https://heylink.me/healthsmood

MCAT: Orientation Seminars: 14th to 20th July, 2014 Availability & Submission of Entrance Test Kits (Admission Forms): 4th to 9th August, 2014 Issuance of Admittance Cards: 11th to 16th August, 2014 (excluding 14 Aug) Entrance Test: 31st August, 2014 ECAT: Registration Dates: 14-07-2014  to 08-08-2014 Test Date: SUNDAY, 24-08-2014 at 10.00 A.M Result Date: 31-08-2014  (Midnight) 

Closing Merits of IST in 2013 Field Closing Merit Aerospace Engineering 80.36% Mechanical Engineering 78.62% Electrical Engineering 77.93% Material Engineering 75.62% Space Science 73.53%

MCQs of NET-3 1) The electric effects of friction are due to: a) electrostatics b) electromagnetics c) electrodynamics d) All 2) Heat energy absorbed by a body does not depend upon: a) mass of the body b) specific heat c) change in temperature d) duration of heating 3) If electron jumps from 3rd to 1st orbit, its speed : a) decreases 9 times b) increases 9 times c) decreases 3 times d) increases 3 times 4) Which of the following has unit kgm/s a) Impulse b) force c) momentum d) impulse and momentum 5) If the amplitude of oscillating body is halved, its frequency becomes: a) 2f b) f/2 c) 0 d) 1.5f 6) A mass spring system oscillates between 10cm and 50cm. It has max. P.E. at a) 10cm b) 20cm c) 30cm d) 40cm 7) Which of the following is soluble in hot water: a) keratin b) silk fibres c) fibrin d) none 8) Choose the Odd one out a) magazine b) dictionary c) novel d) thesis 9) If RTTHHX encodes for purify, then for blanch a) DKDSHG b) AOCGTY c) DKCMEG d) EMGCKD 10) Al evolves Hydrogen by reac

  Relation between Momentum and K.E Question: If Momentum of a body is increased by 20%, what is percentage change in the K.E of the body? a) 40% b) 44% c) 20% d) 10% Answer: 44% Solution: Let the original momentum of the body be P, Then: P=mv Momentum is increased by 20%, so new momentum will be: P' = P + 20%P P' = P + 0.2P = 1.2P K.E = 0.5 mv^2         =0.5 m^2v^2/m                          (multiply and divide by m) K.E  = 0.5 P^2 / m                               ---------- (i) New K.E = K.E'                 = 0.5 (P')^2/m                 = 0.5 (1.2P)^2 / m                 = 0.5  (1.44) P^2 / m                 = 1.44 K.E                              (using i) Percentage Change in K.E = (New K.E - Old K.E) /Old K.E x 100%                                          = (1.44K.E - K.E)/K.E x 100%                                          = 44% Second Method: Momentum of a body is given by : P = mv P can be varied by varying either m or v. Since mass of a body is a constant