JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A uniform cylinder of mass \(M\) and radius \(R\) is to be pulled over a step of height \(a (a < R\)) by applying a force \(F\) at its centre \('O'\) perpendicular to the plane through the axes of the cylinder on the edge of the step ( see figure). The minimum value of \(F\) required is

- A \(Mg \sqrt{1-\frac{ a ^{2}}{ R ^{2}}}\)
- B \(Mg \sqrt{\left(\frac{ R }{ R - a }\right)^{2}-1}\)
- C \(Mg \frac{ a }{ R }\)
- D \(M g \sqrt{1-\left(\frac{R-a}{R}\right)^{2}}\)
Answer & Solution
Correct Answer
(D) \(M g \sqrt{1-\left(\frac{R-a}{R}\right)^{2}}\)
Step-by-step Solution
Detailed explanation
To Step up, \(F \times R \geq M g \times X\) \(\Rightarrow F_{\min }=\frac{M g}{R} \times \sqrt{R^{2}-(R-a)^{2}}\) \(\quad\quad\quad\,=M g \sqrt{1-\left(\frac{R-a}{R}\right)^{2}}\)
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