InterviewSolution
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InterviewSolution
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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 - 5 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 xx 10^(3)) xx 10` `= 15 N`. The upthrust when the liquid level has decended by `x cm`. `= 100 (10 - x). 10^(-6). 10^(3). 10` `= 10 - x` Newton. `:.` Net downward force on the block `= 15 - (10 - x) = 5 + x = "Tansion" T` `:.` The wire will break when `5 + x = 7`. i.e. when `x = 2cm` Let the level decends by `2 cm` in `t` time then `2t = (200 - 100).2` `t = 100 sec`. ........Ans. `20x = 10 :. x = 5` |
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