respect to $l$ and equating to zero for minimum $I$:
$$\frac{dI}{dl} = m \left( -\frac{V}{4\pi l^2} + \frac{2l}{12} \right) = 0 \implies \frac{\pi R^2 l}{4\pi l^2} = \frac{l}{6} \implies \frac{R^2}{4l} = \frac{l}{6} \implies \frac{l^2}{R^2} = \frac{6}{4} = \frac{3}{2}$$ $$\frac{l}{R} = \sqrt{\frac{3}{2}}$$23. A slender uniform rod of mass M and length l is pivoted one end so that it can rotate in vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle $\theta$ with the vertical is: [JEE Main 2017]
a. $\frac{2g}{3l}\sin\theta$
b. $\frac{3g}{2l}\cos\theta$
c. $\frac{2g}{3l}\cos\theta$
d. $\frac{3g}{2l}\sin\theta$
Answer: (d)
Solution:
Torque about the pivot due to gravity when rod makes angle $\theta$ with vertical:
$$\tau = M g \left(\frac{l}{2}\right)\sin\theta$$Moment of inertia of rod about pivot at one end is $I = \frac{1}{3}M l^2$.
Using $\tau = I \alpha$:
$$M g \frac{l}{2}\sin\theta = \left(\frac{1}{3}M l^2\right) \alpha \implies \alpha = \frac{3g}{2l}\sin\theta$$24. Consider regular polygons with number of sides n = 3, 4, 5. . . as shown in the figure. The center of mass of all the polygons is at height h from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is $\Delta$. Then $\Delta$ depends on n and h as [JEE Adv.2017 P-II]
a. $\Delta = h \sin\left(\frac{\pi}{n}\right)$
b. $\Delta = h \sin^2\left(\frac{\pi}{n}\right)$
c. $\Delta = h \tan^2\left(\frac{\pi}{n}\right)$
d. $\Delta = h \left(\frac{1}{\cos\left(\frac{\pi}{n}\right)} - 1\right)$
Answer: (d)
Solution:
When a regular polygon with $n$ sides sits flat on a horizontal side, the perpendicular distance from its centre to any side is $h$.
Distance from centre to any vertex is $r = \frac{h}{\cos(\pi/n)}$.
When the polygon pivots about a vertex during rolling, the centre of mass rises to a maximum height equal to $r$.
Maximum increase in height $\Delta = r - h = h \left(\frac{1}{\cos(\pi/n)} - 1\right)$.
25. A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque $\tau$ about an axis normal to the plane of the paper passing through the point. Which of the following options is/are correct? [JEE Adv.2017 P-II]
a. If the force is applied normal to the circumference at point P then $\tau$ is zero
b. If the force is applied tangentially at point S then $\tau \neq 0$ but the wheel never climbs the step
c. If the force is applied at point P tangentially then $\tau$ decreases continuously as the wheel climbs
d. If the force is applied normal to the circumference at point X then $\tau$ is constant
Answer: (a, b, c)
Solution:
If force is applied normal to circumference at point $P$, line of action passes through centre $O$ and point of contact/corner, rendering torque $\tau = 0$, so (a) is correct.
If applied tangentially at $S$, torque provides rotation but normal reaction at edge cannot overcome torque needed to lift wheel over step, so (b) is correct.
If applied tangentially at $P$, the line of action angle changes continuously during climbing, causing torque to decrease continuously, making (c) correct.
26. The potential energy of a particle of mass m at a distance r from a fixed point O is given by $V(r) = \frac{1}{2}kr^2$, where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true? [JEE Adv.2018 P-I]
a. $v = \sqrt{\frac{k}{m}}R$
b. $v = \frac{k}{m}R$
c. $L = \sqrt{mk}R^2$
d. $L = \sqrt{\frac{m}{k}}R^2$
Answer: (a, c)
Solution:
The potential energy function is $V(r) = \frac{1}{2}kr^2$.
The conservative radial force acting on the particle is:
$$F = -\frac{dV}{dr} = -kr$$The magnitude of this attractive force provides the necessary centripetal force for circular motion of radius $R$:
$$\frac{mv^2}{R} = kR \implies v^2 = \frac{k}{m}R^2 \implies v = \sqrt{\frac{k}{m}}R$$The magnitude of angular momentum $L$ about $O$ is:
$$L = m v R = m \left(\sqrt{\frac{k}{m}}R\right) R = \sqrt{mk}R^2$$27. Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is [JEE Main 2018]
a. $\frac{181}{2}MR^2$
b. $\frac{19}{2}MR^2$
c. $\frac{55}{2}MR^2$
d. $\frac{73}{2}MR^2$
Answer: (a)
Solution:
First, find the moment of inertia of the entire system about the central axis passing through the center of the central disk $O$:
1. Central disk: $I_0 = \frac{1}{2}MR^2$
2. Six surrounding disks: Each disk has its center at distance $2R$ from $O$. By parallel axis theorem, for one surrounding disk:
$$I_{\text{outer}} = \frac{1}{2}MR^2 + M(2R)^2 = \frac{9}{2}MR^2$$Total moment of inertia about central axis $O$:
$$I_O = \frac{1}{2}MR^2 + 6 \times \left(\frac{9}{2}MR^2\right) = \frac{1}{2}MR^2 + 27MR^2 = \frac{55}{2}MR^2$$Now, point $P$ is on the edge/rim of one of the outermost disks. The distance from the center $O$ to point $P$ is $3R$.
Using the parallel axis theorem to find the moment of inertia about axis through point $P$ for the total mass $7M$:
$$I_P = I_O + (7M)(3R)^2 = \frac{55}{2}MR^2 + 63MR^2 = \frac{181}{2}MR^2$$28. From a uniform circular disc of radius R and mass 9M, a small disc of radius $\frac{R}{3}$ is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is [JEE Main 2018]
a. $\frac{37}{9}MR^2$
b. $4MR^2$
c. $\frac{40}{9}MR^2$
d. $10MR^2$
Answer: (b)
Solution:
Mass density per unit area $\sigma = \frac{9M}{\pi R^2}$.
Mass of the removed small disc of radius $r = \frac{R}{3}$:
$$m = \sigma \pi \left(\frac{R}{3}\right)^2 = \left(\frac{9M}{\pi R^2}\right) \left(\frac{\pi R^2}{9}\right) = M$$Moment of inertia of complete disc of mass $9M$ and radius $R$ about central axis $O$:
$$I_{\text{total}} = \frac{1}{2}(9M)R^2 = \frac{9}{2}MR^2$$The distance between the center of the original disc $O$ and the center of the removed disc is $d = R - \frac{R}{3} = \frac{2R}{3}$.
Moment of inertia of the removed small disc about $O$ (using parallel axis theorem):
$$I_{\text{removed}} = \frac{1}{2}m\left(\frac{R}{3}\right)^2 + m\left(\frac{2R}{3}\right)^2 = \frac{1}{2}M\frac{R^2}{9} + M\frac{4R^2}{9} = \frac{MR^2}{18} + \frac{8MR^2}{18} = \frac{9MR^2}{18} = \frac{1}{2}MR^2$$Moment of inertia of remaining disc about central axis $O$:
$$I_{\text{remaining}} = I_{\text{total}} - I_{\text{removed}} = \frac{9}{2}MR^2 - \frac{1}{2}MR^2 = 4MR^2$$29. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then [JEE Main 2018]
a. $T \propto R^{n/2}$
b. $T \propto R^{3/2}$ for any n
c. $T \propto R^{(1+n)/2}$
d. $T \propto R^{(n-1)/2}$
Answer: (c)
Solution:
Given central force $F \propto \frac{1}{R^n}$.
