An alphabet ' $T$ ' made of two similar thin uniform metal plates of each length ' $L$ ' and width ' $a$ ' is placed on a horizontal surface as shown in the figure. If the alphabet is vertically inverted, the shift in the position of its centre of mass from the horizontal surface is

• Correct answer
  A

  $\frac{L-a}{2}$
• B

  $\frac{a-L}{2}$
• C

  $L-\frac{a}{2}$
• D

  $\frac{L}{2}-a$

Solution

$\begin{aligned} & \text { (a) } A_1=L a, A_2=L a \\ & y_1=\frac{L}{2}, y_2=L+\frac{a}{2} \\ & \therefore y_{c m}=\frac{A_1 y_1+A_2 A_2}{A_1+A_2} \\ & =\frac{\frac{L}{2}+L+\frac{a}{2}}{2} \\ & =\left(\frac{3 L}{4}+\frac{a}{4}\right) \end{aligned}$
Centre of mass from top, $\begin{aligned} & \mathrm{y}_{\mathrm{cm}}^{\prime}=(\mathrm{L}+\mathrm{a})-\left(\frac{3 \mathrm{~L}}{4}+\frac{\mathrm{a}}{4}\right) \\ & =\left(\frac{\mathrm{L}}{4}+\frac{3 \mathrm{a}}{4}\right) \end{aligned}$ $\therefore$ Shift in centre of mass, $\begin{aligned} & \Delta \mathrm{y}_{\mathrm{cm}}=\mathrm{y}_{\mathrm{cm}}-\mathrm{y}_{\mathrm{cm}}^{\prime}=\left(\frac{3 \mathrm{~L}}{4}+\frac{\mathrm{a}}{4}\right)-\left(\frac{\mathrm{L}}{4}+\frac{3 \mathrm{a}}{4}\right) \\ & =\frac{\mathrm{L}}{2}-\frac{\mathrm{a}}{2}=\frac{\mathrm{L}-\mathrm{a}}{2} \end{aligned}$

Hence, the correct answer is option: 'A'.