Rotational Motion
JEE (Main) 2023
Q1
Allen
Rotational Motion
Numerical
A ring and a solid sphere rotating about an axis passing through their centers have same radii of gyration. The axis of rotation is perpendicular to plane of ring. The ratio of radius of ring to that of sphere is $\sqrt{\frac{2}{x}}$ . The value of x is ____.
Correct Answer: 5
Explanation:
5
JEE (Main) 2023
Q2
Allen
Rotational Motion
Numerical
For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is $\frac{x}{5}$ . The value of x is ____.
Correct Answer: 2
Explanation:
2
JEE (Main) 2023
Q3
Allen
Rotational Motion
Numerical
A uniform solid cylinder with radius R and length L has moment of inertia $I_{1}$ , about the axis of cylinder. A concentric solid cylinder of radius $R'=\frac{R}{2}$ and length $L'=\frac{L}{2}$ is caned out of the original cylinder. If $I_{2}$ is the moment of inertia of the carved-out portion to the cylinder then $\frac{I_{1}}{I_{2}}=$ ____. (Both $I_{1}$ and $I_{2}$ are about the axis of the cylinder)
Correct Answer: 32
Explanation:
32
JEE (Advanced) 2023
Q4
Allen
JEE Advanced PYQ
Numerical
Two point-like objects of masses 20 gm and 30 gm are fixed at the two ends of a rigid massless rod of length 10 cm. This system is suspended vertically from a rigid ceiling using a thin wire attached to its center of mass, as shown in the figure. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is $1.2 \times 10^{-8}$ N m rad $^{-1}$ . The angular frequency of the oscillations is $n \times 10^{-3}$ rad s $^{-1}$ . The value of n is ____.
Correct Answer: 10
Explanation:
Ans. (10)
$T = \frac {2 \pi}{\omega} = 2 \pi \sqrt {\frac {I}{C}}$
$\Rightarrow \omega = \sqrt {\frac {C}{I}}$
Where I = moment of inertia
$I = (3 0) (4) ^ {2} + (2 0) (6) ^ {2}$
$= 1 2 0 0 g m \cdot c m ^ {2}$
$= 1. 2 \times 1 0 ^ {- 4} k g \cdot m ^ {2}$
$\Rightarrow \omega = \sqrt {\frac {1 . 2 \times 1 0 ^ {- 8}}{1 . 2 \times 1 0 ^ {- 4}}}$
$\Rightarrow \omega = \sqrt {1 0 ^ {- 4}}$
$\omega = (1 0 ^ {- 2})$
$n \times 1 0 ^ {- 3} = 1 0 ^ {- 2} \Rightarrow n = 1 0$
$T = \frac {2 \pi}{\omega} = 2 \pi \sqrt {\frac {I}{C}}$
$\Rightarrow \omega = \sqrt {\frac {C}{I}}$
Where I = moment of inertia
$I = (3 0) (4) ^ {2} + (2 0) (6) ^ {2}$
$= 1 2 0 0 g m \cdot c m ^ {2}$
$= 1. 2 \times 1 0 ^ {- 4} k g \cdot m ^ {2}$
$\Rightarrow \omega = \sqrt {\frac {1 . 2 \times 1 0 ^ {- 8}}{1 . 2 \times 1 0 ^ {- 4}}}$
$\Rightarrow \omega = \sqrt {1 0 ^ {- 4}}$
$\omega = (1 0 ^ {- 2})$
$n \times 1 0 ^ {- 3} = 1 0 ^ {- 2} \Rightarrow n = 1 0$
JEE (Advanced) 2023
Q5
Allen
JEE Advanced PYQ
Numerical
A thin circular coin of mass 5 gm and radius 4/3 cm is initially in a horizontal xy-plane. The coin is tossed vertically up (+z direction) by applying an impulse of $\sqrt{\frac{\pi}{2}} \times 10^{-2} N - s$ at a distance 2/3 cm from its center. The coin spins about its diameter and moves along the +z direction. By the time the coin reaches back to its initial position, it completes n rotations. The value of n is _. [Given: The acceleration due to gravity g = 10 ms $^{-2}$ ]
Correct Answer: 30
Explanation:
Ans. (30)
J = m v
$J r = I _ {\mathrm{c}} \omega$
... (1)
$J _ {C} = \frac {1}{4} m R ^ {2}$
... (2)
... (3)
$t = \frac {2 v}{g}$
... (4)
$\theta = 2 \pi N = \omega t$