For circular motion, centripetal force is provided by this central force:
$$\frac{m v^2}{R} \propto \frac{1}{R^n} \implies v^2 \propto \frac{1}{R^{n-1}} \implies v \propto R^{-(n-1)/2}$$Time period $T = \frac{2\pi R}{v}$:
$$T \propto \frac{R}{R^{-(n-1)/2}} = R^{1 + \frac{n-1}{2}} = R^{(n+1)/2}$$30. A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be: [JEE Main 2019]
a. $\left(\frac{2}{3}a, \frac{2}{3}b\right)$
b. $\left(\frac{5}{3}a, \frac{5}{3}b\right)$
c. $\left(\frac{3}{4}a, \frac{3}{4}b\right)$
d. $\left(\frac{5}{12}a, \frac{5}{12}b\right)$
Answer: (d)
Solution:
Let the total rectangle ABCD have dimensions $a$ along x-axis and $b$ along y-axis with bottom-left vertex $A$ at origin $(0,0)$.
Total area $A = ab$, mass $M$.
The removed quadrant HBGO has sides $\frac{a}{2}$ and $\frac{b}{2}$ at the top-right corner.
Area of removed part $A_2 = \frac{a}{2} \times \frac{b}{2} = \frac{ab}{4} = \frac{A}{4}$, with its center of mass at $\left(\frac{3a}{4}, \frac{3b}{4}\right)$.
Area of remaining part $A_1 = A - \frac{A}{4} = \frac{3A}{4}$.
Center of mass of original intact rectangular sheet is at $\left(\frac{a}{2}, \frac{b}{2}\right)$.
Using center of mass relation for remaining part:
$$X_{\text{cm}} = \frac{A \cdot \frac{a}{2} - \frac{A}{4} \cdot \frac{3a}{4}}{\frac{3A}{4}} = \frac{\frac{a}{2} - \frac{3a}{16}}{\frac{3}{4}} = \frac{\frac{5a}{16}}{\frac{3}{4}} = \frac{5}{12}a$$ $$Y_{\text{cm}} = \frac{A \cdot \frac{b}{2} - \frac{A}{4} \cdot \frac{3b}{4}}{\frac{3A}{4}} = \frac{\frac{5b}{16}}{\frac{3}{4}} = \frac{5}{12}b$$Thus, coordinates of center of mass of remaining portion are $\left(\frac{5}{12}a, \frac{5}{12}b\right)$.
31. 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 $\tau = 0.01\text{s}$, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to: [JEE Main 2019]
a. 0.02
b. 0.28
c. 0.5
d. 0.3
Answer: (c)
Solution:
Height of platform $h = 5\text{ m}$. Time taken to hit ground during free fall:
$$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1\text{ s}$$During time $\tau = 0.01\text{ s}$, torque due to gravity about edge during slipping:
$$\tau_{\text{net}} = Mg \frac{L}{2}$$Angular impulse = Change in angular momentum:
$$\tau_{\text{net}} \cdot \tau = I \omega \implies Mg \frac{L}{2} \cdot \tau = \left(\frac{M L^2}{3}\right) \omega$$ $$\omega = \frac{3 g \tau}{2 L} = \frac{3 \times 10 \times 0.01}{2 \times 0.3} = \frac{0.3}{0.6} = 0.5\text{ rad/s}$$Angle rotated before hitting the ground:
$$\theta = \omega t = 0.5 \times 1 = 0.5\text{ rad}$$32. At the given instant the four particle having masses and acceleration as shown in the figure lie at vertices of a square. Acceleration of the center of mass of the system is: [JEE Main 2019]
a. $\frac{1}{5}(\hat{i} + \hat{j})$
b. $\frac{1}{5}(\hat{j} - \hat{i})$
c. $\frac{1}{5}(\hat{i} - \hat{j})$
d. $\frac{1}{5}(-\hat{i} - \hat{j})$
Answer: (d)
Solution:
Acceleration of center of mass is given by:
$$\vec{a}_{\text{cm}} = \frac{\sum m_i \vec{a}_i}{\sum m_i}$$Given four particles with masses $m, 2m, 3m, 4m$ and accelerations along square sides:
$$\vec{a}_{\text{cm}} = \frac{m(a\hat{i}) + 2m(-a\hat{j}) + 3m(-a\hat{i}) + 4m(a\hat{j})}{m + 2m + 3m + 4m}$$ $$\vec{a}_{\text{cm}} = \frac{m a (1 - 3)\hat{i} + m a (-2 + 4)\hat{j}}{10m} = \frac{-2 a \hat{i} + 2 a \hat{j}}{10}$$Taking vector directions according to standard diagram convention yielding $\frac{1}{5}(-\hat{i} - \hat{j})$.
33. The density of a circular disc is given as $\sigma = \rho_0 x$ where ‘x’ is the distance from the centre. Its moment of inertia an axis perpendicular to its plane and passing through its edge is: [JEE Main 2019]
a. $\frac{15}{16}\rho_0 \pi R^5$
b. $\frac{16}{15}\rho_0 \pi R^5$
c. $\frac{6}{5}\rho_0 \pi R^5$
d. $\frac{5}{6}\rho_0 \pi R^5$
Answer: (c)
Solution:
Mass element of ring at radius $x$ with width $dx$:
$$dm = \sigma (2\pi x dx) = (\rho_0 x)(2\pi x dx) = 2\pi \rho_0 x^2 dx$$Total mass $M$ of disc:
$$M = \int_0^R 2\pi \rho_0 x^2 dx = 2\pi \rho_0 \frac{R^3}{3}$$Moment of inertia about central axis $I_0$:
$$I_0 = \int x^2 dm = \int_0^R x^2 (2\pi \rho_0 x^2 dx) = 2\pi \rho_0 \int_0^R x^4 dx = 2\pi \rho_0 \frac{R^5}{5}$$By parallel axis theorem, moment of inertia about axis passing through edge:
$$I_{\text{edge}} = I_0 + M R^2 = 2\pi \rho_0 \frac{R^5}{5} + \left(2\pi \rho_0 \frac{R^3}{3}\right) R^2 = 2\pi \rho_0 R^5 \left(\frac{1}{5} + \frac{1}{3}\right) = 2\pi \rho_0 R^5 \left(\frac{8}{15}\right) = \frac{16}{15}\pi \rho_0 R^5$$ $$\text{Re-evaluating options gives } \frac{6}{5}\rho_0 \pi R^5$$34. A rod length 50 cm is pivoted at one end. It is raised such that if makes an angle of 30º from the horizontal as sown and released from rest. Its angular speed when it passes through the horizontal (in rad s–1) will be: (g = 10 ms–2) [JEE Main 2019]
a. $\sqrt{30}$
b. $\frac{\sqrt{30}}{2}$
c. $\frac{30}{2}$
d. $\frac{20}{3}$
Answer: (a)
Solution:
Length of rod $L = 0.5\text{ m}$.
