Question

A cylindrical object of outer diameter 20 cm and mass 2 kg floats in water with its axis vertical. If it is slightly depressed

Answer 1

please let us know when you get here I will try to call you later today and will not be in school on Friday and Saturday and I am not able or willing and call you tomorrow to get it done by Friday I will try and get a good night and a very.

Answer 2

$<b><i>As per as the Principle of the Flotation,</b>$

$<b>Weight of the cylindrical object = Weight of the liquid displaced.</b>$

$<b>Weight of the cylindrical object = mass × g</b>$

= $<b>2 × g</b>$

$<b>Weight of the Fluid Displaced = Upthrust</b>$

= Volume of of solid immersed × density of water × g

=  πr²h × 1 × g

$<b>where, h is the height and r is the radius of the solid immersed in a water.</b>$

∴ Weight of the fluid displaced =  π(20)²h × 1 × g

2000 × g = π(10)²h × 1 × g

$<b>2000 = π(10)²h</b>$

2000 = π × 100 × h

∴ $<b>h = 20/π cm.</b>$

$<b>Now, Using the formula of the Time Period,</b>$

 T = 2π√(h/g)

T = 2 × 22/7 × √(20/π × 980)

∴ $<b>T = 0.51 seconds.</b>$

$<b>$Hence, the time period of the S.H.M. is 0.51 seconds.

Hope it helps.