... (5) $\therefore N = 30$
## JEE (Main) Practice Paper
J = m v
$J r = I _ {\mathrm{c}} \omega$
... (1)
$J _ {C} = \frac {1}{4} m R ^ {2}$
... (2)
... (3)
$t = \frac {2 v}{g}$
... (4)
$\theta = 2 \pi N = \omega t$
... (5) $\therefore N = 30$
## JEE (Main) Practice Paper
JEE (Main) 2022
Q6
Allen
Rotational Motion
Numerical
A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 10 g are put one on the top of the other at the 10.0 cm mark the scale is found to be balanced at 40.0 cm mark. The mass of the metre scale is found to be $x \times 10^{-2}$ kg. The value of x is
Correct Answer: 2
Explanation:
Ans. (2)
Say the distance of central line from instantaneous axis of rotation is r. Then r from the point on left becomes lesser than that for right.
So, $v_{left point} = \omega r' < \omega r = v_{right point}$ .
So, the roller will turn to left.
Say the distance of central line from instantaneous axis of rotation is r. Then r from the point on left becomes lesser than that for right.
So, $v_{left point} = \omega r' < \omega r = v_{right point}$ .
So, the roller will turn to left.
JEE (Advanced) 2022
Q7
Allen
JEE Advanced PYQ
Numerical
At time t = 0, a disk of radius 1 m starts to roll without slipping on a horizontal plane with an angular acceleration of $\alpha = \frac{2}{3}$ rad s $^{-2}$ . A small stone is stuck to the disk. At t = 0, it is at the contact point of the disk and the plane. Later, at time $t = \sqrt{\pi} s$ , the stone detaches itself and flies off tangentially from the disk. The maximum height (in m) reached by the stone measured from the plane is $\frac{1}{2} + \frac{x}{10}$ . The value of x is ____. [Take g = 10 m s $^{-2}$ .]
Correct Answer: 0.52
Explanation:
Ans. (3) $\frac{mR^2}{2}\omega = \left(\frac{mR^2}{2} + mR^2\right)\omega'$ $\therefore \omega' = \frac{\omega}{3}$
JEE (Advanced) 2022
Q8
Allen
JEE Advanced PYQ
Numerical
A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination $\theta = 30^{\circ}$ from the horizontal. Two forces of magnitude 1 N each, parallel to the incline, act on the sphere, both at distance r = 0.5 m from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is ____ $ms^{-2}$ . (Take $g = 10 \, m s^{-2}$ .)
Correct Answer: 2.85 to 2.86
Explanation:
Ans. (A-R), (B-S), (C-P), (D-Q)
(A) $V_{cm} = l$ and $\omega R = 1$ (B) $v_{rel\ 1/3} \neq 0$ ā“ Not a rigid body
(C) $v_{2} = \hat{i}, v_{3} = 0$ i.e. rotating at same position only (p).
(D) $V_{i} = -\hat{i} - \hat{j}$ $V_{4} = 0$ (Lowest point) so (Q).
## Important Notes
The image contains only background horizontal lines, which are stylistic or background elements and not text placeholders. According to the correction guidelines, such lines must be ignored. Therefore, the correct OCR output is empty.
(no text)
## Important Notes
The image contains only background horizontal lines, which are stylistic or background elements and not text placeholders. According to the correction guidelines, such lines must be ignored. Therefore, the correct OCR output is empty.
(no text)
(A) $V_{cm} = l$ and $\omega R = 1$ (B) $v_{rel\ 1/3} \neq 0$ ā“ Not a rigid body
(C) $v_{2} = \hat{i}, v_{3} = 0$ i.e. rotating at same position only (p).
(D) $V_{i} = -\hat{i} - \hat{j}$ $V_{4} = 0$ (Lowest point) so (Q).
## Important Notes
The image contains only background horizontal lines, which are stylistic or background elements and not text placeholders. According to the correction guidelines, such lines must be ignored. Therefore, the correct OCR output is empty.
(no text)
## Important Notes
The image contains only background horizontal lines, which are stylistic or background elements and not text placeholders. According to the correction guidelines, such lines must be ignored. Therefore, the correct OCR output is empty.
(no text)
JEE (Advanced) 2022
Q9
Allen
JEE Advanced PYQ
Numerical
A particle of mass 1 kg is subjected to a force which depends on the position as $\vec{F} = -k(x\hat{i} + y\hat{j})kgms^{-2}$ with $k = 1 kgs^{-2}$ . At time $t = 0$ , the particle's position $\vec{r} = \left(\frac{1}{\sqrt{2}}\hat{i} + \sqrt{2}\hat{j}\right)m$ and its velocity $\vec{v} = \left(-\sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \frac{2}{\pi}\hat{k}\right)ms^{-1}$ . Let $v_x$ and $v_y$ denote the $x$ and the $y$ components of the particle's velocity, respectively. Ignore gravity. When $z = 0.5$ m, the value of $(x v_y - y v_x)$ is ____ $m^2 s^{-1}$ .
Correct Answer: 3
Explanation:
Ans. (3)
Torque about origin is zero
So angular momentum about origin remains conserved.
$\left| \begin{array}{c c c} \hat {i} & \hat {j} & \hat {k} \\ \frac {1}{\sqrt {2}} & \sqrt {2} & 0 \\ - \sqrt {2} & \sqrt {2} & \frac {2}{\pi} \end{array} \right| = \left| \begin{array}{c c c} \hat {i} & \hat {j} & \hat {k} \\ x & y & 0. 5 \\ v _ {x} & v _ {y} & \frac {2}{\pi} \end{array} \right|$
$\hat {\imath} \left[ \sqrt {2} \times \frac {2}{\pi} \right] - \hat {\jmath} \left[ \frac {\sqrt {2}}{\pi} \right] + \hat {k} [ 1 + 2 ]$
$= \hat {\imath} \left[ \frac {y \times 2}{\pi} - 0. 5 v _ {y} \right] - \hat {\jmath} \left[ \frac {x \times 2}{\pi} - 0. 5 v _ {x} \right] + \hat {k} \big [ x v _ {y} - y v _ {x} \big ]$
$x _ {v y} - y _ {v x} = 3$
Torque about origin is zero
So angular momentum about origin remains conserved.
$\left| \begin{array}{c c c} \hat {i} & \hat {j} & \hat {k} \\ \frac {1}{\sqrt {2}} & \sqrt {2} & 0 \\ - \sqrt {2} & \sqrt {2} & \frac {2}{\pi} \end{array} \right| = \left| \begin{array}{c c c} \hat {i} & \hat {j} & \hat {k} \\ x & y & 0. 5 \\ v _ {x} & v _ {y} & \frac {2}{\pi} \end{array} \right|$
$\hat {\imath} \left[ \sqrt {2} \times \frac {2}{\pi} \right] - \hat {\jmath} \left[ \frac {\sqrt {2}}{\pi} \right] + \hat {k} [ 1 + 2 ]$
$= \hat {\imath} \left[ \frac {y \times 2}{\pi} - 0. 5 v _ {y} \right] - \hat {\jmath} \left[ \frac {x \times 2}{\pi} - 0. 5 v _ {x} \right] + \hat {k} \big [ x v _ {y} - y v _ {x} \big ]$
$x _ {v y} - y _ {v x} = 3$
JEE (Main) 2021
Q10
Allen
Rotational Motion
Numerical
A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass 'm' travelling along the surface hits at one end of the rod with a velocity 'u' in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses $\left(\frac{m}{M}\right)$ is $\frac{1}{x}$ . The value of 'x' will be ____.
