Question

Please explain me the answer! Question is shown in attacent..

Answer 1

according to attachment option (4) is correct

➡because according to Archimedes principle,loss in weight of body when emersed underwater is equal to up-thrust of water. so,

w1-w2=vρg.............(1)

where v is volume,ρ is density and g is acceleration due to gravity.

➡body float with some emersed part under water. let V be volume of body and v be volume outside water.

• volume of body underwater=V-v
• weight of liquid displace=(V-v)ρg
• weight of body=Vσg

for float,

weight of liquid displace by submerged part=weight of body

(V-v)ρg=Vσg........... (2)

from equation (1) and (2) if density of ball and liquid is not same then total reading of weighting is sum of weight of ball and liquid and container