A metre stick, of mass 400 g, is pivoted at one end displaced through an angle of 60° . The increase in its potential energy is

The correct option is 1 J.

To find the increase in the potential energy of the metre stick, we need to determine how much its center of mass is raised when the stick is displaced.

Step 1: Identify the given values and convert units
- Length of the metre stick, $L=1\text{m}$
- Mass of the metre stick, $m=400\text{g}=0.4\text{kg}$
- Angle of displacement, $\theta =\mathrm{60°}$
- Acceleration due to gravity, $g\approx 10{\text{m/s}}^2$

Step 2: Find the position of the center of mass
For a uniform metre stick, the center of mass is located at its midpoint. Since the stick is pivoted at one end, the distance of the center of mass from the pivot is:
$d=\frac{L}{2}=\frac{1\text{m}}{2}=0.5\text{m}$

Step 3: Calculate the vertical rise of the center of mass
Initially, the stick hangs vertically downward, so the center of mass is at a depth of $d$ below the pivot.
When the stick is pivoted and displaced through an angle $\theta$, the vertical depth of the center of mass below the pivot becomes $d\cos \theta$.
The increase in vertical height ($\Delta h$) of the center of mass is:
$\Delta h=d-d\cos \theta =d(1-\cos \theta )$
Substituting the values:
$\Delta h=0.5\times (1-\cos \mathrm{60°})$
Since $\cos \mathrm{60°}=0.5$:
$\Delta h=0.5\times (1-0.5)=0.5\times 0.5=0.25\text{m}$

Step 4: Calculate the increase in potential energy
The increase in gravitational potential energy ($\Delta U$) is given by:
$\Delta U=mg\Delta h$
Substituting the values:
$\Delta U=0.4\times 10\times 0.25=1\text{J}$
Thus, the increase in potential energy is 1 J.