RankLab
Back to Questions
PhysicsMediumMCQ2021 · 20 Jul Shift 2

Q10.Two small drops of mercury each of radius R coalesce to form a single large drop. The ratio of total surface energy before and after the change is (1) 2 13 : 1 (2) 1 : 2 31 (3) 2 : 1 (4) 1 : 2

What This Question Tests

This question involves calculating the surface energy of mercury drops before and after coalescence, using the principle of volume conservation to find the radius of the larger drop.

Concepts Tested

Surface energySurface tensionConservation of volume

Formulas Used

E = T × Area

Volume of sphere = 4/3 πR^3

📚 NCERT Sections This Tests

13.4 — Obtain Approximately The Ratio Of The Nuclear Radii Of The Gold Isotope

Physics Class 12 · Chapter 13

74% match

13.4 Obtain approximately the ratio of the nuclear radii of the gold isotope 197 79 Au and the silver isotope 10747 Ag .

13.2 — Obtain The Binding Energy Of The Nuclei 5626Fe And 20983 Bi In Units Of

Physics Class 12 · Chapter 13

73% match

13.2 Obtain the binding energy of the nuclei 5626Fe and 20983 Bi in units of MeV from the following data: m ( 5626Fe ) = 55.934939 u m ( 20983 Bi ) = 208.980388 u

13.5 — The Q Value Of A Nuclear Reaction A + B ® C + D Is Defined By

Physics Class 12 · Chapter 13

73% match

13.5 The Q value of a nuclear reaction A + b ® C + d is defined by Q = [ mA + mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 11 H+13 H →12 H+12 H (ii) 126 C+126 C →1020 Ne+ 24 He Atomic masses are given to be m ( 12 H ) = 2.014102 u m ( 13 H) = 3.016049 u m ( 126 C ) = 12.000000 u m ( 1020 Ne ) = 19.992439 u