It takes a noun and a verb to make a complete sentence.
There isn't a single verb in a), b), or c).
"Affords" is the verb (predicate) in d)., the only complete sentence.
The bearing could be the below:
oppositely charged, same initial direction
same charge, opposite initial direction
You can decide by utilizing your correct hand and put your fingers toward the attractive field (North to South). Thumb toward present or charged molecule. The course of your palm will demonstrate the heading of compelling set on a decidedly charged molecule and the bearing of the back of your hand will demonstrate the bearing of a contrarily charged molecule.
The answer to this question lies in the definition of density. One material will just float over another if its density is smaller. If one material is denser than the other, it will sink.
Density can be defined as the mass per unit volume of a substance at a given pressure and temperature.
Thus, for a material to float in water, it does not depend on the weight, or rather on the mass, but on the distribution of the mass by the volume occupied, that is, of the density. The more distributed the mass, that is, the larger its volume, the less dense the object and it will float.
Object C has the lowest density<span>
65 N or 6.5 Kg ------------ 6 N or 6 Kg
This effective mass under water will be its actual mass minus the mass of the fluid displaced.
The buoyant force on a submerged object is equal to the weight of the fluid displaced.
Weight of object - buoyant force on object (the mass of the fluid displaced)
6Kg - 6.5Kg= - 0.5Kg
</span>Answer: C. object C
Answer:
i) 21 cm
ii) At infinity behind the lens.
iii) A virtual, upright, enlarged image behind the object
Explanation:
First identify,
object distance (u) = 42 cm (distance between object and lens, 50 cm - 8 cm)
image distance (v) = 42 cm (distance between image and lens, 92 cm - 50 cm)
The lens formula,
Then applying the new Cartesian sign convention to it,
Where f is (-), u is (+) and v is (-) in all 3 cases. (If not values with signs have to considered, this method that need will not arise)
Substituting values you get,
i) 
f = 21 cm
ii) u =21 cm, f = 21 cm v = ?
Substituting in same equation
v ⇒ ∞ and image will form behind the lens
iii) Now the object will be within the focal length of the lens. So like in the attachment, a virtual, upright, enlarged image behind the object.
