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Figure shows a cubical block of side 10 cm and relative density 1.5 suspended by a wire of cross sectional area `10^(-6) m^(2)`. The breaking stress of the wire is `7 xx 10^(6) N//m^(2)`. The block is placed in a beaker of base area `200 cm^(2)` and initially i.e. at t = 0, the top surface of water & the block coincide. There is a pump at the bottom corner which ejects `2 cm^(3)` of water per sec. Find the time at which the wire will break. |
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Answer» Correct Answer - 100 sec Breaking force = Breaking stress `xx` area of wire `=7 xx 10^(6) xx 10^(-6) = 7N` weight of block = vol `xx` density `xx g` `=10^(-3) xx(1.5 xx10^(3)) xx10 = 15 N` The upthrust when the liquid level has decended by `xx` cm `100 (10-x)10^(-6)10^(3)10=10-x` Newton `:.` Net downward force on the block `= 15 -(10 -x)=5 + x` = Tension T `:.` The wire will break when `5 + x = 7` i.e. when x = 2 cm Let the level descends by 2 cm in t time then `2t =(200-100) 2 t = 100 sec`. |
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