
Two blocks of mass $m = 5Kg$ and $M = 10Kg$ are connected by a string passing over a pulley B as shown. Another string connects the center of pulley B to the floor and passes over another pulley A as shown. An upward force F is applied at the center of pulley A. Both the pulleys are massless. An upward force F is applied in the pulley A. Find the acceleration of blocks m and M if F is $100N$.
(A)$0\dfrac{m}{{{s^2}}};0\dfrac{m}{{{s^2}}}$
(B)$5\dfrac{m}{{{s^2}}};2.5\dfrac{m}{{{s^2}}}$
(C)$10\dfrac{m}{{{s^2}}};5\dfrac{m}{{{s^2}}}$
(D)$2.5\dfrac{m}{{{s^2}}};5\dfrac{m}{{{s^2}}}$
Answer
134.7k+ views
Hint For lifting a body in an upward direction the upward directing force must be greater than the force acting in the downward direction. Tension in a massless string is the same along its length. We will use this property to solve this question.
Complete Step-by-step solution
Let,
${T_0}$= tension in the string passing over A
${T_1}$= tension in the string passing over B
Free body diagram:

$\therefore 2{T_0}= F$ and $2{T_1} = {T_0}$(balancing forces in the free body diagram drawn above)
$ \Rightarrow {T_1} = \dfrac{F}{4}$ (It’s given in the question that $F=100N$)
$ \therefore {T_1} = 25N$
The upward force on block m and M is the same as that force is due to tension in the string which is the same across its length as tension in a massless string is the same along its length.
Weights of blocks are,
$M=10Kg$ and $m=5Kg$ (given in question)
$mg = 50N$ and $Mg = 100N$ (Weight is the downward force acting on the block)
As${T_1}$ is less than mg and Mg hence the blocks will remain stationary on the floor as for lifting a body in an upward direction the upward directing force must be greater than the force acting in the downward direction. So, the acceleration of both the blocks will be zero as they are at rest.
$\therefore $ Hence, the correct answer will be (A)$0\dfrac{m}{{{s^2}}};0\dfrac{m}{{{s^2}}}$.
Note As discussed earlier tension along the same massless string will be the same this is an important point to be noted as many students consider tension in both strings different and end up making mistakes due to this mistake there time also got wasted in the examination hall and could get negative marks in the question.
Complete Step-by-step solution
Let,
${T_0}$= tension in the string passing over A
${T_1}$= tension in the string passing over B
Free body diagram:

$\therefore 2{T_0}= F$ and $2{T_1} = {T_0}$(balancing forces in the free body diagram drawn above)
$ \Rightarrow {T_1} = \dfrac{F}{4}$ (It’s given in the question that $F=100N$)
$ \therefore {T_1} = 25N$
The upward force on block m and M is the same as that force is due to tension in the string which is the same across its length as tension in a massless string is the same along its length.
Weights of blocks are,
$M=10Kg$ and $m=5Kg$ (given in question)
$mg = 50N$ and $Mg = 100N$ (Weight is the downward force acting on the block)
As${T_1}$ is less than mg and Mg hence the blocks will remain stationary on the floor as for lifting a body in an upward direction the upward directing force must be greater than the force acting in the downward direction. So, the acceleration of both the blocks will be zero as they are at rest.
$\therefore $ Hence, the correct answer will be (A)$0\dfrac{m}{{{s^2}}};0\dfrac{m}{{{s^2}}}$.
Note As discussed earlier tension along the same massless string will be the same this is an important point to be noted as many students consider tension in both strings different and end up making mistakes due to this mistake there time also got wasted in the examination hall and could get negative marks in the question.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

What are examples of Chemical Properties class 10 chemistry JEE_Main

JEE Main 2025 Session 2 Schedule Released – Check Important Details Here!

JEE Main 2025 Session 2 Admit Card – Release Date & Direct Download Link

JEE Main 2025 Session 2 Registration (Closed) - Link, Last Date & Fees

JEE Mains Result 2025 NTA NIC – Check Your Score Now!

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Elastic Collisions in One Dimension - JEE Important Topic

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3
