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A bar of length 1 m is supported at its two ends. The breadth and depth of the bar are 5 cm and 0.5 cm., respectively , A body of mass 0.1 is suspended at the centre of the bar. Calculate the depression produced in the bar. |
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Answer» Solution :Data : L = 1 m, b = 5 cm= ` 5 xx 10^(-2) m .d = 0.5 cm = 5 xx 10^(-3) ` m `M = 0.1 KG,Y = 2 xx 10^(11) N//m^(2)` Load, W = MG = ` 0.1 xx 9.8 = 0.98 N` The DEPRESSION (sag) produced in the bar. ` delta = W/(4by) (l/d)^(3) = 0.98/(4 (5 xx 10^(-2)) (2xx 10^(11)) . (1/(5 xx 10^(-3)))^(3)` ` = 0.98/(4xx 10^(16) . (2 xx 10^(2))^(3) = (0.98 xx 8 xx 10^(6))/( 4 xx 10^(10)) = 1.96 xx 10^(-4) m = 0.196 mm` |
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