Answer
(a) $13.2N$
(b) $12.9N$
Work Step by Step
(a) We can find the reading on the scale before the string breaks as follows:
$W_1=(m_1+m_2)g$
We plug in the known values to obtain:
$W_1=(1.2Kg+0.15Kg)(9.8m/s^2)$
$W_1=13.2N$
(b) The reading on the scale after the string breaks is given as follows:
Due to the presence of the liquid, the retarding force exerted on the ball is $m_2(g-\frac{g}{4})=\frac{3}{4}m_2g$
$W_2=m_1g+\frac{3}{4}m_2g$
$\implies W_2=(1.2Kg)(9.8m/s^2)+\frac{3}{4}(0.15Kg)(9.8m/s^2)$
$W_2=12.9N$
