Question

A solid cube of density 3.5 g/cm3 and side length
7 cm is dipped completely into water. Its magnitude
of apparent weight, when completely immersed in
water, will be [Take g = 10 m/s2]
(1) Greater than 12.005 N
(2) Less than 12.005 N
(3) Equal to 12.005 N
(4) 35 N

Answer 1

Answer:

Option (2) less than 12.005N

Explanation:

Since weight of cube is 12.005N and there is a Buyounce Force of water is acting on cube against gravity therefore it's weight will get reduced.

Answer 2

Answer:

(2) Less than 12.005 N

Explanation:

Given that

density ,ρ=3.5 g/cm³

length , a= 7 cm

V = a ³ = 7³ = 343 cm ³

g = 10 m/s²

When the block completely dipped in the water then

Lets take apparent weight = W'

The actual weight =W

W = m g =  ρ  V g

W= 3.5 x 343 x 10 x 100 = 1.2005 N                      ( 1 kg = 1000 g)

Bouncy force will act up ward

$F_b =\rho_wV.g$

Now by balancing force

Fb+ W ' = W

W' = W - Fb

W' = 1.2005 - Fb

So apparent weight is less than 1.2005 N.

(2) Less than 12.005 N