An object of mass 400g and density 600kgm^-3 is suspended with a sting so that half of it is immersed in paraffin of density 900kgm^-3. The tension in the string is

Efranklyn

4 years ago

The resolves forces on the body is given as
T + U = Mg
T = Mg - U
But the Upthrust is given by
U = mfd ×g = d × Vfd × g
Then the volume of the body
V = 0.4/600 = 0.000666667m3
But from the question half is immersed meaning that
Vfd = 0.5× V
Vfd = 0.5 × 0.000666667m3
Vfd = 0.000333333m3
Now
U = 900× 0.00033333 ×gravity
U = 3N
Tension = 0.4×10 - 3
Tension = 1N
I hope I have cleared myschool

Somysoph

5 years ago

The answer should be 1N because tension = Wair-Wparaffin. =4- 3.0015=0.9985 approximately=1N

chijioke7

7 years ago

If the body weighs 0.4kg which is 4N, shouldnt the tension on the string be 1N?

ozioma122

4 months ago

All these density and or equilibrium in liquids sef tire me maybe bekos dey no teach am for the school wey i dey attend

ozioma122

4 months ago

pls who can put me though this topic because i am really lost

OdegokeIfeoluwa

2 months ago

The formula is Actual weight= Apparent weight (tension) + Upthrust. The upthrust which is gotten as 3N using density of liquid * volume of solid* acceleration due to gravity. The actual weight is simply mass * acceleration due to gravity which is 4N. Then 4N - 3N= 1N which is the correct value of tension.

Annabel

2 months ago

A is the correct answer not B