JEE Exam > JEE Questions > Masses 1 kg, 1.5 kg, 2 kg, and “M&rdquo...
Start Learning for Free
Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?
- a)1 kg
- b)1.5 kg
- c)2 kg
- d)2.5 kg
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1...
Sum of masses = 1 + 1.5 + 2 + M = 4.5 + M
x-coordinate;
(1*2 + 1.5*1 + 2*2 – M)/(4.5 + M) = 1
4.5 + M = 7.5 – M
2M = 3
M = 1.5 kg.
x-coordinate;
(1*2 + 1.5*1 + 2*2 – M)/(4.5 + M) = 1
4.5 + M = 7.5 – M
2M = 3
M = 1.5 kg.
Free Test
FREE
| Start Free Test |
Community Answer
Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1...
3 kg are placed on a frictionless surface connected by a string over a pulley. The 3 kg mass is hanging vertically downward, and the other masses are on the surface. Assuming the pulley is massless and there is no energy loss due to friction, find:
a) The acceleration of the system
b) The tension in the string
Solution:
a) The acceleration of the system can be found using Newton's second law, which states that the net force on an object is equal to its mass times its acceleration. Since the masses are connected by a string, they will move together with the same acceleration. Therefore, we can write:
net force = (mass of the system) x (acceleration of the system)
The net force on the system is the difference between the weight of the hanging mass and the weight of the other masses, which are balanced by the tension in the string. The weight of a mass is given by:
weight = (mass) x (acceleration due to gravity)
In this case, the acceleration due to gravity is 9.8 m/s^2. Therefore, we can write:
net force = (3 kg)(9.8 m/s^2) - [(1 kg)(9.8 m/s^2) + (1.5 kg)(9.8 m/s^2) + (2 kg)(9.8 m/s^2)]
net force = 29.4 N - 44.1 N
net force = -14.7 N
The negative sign indicates that the net force is in the opposite direction to the hanging mass. Therefore, we can write:
-14.7 N = (mass of the system) x (acceleration of the system)
mass of the system = 1 kg + 1.5 kg + 2 kg + 3 kg = 7.5 kg
acceleration of the system = (-14.7 N) / (7.5 kg) = -1.96 m/s^2
The negative sign indicates that the system is accelerating in the opposite direction to the hanging mass, which is upward.
b) The tension in the string can be found using Newton's second law again. In this case, we can focus on one of the masses on the surface, say the 1.5 kg mass. The tension in the string is the force pulling it in the direction of the hanging mass. Therefore, we can write:
net force = (mass) x (acceleration)
The net force on the 1.5 kg mass is the tension in the string minus its weight. The weight is given by:
weight = (mass) x (acceleration due to gravity)
weight = (1.5 kg) x (9.8 m/s^2) = 14.7 N
Therefore, we can write:
net force = (tension) - (weight)
net force = (1.5 kg) x (-1.96 m/s^2) = -2.94 N
Substituting the weight and net force, we get:
(tension) - (14.7 N) = -2.94 N
tension = 11.76 N
Therefore, the tension in the string is 11.76 N.
a) The acceleration of the system
b) The tension in the string
Solution:
a) The acceleration of the system can be found using Newton's second law, which states that the net force on an object is equal to its mass times its acceleration. Since the masses are connected by a string, they will move together with the same acceleration. Therefore, we can write:
net force = (mass of the system) x (acceleration of the system)
The net force on the system is the difference between the weight of the hanging mass and the weight of the other masses, which are balanced by the tension in the string. The weight of a mass is given by:
weight = (mass) x (acceleration due to gravity)
In this case, the acceleration due to gravity is 9.8 m/s^2. Therefore, we can write:
net force = (3 kg)(9.8 m/s^2) - [(1 kg)(9.8 m/s^2) + (1.5 kg)(9.8 m/s^2) + (2 kg)(9.8 m/s^2)]
net force = 29.4 N - 44.1 N
net force = -14.7 N
The negative sign indicates that the net force is in the opposite direction to the hanging mass. Therefore, we can write:
-14.7 N = (mass of the system) x (acceleration of the system)
mass of the system = 1 kg + 1.5 kg + 2 kg + 3 kg = 7.5 kg
acceleration of the system = (-14.7 N) / (7.5 kg) = -1.96 m/s^2
The negative sign indicates that the system is accelerating in the opposite direction to the hanging mass, which is upward.
b) The tension in the string can be found using Newton's second law again. In this case, we can focus on one of the masses on the surface, say the 1.5 kg mass. The tension in the string is the force pulling it in the direction of the hanging mass. Therefore, we can write:
net force = (mass) x (acceleration)
The net force on the 1.5 kg mass is the tension in the string minus its weight. The weight is given by:
weight = (mass) x (acceleration due to gravity)
weight = (1.5 kg) x (9.8 m/s^2) = 14.7 N
Therefore, we can write:
net force = (tension) - (weight)
net force = (1.5 kg) x (-1.96 m/s^2) = -2.94 N
Substituting the weight and net force, we get:
(tension) - (14.7 N) = -2.94 N
tension = 11.76 N
Therefore, the tension in the string is 11.76 N.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
Question Description
Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?a)1 kgb)1.5 kgc)2 kgd)2.5 kgCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?a)1 kgb)1.5 kgc)2 kgd)2.5 kgCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?a)1 kgb)1.5 kgc)2 kgd)2.5 kgCorrect answer is option 'B'. Can you explain this answer?.
Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?a)1 kgb)1.5 kgc)2 kgd)2.5 kgCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?a)1 kgb)1.5 kgc)2 kgd)2.5 kgCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?a)1 kgb)1.5 kgc)2 kgd)2.5 kgCorrect answer is option 'B'. Can you explain this answer?.
Solutions for Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?a)1 kgb)1.5 kgc)2 kgd)2.5 kgCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?a)1 kgb)1.5 kgc)2 kgd)2.5 kgCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?a)1 kgb)1.5 kgc)2 kgd)2.5 kgCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?a)1 kgb)1.5 kgc)2 kgd)2.5 kgCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?a)1 kgb)1.5 kgc)2 kgd)2.5 kgCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?a)1 kgb)1.5 kgc)2 kgd)2.5 kgCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup