A monkey of mass 40 kg climbs on a rope | Class 11 Physics Chapter Laws of Motion, Laws of Motion NCERT Solutions

Current NCERT chapter

Academic session 2026-27, Chapter 4 in the current curriculum.

Welcome to the NCERT Solutions for Class 11 Physics - Chapter Laws of Motion. This page offers a step-by-step solution to the specific question from Exercise 1, Question 33:

A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey

(a) climbs up with an acceleration of 6 m s-2

(b) climbs down with an acceleration of 4 m s-2

(c) climbs up with a uniform speed of 5 m s-1

(d) falls down the rope nearly freely under gravity?

(Ignore the mass of the rope).

. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.

Question 33:

A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey

(a) climbs up with an acceleration of 6 m s-2

(b) climbs down with an acceleration of 4 m s-2

(c) climbs up with a uniform speed of 5 m s-1

(d) falls down the rope nearly freely under gravity?

(Ignore the mass of the rope).

Answer:

Case (a)

Mass of the monkey, m= 40 kg

Acceleration due to gravity, g= 10 m/s

Maximum tension that the rope can bear, Tmax= 600 N

Acceleration of the monkey, a= 6 m/s2upward

UsingNewton's second law of motion, we can write the equation of motion as:

T- mg= ma

∴T= m(g+ a)

= 40 (10 + 6)

= 640 N

Since T > Tmax, the rope will break in this case.

Case (b)

Acceleration of the monkey, a= 4 m/s2downward

UsingNewton's second law of motion, we can write the equation of motion as:

mg - T = ma

∴T= m (g- a)

= 40(10 - 4)

= 240 N

Since T < Tmax, the rope will not break in this case.

Case (c)

The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., a= 0.

UsingNewton's second law of motion, we can write the equation of motion as:

T- mg = ma

T- mg = 0

∴T = mg

= 40 x 10

= 400 N

Since T < Tmax, the rope will not break in this case.

Case (d)

When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e., a = g

UsingNewton's second law of motion, we can write the equation of motion as:

mg - T= mg

∴T= m(g- g) = 0

Since T < Tmax, the rope will not break in this case.


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Laws of Motion.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 33: A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In w....

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