WebIf ρ is the density of a small drop and r its radius, then the mass of each small drop is m = 4 π 3 r 3 ρ. If n such drops coalesce to form a big drop of radius R, then the mass of the big drop is n m= 4 π 3 R 3 ρ. Now, by conservation of mass: n* 4 π 3 r 3 ρ. = 4 π 3 R 3 ρ. Hence, R= n 1 3 r. Now, the capacitance of a sphere is ... WebClick here👆to get an answer to your question ️ V 24 V 33 V 4)2 V 35. n drops each of radius' r' and carrying a charge 'q' are combined to form a bigger drop. The ratio of potentials of bigger to that of smaller is TH con d haped ne speed 2 Canectance: 36. A sphere A of radius 5 em contains a charge of 120 nC. It is connected to two other …
Eight small drops, each of radius \\( r \\) and having same …
WebAug 8, 2024 · 1000 small water drops, each of radius r and charge q, coalesce to form a single big drop. The potential of big drop, asked Mar 29, 2024 in Electric Potential by Abhinay (62.9k points) electric potential; class-12; 0 votes. 1 answer. n small drops of the same size of the same size are charged to V volt each. They coalesce to form a big drop. WebEight small drops, each of radius r and having same charge q are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is. Q. 8 charged drops of water each of radius 1mm and having a charge of 10^-10C , combine to form a bigger drop . determine the potential of bigger drop. ... univ. of south florida - alec braynen
Eight small drops, each of radius r and having same charge q are ...
WebMar 20, 2024 · Ratio between potentials of the bigger drop and samller drop = 8. Step by step: Let the charge on big drop = Q. Redius of big drop = R. Number of small drops = n = 8. Redius of small drop = r. Charge on small drop = q. Now, Charges gets added when multiple drops are combined. Charge on big drop, is the addition of cahrges o smaller … WebApr 4, 2024 · $\Rightarrow V = \dfrac{4}{3}\pi {r^3} $ where $ V $ is the volume of a sphere, $ r $ is the radius of the sphere. Complete step by step answer One of the 64 small drops has a surface area of $ A = 4\pi {r^2} $ $ r $ is its radius. Now these small drops coalesce into a large drop, the total volume hence must be the same as in $ V = 64 \times ... receiving objects很慢