Question

A wooden cube just floats inside the water when a 200 gm mass is placed on it. When the mass is removed the cube is 2 cm above the water level. The side of the cube is:
(a) 5 cm (b) 10 cm (c) 15 cm (d) 20 cm

Answer 1

Let the side of the cube be S, density of water be d, mass of cube be m, acceleration due to gravity be g.

When the weight is on the block,

Using Archimedes’ Principle ,

Weight of the block + Weight of the mass = Bouyant force (B1)

(Since B1 = volume of water displaced = S3×d×gS3×d×g )

=> m×g+0.2×g=S3×d×gm×g+0.2×g=S3×d×g

=> m=S3×d+0.2m=S3×d+0.2 …. 1

When the mass is removed,

Weight of the block = Bouyant force (B2)

(Since B2 = (S−0.02)S2×d×g(S−0.02)S2×d×g)

=> m×g=(S−0.02)S2×d×gm×g=(S−0.02)S2×d×g

=> m=(S−0.02)S2×dm=(S−0.02)S2×d … 2

From 1 and 2,

S3×d+0.2=(S−0.02)S2×dS3×d+0.2=(S−0.02)S2×d

=>0.2=0.02×S2×d0.2=0.02×S2×d

Taking d = 1000kg/m31000kg/m3

S2=0.01S2=0.01

=> S = 0.1 m = 10cm

Answer 2

heyy bro ur answer..........

Let the side of the cube be S, density of water be d, mass of cube be m, acceleration due to gravity be g.

When the weight is on the block,

Using Archimedes’ Principle ,

Weight of the block + Weight of the mass = Bouyant force (B1)

(Since B1 = volume of water displaced = S3×d×gS3×d×g )

=> m×g+0.2×g=S3×d×gm×g+0.2×g=S3×d×g

=> m=S3×d+0.2m=S3×d+0.2 …. 1

When the mass is removed,

Weight of the block = Bouyant force (B2)

(Since B2 = (S−0.02)S2×d×g(S−0.02)S2×d×g)

=> m×g=(S−0.02)S2×d×gm×g=(S−0.02)S2×d×g

=> m=(S−0.02)S2×dm=(S−0.02)S2×d … 2

From 1 and 2,

S3×d+0.2=(S−0.02)S2×dS3×d+0.2=(S−0.02)S2×d

=>0.2=0.02×S2×d0.2=0.02×S2×d

Taking d = 1000kg/m31000kg/m3

S2=0.01S2=0.01

=> S = 0.1 m = 10cm...............

hope this will help u..............@kundan