\[{T_1}\] and \[{T_2}\] in the given figure are

(A) 28 N, 48 N
(B) 48 N, 28 N
(C) 96 N, 56 N
(D) 56 N, 96 N
Answer
249.9k+ views
Hint: We need to identify the force acting on each block. Perform a Newton’s second law analysis on each of the blocks separately. The acceleration of each object is equal to the acceleration of the system.
Formula used: In this solution we will be using the following formulae;
\[{F_{NET}} = ma\] where \[{F_{NET}}\] is the net force acting on a body, \[m\] is the mass of the body, and \[a\] is the acceleration of the body.
Complete Step-by-Step Answer:
A particular amount of force is acting upon a chain of objects, the tensions in the cords are asked to be determined. To do this, we must analyse the forces on each block independently and link them with the acceleration. The acceleration of the system is the acceleration of the blocks, hence, they are equal.
Newton’s second law can be written as
\[{F_{NET}} = ma\] where \[{F_{NET}}\] is the net force acting on a body, \[m\] is the mass of the body, and \[a\] is the acceleration of the body.
Applying on the 3kg body, we have
\[120 - {T_1} = ma\]
\[ \Rightarrow 120 - {T_1} = 3a\]
Similarly, on the 5kg block
\[{T_1} - {T_2} = 5a\]
And on the 7kg block
\[{T_2} = 7a\]
We shall solve this three equations simultaneously, hence,
Insert \[{T_2} = 7a\] into \[{T_1} - {T_2} = 5a\], then,
\[{T_1} - 7a = 5a\]
\[ \Rightarrow {T_1} = 12a\]
Inserting this into \[120 - {T_1} = 3a\], we have
\[120 - 12a = 3a\]
\[ \Rightarrow 120 = 15a\]
Dividing both sides by 10, we get
\[a = 8m{s^{ - 2}}\]
Then, \[{T_1}\] can be calculated from
\[120 - {T_1} = 3 \times 8\]
\[ \Rightarrow {T_1} = 120 - 24 = 96N\]
Then the tension in the second cord \[{T_2}\] can be calculated from
\[{T_2} = 7a\] as
\[{T_2} = 7 \times 8 = 56N\]
Hence, the correct option is C
Note: Alternatively, one could take the entire system as one block which will be the sum of all three blocks and calculate the acceleration as thus
\[a = \dfrac{{120}}{M} = \dfrac{{120}}{{3 + 5 + 7}} = \dfrac{{120}}{{15}}\]
Hence, by computation,
\[a = 8m{s^{ - 2}}\]
Formula used: In this solution we will be using the following formulae;
\[{F_{NET}} = ma\] where \[{F_{NET}}\] is the net force acting on a body, \[m\] is the mass of the body, and \[a\] is the acceleration of the body.
Complete Step-by-Step Answer:
A particular amount of force is acting upon a chain of objects, the tensions in the cords are asked to be determined. To do this, we must analyse the forces on each block independently and link them with the acceleration. The acceleration of the system is the acceleration of the blocks, hence, they are equal.
Newton’s second law can be written as
\[{F_{NET}} = ma\] where \[{F_{NET}}\] is the net force acting on a body, \[m\] is the mass of the body, and \[a\] is the acceleration of the body.
Applying on the 3kg body, we have
\[120 - {T_1} = ma\]
\[ \Rightarrow 120 - {T_1} = 3a\]
Similarly, on the 5kg block
\[{T_1} - {T_2} = 5a\]
And on the 7kg block
\[{T_2} = 7a\]
We shall solve this three equations simultaneously, hence,
Insert \[{T_2} = 7a\] into \[{T_1} - {T_2} = 5a\], then,
\[{T_1} - 7a = 5a\]
\[ \Rightarrow {T_1} = 12a\]
Inserting this into \[120 - {T_1} = 3a\], we have
\[120 - 12a = 3a\]
\[ \Rightarrow 120 = 15a\]
Dividing both sides by 10, we get
\[a = 8m{s^{ - 2}}\]
Then, \[{T_1}\] can be calculated from
\[120 - {T_1} = 3 \times 8\]
\[ \Rightarrow {T_1} = 120 - 24 = 96N\]
Then the tension in the second cord \[{T_2}\] can be calculated from
\[{T_2} = 7a\] as
\[{T_2} = 7 \times 8 = 56N\]
Hence, the correct option is C
Note: Alternatively, one could take the entire system as one block which will be the sum of all three blocks and calculate the acceleration as thus
\[a = \dfrac{{120}}{M} = \dfrac{{120}}{{3 + 5 + 7}} = \dfrac{{120}}{{15}}\]
Hence, by computation,
\[a = 8m{s^{ - 2}}\]
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

