Archimedes' principle tells us that the upwards force on an object immersed in a fluid is equal to the weight of fluid displaced.
So let's take your initial experiment. You don't tell us the volume of your object, but you do give its weight, $W_g = 100$ g, and density, $\rho_g = 2.6$ g/cm$^3$, so the volume is:
$$ V = \frac{W_g}{\rho_g} = 38.46 cm^3 $$
When you put this in water the volume of water displaced is the same as the above volume, so the mass of water displaced is:
$$ M_w = V\rho_w = \frac{W_g}{\rho_g} \rho_w = 38.46g $$
and hence the effective mass of your glass object is 100 - 38.46 = 61.54g as you say.
If the density of water changes to 1.010 g/cm$^{3}$ just use the equation above but set $\rho_w = 1.010$, and you will indeed get the effective weight of the glass equal to 61.15g. Though if you want to be really accurate you should note that changing the temperature will change the density of the glass as well, so $\rho_g$ would be slightly different as well.
Okay. Firstly, I would like to point out that you are mixing two very different concepts here:
(1) Variation in the value of gravity $g$ as the distance from the surface of the earth changes.
(2) True and apparent weight
(1) Variation in the value of gravity
Alright. Variation in gravity. Firstly, lets get clear on the value of $g$. What exactly is $g$? It's like this: Suppose you are somewhere. Maybe sitting somewhere having pizza or flying in the sky. The earth applies a force on you. Let's call this force $F$. Then the value of $g$ is simply defined as $F/m$. That's it.
Now suppose the radius of earth is $R$ and you are at distance $d$ from the surface. (Note, from surface of the earth, not the center.) The force applied on you by the earth is
$$F = \cfrac{GM_em}{(R+d)^2}$$
So, now,
$$g = F/m = \cfrac{GM_e}{(R+d)^2}$$
Have a look at it. The value of $g$ indeed depends on $d$, your distance from the surface of the earth. But, near the surface of the earth, $d<<R$, so we can approximate the above expression to
$$g = F/m = \cfrac{GM_e}{R^2}$$
which is independent of $d$. But note that it is valid only for small values of $d$.
(2) True and apparent weight
Okay. Answer to the next part of the question. True and apparent weight. True weight is simply weight. What is your true weight? It's simply $mg$. Mass multiplied by gravity. End of story.
Now, Apparent weight. I'll denote it by $W_A$. It's defined as
$$W_A = N$$
where $N$ is the normal force in the direction opposite to the direction of gravity. That is away from the center of the earth. You may be standing and someone may be trying to push you horizontally. That normal reaction force doesn't count. Only the vertical Normal Force counts.
So suppose you jump from the top of the building because your dog died. You are falling. Your '(True) Weight' is simply $mg$. Your Apparent weight is $0$. Because there is no normal force applied on you currently. (Offcourse the ground will apply one hell of a normal force when you finally reach it.)
Now suppose you are standing in an elevator at rest. True weight, offcourse is $mg$. But Apparent weight is also $mg$. Because you are at rest, $N = mg$.
Elevator moving with constant speed: $N = mg$
Suppose the magnitude of elevator's acceleration is $|a|$.
Elevator moving upwards, and slowing down: $N = mg - m|a|$
Elevator moving upwards, and increasing speed: $N = mg + m|a|$
Elevator moving downwards, and slowing down: $N = mg + m|a|$
Elevator moving downwards, and increasing speed: $N = mg - m|a|$
So, why did they introduce the concept of Apparent Weight. Apparent weight is the weight you 'feel'. Think about it! When you are falling, you feel weightlessness. Hence Apparent Weight is $0$. When in an elevator with moving upwards with increasing speed, you feel heavier. Hence more is the Apparent Weight!
(3) $d/R$ ratio
The ratio $d/R$ where weight would be 1% lesser:
$$\cfrac{GM_e}{(R+d_1)^2} = 0.99 \cfrac{GM_e}{R^2}$$
Solve it for $d_1$. That's your answer!
Best Answer
No, there is no universal formula to calculate the apparent weight. It all depends on the situation and how you set the definition for the apparent weight.
The problem is that, to the best of my knowledge, there is no formal definition for the apparent weight. See related discussion: Is there a formal definition for apparent weight?
Not true. The net force on the object standing still on the ground is zero. Your definition implies its’s (apparent) weight is zero, which is not true.
The most important thing is to set the definition for the apparent weight. I will discuss here three different scenarios, each with different definition. Once we set the definition, the apparent weight is easily calculated from the free-body diagram.
Let's consider scenarios when object is (i) on the floor, (ii) in free fall, and (iii) in a fluid.
Object on the floor
The apparent weight is usually defined as the force that the floor exerts on the body. This is somewhat intuitive since this is what a scale would show if the object was standing on one.
Now let's consider the most general scenario - an object in an elevator that moves (accelerates) upwards or downwards. For a person standing on the ground (inertial reference frame), the net force acting on the object in the elevator is:
$$F_\text{net} = m \cdot a \tag 1$$
where $m$ is mass of the object. Note that the mass is property of a body which does not depend on the frame of reference.
The free-body diagram would show that there are only two forces exerted on the object: (i) force that elevator floor exerts on the object $F_\text{f/o}$ in upward direction, and (ii) gravitational force that Earth exerts on the object $F_\text{e/o}$ in downward direction:
$$F_\text{f/o} - F_\text{e/o} = F_\text{net}$$
where positive acceleration $a$ is taken to be in upward direction, and the gravitational force is
$$F_\text{e/o} = m \cdot g$$
The apparent weight equals $F_\text{f/o}$ force which is
$$F_\text{f/o} = F_\text{net} + F_\text{e/o} = m \cdot (g + a)$$
When the elevator accelerates:
Object in free fall
The apparent weight is usually defined such that the apparent weight of an object in free fall is zero. Note that this is very loose definition when we consider the free fall with drag!
When the body just starts falling, the velocity is zero and the free body diagram would show only one force acting on the object - gravitational force that pulls the object down. The velocity starts increasing and the drag starts acting on the object which will eventually be equal in magnitude an opposite in direction to the gravitational force. When this happens, the object's acceleration drops to zero and the object remains falling at the terminal velocity. See related discussion: https://physics.stackexchange.com/a/681282/149541
What is the apparent weight of the object when it reaches the terminal velocity? Unfortunately, the above definition does not cover this case! You could say it is zero or redefine the apparent weight.
Object in a fluid
The apparent weight is usually defined as the gravitational force minus the buoyancy. When the body floats its apparent weight is said to be zero. Note that this is the definition found in most textbooks.
The closest you can get to the intuition is to imagine holding an object submerged in a fluid. The force you need to apply to prevent the object from sinking equals gravitational force minus buoyancy.
What is the apparent weight of a sinking body? This example is very similar to the freely falling object, i.e. the free-body diagram would have three forces: gravitational force, buoyancy, and drag. Note that buoyancy and drag are much larger in fluid than in air. The definition above does not cover the sinking body case, i.e. by definition it is still gravitational force minus buoyancy. Force required to stop the body from sinking further still equals gravitational force minus buoyancy.
To conclude, when asked to find the apparent weight, you should always ask for its definition and then apply free-body diagram.