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16 Aug 2011 20:33:58 IST
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An insect is crawling up a hemispherical bowl of radius R. If the coefficient of friction is 1/3, then upto what height will it be able to crawl? (a)R/5 (b)R/10 (c)R/20 (d)R/30



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hemang's Avatar

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Joined: 27 Dec 2010
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26 Aug 2011 11:07:37 IST
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let the insect crawl upto a height h.

let the radius be r.

also let the point on the surface of the hemishphere at the height h be P.

the center of circular upper portion be O and the point directly below it on the hemisphere be S.

then, there must be a point N on OS such that NS = h. such that ON = (r-h).

also a triangle is formed by O , N and P such that OP = r. let the vertical angle be x in this rigth angled triangle. that is angle PON is x. and ONP = 90 degrees.

also since the insect at the heighest point will be in equilibrium so, there Wsinx = k(Wcosx) where x is the coefficient of friction or 1/3.

so, on solving we get, tanx = k = 1/3

but tanx = (PN)/(ON)

but, if ON = r-h and PN = sqrt[r2 - (r-h)2]

on putting the values and solving we get,

9[r2-(r-h)2] = (r-h)2

or, 3r = sqrt[10](r-h)

or, (sqrt10r - 3r)/sqrt[10] = h

so, h = 0.05r

or, h = r/20, answer.

 

 

 

Yagyadutt Mishra's Avatar

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Joined: 19 Feb 2009
Posts: 1958
26 Aug 2011 18:37:00 IST
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A good Solution !



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