Question

A 10.0-kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended from a scale and immersed in water as shown. The 12.0-cm dimension is vertical, and the top of the block is 5.00 cm below the surface of the water. (a) What are the magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water? (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.

Answer 1

Answer:

First find volume of block

$v = 12 \times 10 \times 10 = 1200 {cm}^{3}$

Now density

$\rho = \frac{10000}{1200} = \frac{25}{3} \frac{gm}{cm {}^{3} }$

Now volume of block inside water

$= 10 \times 10 \times 5 \\ = 500 {cm}^{3}$

Buoyancy force

$b= \rho \: v(inside)g = \frac{25}{3} .1200.981 \: dyen$

force below is gravition

= mg

Spring reading will be

f = mg - b