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
Answers
Answered by
0
ans: option2..... L/2
mass of rod = mass of water displaced
d * 2L * A = h * d_w * A
=> h = 2L * d/d_w = 2L * 0.75
height of rod outside water = 2L - h = L/2
mass of rod = mass of water displaced
d * 2L * A = h * d_w * A
=> h = 2L * d/d_w = 2L * 0.75
height of rod outside water = 2L - h = L/2
Similar questions