Eight mercury droplets having a radius of 1mm
Webr = 0.1 mm = 0.1 × 10-3 m, T = 0.072 N/m. To find: The change in surface energy. Solution: Let R be the radius of the single drop formed due to the coalescence of 27 droplets of mercury. Volume of 27 droplets = volume of the single drop as the volume of the liquid remains constant. ∴ `27 xx 4/3pi"r"^3 = 4/3pi"R"^3` ∴ `27"r"^3 = "R"^3` ∴ ... WebApr 5, 2024 · We can consider a big drop of mercury of radius R. It can split into 1000 droplets of same sized mercury. The radius of the small drop is r. We can find out the …
Eight mercury droplets having a radius of 1mm
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WebFeb 15, 2024 · Data : r = 1 mm = 1 × 10-3 m, T = 0.472 J/m 2. Let R be the radius of the single drop formed due to the coalescence of 8 droplets of mercury. Volume of 8 droplets = volume of the single drop as the volume of the liquid remains constant. ∴ 8 × \(\frac43\) πr 3 = \(\frac43\) πr 3. ∴ 8r 3 = R 3. ∴ 2r = R. Surface area of 8 droplets = 8 ... Web2. A drop with an initial radius of 100 μm falls through a cloud containing 100 droplets per cubic centimeter that it collects in a continuous manner with a collection efficiency of 0.8. If all the cloud droplets have a radius of 10 μm, how long will it take for the drop to reach a radius of 1 mm?
WebA drop of mercury of radius 0.2 cm is broken into 8 droplets of the same size. Find the work done if the surface tension of mercury is 435.5 dyn/cm. Maharashtra State Board HSC Science (Computer Science) 12th Board Exam. Question Papers 222. … WebQuestion. Eight mercury droplets having a radius of 1 mm and a charge of 0.066 PC each merge to form droplet. Its potential is (A) 2.4 V 0 (B) 1.2 V (C) 3.6 v (D) 4.8 V: …
WebEight droplets of mercury each of radius 1 mm thickness 0.5cm is held vertically with its lower end coalesce into a single drop. Find change in surface immersed in water.Find the downward pull on the energy. Surface tension of mercury is 0.465 J/m2. tube due to surface tension. S. T. of water is 0.074 ... WebApr 2, 2024 · A mercury drop of radius 1 cm in broken into ${10^6}$ droplets of equal size The work done is $\rho = 35 \times {10^{ - 2}}N/m$ A). $4.35 \times {10^{ - 2}}J$ B). $4.35 \times {10^{ - 3}}J$ C). $4.35 \times {10^{ - 6}}J$ D). $4.35 \times {10^{ - 8}}J$
WebQuestion: A drop with an initial radius of 100 μm falls through a cloud containing 100 droplets cm-3, which it collects in a continuous manner with a collection efficiency of 0.8. If all these droplets have a radius of 10 μm, how long will it take for the drop to reach a radius of 1 mm? Assume that the smaller droplets are stationary and that the larger …
Web1. The radius of a soap bubble is increased from R to 2 R. Work done in this process (T = surface tension) is: 1. 24 πR2T 24 πR 2 T. 2. 48 πR2T 48 πR 2 T. 3. 12 πR2T 12 πR 2 T. 4. 36 πR2T 36 πR 2 T. 2. Water flowing from a hose pipe … jen caraubalWebNov 22, 2024 · A drop of mercury of radius 0.1 cm is broken into 8 droplets of the same size. Find the work done if the surface tension of mercury is 540 dyne/cm. Given: Initial … lake haven apartments kemah txWebJun 22, 2024 · Solution For Eight mercury droplets having a radius of 1 mm and a charge of 0.066pC each merge to form one droplet. Its potential is Eight mercury droplets having a radius of 1 mm and a charge of 0.066pC ea.. lake havasu state park camping cabinsWebSep 5, 2024 · Answer: eight droplets of Mercury each of radius 1mm coalesces into a single drop. Find change in the surface energy. Surface tension of Mercury is 0.456J/m^2 answer is 23.316 × 10 ^-6 J jen cardonaWebApr 18, 2024 · Eight droplets of mercury, each of radius 1 mm, coalesce to form a single drop. Find the change in the surface energy. asked Feb 15, 2024 in Physics by ShubhamMahanti (33.9k points) mechanical … lake haven kemah texasWebr = 0.1 mm = 0.1 × 10-3 m, T = 0.072 N/m. To find: The change in surface energy. Solution: Let R be the radius of the single drop formed due to the coalescence of 27 droplets of … lake havasu yamaha dealerWebRadius of larger mercury drop, R = 1 cm = 10 − 2 m Surface Tension, T = 32 × 10 − 2 N/m Number of smaller mercury drops, n = 10 6 Let us suppose, radius of smaller mercury … jen card game