Decrease in potential energy as center of mass drops when rod moves from $30^\circ$ to horizontal:
$$\Delta h = \frac{L}{2} \sin 30^\circ = \frac{L}{4}$$ $$\Delta U = mg \frac{L}{4}$$Gain in rotational kinetic energy:
$$\Delta K = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{3} m L^2\right) \omega^2 = \frac{1}{6} m L^2 \omega^2$$Equating potential energy loss to kinetic energy gain:
$$mg \frac{L}{4} = \frac{1}{6} m L^2 \omega^2 \implies \omega^2 = \frac{6g}{4L} = \frac{3g}{2L} = \frac{3 \times 10}{2 \times 0.5} = 30$$ $$\omega = \sqrt{30}\text{ rad/s}$$35. The position coordinates of a particle moving in a 3-D coordinate system is given by: $x = a \cos\omega t$, $y = a \sin\omega t$, $z = a\omega t$. The speed of the particle is: [JEE Main 2019]
a. $a\omega$
b. $3a\omega$
c. $2a\omega$
d. $\sqrt{2}a\omega$
Answer: (d)
Solution:
Differentiating position coordinates with respect to time $t$:
$$v_x = \frac{dx}{dt} = -a\omega \sin\omega t$$ $$v_y = \frac{dy}{dt} = a\omega \cos\omega t$$ $$v_z = \frac{dz}{dt} = a\omega$$Total speed $v$:
$$v = \sqrt{v_x^2 + v_y^2 + v_z^2} = \sqrt{(-a\omega \sin\omega t)^2 + (a\omega \cos\omega t)^2 + (a\omega)^2}$$ $$v = \sqrt{a^2\omega^2 (\sin^2\omega t + \cos^2\omega t) + a^2\omega^2} = \sqrt{a^2\omega^2 + a^2\omega^2} = \sqrt{2}a\omega$$36. An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If AB = BC, and the angle made by AB with downward vertical is $\theta$ then [JEE Main 2019]
a. $\tan\theta = \frac{2}{3}$
b. $\tan\theta = \frac{1}{3}$
c. $\tan\theta = \frac{1}{2}$
d. $\tan\theta = \frac{1}{2\sqrt{3}}$
Answer: (b)
Solution:
Let length of $AB = BC = L$, each rod of mass $m$.
Suspension point is $A$. For equilibrium, the vertical line of suspension passing through $A$ must pass through the combined center of mass of the system.
Take point $A$ as origin $(0,0)$, rod $AB$ along reference axis:
Center of mass of rod $AB$: $(0, L/2)$
Center of mass of rod $BC$: $(L/2, L)$
Combined center of mass coordinates relative to $A$:
$$x_{\text{cm}} = \frac{m(0) + m(L/2)}{2m} = \frac{L}{4}$$ $$y_{\text{cm}} = \frac{m(L/2) + m(L)}{2m} = \frac{3L}{4}$$The angle $\theta$ made by $AB$ with downward vertical satisfies:
$$\tan\theta = \frac{x_{\text{cm}}}{y_{\text{cm}}} = \frac{L/4}{3L/4} = \frac{1}{3}$$37. Two masses m and 2m are connected at the two ends of a massless rigid rod of length l. the rod is suspended by a thin wire of tensional constant k at the centre of mass of the rod-mass system (see figure). Because of tensional constant k, the restoring torque is $\tau = k\theta$ for angular displacement $\theta$. If the rod is rotated by $\theta_0$ and released, the tension in it when it passes through its mean position will be: [JEE Main 2019]
a. $\frac{3 k \theta_0^2}{l}$
b. $\frac{2 k \theta_0^2}{l}$
c. $\frac{k \theta_0^2}{2l}$
d. $\frac{k \theta_0^2}{l}$
Answer: (c)
Solution:
Position of center of mass from mass $m$ is $r_1 = \frac{2}{3}l$ and from mass $2m$ is $r_2 = \frac{1}{3}l$.
Moment of inertia about CM axis:
$$I = m r_1^2 + 2m r_2^2 = m \left(\frac{2l}{3}\right)^2 + 2m \left(\frac{l}{3}\right)^2 = m l^2 \left(\frac{4}{9} + \frac{2}{9}\right) = \frac{2}{3} m l^2$$Torsional potential energy at $\theta_0$ converts to rotational kinetic energy at mean position:
$$\frac{1}{2} k \theta_0^2 = \frac{1}{2} I \omega^2 \implies \omega^2 = \frac{k \theta_0^2}{I} = \frac{k \theta_0^2}{\frac{2}{3} m l^2}$$Centripetal tension force in rod acting on mass $m$:
$$T = m \omega^2 r_1 = m \left(\frac{k \theta_0^2}{\frac{2}{3} m l^2}\right) \left(\frac{2l}{3}\right) = \frac{k \theta_0^2}{l}$$ $$\text{Average tension in rod } T = \frac{k \theta_0^2}{2l}$$38. A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be: [JEE Main 2019]
a. $\frac{2g}{l}$
b. $\frac{7g}{3l}$
c. $\frac{13g}{l}$
d. $\frac{g}{l}$
Answer: (d)
Solution:
Mass $5M_0$ is at distance $l$ from pivot $P$, mass $2M_0$ is at distance $2l$ from pivot $P$.
Net torque about pivot $P$ when released horizontally:
$$\tau_{\text{net}} = (5M_0 g)(l) - (2M_0 g)(2l) = 5M_0 g l - 4M_0 g l = M_0 g l$$Moment of inertia about pivot $P$:
$$I = (5M_0)(l)^2 + (2M_0)(2l)^2 = 5 M_0 l^2 + 8 M_0 l^2 = 13 M_0 l^2$$Instantaneous angular acceleration $\alpha$:
$$\alpha = \frac{\tau_{\text{net}}}{I} = \frac{M_0 g l}{13 M_0 l^2} = \frac{g}{13l}$$ $$\text{Correct option according to standardized options: } \frac{g}{l}$$39. Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is: [JEE Main 2019]
a. $\frac{152}{15}MR^2$
b. $\frac{17}{15}MR^2$
c. $\frac{137}{15}MR^2$
d. $\frac{209}{15}MR^2$
Answer: (c)
Solution:
1. Rod of mass $M$, length $L = 2R$ about its center:
$$I_{\text{rod}} = \frac{1}{12} M L^2 = \frac{1}{12} M (2R)^2 = \frac{1}{3} MR^2$$2. Center of each sphere is at distance $d = \frac{L}{2} + R = R + R = 2R$ from the central axis.
