CBSE Class 11-science Answered
a block of mass 2 kg and specific gravity 5/2 is attached with spring of spring constant=100n/mand is half dipped in water if extension is 1cm what is the force exerted by bottom of tank on block
Asked by SELVA | 30 May, 2019, 04:36: PM
Expert Answer
Net force stretching the spring F = weight of block - upward buoyant force = mg - FB ............... (1)
Spring constant = 100 N/m = 1 N/cm
Hence if extension of spring is 1 cm, Net force = 1 N
Hence from eqn.(1), we get upward buoyant force FB = mg +1 = 2×9.8+1 = 20.6 N
Answered by Thiyagarajan K | 30 May, 2019, 05:00: PM
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