Choice-B is the true one.
Answer:

and

Explanation:
Given:
- first charge,

- second charge,

- position of first charge,

- position of second charge,

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.
<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

- since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.





Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

and

Answer:
E = 3.8 kJ
Explanation:
Given that,
The mass of the object, m = 10 g = 0.01 kg
The heat of fusion of aluminum is 380 kJ/kg
We need to find the energy required to melt the mass of the aluminium. It can be calculated as follows:
E = mL
So,
E = 0.01 × 380
E = 3.8 kJ
So, the energy required to melt the mass is equal 3.8 kJ.
The answer they are looking for is the last one. However the last two are technically correct but the third one would result in negative work.
During the daytime, I have mostly line symmetry.
During the night, I often have almost spherical symmetry.