Answer
64.8k+ views
Hint: In this solution, we will draw the free body diagram of the string pulley system. Then we will calculate the tension in the strings to calculate the acceleration in the two blocks.
Complete step by step answer:
Let us start by drawing a free body diagram of the tension in the different strings.
![](https://www.vedantu.com/question-sets/8000f56e-fee3-4b63-bd5f-5caf0c70f3c43289482965257672784.png)
AS we can see, for the mass ${m_1}$ the mass it experiences is associated with only the bottom left pulley. Hence, we can write the equation of motion for the first mass as:
$T = m{a_1}$
Now the string connecting the lower two pulleys is the same so the net displacement will be zero. Also, we can see that for the pulley in the bottom right, the tension in the two strings will be $2T$ as can be seen from the diagram. Also, the tension in the string can be represented as the acceleration of the object which will be
$T = 2m{a_2}$
Since the tension will be the same in the string for both the bottom pulleys, we have
$m{a_1} = 2m{a_2}$
So, the relation of the acceleration of the two blocks will be
${a_1} = 2{a_2}$
Hence the correct choice will be choice (A).
Note: We can intuitively expect the first mass to have higher acceleration since it will have to compensate for the tension exerted by the other mass as it is connected by two pulleys. As a result, the displacement of the first block will always be higher than the displacement of the second block. We shouldn’t worry about the movement of individual pulleys but only the tension in the strings as it simplifies our calculations.
Complete step by step answer:
Let us start by drawing a free body diagram of the tension in the different strings.
![](https://www.vedantu.com/question-sets/8000f56e-fee3-4b63-bd5f-5caf0c70f3c43289482965257672784.png)
AS we can see, for the mass ${m_1}$ the mass it experiences is associated with only the bottom left pulley. Hence, we can write the equation of motion for the first mass as:
$T = m{a_1}$
Now the string connecting the lower two pulleys is the same so the net displacement will be zero. Also, we can see that for the pulley in the bottom right, the tension in the two strings will be $2T$ as can be seen from the diagram. Also, the tension in the string can be represented as the acceleration of the object which will be
$T = 2m{a_2}$
Since the tension will be the same in the string for both the bottom pulleys, we have
$m{a_1} = 2m{a_2}$
So, the relation of the acceleration of the two blocks will be
${a_1} = 2{a_2}$
Hence the correct choice will be choice (A).
Note: We can intuitively expect the first mass to have higher acceleration since it will have to compensate for the tension exerted by the other mass as it is connected by two pulleys. As a result, the displacement of the first block will always be higher than the displacement of the second block. We shouldn’t worry about the movement of individual pulleys but only the tension in the strings as it simplifies our calculations.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)