SHMHard
Question
A cylindrical block of wood (density = 650 kg m-3), of base area 30cm2 and height 54 cm, floats in a liquid of density 900 kg m-3. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly) :
Options
A.52 cm
B.65 cm
C.39 cm
D.26 cm
Solution
mg = FB
h = Length of block immerged in water
650 × A × 54 × 10-2g = 900 × A × hg
h = 0.39m = 39 cm. Option (3)
Create a free account to view solution
View Solution FreeMore SHM Questions
The displacement of a particle in S.H.M. is indicated by equation y = 10 sin(20t + π/3) where y is in metres. The v...The phase of a particle in S.H.M at time t is . The following inference is drawn from this...The figure shown the displacement time graph of a executing S.H.M if the time period of oscillation is 2 s the equation ...In the compound pendulum, the minimum period of oscillation will be :...An infinite number of springs having force constants as K, 2K, 4K, 8K, ..... ∞ respectively are connected in serie...