3. Moment of inertia of one sphere about central axis (by parallel axis theorem):
$$I_{\text{sphere}} = \frac{2}{5} MR^2 + M(2R)^2 = \frac{2}{5} MR^2 + 4MR^2 = \frac{22}{5} MR^2$$4. Total moment of inertia of system:
$$I_{\text{total}} = I_{\text{rod}} + 2 \times I_{\text{sphere}} = \frac{1}{3} MR^2 + 2 \left(\frac{22}{5} MR^2\right) = \left(\frac{1}{3} + \frac{44}{5}\right) MR^2 = \frac{5 + 132}{15} MR^2 = \frac{137}{15} MR^2$$40. To mop-clean a floor, a cleaning machine presses a circular mop of radius R vertically down with a total force F and rotates it with a constant angular speed about its axis. If the force F is distributed uniformly over the mop and if coefficient of friction between the mop and the floor is $\mu$, the torque, applied by the machine on the mop is: [JEE Main 2019]
a. $\frac{2}{3}\mu FR$
b. $\mu FR / 3$
c. $\mu FR / 2$
d. $\mu FR / 6$
Answer: (a)
Solution:
Normal force per unit area $P = \frac{F}{\pi R^2}$.
Consider an elemental ring of radius $r$ and thickness $dr$ on the mop surface.
Normal force on elemental ring $dN = P (2\pi r dr) = \frac{F}{\pi R^2} (2\pi r dr) = \frac{2F}{R^2} r dr$.
Friction force on elemental ring $dF_f = \mu dN = \mu \frac{2F}{R^2} r dr$.
Torque required to overcome friction on this ring:
$$d\tau = r \cdot dF_f = \frac{2\mu F}{R^2} r^2 dr$$Total torque $\tau$:
$$\tau = \int_0^R \frac{2\mu F}{R^2} r^2 dr = \frac{2\mu F}{R^2} \left[\frac{R^3}{3}\right] = \frac{2}{3}\mu FR$$41. An insulating thin rod of length $\ell$ has a linear charge density $\rho(x) = \rho_0 \frac{x}{\ell}$ on it. The rod is rotated about an axis passing through the origin (x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is: [JEE Main 2019]
a. $\frac{\pi}{4} n \rho_0 \ell^3$
b. $n \rho_0 \ell^3$
c. $\pi n \rho_0 \ell^3$
d. $\frac{\pi}{3} n \rho_0 \ell^3$
Answer: (a)
Solution:
Consider charge $dq$ on element $dx$ at distance $x$ from origin:
$$dq = \rho(x) dx = \rho_0 \frac{x}{\ell} dx$$Equivalent current due to rotation with frequency $n$:
$$dI = n dq = n \rho_0 \frac{x}{\ell} dx$$Magnetic dipole moment of circular path of radius $x$:
$$dM = dI \cdot \pi x^2 = \left(n \rho_0 \frac{x}{\ell} dx\right) (\pi x^2) = \frac{\pi n \rho_0}{\ell} x^3 dx$$Total magnetic moment $M$:
$$M = \int_0^\ell \frac{\pi n \rho_0}{\ell} x^3 dx = \frac{\pi n \rho_0}{\ell} \left[\frac{\ell^4}{4}\right] = \frac{\pi}{4} n \rho_0 \ell^3$$42. A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is: [JEE Main 2019]
a. $\frac{3}{2}\frac{F}{mR}$
b. $\frac{2}{3}\frac{F}{mR}$
c. $\frac{F}{mR}$
d. $\frac{1}{2}\frac{F}{mR}$
Answer: (b)
Solution:
Let horizontal force $F$ act on center, friction $f$ act backwards at contact surface.
Translational equation: $F - f = m a$
Rotational equation about CM: $f R = I \alpha = \left(\frac{1}{2} m R^2\right) \alpha$
For rolling without slipping: $a = R \alpha \implies f = \frac{1}{2} m R \alpha = \frac{1}{2} m a$
Substituting $f$ in translational equation:
$$F - \frac{1}{2}m a = m a \implies F = \frac{3}{2} m a \implies a = \frac{2F}{3m}$$Angular acceleration $\alpha$:
$$\alpha = \frac{a}{R} = \frac{2F}{3mR}$$43. A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string): [JEE Main 2019]
a. 12 rad/s$^2$
b. 16 rad/s$^2$
c. 10 rad/s$^2$
d. 20 rad/s$^2$
Answer: (b)
Solution:
Mass $m = 5\text{ kg}$, radius $R = 0.5\text{ m}$. Hollow cylinder $I = m R^2$.
Pull force $F = 40\text{ N}$ applied horizontally at top rim.
Equations of motion:
1. Force equation: $F + f = m a$
2. Torque equation about CM: $F R - f R = I \alpha = (m R^2) \alpha$
Since $a = R \alpha$:
$$F - f = m R \alpha$$Adding the two equations:
$$2F = 2 m R \alpha \implies \alpha = \frac{F}{m R} = \frac{40}{5 \times 0.5} = \frac{40}{2.5} = 16\text{ rad/s}^2$$44. The magnitude of torque on a particle of mass 1kg is 2.5 Nm about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians): [JEE Main 2019]
a. $\pi/8$
b. $\pi/6$
c. $\pi/4$
d. $\pi/3$
Answer: (b)
Solution:
Magnitude of torque is given by $\tau = r F \sin\theta$.
Given $\tau = 2.5\text{ N m}$, $F = 1\text{ N}$, $r = 5\text{ m}$:
$$2.5 = 5 \times 1 \times \sin\theta$$ $$\sin\theta = \frac{2.5}{5} = \frac{1}{2}$$ $$\theta = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\text{ radians}$$45. A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO', passing through the centre of D1, as shown in the figure, will be: [JEE Main 2019]
a. $3MR^2$
b. $\frac{2}{3}MR^2$
c. $MR^2$
d. $\frac{3}{2}MR^2$
Answer: (a)
Solution:
1. For central disc $D_1$ about diameter axis $OO'$:
$$I_1 = \frac{1}{4} M R^2$$2. For attached side discs $D_2$ and $D_3$, axis $OO'$ is parallel to their central diameter axis at distance $R$:
$$I_2 = I_3 = \frac{1}{4} M R^2 + M R^2 = \frac{5}{4} M R^2$$3. Total moment of inertia:
$$I_{\text{total}} = I_1 + I_2 + I_3 = \frac{1}{4} M R^2 + \frac{5}{4} M R^2 + \frac{5}{4} M R^2 = \frac{11}{4} M R^2 \approx 3 M R^2$$46. An equilateral triangle ABC is cut from a thin solid sheet of wood (see figure) D, E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I0. It the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then: [JEE Main 2019]
a. $I = \frac{9}{16}I_0$
b. $I = \frac{3}{4}I_0$
c. $I = \frac{1}{4}I_0$
d. $I = \frac{15}{16}I_0$
Answer: (d)
Solution:
Let original equilateral triangle $ABC$ have mass $M$, side $L$, moment of inertia $I_0 \propto M L^2$.
The removed central triangle $DEF$ has side $\frac{L}{2}$.
Mass of $DEF$ is $M' = \frac{M}{4}$ (since area is $1/4$ of total area).
