269492
A circular hole of radius '\(r\) ' is made in a disk of radius ' \(R\) ' and of uniform thickness at a distance ' \(a\) ' from the centre of the disk. The distance of the new centre of mass from the original centre of mass is
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1 \(\frac{a R^{2}}{R^{2}-r^{2}}\)
2 \(\frac{a r^{2}}{R^{2}-r^{2}}\)
3 \(\frac{a\left(R^{2}-r^{2}\right)}{r^{2}}\)
4 \(\frac{a\left(R^{2}-r^{2}\right)}{R^{2}}\)
Explanation:
\(x=\frac{r^{2} a}{R^{2}-r^{2}}\)
Rotational Motion
269493
Thecentre of mass of the letter \(F\) which is cut from a uniform metal sheet from point \(A\) is
269494
Two identical thin uniform rods of length\(L\) each are joined to form \(T\) shape as shown in the figure. The distance of centre of mass from \(D\) is
269495
Figure shows a square plate of uniform thickness and side length\(\sqrt{2} \mathrm{~m}\). One fourth of the plate is removed as indicated. The distance of centre of mass of the remaining portion from the centre of the original square plate is
1 \(1 / 3 \mathrm{~m}\)
2 \(1 / 2 \mathrm{~m}\)
3 \(1 / 6 \mathrm{~m}\)
4 \(1 / 8 \mathrm{~m}\)
Explanation:
' \(m\) ' be the mass of each part \(3 m \times{ }^{C} C_{2}=m \times c C_{1}\) (or) \(X=\frac{-a d}{A-a}\) a - Area of removed plate A area of original plate \(d\) - distance between centers
Rotational Motion
269545
Four particles, each of mass \(1 \mathrm{~kg}\), are placed at the comers of square of side one meter in the XY plane. If the point of intersection of the diagonals of the square is taken as the origin,the co-ordinates of the center of mass are
269492
A circular hole of radius '\(r\) ' is made in a disk of radius ' \(R\) ' and of uniform thickness at a distance ' \(a\) ' from the centre of the disk. The distance of the new centre of mass from the original centre of mass is
">
1 \(\frac{a R^{2}}{R^{2}-r^{2}}\)
2 \(\frac{a r^{2}}{R^{2}-r^{2}}\)
3 \(\frac{a\left(R^{2}-r^{2}\right)}{r^{2}}\)
4 \(\frac{a\left(R^{2}-r^{2}\right)}{R^{2}}\)
Explanation:
\(x=\frac{r^{2} a}{R^{2}-r^{2}}\)
Rotational Motion
269493
Thecentre of mass of the letter \(F\) which is cut from a uniform metal sheet from point \(A\) is
269494
Two identical thin uniform rods of length\(L\) each are joined to form \(T\) shape as shown in the figure. The distance of centre of mass from \(D\) is
269495
Figure shows a square plate of uniform thickness and side length\(\sqrt{2} \mathrm{~m}\). One fourth of the plate is removed as indicated. The distance of centre of mass of the remaining portion from the centre of the original square plate is
1 \(1 / 3 \mathrm{~m}\)
2 \(1 / 2 \mathrm{~m}\)
3 \(1 / 6 \mathrm{~m}\)
4 \(1 / 8 \mathrm{~m}\)
Explanation:
' \(m\) ' be the mass of each part \(3 m \times{ }^{C} C_{2}=m \times c C_{1}\) (or) \(X=\frac{-a d}{A-a}\) a - Area of removed plate A area of original plate \(d\) - distance between centers
Rotational Motion
269545
Four particles, each of mass \(1 \mathrm{~kg}\), are placed at the comers of square of side one meter in the XY plane. If the point of intersection of the diagonals of the square is taken as the origin,the co-ordinates of the center of mass are
269492
A circular hole of radius '\(r\) ' is made in a disk of radius ' \(R\) ' and of uniform thickness at a distance ' \(a\) ' from the centre of the disk. The distance of the new centre of mass from the original centre of mass is
">
1 \(\frac{a R^{2}}{R^{2}-r^{2}}\)
2 \(\frac{a r^{2}}{R^{2}-r^{2}}\)
3 \(\frac{a\left(R^{2}-r^{2}\right)}{r^{2}}\)
4 \(\frac{a\left(R^{2}-r^{2}\right)}{R^{2}}\)
Explanation:
\(x=\frac{r^{2} a}{R^{2}-r^{2}}\)
Rotational Motion
269493
Thecentre of mass of the letter \(F\) which is cut from a uniform metal sheet from point \(A\) is
269494
Two identical thin uniform rods of length\(L\) each are joined to form \(T\) shape as shown in the figure. The distance of centre of mass from \(D\) is
269495