Correct Answer: 4
Explanation:
Ans. (4)
Different
Different
JEE (Main) 2021
Q11
Allen
Rotational Motion
Numerical
In the given figure, two wheels P and Q are connected by a belt B. The radius of P is three times as that of Q. In case of same rotational kinetic energy, the ratio of rotational inertias $\left(\frac{I_{1}}{I_{2}}\right)$ will be x:1. The value of x will be ____.
Correct Answer: 4
Explanation:
Ans. (4)
$\begin{array}{r l} & {\vec {r _ {c}} = \frac {1 \times 1 (\cos 6 0 \hat {i} + \sin 6 0 \hat {j}) + 3 \hat {i}}{6}} \\ & {\quad = \frac {3 . 5 \hat {i} + \frac {\sqrt {3}}{2} \hat {j}}{6}} \end{array}$
$I = I _ {\mathrm{C}} + m r _ {\mathrm{C}} ^ {2} = 1 \times 1 2 + 3 \times 1 2$
$\begin{array}{r l} I _ {\mathrm{C}} & = 4 - \frac {6 \times \left(\frac {4 9}{4} + \frac {3}{4}\right)}{6 ^ {2}} \\ & = 4 - \frac {5 2}{2 4} \\ & = \frac {2 1}{6} k g - m ^ {2} = 3. 5 k g \cdot m ^ {2} \end{array}$
$\begin{array}{r l} & {\vec {r _ {c}} = \frac {1 \times 1 (\cos 6 0 \hat {i} + \sin 6 0 \hat {j}) + 3 \hat {i}}{6}} \\ & {\quad = \frac {3 . 5 \hat {i} + \frac {\sqrt {3}}{2} \hat {j}}{6}} \end{array}$
$I = I _ {\mathrm{C}} + m r _ {\mathrm{C}} ^ {2} = 1 \times 1 2 + 3 \times 1 2$
$\begin{array}{r l} I _ {\mathrm{C}} & = 4 - \frac {6 \times \left(\frac {4 9}{4} + \frac {3}{4}\right)}{6 ^ {2}} \\ & = 4 - \frac {5 2}{2 4} \\ & = \frac {2 1}{6} k g - m ^ {2} = 3. 5 k g \cdot m ^ {2} \end{array}$
JEE (Main) 2021
Q12
Allen
Rotational Motion
Numerical
Consider a badminton racket with length scales as shown in the figure.
If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, $\frac{r}{2}$ distance from the end A of the handle will be ..... $Mr^{2}$ .
If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, $\frac{r}{2}$ distance from the end A of the handle will be ..... $Mr^{2}$ .
Correct Answer: 1
Explanation:
Ans. (1)
Let mass and side of cube be $M'$ and $\sqrt{3}$ a = 2R
$M ^ {\prime} = \frac {M}{\frac {4}{3} \pi R ^ {3}} a ^ {3}$
Moment of Inertia of cube = $\frac{M'a^{2}}{6}$
$= \left(\frac {M}{\frac {4}{3} \pi R ^ {3}} a ^ {3}\right) \frac {a ^ {2}}{6} = \boxed {\frac {4 M R ^ {2}}{9 \sqrt {3} \pi}}$
Let mass and side of cube be $M'$ and $\sqrt{3}$ a = 2R
$M ^ {\prime} = \frac {M}{\frac {4}{3} \pi R ^ {3}} a ^ {3}$
Moment of Inertia of cube = $\frac{M'a^{2}}{6}$
$= \left(\frac {M}{\frac {4}{3} \pi R ^ {3}} a ^ {3}\right) \frac {a ^ {2}}{6} = \boxed {\frac {4 M R ^ {2}}{9 \sqrt {3} \pi}}$
JEE (Advanced) 2021
Q13
Allen
JEE Advanced PYQ
Numerical
A thin rod of mass M and length a is free to rotate in horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass M and of radius a/4 is pivoted on this rod with its center at a distance a/4 from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity $\Omega$ and the disc rotating about its vertical axis with angular velocity $4\Omega$ . The total angular momentum of the system about the point O is $\left(\frac{Ma^{2}\Omega}{48}\right)n$ . The value of n is ____.
Correct Answer: 49
Explanation:
Ans. (5)
$4 0 - T _ {A} = 4 a\tag{... (i}$
$T _ {B} - 2 0 = 2 a$
$(T _ {A} - T _ {B}) \times 0. 1 \frac {M (0 . 1) ^ {2}}{2} \alpha\tag{... (ii}$
... (iii)
$T _ {A} - T _ {B} = 2 \alpha\tag{... (vi}$
$a = \alpha R \Rightarrow a = \frac {\alpha}{1 0}\tag{... (iv}$
$\text { Now } (1) + (2) \Rightarrow T _ {B} - T _ {A} + 2 0 = 6 a$
$\therefore - 2 \alpha + 2 0 = 6 a$
$\therefore \quad - 2 \alpha + 2 0 = \frac {6 \alpha}{1 0} \Rightarrow 2 0 \frac {2 6 \alpha}{1 0}$
$\therefore \alpha = \frac {2 0 0}{2 6}$
$\therefore \alpha = \frac {1 0 0}{1 3}$
$\therefore \quad a = \frac {\alpha}{1 0} = \frac {1 0}{1 3} m / s ^ {2}$
$\frac {1}{2} \alpha t ^ {2} = \frac {1}{2} \times \frac {1 0 0}{1 3} \times \frac {1 0 0}{2 \pi} = \frac {5 0 0 0}{2 6 \pi} r \ell v$
$4 0 - T _ {A} = 4 a\tag{... (i}$
$T _ {B} - 2 0 = 2 a$
$(T _ {A} - T _ {B}) \times 0. 1 \frac {M (0 . 1) ^ {2}}{2} \alpha\tag{... (ii}$
... (iii)
$T _ {A} - T _ {B} = 2 \alpha\tag{... (vi}$
$a = \alpha R \Rightarrow a = \frac {\alpha}{1 0}\tag{... (iv}$
$\text { Now } (1) + (2) \Rightarrow T _ {B} - T _ {A} + 2 0 = 6 a$
$\therefore - 2 \alpha + 2 0 = 6 a$
$\therefore \quad - 2 \alpha + 2 0 = \frac {6 \alpha}{1 0} \Rightarrow 2 0 \frac {2 6 \alpha}{1 0}$
$\therefore \alpha = \frac {2 0 0}{2 6}$
$\therefore \alpha = \frac {1 0 0}{1 3}$
$\therefore \quad a = \frac {\alpha}{1 0} = \frac {1 0}{1 3} m / s ^ {2}$
$\frac {1}{2} \alpha t ^ {2} = \frac {1}{2} \times \frac {1 0 0}{1 3} \times \frac {1 0 0}{2 \pi} = \frac {5 0 0 0}{2 6 \pi} r \ell v$
JEE (Main) 2020
Q14
Allen
JEE Main PYQ
Numerical
ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid points of AB, AC and G is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is $I_{0}$ . If part ADE is removed, the moment of inertia of the remaining part about the same axis is $\frac{NI_{0}}{16}$ where N is an integer. Value of N is ____.
Correct Answer: 2
Explanation:
Ans. (2)
$m_{a}g\ell_{1} + m_{s}x = m_{b}g\ell_{2}$ and $m_{a}g\frac{\ell_{1}}{2} +m_{s}$
$x > \frac {m _ {b} g \ell_ {2}}{2}$
$m_{a}g\ell_{1} + m_{s}x = m_{b}g\ell_{2}$ and $m_{a}g\frac{\ell_{1}}{2} +m_{s}$
$x > \frac {m _ {b} g \ell_ {2}}{2}$
JEE (Main) 2020
Q15
Allen
Rotational Motion
Numerical
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 $100 \times \frac{b}{a}$ for a box not to topple before moving is ____.
