Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • I need help with Three identical blocks of masses m = 2 kg are drawn by a force F = 10.2 N with an acceleration of 0.6 ms-2 on a frictionless surface, then what is the tension (in N) in the string between the blocks B and C ?

Three identical blocks of masses  m = 2 kg are drawn by a force F = 10.2 N  with an acceleration of 0.6 ms-2 on a  frictionless surface, then what is the tension (in N) in the string between the blocks B and C ?

  • Option 1)

    9.2

  • Option 2)

    7.8

  • Option 3)

    4

  • Option 4)

    9.8

 

Answers (1)

best_answer

As we learnt in

3 block connected with string -

- wherein

a=\frac{F}{m_{1}+m_{2}+m_{3}}


T_{1}=\frac{(m_{2}+m_{3})F}{m_{1}+m_{2}+m_{3}}


T_{2}=\frac{m_{3}F}{m_{1}+m_{2}+m_{3}}

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Since,     Force  =  mass × acceleration

\therefore    F - TAB = ma  and TAB - TBC = ma

\therefore    TBC  =  F - 2 ma

or    TBC  =  10.2 - ( 2 x 2 x 0.6)

or    TBC  =  7.8 N

Correct option is 2.


Option 1)

9.2

This is an incorrect option.

Option 2)

7.8

This is the correct option.

Option 3)

4

This is an incorrect option.

Option 4)

9.8

This is an incorrect option.

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 session 2 result date announced; Steps to check NTA session 2 scores
anam-khan

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today

Latest Question