A SLENDER HOMOGENEOUS ROD OF LENGTH 2 L FLOATS PARTLY IMMERSED IN WATER BEING SUPPORTED BY A STRING FASTENED TO ONE OF ITS END . THE SPECIFIC GRAVITY OF THE ROD IS 0.75 .THE LENGTH OF THE THAT EXERTS OUT OF WATER IS 1. L 2. L/2 3. L/4 4. L/3
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Option 2... L/2
let h be height of rod inside water. A = area of cross section.
d = density of rod, d_w density of water
mass of rod = mass of water displaced
d * A * 2L = d_w * h * A
h = 2L * d/d_w = 2L * 0.75 = 3L/2
Height of the rod outside water = L/2.
let h be height of rod inside water. A = area of cross section.
d = density of rod, d_w density of water
mass of rod = mass of water displaced
d * A * 2L = d_w * h * A
h = 2L * d/d_w = 2L * 0.75 = 3L/2
Height of the rod outside water = L/2.
kvnmurty:
click on red heart thanks above
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