If there was a bubble of air trapped in an object, how would it affect the density? Would the density increase or decrease. Explain

Suppose I'm finding the mass of a beaker of water, and I have it on the balance and are ready to record the mass. My partner dips the tip of his or her pen in the water, but does not touch the beaker or the balance. Does the presence of the pen affect the reading on the balance? If so how? If not why?

Woah - two completely different questions.. .

How will a bubble of air affect the density of an object - well, density is mass divided by volume. Adding a bubble of air is increasing the volume, but it's also increasing the mass. Whether the average density of the object increases or decreases depends on the relative density of the original object and the density of the air bubble. If the original material was denser than air, then the bubble will make the average density go down. Vice versa as well.

As for your second question -

Note that what the balance scale actually measures is force. Usually, it's the force of gravity on an object, which is called weight. It's the mass of the object times acceleration due to gravity, so you can derive the mass from the simple force on the object. Now, think about what happens to the forces when your partner dips their pen into the water:

The water exerts an upwards buoyancy force on the pen, equal to the weight of the displaced water. In order to do this, there has to be an equal and opposite force downwards. Now, the force registered on the scale will be weight of the water, plus the weight of the water displaced by the pen. The reading on the balance will increase.

It doesn't matter that the pen does not touch the sides of the container. Imagine if you'd put a small toy boat in the beaker. The weight of the boat will measure on the scale even if it doesn't touch the sides of the beaker.