Starting from rest, a body slides down a 45° inclined plane in twice the time it takes to slide the same distance in the absence of friction. What is the coefficient of friction between the body and the inclined plane ?

For a frictionless surface, o =g sin 45° = g/√2

$\therefore \mathrm{l}=\frac{1}{2}\times \frac{\mathrm{g}}{\sqrt{2}}\times {\mathrm{t}}_2^2$   …(1)

In presence of friction.

$\mathrm{a}=\frac{\mathrm{g}}{\sqrt{2}}-\frac{\mathrm{\mu g}}{\sqrt{2}}=\frac{\mathrm{g}}{\sqrt{2}}(1-\mathrm{\mu })$

$\therefore \mathrm{l}=\frac{1}{2}\times \left[\frac{\mathrm{g}}{\sqrt{2}}(1-\mathrm{\mu })\right]\times {\mathrm{t}}_1^2$   …(2)

Dividing eq. (2) by eq. (1), we get

$1=(1-\mathrm{\mu })\frac{{{\mathrm{t}}_1}^2}{{{\mathrm{t}}_2}^2}=(1-\mathrm{\mu })\frac{{\left(2{\mathrm{t}}_2\right)}^2}{{{\mathrm{t}}_2}^2}=(1-\mathrm{\mu })\times 4$

$\text{ or }(1-\mathrm{\mu })=1/4\text{ or }\mathrm{\mu }=3/4$.