Question

a block of mass 4kg and volume 5×10^-4m3 is suspended by a spring balance in a lift which is accelerating. the apparent weight shown by the spring balance is 3 kg .now the block is immersed in water in a container inside the lift .the apparent weight in kg shown by the spring balance is

Answer 1

apparent weight is less than actual weight that mean , the lift is moving downward with acceleration a.
therefore,
apparent weight = m(g-a)
3g = 4 (g-a)
now,
g-a = 3g/4

now when the body is immersed in water then buoyant force will further decrease it's weight
therefore,
buoyant force = vol. × density of water× acceleration
buoyant force= 5 ×
${10 }^{ - 4}$
×
${10}^{3}$
× (g-a)

=5×
${10}^{ - 1}$
×3g/4

=0.375 kg

now the total weight loss by it is
= 3 - 0.375 = 2.625 kg

thus, apparent weight is 2.625kg

Answer 2

Answer:yaa

Explanation: