One twirls a circular ring (of mass $M$ and radius $R$ ) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is $r$. The finger rotates with an angular velocity $\omega_{0}$. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is $\mu$ and the acceleration due to gravity is $g$.
The total kinetic energy of the ring is
- $M \omega_{0}^{2} R^{2}$
- $\frac{1}{2} M \omega_{0}^{2}(R-r)^{2}$
- $M \omega_{0}^{2}(R-r)^{2}$
- $\frac{3}{2} M \omega_{0}^{2}(R-r)^{2}$
The Correct Option is C
Solution and Explanation
Top Questions on Gravitation
- Given below are two statements: Statement I: A satellite is moving around earth in an orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth.
Statement II: The time period of revolution of the satellite is \[ T = 2\pi \sqrt{\frac{R_e}{g}} \] (for satellite very close to the earth surface), where \(R_e\) is the radius of earth and \(g\) is acceleration due to gravity.
In the light of the above statements, choose the correct answer from the options given below.- JEE Main - 2026
- Physics
- Gravitation
- Initially a satellite of 100 kg is in a circular orbit of radius 1.5\(R_E\). This satellite can be moved to a circular orbit of radius 3\(R_E\) by supplying \(a \times 10^6\) J of energy. The value of a is_________ . (Take Radius of Earth \(R_E = 6 \times 10^6\) m and g = 10 m/s\(^2\)).
- JEE Main - 2026
- Physics
- Gravitation
- The escape velocity from a spherical planet \(A\) is \(10\ \text{km/s}\). The escape velocity from another planet \(B\), whose density and radius are \(10%\) of those of planet \(A\), is _______ m/s.
- JEE Main - 2026
- Physics
- Gravitation
Net gravitational force at the center of a square is found to be \( F_1 \) when four particles having masses \( M, 2M, 3M \) and \( 4M \) are placed at the four corners of the square as shown in figure, and it is \( F_2 \) when the positions of \( 3M \) and \( 4M \) are interchanged. The ratio \( \dfrac{F_1}{F_2} = \dfrac{\alpha}{\sqrt{5}} \). The value of \( \alpha \) is
- JEE Main - 2026
- Physics
- Gravitation
- In the given situation, the force at the center on 1 kg mass is \( F_1 \). Now if \( 4m \) and \( 3m \) are interchanged, the force is \( F_2 \). Given \( \frac{F_1}{F_2} = \frac{2}{\sqrt{\alpha}} \), find \( \alpha \).
- JEE Main - 2026
- Physics
- Gravitation
Questions Asked in JEE Advanced exam
- Let $ x_0 $ be the real number such that $ e^{x_0} + x_0 = 0 $. For a given real number $ \alpha $, define $$ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} $$ for all real numbers $ x $. Then which one of the following statements is TRUE?
- JEE Advanced - 2025
- Fundamental Theorem of Calculus
- A linear octasaccharide (molar mass = 1024 g mol$^{-1}$) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is 58.26 % (w/w) of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is _____.
Use: Molar mass (in g mol$^{-1}$): ribose = 150, 2-deoxyribose = 134, glucose = 180; Atomic mass (in amu): H = 1, O = 16- JEE Advanced - 2025
- Biomolecules
Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?
- JEE Advanced - 2025
- Conic sections
- Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from 10 mg g$^{-1}$ and 16 mg g$^{-1}$ aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be 4 mg g$^{-1}$ and 10 mg g$^{-1}$, respectively. At this temperature, the concentration (in mg g$^{-1}$) of adsorbed phenol from 20 mg g$^{-1}$ aqueous solution of phenol will be ____. Use: $\log_{10} 2 = 0.3$
- JEE Advanced - 2025
- Adsorption
- At 300 K, an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (h) of the solution (density = 1.00 g cm$^{-3}$) where h is equal to 2.00 cm. If the concentration of the dilute solution of the macromolecule is 2.00 g dm$^{-3}$, the molar mass of the macromolecule is calculated to be $X \times 10^{4}$ g mol$^{-1}$. The value of $X$ is ____. Use: Universal gas constant (R) = 8.3 J K$^{-1}$ mol$^{-1}$ and acceleration due to gravity (g) = 10 m s$^{-2}\}$
- JEE Advanced - 2025
- Colligative Properties
Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].