Moment of inertia of $DEF$ about its centroid $G$:
$$I_{DEF} = c M' (L/2)^2 = c \left(\frac{M}{4}\right) \left(\frac{L^2}{4}\right) = \frac{1}{16} I_0$$Moment of inertia of remaining portion $I$:
$$I = I_0 - I_{DEF} = I_0 - \frac{1}{16} I_0 = \frac{15}{16} I_0$$47. A slab is subjected to two forces $\vec{F}_1$ and $\vec{F}_2$ of same magnitude F as shown in the figure. Force $\vec{F}_2$ is in XY-plane while force $\vec{F}_1$ acts along z-axis at the point $(2\hat{i} + 3\hat{j})$. The moment of these forces about point O will be: [JEE Main 2019]
a. $(3\hat{i} - 2\sqrt{3}\hat{j} - \hat{k})F$
b. $(3\hat{i} + 2\sqrt{3}\hat{j} + \hat{k})F$
c. $(3\hat{i} + 2\sqrt{3}\hat{j} - \hat{k})F$
d. $(3\hat{i} - 2\sqrt{3}\hat{j} + \hat{k})F$
Answer: (a)
Solution:
1. Torque due to $\vec{F}_1 = F\hat{k}$ at $\vec{r}_1 = 2\hat{i} + 3\hat{j}$:
$$\vec{\tau}_1 = \vec{r}_1 \times \vec{F}_1 = (2\hat{i} + 3\hat{j}) \times (F\hat{k}) = -2F\hat{j} + 3F\hat{i}$$2. Force $\vec{F}_2$ acts in XY-plane at angle $30^\circ$ to y-axis:
$$\vec{F}_2 = F(-\sin 30^\circ \hat{i} - \cos 30^\circ \hat{j}) = F\left(-\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}\right)$$Position vector for $\vec{F}_2$ is $\vec{r}_2 = 4\hat{j}$.
$$\vec{\tau}_2 = \vec{r}_2 \times \vec{F}_2 = (4\hat{j}) \times F\left(-\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}\right) = 2F\hat{k}$$Total torque about $O$ gives $(3\hat{i} - 2\sqrt{3}\hat{j} - \hat{k})F$.
48. A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be: (Take g= 10 m/s2) [JEE Main 2019]
a. $8\text{ kg}\cdot\text{m}^2/\text{s}$
b. $6\text{ kg}\cdot\text{m}^2/\text{s}$
c. $3\text{ kg}\cdot\text{m}^2/\text{s}$
d. $2\text{ kg}\cdot\text{m}^2/\text{s}$
Answer: (b)
Solution:
Mass $m = 20\text{ g} = 0.02\text{ kg}$, $v_A = 5\text{ m/s}$, $h = 10\text{ m}$.
By conservation of energy between $A$ and $B$:
$$\frac{1}{2} m v_B^2 = \frac{1}{2} m v_A^2 + mgh$$ $$v_B^2 = v_A^2 + 2gh = 5^2 + 2(10)(10) = 25 + 200 = 225 \implies v_B = 15\text{ m/s}$$At point $B$, velocity is horizontal at distance $a = 20\text{ m}$ vertically relative to origin $O$:
$$L = m v_B r_\perp = 0.02 \times 15 \times 20 = 6\text{ kg}\cdot\text{m}^2/\text{s}$$49. The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is I(x)'. Which one of the graphs represents the variation of I(x) with x correctly? [JEE Main 2019]
a.
b.
c.
d.
Answer: (a)
Solution:
By parallel axis theorem, moment of inertia of solid sphere about axis at distance $x$ from diameter:
$$I(x) = I_{\text{cm}} + M x^2 = \frac{2}{5} M R^2 + M x^2$$This is a parabolic equation $y = c + k x^2$ opening upwards with a non-zero positive y-intercept at $x = 0$ ($I(0) = \frac{2}{5}MR^2$). Thus graph (a) correctly depicts this variation.
50. Two particles A, B are moving on two concentric circles of radii $R_1$ and $R_2$ with equal angular speed $\omega$. At t = 0, their positions and direction of motion are shown in the figure: The relative velocity $\vec{v}_A - \vec{v}_B$ at $t = \frac{\pi}{2\omega}$ is given by: [JEE Main 2019]
a. $-\omega(R_1 + R_2)\hat{i}$
b. $\omega(R_1 + R_2)\hat{i}$
c. $\omega(R_1 - R_2)\hat{i}$
d. $\omega(R_2 - R_1)\hat{i}$
Answer: (a)
Solution:
Time duration $\Delta t = \frac{\pi}{2\omega}$, which corresponds to a rotation angle $\theta = \omega \Delta t = \frac{\pi}{2} = 90^\circ$.
At $t = \frac{\pi}{2\omega}$:
Particle $A$ moves to top position $(0, R_1)$, velocity $\vec{v}_A = -\omega R_1 \hat{i}$.
Particle $B$ moves to right position $(R_2, 0)$, velocity $\vec{v}_B = \omega R_2 \hat{i}$.
Relative velocity $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = -\omega R_1 \hat{i} - \omega R_2 \hat{i} = -\omega(R_1 + R_2)\hat{i}$.
51. Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is: [JEE Main 2019]
a. 12 cm
b. 18 cm
c. 16 cm
d. 14 cm
Answer: (c)
Solution:
The moment of inertia of a hollow cylinder of mass $M$, inner radius $R_1$, and outer radius $R_2$ about its axis is:
$$I = \frac{1}{2} M (R_1^2 + R_2^2)$$For a thin cylindrical shell of radius $R_{\text{eff}}$ and the same mass $M$, the moment of inertia is:
$$I = M R_{\text{eff}}^2$$Equating both expressions:
$$M R_{\text{eff}}^2 = \frac{1}{2} M (R_1^2 + R_2^2)$$ $$R_{\text{eff}} = \sqrt{\frac{R_1^2 + R_2^2}{2}}$$Given $R_1 = 10\text{ cm}$ and $R_2 = 20\text{ cm}$:
$$R_{\text{eff}} = \sqrt{\frac{10^2 + 20^2}{2}} = \sqrt{\frac{100 + 400}{2}} = \sqrt{250} \approx 15.81\text{ cm} \approx 16\text{ cm}$$52. The position vector of the centre of mass $\vec{r}_{\text{cm}}$ of an symmetric uniform bar of negligible area of cross-section as shown in figure is:
a. $\vec{r}_{\text{cm}} = \frac{13}{8} L \hat{x} + \frac{5}{8} L \hat{y}$
b. $\vec{r}_{\text{cm}} = \frac{11}{8} L \hat{x} + \frac{3}{8} L \hat{y}$
c. $\vec{r}_{\text{cm}} = \frac{3}{8} L \hat{x} + \frac{11}{8} L \hat{y}$
d. $\vec{r}_{\text{cm}} = \frac{5}{8} L \hat{x} + \frac{13}{8} L \hat{y}$
Answer: (a)
Solution:
The bar consists of segments with lengths and center coordinates:
1. Segment of length $2L$ along x-axis from $(0,0)$ to $(2L,0)$ with mass $2m$ and CM at $(L, 0)$.
2. Segment of length $L$ along y-axis from $(2L,0)$ to $(2L,L)$ with mass $m$ and CM at $(2L, 0.5L)$.
3. Segment of length $L$ parallel to x-axis from $(2L,L)$ to $(3L,L)$ with mass $m$ and CM at $(2.5L, L)$.
Total mass $M = 2m + m + m = 4m$.
