Answer:
The charge carried by the droplet is 
Explanation:
Given that,
Distance =8.4 cm
Time = 0.250 s
Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude
points straight down and if the mass of the droplet is 
We need to calculate the acceleration
Using equation of motion

Put the value into the formula



We need to calculate the charge carried by the droplet
Using formula of electric filed


Put the value into the formula


Hence, The charge carried by the droplet is 
A is the answer to the question
Answer:
The distance between the slits is given by 1.3 ×
m
Given:

D = 4 m
y = 
m = 1
To find:
distance between slits, d = ?
Formula used:
y = 
y = distance of first bright band from central maxima
D = distance between screen and source
d = distance between slits
= wavelength
Solution:
distance of first bright band from central maxima is given by,
y = 
y = distance of first bright band from central maxima
D = distance between screen and source
d = distance between slits
= wavelength
Thus,
d = 
d = 
d = 1.28 × 
d = 1.3 ×
m
The distance between the slits is given by 1.3 ×
m
Answer:
So to increase current of the circuit what you can do is :
1. Use conductor of low resistivity, ¶.
2. Use conductor of small length.
3. Use thick wire.
4. Decrease the temperature of the circuit.
5. If operating temprature is high than use semiconductor, because it have negative temprature coefficient.
6. Minimise the circuit losses.