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                                    A composite inclined plane has three different inclined surfaces `AB, BC` and `CD` of heights `1 m` each and coefficients of friction `(1)/(sqrt(3)), (1)/(sqrt(8))` and `(1)/(sqrt(15))` respectively. A particle given an initial velocity at `A` along `AB` transverses the inclined surfaces with uniform speed, reaches `D` in `5s`. The initial speed given is `("in" m//s)` A. `1.6`B. 1.8C. 2.4D. 3 | 
                            
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Answer» Correct Answer - B We know that if the angle of inclination is `tan^(-1) mu` no resultant force acts on the body . `therefore " " AB = (1)/("sin" (tan^(-1) (1)/(sqrt3))) = 2m` `BC = (1)/("sin" (tan^(-1) (1)/(sqrt8))) = 3m` CD = `(1)/("sin" (tan^(-1) (1)/(sqrt15))) = 4m ` `therefore` ABCD = 9m , T = 5 s , v = 1.8 m/s  | 
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