Question

Urgent!

Answer 1

mass of iron block given as

$m_1 = 1.90 kg$

density of iron block is

$\rho = 7860 kg/m^3$

now the volume of the iron piece is given as

$V = \frac{m}{\rho}$

$V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3$

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

$F_b = \rho_L V g$

here we know that

$\rho_L$ = density of liquid = 916 kg/m^3

$F_b = 916* 2.42 * 10^{-4} * 9.8$

$F_b = 2.17 N$

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

$F_s + F_b = mg$

$F_s + 2.17 = 1.90* 9.8$

$F_s = 16.45 N$

So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

$F_n = F_g + F_b$

$F_n = (1 + 2.50)*9.8 + 2.17$

$F_n = 34.3 + 2.17 = 36.47 N$

So the other scale will read 36.47 N