Figure shows a square plate of uniform thickness and side length\(\sqrt{2} \mathrm{~m}\). One fourth of the plate is removed as indicated. The distance of centre of mass of the remaining portion from the centre of the original square plate is
1 \(1 / 3 \mathrm{~m}\)
2 \(1 / 2 \mathrm{~m}\)
3 \(1 / 6 \mathrm{~m}\)
4 \(1 / 8 \mathrm{~m}\)
Explanation:
' \(m\) ' be the mass of each part \(3 m \times{ }^{C} C_{2}=m \times c C_{1}\) (or) \(X=\frac{-a d}{A-a}\) a - Area of removed plate A area of original plate \(d\) - distance between centers
Rotational Motion
269545
Four particles, each of mass \(1 \mathrm{~kg}\), are placed at the comers of square of side one meter in the XY plane. If the point of intersection of the diagonals of the square is taken as the origin,the co-ordinates of the center of mass are
269492
A circular hole of radius '\(r\) ' is made in a disk of radius ' \(R\) ' and of uniform thickness at a distance ' \(a\) ' from the centre of the disk. The distance of the new centre of mass from the original centre of mass is
">
1 \(\frac{a R^{2}}{R^{2}-r^{2}}\)
2 \(\frac{a r^{2}}{R^{2}-r^{2}}\)
3 \(\frac{a\left(R^{2}-r^{2}\right)}{r^{2}}\)
4 \(\frac{a\left(R^{2}-r^{2}\right)}{R^{2}}\)
Explanation:
\(x=\frac{r^{2} a}{R^{2}-r^{2}}\)
Rotational Motion
269493
Thecentre of mass of the letter \(F\) which is cut from a uniform metal sheet from point \(A\) is
269494
Two identical thin uniform rods of length\(L\) each are joined to form \(T\) shape as shown in the figure. The distance of centre of mass from \(D\) is
269495
Figure shows a square plate of uniform thickness and side length\(\sqrt{2} \mathrm{~m}\). One fourth of the plate is removed as indicated. The distance of centre of mass of the remaining portion from the centre of the original square plate is
1 \(1 / 3 \mathrm{~m}\)
2 \(1 / 2 \mathrm{~m}\)
3 \(1 / 6 \mathrm{~m}\)
4 \(1 / 8 \mathrm{~m}\)
Explanation:
' \(m\) ' be the mass of each part \(3 m \times{ }^{C} C_{2}=m \times c C_{1}\) (or) \(X=\frac{-a d}{A-a}\) a - Area of removed plate A area of original plate \(d\) - distance between centers
Rotational Motion
269545
Four particles, each of mass \(1 \mathrm{~kg}\), are placed at the comers of square of side one meter in the XY plane. If the point of intersection of the diagonals of the square is taken as the origin,the co-ordinates of the center of mass are
269492
A circular hole of radius '\(r\) ' is made in a disk of radius ' \(R\) ' and of uniform thickness at a distance ' \(a\) ' from the centre of the disk. The distance of the new centre of mass from the original centre of mass is
">
1 \(\frac{a R^{2}}{R^{2}-r^{2}}\)
2 \(\frac{a r^{2}}{R^{2}-r^{2}}\)
3 \(\frac{a\left(R^{2}-r^{2}\right)}{r^{2}}\)
4 \(\frac{a\left(R^{2}-r^{2}\right)}{R^{2}}\)
Explanation:
\(x=\frac{r^{2} a}{R^{2}-r^{2}}\)
Rotational Motion
269493
Thecentre of mass of the letter \(F\) which is cut from a uniform metal sheet from point \(A\) is
269494
Two identical thin uniform rods of length\(L\) each are joined to form \(T\) shape as shown in the figure. The distance of centre of mass from \(D\) is
269495
Figure shows a square plate of uniform thickness and side length\(\sqrt{2} \mathrm{~m}\). One fourth of the plate is removed as indicated. The distance of centre of mass of the remaining portion from the centre of the original square plate is
1 \(1 / 3 \mathrm{~m}\)
2 \(1 / 2 \mathrm{~m}\)
3 \(1 / 6 \mathrm{~m}\)
4 \(1 / 8 \mathrm{~m}\)
Explanation:
' \(m\) ' be the mass of each part \(3 m \times{ }^{C} C_{2}=m \times c C_{1}\) (or) \(X=\frac{-a d}{A-a}\) a - Area of removed plate A area of original plate \(d\) - distance between centers
Rotational Motion
269545
Four particles, each of mass \(1 \mathrm{~kg}\), are placed at the comers of square of side one meter in the XY plane. If the point of intersection of the diagonals of the square is taken as the origin,the co-ordinates of the center of mass are
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