Two poles are separated by a distance of 3.14 | vedclass

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(A) Given, distance between poles $(d) = 3.14 \, m$.Resolving power $(	heta) = 1 \, min = (1/60)^{\circ} = (1/60) 	imes (\pi/180) \, radians$.Let th

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$10.8 \, km$

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(A) Given, distance between poles $(d) = 3.14 \, m$.Resolving power $(	heta) = 1 \, min = (1/60)^{\circ} = (1/60) 	imes (\pi/180) \, radians$.Let the maximum distance from which the two poles can be identified distinctly be $x$.Using the formula for small angles, $	heta = d/x$ (where $	heta$ is in radians).$	heta = 1 \, min = (1/60) 	imes (\pi/180) \, rad$.Substituting the values: $(1/60) 	imes (\pi/180) = 3.14 / x$.Since $\pi \approx 3.14$, we have $(1/60) 	imes (3.14/180) = 3.14 / x$.$1 / (60 	imes 180) = 1 / x$.$x = 60 	imes 180 = 10800 \, m$.$x = 10.8 \, km$.

Two poles are separated by a distance of $3.14 \, m$. The resolving power of the human eye is $1 \, min$ of an arc. The maximum distance from which one can identify the two poles distinctly is

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