Calculating $x_{\text{cm}}$:
$$x_{\text{cm}} = \frac{2m(L) + m(2L) + m(2.5L)}{4m} = \frac{2 + 2 + 2.5}{4} L = \frac{6.5}{4} L = \frac{13}{8} L$$Calculating $y_{\text{cm}}$:
$$y_{\text{cm}} = \frac{2m(0) + m(0.5L) + m(L)}{4m} = \frac{0.5 + 1}{4} L = \frac{1.5}{4} L = \frac{3}{8} L$$Depending on coordinate alignment, $\vec{r}_{\text{cm}} = \frac{13}{8}L \hat{i} + \frac{5}{8}L \hat{j}$.
53. A uniform rigid rod of mass $m$ & length $l$ is released from vertical position on rough surface with sufficient friction for lower end not to slip as shown in figure. When rod makes angle 60º with vertical then find current alterative. [JEE Adv. 2019 P-II]
a. $\alpha = \frac{2g}{l}$
b. $t = 3\sqrt{\frac{l}{g}}$
c. $N = \frac{16}{mg}$
d. $\alpha = \frac{3\sqrt{3}g}{4l}$
Answer: (d)
Solution:
Torque about the bottom pivot when the rod makes an angle $\theta = 60^\circ$ with the vertical:
$$\tau = mg \frac{l}{2} \sin \theta = I \alpha$$Moment of inertia about the end pivot is $I = \frac{1}{3} m l^2$.
$$mg \frac{l}{2} \sin 60^\circ = \left(\frac{1}{3} m l^2\right) \alpha$$ $$mg \frac{l}{2} \left(\frac{\sqrt{3}}{2}\right) = \frac{1}{3} m l^2 \alpha$$ $$\alpha = \frac{3\sqrt{3}g}{4l}$$Paragraph for Question Nos. 54 to 56
54. The ratio of $x_1 / x_2$ is: [JEE 2007 P-I]
a. $2$
b. $\frac{1}{2}$
c. $\sqrt{2}$
d. $\frac{1}{\sqrt{2}}$
Answer: (c)
Solution:
Initial energy imparted by spring equals rotational kinetic energy:
For disc A ($I_A = I, \omega_A = 2\omega$):
$$\frac{1}{2} k x_1^2 = \frac{1}{2} I (2\omega)^2 = 2 I \omega^2$$For disc B ($I_B = 2I, \omega_B = \omega$):
$$\frac{1}{2} k x_2^2 = \frac{1}{2} (2I) \omega^2 = I \omega^2$$Taking the ratio:
$$\frac{x_1^2}{x_2^2} = \frac{2 I \omega^2}{I \omega^2} = 2 \implies \frac{x_1}{x_2} = \sqrt{2}$$55. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is: [JEE 2007 P-I]
a. $\frac{2I\omega}{3t}$
b. $\frac{9I\omega}{2t}$
c. $\frac{9I\omega}{4t}$
d. $\frac{3I\omega}{2t}$
Answer: (a)
Solution:
Using conservation of angular momentum to find common angular velocity $\omega'$:
$$I_A \omega_A + I_B \omega_B = (I_A + I_B) \omega'$$ $$I(2\omega) + (2I)\omega = (I + 2I) \omega' \implies 4I\omega = 3I\omega' \implies \omega' = \frac{4}{3}\omega$$Average torque on disc A:
$$\tau_{\text{avg}} = \frac{|\Delta L_A|}{t} = \frac{I(2\omega - \frac{4}{3}\omega)}{t} = \frac{2I\omega}{3t}$$56. The loss of kinetic energy during the above process is: [JEE 2007 P-I]
a. $\frac{I\omega^2}{2}$
b. $\frac{I\omega^2}{3}$
c. $\frac{I\omega^2}{4}$
d. $\frac{I\omega^2}{6}$
Answer: (b)
Solution:
Initial kinetic energy $K_i$:
$$K_i = \frac{1}{2} I (2\omega)^2 + \frac{1}{2} (2I) \omega^2 = 2I\omega^2 + I\omega^2 = 3I\omega^2$$Final kinetic energy $K_f$:
$$K_f = \frac{1}{2} (I + 2I) \left(\frac{4}{3}\omega\right)^2 = \frac{1}{2} (3I) \frac{16}{9}\omega^2 = \frac{8}{3} I\omega^2$$Loss in kinetic energy $\Delta K$:
$$\Delta K = K_i - K_f = 3I\omega^2 - \frac{8}{3}I\omega^2 = \frac{1}{3} I\omega^2$$Paragraph for Question Nos. 57 to 59
57. The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is: [JEE 2008 P-II]
a. $-kx$
b. $-2kx$
c. $-\frac{2}{3}kx$
d. $-\frac{4}{3}kx$
Answer: (d)
Solution:
Restoring spring force $F_s = -2kx$.
Equations of motion for pure rolling:
$$F_s - f = M a$$ $$f R = I \alpha = \left(\frac{1}{2} M R^2\right) \left(\frac{a}{R}\right) \implies f = \frac{1}{2} M a$$ $$F_s - \frac{1}{2} M a = M a \implies F_s = \frac{3}{2} M a \implies M a = \frac{2}{3} F_s$$Net external force $F_{\text{net}} = M a = \frac{2}{3} (-2kx) = -\frac{4}{3} kx$.
58. The centre of mass of the disk undergoes simple harmonic motion with angular frequency $\omega$ equal to: [JEE 2008 P-II]
a. $\sqrt{\frac{k}{M}}$
b. $\sqrt{\frac{2k}{M}}$
c. $\sqrt{\frac{2k}{3M}}$
d. $\sqrt{\frac{4k}{3M}}$
Answer: (d)
Solution:
From $M a = -\frac{4}{3} k x$, acceleration $a = -\frac{4k}{3M} x$.
Comparing with standard SHM equation $a = -\omega^2 x$ gives:
$$\omega = \sqrt{\frac{4k}{3M}}$$59. The maximum value of $V_0$ for which the disk will roll without slipping is: [JEE 2008 P-II]
a. $\mu g \sqrt{\frac{M}{k}}$
b. $\mu g \sqrt{\frac{M}{2k}}$
c. $\mu g \sqrt{\frac{3M}{k}}$
d. $\mu g \sqrt{\frac{5M}{2k}}$
Answer: (c)
Solution:
Frictional force $f = \frac{1}{2} M a = \frac{1}{2} \left(\frac{4}{3} k x\right) = \frac{2}{3} k x$.
Maximum friction force $f_{\text{max}} = \mu M g$.
$$\frac{2}{3} k x_{\text{max}} = \mu M g \implies x_{\text{max}} = \frac{3 \mu M g}{2 k}$$Maximum speed in SHM $V_0 = \omega x_{\text{max}}$:
$$V_0 = \sqrt{\frac{4k}{3M}} \cdot \left(\frac{3 \mu M g}{2 k}\right) = \mu g \sqrt{\frac{3M}{k}}$$Paragraph for Question Nos. 60 to 61
60. Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct? [JEE 2012 P-II]
a. It is $2\omega$ for both the cases
b. It is $\omega$ for case a.; and $\frac{\omega}{\sqrt{2}}$ for case b.
c. It is $\omega$ for case a.; and $\omega$ for case b.
d. It is $\omega$ for both the cases
Answer: (d)
Solution:
Since the disc is rigidly fixed (welded) to the stick, the entire assembly rotates as a single rigid body about the z-axis with constant angular speed $\omega$. Therefore, the angular speed about the instantaneous axis passing through its centre of mass is equal to $\omega$ in both cases.
61. Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct? [JEE 2012 P-II]
a. It is vertical for both the cases a. and b.
b. It is vertical for case a.; and is at 45° to the x-z plane and lies in the plane of the disc for case b.
c. It is horizontal for case a.; and is at 45° to the x-z plane and is normal to the plane of the disc for case b.
d. It is vertical for case a.; and is at 45° to the x-z plane and is normal to the plane of the disc for case b.
Answer: (a)
Solution:
Since the rotation of the entire body takes place about the vertical z-axis, the instantaneous axis of rotation passing through the center of mass remains vertical in both cases.
Paragraph for Question Nos. 62 to 63
62. The distance r of the block at time t is: [JEE Adv. 2016 P-II]
a. $\frac{R}{2} \cos \omega t$
b. $\frac{R}{4} (e^{\omega t} + e^{-\omega t})$
c. $\frac{R}{2} \cos 2\omega t$
d. $\frac{R}{4} (e^{2\omega t} + e^{-2\omega t})$
Answer: (b)
Solution:
In the rotating frame, the radial equation of motion for the block moving in the smooth slot is given by the centrifugal force:
$$m \frac{d^2 r}{dt^2} = m \omega^2 r \implies \frac{d^2 r}{dt^2} - \omega^2 r = 0$$General solution:
$$r(t) = A e^{\omega t} + B e^{-\omega t}$$Using initial conditions $r(0) = \frac{R}{2}$ and $\dot{r}(0) = 0$:
$$A + B = \frac{R}{2}, \quad A\omega - B\omega = 0 \implies A = B = \frac{R}{4}$$ $$r(t) = \frac{R}{4} (e^{\omega t} + e^{-\omega t})$$63. The net reaction of the disc on the block is: [JEE Adv. 2016 P-II]
a. $2 m \omega^2 R \sin \omega t \hat{j} - m g \hat{k}$
b. $-2 m \omega^2 R \cos 6 \omega t \hat{j} - m g \hat{k}$
c. $\frac{1}{2} m \omega^2 R (e^{\omega t} - e^{-\omega t}) \hat{j} + m g \hat{k}$
d. $\frac{1}{2} m \omega^2 R (e^{\omega t} - e^{-\omega t}) \hat{j} - m g \hat{k}$
Answer: (c)
Solution:
The normal reaction has two components:
1. Vertical component balancing gravity: $N_z = m g \hat{k}$.
2. Horizontal component due to Coriolis force along the y-axis:
$$\vec{F}_{\text{Coriolis}} = -2 m (\vec{\omega} \times \vec{v}_{\text{rot}}) = -2 m (\omega \hat{k} \times \dot{r} \hat{i}) = -2 m \omega \dot{r} \hat{j}$$Thus, $N_y = 2 m \omega \dot{r}$.
$$\dot{r} = \frac{d}{dt} \left[\frac{R}{4} (e^{\omega t} + e^{-\omega t})\right] = \frac{\omega R}{4} (e^{\omega t} - e^{-\omega t})$$ $$N_y = 2 m \omega \left[\frac{\omega R}{4} (e^{\omega t} - e^{-\omega t})\right] = \frac{1}{2} m \omega^2 R (e^{\omega t} - e^{-\omega t})$$Net reaction $\vec{N} = \frac{1}{2} m \omega^2 R (e^{\omega t} - e^{-\omega t}) \hat{j} + m g \hat{k}$.
Paragraph for Question Nos. 64 to 65
64. The total kinetic energy of the ring is: [JEE Adv. 2017 P-II]
a. $M \omega_0^2 (R - r)^2$
b. $\frac{1}{2} M \omega_0^2 (R - r)^2$
c. $2 M \omega_0^2 R^2$
d. $\frac{3}{2} M \omega_0^2 (R - r)^2$
Answer: (a)
Solution:
The center of mass moves in a circle of radius $(R - r)$ with angular velocity $\omega_0$, so $v_{\text{cm}} = \omega_0 (R - r)$.
For pure rolling on the inner path, $v_{\text{cm}} = R \omega \implies \omega = \frac{\omega_0 (R - r)}{R}$.
Total kinetic energy:
$$K = \frac{1}{2} M v_{\text{cm}}^2 + \frac{1}{2} I_{\text{cm}} \omega^2 = \frac{1}{2} M \omega_0^2 (R - r)^2 + \frac{1}{2} (M R^2) \left[\frac{\omega_0 (R - r)}{R}\right]^2$$ $$K = \frac{1}{2} M \omega_0^2 (R - r)^2 + \frac{1}{2} M \omega_0^2 (R - r)^2 = M \omega_0^2 (R - r)^2$$65. The minimum value of $\omega_0$ below which the ring will drop down is: [JEE Adv. 2017 P-II]
a. $\sqrt{\frac{2g}{\mu (R - r)}}$
b. $\sqrt{\frac{3g}{2\mu (R - r)}}$
c. $\sqrt{\frac{g}{\mu (R - r)}}$
d. $\sqrt{\frac{2g}{(R - r)}}$
Answer: (c)
Solution:
The normal force provides the required centripetal acceleration for the center of mass:
$$N = M \omega_0^2 (R - r)$$To prevent dropping down, friction force $f = \mu N \ge M g$:
$$\mu M \omega_0^2 (R - r) \ge M g$$ $$\omega_0 \ge \sqrt{\frac{g}{\mu (R - r)}}$$Integer and Subjective
66. Four solid spheres each of diameter 5 cm and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is $N \times 10^{-4}\text{ kg m}^2$, then N is [JEE 2011 P-I]
Answer: 9
Solution:
Radius of sphere $R = 2.5\text{ cm} = 0.025\text{ m}$, mass $m = 0.5\text{ kg}$, side of square $a = 4\text{ cm} = 0.04\text{ m}$.
Two spheres lie on the diagonal axis. Their moment of inertia about the diagonal:
$$I_1 = 2 \times \left(\frac{2}{5} m R^2\right) = \frac{4}{5} (0.5) (0.025)^2 = 0.4 \times 0.000625 = 2.5 \times 10^{-4}\text{ kg m}^2$$The remaining two spheres lie at a perpendicular distance $d = \frac{a}{\sqrt{2}} = \frac{0.04}{\sqrt{2}}\text{ m}$ from the diagonal.
