575 - Physics Questions Answers
A CHARGED PARTICLE IS PROJECTED PARALLEL TO A UNIFORM FIELD . THE PATH OF D PARTICLE WILL BE
1) PARABOLIC
2) ELLIPTICAL
3) CIRCULAR
4) LINEAR
linear and accelerated.
CHARGES ARE PLACED ON VERTICES OF A SQ. WITH a, b HAS +q & c,d has -q charge . LET E ELECTRIC FIELD & V IS POT. AT CENTRE . IF CHARGES ON a &b R INTERCHANGED WITH THOSE OF c& d RESPTLY THEN E CHANGES WHY??
in this case magnitude will remain same. Only direction of E will change.
a II plate capacitor has an electric field of 105 V/m b/w d plates . if charge on capacitor plate is 1uC d force on each capacitor is??
ans 0.05N
F = σq/2ε0 = Eq/2 = 105*10-6/2 = 0.05
A charge of +2.0 x 10^-8 C is placed on the positive plate and a charge of -1.0 x 10^-8 C on the negative plate of a parallel plate capacitor of capacitance 1.2 x 10^-3 uF. Calculate the potential difference developed between the plates. Please add explanation!
due to induction the charges on the four faces will be
so V = q/C = 1.5 * 10-8/1.2 * 10-9
solve now
2 MOLES OF MONOATOMIC GAS IS MIXED WITH 1 MOLE OF A DIATOMIC GAS. THEN γ FOR THE MIXTURE??
ANS 1.55
total no of freedom for this mixture = (2*3+1*5)/(2+1) = 11/3
so γ = 1+[2/f] = 1 + 2/[11/3] = 1 + [6/11] = 17/11 = 1.55
a thin square plate with each side equal to 10cm , is heated by a blacksmith . the rate radiated energy by the heated plate is 1134W the temp of hot square plate is
( σ = 5.67 X 10-8 Wm2/k4 emissivity of plate =1)
ANS 1000K
according to the Stefans law P = AeσT4
so 1134 = 0.02*1*5.67*10-8*T4 (area of both the faces are considered)
so 1134*1010/11.34 = T4
so T = 1000
TWO PENDULUM HAVE TIME PERIODS T &5T/4. THEY STARTS SHM AT THE SAME TIME FROM THE MEAN POSITION . WHAT WILL BE THE PHASE DIFFERENCE B/W THEM AFTER D BIGGER PENDULUM COMPLETED ONE OSCILLATn ??
ans 900
When bigger pendulum will complete its one oscillation, the smaller one will complete [1+(1/4)] oscillation. so phase difference = 2π*(T/4)/T = π/2
QUES A MAN RUNS AT A SPEED OF 4M/s TO OVERTAKE A STANDING BUS . WHEN HE IS 6m BEHIND THE DOOR (AT t=0) , THE BUS MOVES FORWARD & CONTINUES WITH A CONSTANT ACCELERATn OF 1.2m/s2 . the MAN SHALL GAIN THE DOOR AT TIME t EQUAL TO
A) 5.2s
b) 4.3s
c)2.3s
d) the man shell never gain the door.
ans 4.3s , Sir ,on solving I 'm getting both 4.3 & 2.3 s ( when we solve sq root) so here i want to know why not the ans is 2.3s. AMU 2012
After 2.3 sec. the man overtake the door first, in this time velocity of bus will be less than that of man so it is not possible to catch the bus then after 4.3 sec. the bus door will again reach to the man, in this time velocity of bus and man are same, so relative velocity is zero and catching of bus is possible.
A RADIOACTIVE SUBSTANCE EMITS n BETA PARTICLES IN THE FIRST 2 SECONDS & 0.5 n BETA PARTICLES IN THE NEXT 2 SECONDS . THE MEAN LIFE OF THE SAMPLE IS
ANS 2/(ln2) s
A RADIOACTIVE SUBSTANCE EMITS n BETA PARTICLES IN THE FIRST 2 SECONDS & 0.5 n BETA PARTICLES IN THE NEXT 2 SECONDS. Definitely by definition its half life will be 2 sec.
so mean life = half life / ln2 This is the relation between mean life and half life.
A NUCLEUS OF MASS NO. 220, INITIALLY AT REST , EMITS AN .α PARTICLE . IF THE Q VALUE OF THE Rxn IS 5.5MeV THE ENERGY OF THE EMITTED α PARTICLE WILL BE ??
ANS 5.4 MeV
5.5*216/220 = 216/40 = 5.4 MeV here 216 is the mass of nucleus after ejecting α particle.