Calculate the height upto which an insect can crawl up a fixed bowl in the from of a hemisphere of radius r Given coefficient of friction `= 1 sqrt(3)`
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Explanation

B is botton of the bowl of radius r = OB . The insect can crawl up the bowl From B to P through a height
` BA = h `
As is clear from ` R = W cos alpha `
and `F = W sin alpha ` where W is weight of insect and F is force of friction
`mu = (F)/(R) = (Wsin alpha)/(W cos alpha) = tan alpha (1)/(sqrt3) `
In ` triangleOPA , tan alpha = (PA)/(OA) `
`:. tan alpha = sqrt(r^(2) - Y^(2))/(y)= (1)/sqrt3 `
` (r^(2) - y^(y))/(y^(2)) = (1)/(3) `
` y^(2) = (3 r^(2))/(4) , y = (sqrt3 r)/(2) `
` h = BA = OB = OB - OA = r - y `
` h = r - (sqrt3 r )/(2) = 0 .134 r `
`h = 13 .4 %` of r.