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Ajneesh

HomeAuthor: AjneeshPage 7
(a)  Decreases (b)  Increases (c)   Remains constant (d)  Depends on the molecular motion Ans. c (22)
(a)  – 50 J                                  (b)  20 J (c)   30 J                                     (d)  50 J Ans. b (9)
(a)  Q + W                                 (b)  Q – W (c)   Q                                         (d) (Q – W)/2 Ans. b   (3)
area 0.05 m2. Increase in its potential energy will be (T = 0.2 N/m)    (a)        5...
(a)      24πR2S                        (b) 48 πR2S (c)   ...
(a)  dQ = dU + PdV                 (b) dQ = dU x Pdv (c) ...
(a)  r                                          (b)  0 (c)   Infinity                               (d)  1/2r Ans. c (15)
(a)  3 J                                       (b)  6.5 J (c)  1.5 J                                    (d)  4 J Ans. a (9)
(a)     192π  x   10-4 J          (b) 280π  x   10-4 J (c)      200π  x   10-3...
the increase in surface energy. (Surface tension of mercury is 0.465 J/m2 ) (a)    23.4 μJ       ...
(a)  Released                            (b)  Absorbed (c)   Both (a) and (b)               (d)  None of these Ans. a (13)
108 drops of equal size. The energy expended in joules is (surface tension of Mercury is 460 x 10-3N/m (a) ...
(a) 2 x 10-2Nm-1                          (b) 2 x 10-4Nm-1...
(a)  1                                         (b)  2 (c)   4                                         (d)  6 Ans. c (13)
a ring of area  is (Surface tension of liquid   = 5Nm-1) (a) 0.75 J                                         (b)  1.5 J (c)   2.25 J                                ...
(a)  1 : 10                                  (b)  1 : 15 (c)   1 : 20                                  (d)  1 : 25 Ans. c (7)
(a)  Energy is released (b)  Energy is absorbed (c)   The surface area of the bigger drop is greater than the...
radius is (Surface tension of the soap solution is (3/100) N/m) (a)   75.36 x 10-4 joule (b) 37.68 x 10-4...
size 10 cm x 10 cm is (Surface tension T= 3 x 10-2 Nm (a)6 x 10-4 j                                     (b)...
106 small drops. The energy used will be (surface tension of mercury is 35 x10-3N/cm (a) 4.4 x 10-3 j...
The surface tension of liquid is 0.5 N/m. If a film is held on a ring of area 0.02 m2,...
(a)   Becomes zero (b)  Becomes infinite (c)   is equal to the value at room temperature (d)  is half to the...
is 2 x 10-2 N/m. To blow a bubble of radius 1 cm, the work done is  (a)    4 π...
106 small drops coalesce to make a new larger drop then the drop (a)  Density increases (b)  Density decreases (c)  ...
(a)    R/2 (b)   R/5 (c)   R/6 (d)  R/10 Ans. d (11)
(a) Rn(2/3) σ (b) (n(2/3) – 1) Rσ (c) (n(1/3) – 1) Rσ (d) 4πR2(n(1/3) – 1)σ (e) [1/( n(1/3)...
is 1.9 x 10-2N/m . Work done in blowing a bubble of 2.0 cm diameter will be (a)  7.6 x...
10 cm× 6 cm to 10 cm × 11 cm is 3 ×10-4 joule. The surface tension of the film...
(a)      0o      (b)   90o (c)   Less than 90o                               (d)  Greater than 90o Ans. d (5)
from (1/√π) cm to (2/√π) cm. If the surface tension of water is 30 dynes per cm, then the work done...
(a)  Decrease (b)  Increase (c)  Remains the same (d)  None of the above Ans. b (4)
(a)  8πr2T (b) 2πr2T (c)  4πr2T (d) (4/3)πr2T Ans. a (6)
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