मराठी

A Stone of Mass M Tied to the End of a String Revolves in a Vertical Circle of Radius R. the Net Forces at the Lowest and Highest Points of the Circle Directed Vertically Downwards Are: [Choose the Correct Alternative] - Physics

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प्रश्न

A stone of mass tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]

  Lowest Point Highest Point
a) mg – T1 mg + T2
b) mg + T1 mg – T2
c) `mg + T1 –(m_v_1^2)/R` mg – T2 + (`mv_1^2`)/R
d) `mg – T1 – (mv)/R` mg + T2 + (mv_1^2)/R

 Tand v1 denote the tension and speed at the lowest point. Tand v2 denote corresponding values at the highest point.

उत्तर १

(a)The free body diagram of the stone at the lowest point is shown in the following figure.

According to Newton’s second law of motion, the net force acting on the stone at this point is equal to the centripetal force, i.e.,

`F_"net" = T - mg = (mv_1^2)/R ...(i)`

Where, v1 = Velocity at the lowest point

The free body diagram of the stone at the highest point is shown in the following figure

Using Newton’s second law of motion, we have:

`T + mg = (mv_2^2)/R `...(ii)

Where, v= Velocity at the highest point

It is clear from equations (i) and (ii) that the net force acting at the lowest and the highest points are respectively (T – mg) and (T + mg).

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उत्तर २

The net force at the lowest point is (mg – T1) and the net force at the highest point is (mg + T2). Therefore, alternative (a) is correct. Since mg and T1 are in mutually opposite directions at lowest point and mg and T2are in same direction at the highest point.

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Laws of Motion - Exercises [पृष्ठ ११२]

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एनसीईआरटी Physics [English] Class 11
पाठ 5 Laws of Motion
Exercises | Q 26 | पृष्ठ ११२

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