Correct Answer: 3
Explanation:
Ans. (3)
$\begin{array}{r l} & L = m _ {1} v r + m _ {2} v r + I \omega \\ & = (m _ {1} + m _ {2}) r ^ {2} \omega + I \omega \end{array}$
$\begin{array}{r l} & L = m _ {1} v r + m _ {2} v r + I \omega \\ & = (m _ {1} + m _ {2}) r ^ {2} \omega + I \omega \end{array}$
JEE (Advanced) 2020
Q16
Allen
JEE Advanced PYQ
Numerical
Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00 cm and the right one at 90.00 cm. When you attempt to move both the fingers slowly towards the center, initially only the left finger slips with respect to the scale and the right finger does not. After some distance, the left finger stops and the right one starts slipping. Then the right finger stops at a distance $x_{R}$ from the center (50.00 cm) of the scale and the left one starts slipping again. This happens because of the difference in the frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers and the scale are 0.40 and 0.32, respectively, the value of $x_{R}$ (in cm) is ____.
Correct Answer: 25.60
Explanation:
Ans. (C) At $\ell_{\mathrm{max}}$ , component of velocity along the length of spring is zero.
Q17
Allen
Concept of $w, \alpha, v, a$
Numerical
A uniform bar AB of mass m and a ball of the same mass are released from rest from the same horizontal position. The bar is hinged at end A. There is gravity downwards. What is the distance of the point from point B that has the same acceleration as that of ball, immediately after release?
Correct Answer: $\frac{L}{3}$
Explanation:
$\frac{L}{3}$
Q18
Allen
Concept of $w, \alpha, v, a$
Numerical
A uniform beam of length L and mass m is supported as shown. If the cable at B suddenly breaks, determine; (a) the acceleration of end B. (b) the reaction at the pin support.
Correct Answer: (a) $\downarrow\frac{9g}{7}$ (b) $\uparrow\frac{4mg}{7}$
Explanation:
(a) $\downarrow\frac{9g}{7}$ (b) $\uparrow\frac{4mg}{7}$
Q19
Allen
Concept of $w, \alpha, v, a$
Numerical
A rod of mass m and length $\ell$ is hinged at it's upper end. It can rotate in vertical plane. It is given angular velocity $\omega$ so that it can complete vertical circle. Find (a) $\omega$ (b) Tension at centre of rod at initial moment.
Correct Answer: (a) $\omega = \sqrt{\frac{6g}{\ell}}$ , (b) $T = \frac{11mg}{4}$
Explanation:
(a) $\omega = \sqrt{\frac{6g}{\ell}}$ , (b) $T = \frac{11mg}{4}$
Q20
Allen
Moment of Inertia
Numerical
Three equal masses m are rigidly connected to each other by massless rods of length l forming an equilateral triangle, as shown in the figure. What is the ratio of the moment of inertia of the assembly for an axis through B compared with that for an axis through A (centroid). Both the axis are perpendicular to the plane of triangle.
Correct Answer: 2
Explanation:
2
Q21
Allen
Moment of Inertia
Numerical
Two rods of equal mass m and length l lie along the x axis and y axis with their centres at origin. What is the moment of inertia of the system about the line x = y
Correct Answer: $\frac{ml^{2}}{12}$
Explanation:
$\frac{ml^{2}}{12}$
Q22
Allen
Moment of Inertia
Numerical
Find the MI of a rod about an axis through its centre of mass and perpendicular to the length whose linear density varies as $\lambda = ax$ where a is a constant and x is the position of an element of the rod relative to its left end. The length of the rod is $\ell$ .
Correct Answer: $\frac{a\ell^{4}}{36}$
Explanation:
$\frac{a\ell^{4}}{36}$
Q23
Allen
Moment of Inertia
Numerical
A hole of radius $R / 2$ is cut from a solid sphere of radius $R$ . If the mass of the remaining part of sphere is $M$ , then find moment of inertia of the body about an axis through $O$ .
Correct Answer: $57 / 140MR^{2}$
Explanation:
$57 / 140MR^{2}$
Q24
Allen
Equilibrium
Numerical
In an experiment with a massless beam balance an unknown mass m is balanced by two known masses of 16kg and 4 kg as shown in figure. Find the value of the unknown mass m.
Correct Answer: 8 kg
Explanation:
8 kg
Q25
Allen
Equilibrium
Numerical
Find the force F required to keep the system in equilibrium. The dimensions of the system are d = 0.3 m and a = 0.2 m. Assume the rods to be massless.
Correct Answer: 150(+i)
Explanation:
150(+i)