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A square of side 4 cm and of uniform thickness is divided into four equal squares . If one of them is cut off (OECF) , then the position of the centre of mass of the remaining portion from O is (1) 1/√3 cm (2) 2/√3 cm (3) 5/√3 cm (4) √2/3 cm?
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A square of side 4 cm and of uniform thickness is divided into four eq...
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A square of side 4 cm and of uniform thickness is divided into four eq...
Solution:
Given, a square of side 4 cm is divided into four equal squares.
Area of the square = (side)^2 = 4^2 = 16 cm^2
Area of each small square = (16/4) cm^2 = 4 cm^2
One of the small squares is cut-off as shown below:
[Insert image of the square with one small square cut-off]
The position of the centre of mass of the remaining portion from O is to be found.
Let the coordinates of O be (0,0) and the side of the small square be 'a'.
The centre of mass of the remaining portion can be found by finding the centroid of the remaining portion.
Let the coordinates of the centroid be (x,y).
The coordinates of the centroid of a square of side 'a' are (a/2, a/2).
Therefore, the coordinates of the centroid of the remaining portion are:
x = (2a + 3a/2)/3 = 7a/6
y = (2a + 3a/2)/3 = 7a/6
Now, we need to find the value of 'a' for the given square.
From the given information, we can see that the diagonal of the square is equal to the sum of the diagonals of the small squares.
Diagonal of the square = 4√2 cm
Diagonal of each small square = a√2 cm
Therefore, 4a√2 = 4√2
a = 1 cm
Substituting this value of 'a' in the coordinates of the centroid, we get:
x = 7/6 cm
y = 7/6 cm
Hence, the position of the centre of mass of the remaining portion from O is (7/6, 7/6) cm.
But the question asks for the distance of the centre of mass from O, which can be found using the distance formula:
d = √[(x-0)^2 + (y-0)^2]
d = √[(7/6)^2 + (7/6)^2]
d = √(2/3)
Therefore, the position of the centre of mass of the remaining portion from O is √(2/3) cm.
Hence, the correct option is (4) √(2/3) cm.
Given, a square of side 4 cm is divided into four equal squares.
Area of the square = (side)^2 = 4^2 = 16 cm^2
Area of each small square = (16/4) cm^2 = 4 cm^2
One of the small squares is cut-off as shown below:
[Insert image of the square with one small square cut-off]
The position of the centre of mass of the remaining portion from O is to be found.
Let the coordinates of O be (0,0) and the side of the small square be 'a'.
The centre of mass of the remaining portion can be found by finding the centroid of the remaining portion.
Let the coordinates of the centroid be (x,y).
The coordinates of the centroid of a square of side 'a' are (a/2, a/2).
Therefore, the coordinates of the centroid of the remaining portion are:
x = (2a + 3a/2)/3 = 7a/6
y = (2a + 3a/2)/3 = 7a/6
Now, we need to find the value of 'a' for the given square.
From the given information, we can see that the diagonal of the square is equal to the sum of the diagonals of the small squares.
Diagonal of the square = 4√2 cm
Diagonal of each small square = a√2 cm
Therefore, 4a√2 = 4√2
a = 1 cm
Substituting this value of 'a' in the coordinates of the centroid, we get:
x = 7/6 cm
y = 7/6 cm
Hence, the position of the centre of mass of the remaining portion from O is (7/6, 7/6) cm.
But the question asks for the distance of the centre of mass from O, which can be found using the distance formula:
d = √[(x-0)^2 + (y-0)^2]
d = √[(7/6)^2 + (7/6)^2]
d = √(2/3)
Therefore, the position of the centre of mass of the remaining portion from O is √(2/3) cm.
Hence, the correct option is (4) √(2/3) cm.
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A square of side 4 cm and of uniform thickness is divided into four equal squares . If one of them is cut off (OECF) , then the position of the centre of mass of the remaining portion from O is (1) 1/√3 cm (2) 2/√3 cm (3) 5/√3 cm (4) √2/3 cm?
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A square of side 4 cm and of uniform thickness is divided into four equal squares . If one of them is cut off (OECF) , then the position of the centre of mass of the remaining portion from O is (1) 1/√3 cm (2) 2/√3 cm (3) 5/√3 cm (4) √2/3 cm? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A square of side 4 cm and of uniform thickness is divided into four equal squares . If one of them is cut off (OECF) , then the position of the centre of mass of the remaining portion from O is (1) 1/√3 cm (2) 2/√3 cm (3) 5/√3 cm (4) √2/3 cm? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A square of side 4 cm and of uniform thickness is divided into four equal squares . If one of them is cut off (OECF) , then the position of the centre of mass of the remaining portion from O is (1) 1/√3 cm (2) 2/√3 cm (3) 5/√3 cm (4) √2/3 cm?.
A square of side 4 cm and of uniform thickness is divided into four equal squares . If one of them is cut off (OECF) , then the position of the centre of mass of the remaining portion from O is (1) 1/√3 cm (2) 2/√3 cm (3) 5/√3 cm (4) √2/3 cm? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A square of side 4 cm and of uniform thickness is divided into four equal squares . If one of them is cut off (OECF) , then the position of the centre of mass of the remaining portion from O is (1) 1/√3 cm (2) 2/√3 cm (3) 5/√3 cm (4) √2/3 cm? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A square of side 4 cm and of uniform thickness is divided into four equal squares . If one of them is cut off (OECF) , then the position of the centre of mass of the remaining portion from O is (1) 1/√3 cm (2) 2/√3 cm (3) 5/√3 cm (4) √2/3 cm?.
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