Question

a silver ingot weighing 2.1 kg is held by string so as to be completely immersed in a liquid of relative density 0.8 the relative of silver is 10.5 the tension in the string in kg wt is

Answer 1

Given.

Relative density of the liquid = 0.8

Relative density of the silver = 10.5

Let the volume of the silver ingot be V.

Then 2.1 kg = V × 10³ kg/m³

V = 2.1 × 10⁻³

Now , mass of the liquid displaced = V × ( 0.8) × 10³

= 2.1 × 10⁻³ × 0.8 × 10³

= 2.1 × 0.8

= 1.68 kg.

Hence Buyont force = 1.68 g. = 16.8 N

Now ,

$T + F_B$ = 2 g.

T + 16.8 = 20

T = 3.2 N

The tension in the string is 3.2 N

{Refer to the attachment for diagram}

Hope it Helps :-)

Answer 2

The solution is provided in the attachment.
PLEASE MARK AS BRAINLIEST