Question

An iron cube of side 5 cm and of R.D 7.6 is suspended by a
thread in
brine solution of relative density 1.2. Find Tension in thread.

​

Answer 1

Explanation:

Given An iron cube of side 5 cm and of R.D 7.6 is suspended by a

thread in  brine solution of relative density 1.2. Find Tension in thread.

•  According to question volume of iron cube = 5 cm^3 = 125 x 10^-6 m^3
• Density of iron = 7.6 g/cm^3 = 7600 kg/m^3
• Mass of iron cube m = volume x density
•                                   = 125 x 10^-6 m^3 x 7600 kg/m^3
•                                    = 0.95 kg
• Now weight of iron cube in air = mg
•                                                   = 0.95 x 10
•                                                    = 9.5 N
• So volume of liquid displaced = volume of iron cube = 125 x 10^-6 m^3
• Mass of liquid displaced = volume x density of liquid
•                                          = 125 x 10^-6 m^3 x 1200 kg/m^3
•                                            = 0.15 kg
• Weight of the liquid displaced = 0.15 kg x 10 m/s^2
•                                                   = 1.5 N which is upthrust
• Weight of iron cube in solution = weight of iron cube in air – upthrust
•                                                    = 9.5 – 1.5
•                                                      = 8 N
• Therefore tension in the thread is 8 N

Reference link will be

https://brainly.in/question/2707215

Answer 2

Answer:

Volume of liquid displaced = Volume of metal cube = 125 x 10−6 m3. = 125 x 10−6 m3. 1200 kg/m3. Therefore tension in the string = 9.75 N.

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