Questions

# Half of the uniform rectangular plate of Length ‘L’ is made up of material of density $\text{'}{\mathrm{d}}_{1}\text{'}$ and the other half  with density ${d}_{2}$ . The perpendicular distance of center of mass from AB is

a
$\frac{2{\mathrm{d}}_{1}+3{\mathrm{d}}_{2}}{{\mathrm{d}}_{1}+{\mathrm{d}}_{2}}×\frac{\mathrm{L}}{4}$
b
$\frac{{\mathrm{d}}_{1}+3{\mathrm{d}}_{2}}{{\mathrm{d}}_{1}+{\mathrm{d}}_{2}}×\frac{\mathrm{L}}{4}$
c
$\frac{3{\mathrm{d}}_{1}}{{\mathrm{d}}_{1}+{\mathrm{d}}_{2}}×\frac{\mathrm{L}}{4}$
d
$\frac{3{\mathrm{d}}_{2}}{{\mathrm{d}}_{1}+{\mathrm{d}}_{2}}×\frac{\mathrm{L}}{4}$

detailed solution

Correct option is B

${\mathrm{m}}_{1}=\frac{\mathrm{L}}{2}×{\mathrm{d}}_{1}$

${X}_{1}=\frac{L}{4}$

${x}_{2}=\frac{3L}{4}$

$=\frac{\frac{\mathrm{L}}{2}×{\mathrm{d}}_{1}×\frac{\mathrm{L}}{4}+\frac{\mathrm{L}}{2}×{\mathrm{d}}_{2}×\frac{3\mathrm{L}}{4}}{\frac{\mathrm{L}}{2}{\mathrm{d}}_{1}+\frac{\mathrm{L}}{2}{\mathrm{d}}_{2}}$

=$\frac{\frac{{d}_{1}L}{4}+\frac{3{d}_{2}L}{4}}{\frac{{d}_{1}L}{2}+\frac{{d}_{2}L}{2}}$

$=\frac{{\mathrm{d}}_{1}+3{\mathrm{d}}_{2}}{{\mathrm{d}}_{1}+{\mathrm{d}}_{2}}×\frac{\mathrm{L}}{4}$