A uniform pole AB of weight 5w and length 8a is suspended horizontally by tho strings attached to it at C and D where AC=DB=a. A
body of weight 9w hangs vertically from the pole at E where ED
=2a.
Calculate the tension in each string.
=2a.
Calculate the tension in each string.
1 answer:
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Answer:
a) 9.72 mm
b) 4.86 mm
Explanation:
wave length of light λ is 580 nm = 580 \times 10⁻⁹ m
Width of slit d = 0.215\times 10⁻³ m
Distance of screen D = 1.8 m.
Width of one fringe =
Putting the values we get fringe width
=
=4.86 mm.
a) Width of central maxima = 2 times width of one fringe
= 2 times 4.86
=9.72 mm
b) width of each fringe except central fringe is same , no matter what the order is.Only brightness changes .
So width of either of the two first order bright fringe will be same and it will be
= 4.86 mm.