Given a 12 m long wire and a cube with each edge measuring 1 m, the wire must be divided into 12 pieces, each 1 m long.
Step 1: Each 1 m piece corresponds to one edge of the cube.
Step 2: If we are to minimize the number of cuts, strategically:
Make 1 cut to get 2 pieces of 6 m each.
Cut each 6 m piece into two 3 m pieces (2 cuts total so far).
Finally, cut each 3 m piece into three 1 m pieces (4 cuts in total, as each 3 m cut into three 1 m pieces adds 2 cuts).
Step 3: This method requires a total of 4 cuts. Therefore, the minimum number of cuts required is 4.
A regular dodecagon (12-sided regular polygon) is inscribed in a circle of radius \( r \) cm as shown in the figure. The side of the dodecagon is \( d \) cm. All the triangles (numbered 1 to 12 in the figure) are used to form squares of side \( r \) cm, and each numbered triangle is used only once to form a square. The number of squares that can be formed and the number of triangles required to form each square, respectively, are:
In the given figure, the numbers associated with the rectangle, triangle, and ellipse are 1, 2, and 3, respectively. Which one among the given options is the most appropriate combination of \( P \), \( Q \), and \( R \)?
A wheel of mass \( 4M \) and radius \( R \) is made of a thin uniform distribution of mass \( 3M \) at the rim and a point mass \( M \) at the center. The spokes of the wheel are massless. The center of mass of the wheel is connected to a horizontal massless rod of length \( 2R \), with one end fixed at \( O \), as shown in the figure. The wheel rolls without slipping on horizontal ground with angular speed \( \Omega \). If \( \vec{L} \) is the total angular momentum of the wheel about \( O \), then the magnitude \( \left| \frac{d\vec{L}}{dt} \right| = N(MR^2 \Omega^2) \). The value of \( N \) (in integer) is: