A weight hangs on the hook of the spring dynamometer. The volume of the weight is 130 cubic cm, the weight is 1 kg.

A weight hangs on the hook of the spring dynamometer. The volume of the weight is 130 cubic cm, the weight is 1 kg. What will the dynamometer show if the weight is immersed in water? into gasoline?

A body in a fluid will be acted upon by the force of gravity P = mg (down) and the buoyant force Fа = rVg (up), where V is the volume of the body, r is the specific density of the fluid, and g is the acceleration of gravity.
The response of the dynamometer will be equal to the sum of the action of these forces:
F = mg – grV = g (m-rV).
Indication for weight in water (r = 1000 kg / m ^ 3):
Fw = 10 m / s ^ 2 * (1 kg – 0.00013 m ^ 3 * 1000 kg / m ^ 3) = 10 N – 1.3 N = 8.7 N.
For a weight in gasoline (r = 750 kg / m ^ 3):
Fb = 10 m / s ^ 2 * (1 kg – 0.00013 m ^ 3 * 750 kg / m ^ 3) = 10 N – 0.97 N = 9.03 N.
Answer: weight in water – 8.7 N; weight in gasoline – 9.03 N.