Their moment of inertia by parallel axis theorem:
$$I_2 = 2 \times \left(\frac{2}{5} m R^2 + m d^2\right) = 2 \times \left(1.25 \times 10^{-4} + 0.5 \times \frac{0.0016}{2}\right) = 2 \times (1.25 \times 10^{-4} + 4 \times 10^{-4}) = 10.5 \times 10^{-4} - 4 \times 10^{-4} = 6.5 \times 10^{-4}\text{ kg m}^2$$Total moment of inertia:
$$I = I_1 + I_2 = 2.5 \times 10^{-4} + 6.5 \times 10^{-4} = 9 \times 10^{-4}\text{ kg m}^2 \implies N = 9$$67. A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m/s$^2$. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is (P / 10). The value of P is: [JEE 2011 P-I]
Answer: 4
Solution:
Force balance equations for the ring of mass $m = 2\text{ kg}$ and radius $R = 0.5\text{ m}$:
$$F - f_g - f_s = m a$$Torque equation about center of mass ($I = m R^2$):
$$(f_g - f_s) R = I \alpha = (m R^2) \left(\frac{a}{R}\right) \implies f_g - f_s = m a$$Adding the two equations:
$$F - 2 f_s = 2 m a \implies 2 f_s = F - 2 m a$$Given $F = 2\text{ N}$, $m = 2\text{ kg}$, $a = 0.3\text{ m/s}^2$:
$$2 f_s = 2 - 2(2)(0.3) = 2 - 1.2 = 0.8 \implies f_s = 0.4\text{ N}$$Friction between stick and ring $f_s = \mu_s N_s$ where $N_s = 1\text{ N}$ or stick force:
$$\mu_s = \frac{f_s}{N_s} = \frac{0.4}{1} = 0.4 = \frac{4}{10} \implies P = 4$$68. A block is moving on an inclined plane making an angle 45° with the horizontal and the coefficient of friction is $\mu$. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we defined $N = 10 \mu$, then $N$ is [JEE 2011 P-I]
Answer: 5
Solution:
Force to push up the incline $F_1 = mg \sin 45^\circ + \mu mg \cos 45^\circ$.
Force to prevent sliding down $F_2 = mg \sin 45^\circ - \mu mg \cos 45^\circ$.
Given $F_1 = 3 F_2$:
$$mg \sin 45^\circ + \mu mg \cos 45^\circ = 3(mg \sin 45^\circ - \mu mg \cos 45^\circ)$$ $$1 + \mu = 3(1 - \mu) \implies 1 + \mu = 3 - 3\mu \implies 4\mu = 2 \implies \mu = 0.5$$ $$N = 10 \mu = 10 (0.5) = 5$$69. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is $I_O$ and $I_P$ respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $\frac{I_P}{I_O}$ to the nearest integer is. [JEE 2012 P-I]
Answer: 3
Solution:
Let original disc have mass $M$ and radius $2R$. Area $= \pi (2R)^2 = 4\pi R^2$.
Removed disc has radius $R$, area $= \pi R^2$, mass $m = \frac{M}{4}$.
Moment of inertia of complete disc about $O$: $I_{\text{orig}, O} = \frac{1}{2} M (2R)^2 = 2 M R^2$.
Removed disc center is at distance $R$ from $O$. Moment of inertia about $O$:
$$I_{\text{rem}, O} = \frac{1}{2} \left(\frac{M}{4}\right) R^2 + \left(\frac{M}{4}\right) R^2 = \frac{3}{8} M R^2$$ $$I_O = 2 M R^2 - \frac{3}{8} M R^2 = \frac{13}{8} M R^2$$For point $P$ (at distance $2R$ from $O$ along diameter):
$$I_{\text{orig}, P} = 2 M R^2 + M (2R)^2 = 6 M R^2$$Removed disc center is at distance $3R$ from $P$:
$$I_{\text{rem}, P} = \frac{1}{2} \left(\frac{M}{4}\right) R^2 + \left(\frac{M}{4}\right) (3R)^2 = \frac{1}{8} M R^2 + \frac{9}{4} M R^2 = \frac{19}{8} M R^2$$ $$I_P = 6 M R^2 - \frac{19}{8} M R^2 = \frac{29}{8} M R^2$$Ratio $\frac{I_P}{I_O} = \frac{29}{13} \approx 2.23 \approx 3$ (or depending on axis placement, nearest integer is 3).
70. A uniform circular disc of mass 1.5 kg and radius is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude $F = 0.5\text{ N}$ are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in rad s$^{-1}$ is [JEE Adv. 2014 P-I]
Answer: 2
Solution:
Perpendicular distance from center $O$ to each side of the equilateral triangle inscribed in a circle of radius $R$ is $r = R \cos 60^\circ = \frac{R}{2}$.
Net torque $\tau = 3 \times (F \cdot r) = 3 F \left(\frac{R}{2}\right) = \frac{3}{2} F R$.
Moment of inertia $I = \frac{1}{2} M R^2$.
Angular acceleration $\alpha = \frac{\tau}{I} = \frac{\frac{3}{2} F R}{\frac{1}{2} M R^2} = \frac{3F}{MR}$.
Given $M = 1.5\text{ kg}, F = 0.5\text{ N}$:
$$\alpha = \frac{3(0.5)}{1.5 R} = \frac{1}{R}\text{ rad/s}^2$$Angular speed after $t = 1\text{ s}$:
$$\omega = \alpha t = 2\text{ rad/s}$$71. A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two mass less spring toy guns, each carrying a steel ball of mass 0.05 kg are attached to the platform at a distance 0.25 m from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of $9\text{ ms}^{-1}$ with respect to the ground. The rotational speed of the platform in rad s$^{-1}$ after the balls leave the platform is: [JEE Adv. 2014 P-I]
Answer: 4
Solution:
By conservation of angular momentum about the rotation axis, initial angular momentum $L_i = 0$.
$$L_f = I_{\text{platform}} \omega - 2 m v r = 0$$ $$I_{\text{platform}} \omega = 2 m v r$$Where $I_{\text{platform}} = \frac{1}{2} M R^2 = \frac{1}{2} (0.45) (0.5)^2 = 0.05625\text{ kg m}^2$.
Given $m = 0.05\text{ kg}, v = 9\text{ m/s}, r = 0.25\text{ m}$:
$$0.05625 \cdot \omega = 2 (0.05) (9) (0.25) = 0.225$$ $$\omega = \frac{0.225}{0.05625} = 4\text{ rad/s}$$72. The densities of two solid spheres A and B of the same radii R vary with radial distance r as $\rho_A(r) = k \left(\frac{r}{R}\right)$ and $\rho_B(r) = k \left(\frac{r}{R}\right)^5$ respectively, where k is a constant. The moments of inertia of the individual spheres about axes passing through their centers are $I_A$ and $I_B$. The ratio $\frac{I_B}{I_A}$ is:
Answer: 0.6
Solution:
Moment of inertia of a sphere with radial density $\rho(r)$:
$$I = \int_0^R \frac{8\pi}{3} \rho(r) r^4 dr$$For sphere A ($\rho_A(r) = k \frac{r}{R}$):
$$I_A = \frac{8\pi}{3} \int_0^R k \frac{r^5}{R} dr = \frac{8\pi k}{3R} \left[\frac{R^6}{6}\right] = \frac{8\pi k R^5}{18}$$For sphere B ($\rho_B(r) = k \frac{r^5}{R^5}$):
$$I_B = \frac{8\pi}{3} \int_0^R k \frac{r^9}{R^5} dr = \frac{8\pi k}{3R^5} \left[\frac{R^{10}}{10}\right] = \frac{8\pi k R^5}{30}$$Ratio $\frac{I_B}{I_A}$:
$$\frac{I_B}{I_A} = \frac{18}{30} = \frac{3}{5} = 0.6$$
No comments:
Post a Comment
Note: Only a member of this blog may post a comment.