Answer:
The current on the water layer = 1.64×10^-3A
Explanation:
Let's assume that the radius given for the string originates from the centre of the string. The equation for determining the current in the water layer is given by:
I = V × pi[(Rwater + Rstring)^2 - (Rstring)^2/ ( Resitivity × L)
I =[ 166×10^6 ×3.142[(0.519×10^-4) + (2.15×10^-3])^2 - ( 2.15×10^-3)^2] / ( 183 × 831)
I =[ 521572000(4.848×10^6)- 4.623×10^-6]/ 154566
I = 252.83 -(4.623×10^-6)/ 154566
I = 252.83/154566
I = 1.64× 10^-3A
Answer:
option B
Explanation:
given,
Satellite B has an orbital radius nine times that of satellite A.
R' = 9 R
now, orbital velocity of the satellite A
........(1)
now, orbital velocity of satellite B
from equation 1
hence, the correct answer is option B
Place the object in an electronic balance and measure its mass.
Place a measured amount of water in the cylinder.
Place the object in the cylinder so that it’s fully submerged.
Measure the new level of the liquid and subtract the original level. This is equal to the volume of the object.
Density = mass / volume.
Π = c/ f
π = 3×10^8 / 1.5×10^13
π = 3÷1.5 ×10^(8-13)
π = 2×10^-5 m
