Answer
Verified
455.7k+ views
Hint: The given problem is an example of connected motion in which two bodies of different masses \[\mathop m\nolimits_1 \]and \[\mathop m\nolimits_2 \] are connected at the ends of an inextensible string and considered that this string passes over light and frictionless pulley.
Complete step by step solution:
Step 1:
In the question it is given that one body is heavier than the other. So, let \[{m_1} > {m_2}\]where \[\mathop m\nolimits_1 = \]10kg and \[\mathop m\nolimits_2 = \]6kg.
Let \[a\]be the common acceleration of the system of two bodies as shown in the above figure.
Step 2: Since the pulley is light and frictionless, therefore, the tension in the string shall be the same on the both sides of the pulley as shown in the above figure.
Now we have to take the consideration of the forces working on the two bodies and these forces are as given below-
On the heavier body, the forces are:
Its weight \[{m_1}g\] acting downwards,
The tension \[T\] in the string working upwards.
As the body moves downwards with the acceleration \[a\], the net downward force on the body is \[{m_1}a\].
So, \[{m_1}g - T = {m_1}a\] …………………...(1)
On the lighter body, the forces are:
Its weight \[{m_2}g\] acting downwards,
The tension \[T\] in the string acting upwards.
As the body moves upwards with the acceleration \[a\], the net upward force on the body is \[{m_2}a\].
So, \[T - {m_2}g = {m_2}a\]...................... (2)
Step 3: Tension \[T\] in the string can be calculated by dividing equation (1) by (2),
We get, \[\dfrac{{{m_1}g - T}}{{T - {m_2}g}} = \dfrac{{{m_1}a}}{{{m_2}a}}\]
\[\dfrac{{{m_1}g - T}}{{T - {m_2}g}} = \dfrac{{{m_1}}}{{{m_2}}}\]
\[{m_1}{m_2}g - {m_2}T = {m_1}T - {m_2}{m_1}g\]
\[{m_1}{m_2}g + {m_1}{m_2}g = {m_1}T + {m_2}T\]
\[2{m_1}{m_2}g = ({m_1} + {m_2})T\]
\[T = \dfrac{{2{m_1}{m_2}g}}{{({m_1} + {m_2})}}\] ……………... (3)
From the question, values of masses of the bodies and the acceleration due to gravity i.e. \[g = \]10m/s can be kept in the equation (3) and tension \[T\] can be calculated as-
\[T = \dfrac{{2 \times 10 \times 6 \times 10}}{{(10 + 6)}}\]
\[T = \dfrac{{1200}}{{16}}\]
\[T = 75\]N.
The correct option is (A).
Note:
(i) While solving these types of questions balancing the force equations is a must. The force in which direction the body is moving is higher than the opposite force working on the body.
(ii) If there is nothing mentioned about the weight and friction of the pulley then it should be considered light and frictionless respectively.
(iii) And if the pulley is lightweight and frictionless then tension \[T\] will be the same on both sides of the pulley.
Complete step by step solution:
Step 1:
In the question it is given that one body is heavier than the other. So, let \[{m_1} > {m_2}\]where \[\mathop m\nolimits_1 = \]10kg and \[\mathop m\nolimits_2 = \]6kg.
Let \[a\]be the common acceleration of the system of two bodies as shown in the above figure.
Step 2: Since the pulley is light and frictionless, therefore, the tension in the string shall be the same on the both sides of the pulley as shown in the above figure.
Now we have to take the consideration of the forces working on the two bodies and these forces are as given below-
On the heavier body, the forces are:
Its weight \[{m_1}g\] acting downwards,
The tension \[T\] in the string working upwards.
As the body moves downwards with the acceleration \[a\], the net downward force on the body is \[{m_1}a\].
So, \[{m_1}g - T = {m_1}a\] …………………...(1)
On the lighter body, the forces are:
Its weight \[{m_2}g\] acting downwards,
The tension \[T\] in the string acting upwards.
As the body moves upwards with the acceleration \[a\], the net upward force on the body is \[{m_2}a\].
So, \[T - {m_2}g = {m_2}a\]...................... (2)
Step 3: Tension \[T\] in the string can be calculated by dividing equation (1) by (2),
We get, \[\dfrac{{{m_1}g - T}}{{T - {m_2}g}} = \dfrac{{{m_1}a}}{{{m_2}a}}\]
\[\dfrac{{{m_1}g - T}}{{T - {m_2}g}} = \dfrac{{{m_1}}}{{{m_2}}}\]
\[{m_1}{m_2}g - {m_2}T = {m_1}T - {m_2}{m_1}g\]
\[{m_1}{m_2}g + {m_1}{m_2}g = {m_1}T + {m_2}T\]
\[2{m_1}{m_2}g = ({m_1} + {m_2})T\]
\[T = \dfrac{{2{m_1}{m_2}g}}{{({m_1} + {m_2})}}\] ……………... (3)
From the question, values of masses of the bodies and the acceleration due to gravity i.e. \[g = \]10m/s can be kept in the equation (3) and tension \[T\] can be calculated as-
\[T = \dfrac{{2 \times 10 \times 6 \times 10}}{{(10 + 6)}}\]
\[T = \dfrac{{1200}}{{16}}\]
\[T = 75\]N.
The correct option is (A).
Note:
(i) While solving these types of questions balancing the force equations is a must. The force in which direction the body is moving is higher than the opposite force working on the body.
(ii) If there is nothing mentioned about the weight and friction of the pulley then it should be considered light and frictionless respectively.
(iii) And if the pulley is lightweight and frictionless then tension \[T\] will be the same on both sides of the pulley.
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
A rainbow has circular shape because A The earth is class 11 physics CBSE
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What was the Metternich system and how did it provide class 11 social science CBSE
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE