575 - Physics Questions Answers
A SIMPLE ELECTRIC MOTOR HAS AN ARMATURE RESISTANCE OF 1 ohm & RUNS FROM A D.C SOURCE OF 12V . IT DRAWS A CURRENT OF 2A WHEN UNLOADED . WHEN A CERTAIN LOAD IS CONNECTED TO IT ,IT SPEED REDUCES BY 10% OF ITS INITIAL VALUE ,THE CURRENT DRAWN BY THE UNLOADED MOTOR IS??
ANS 3A
R=1ohm , E=12V , I=2A
I=(E-e)/R => e=10 ON PUTTING D VALUES
NOW AS IT IS REDUCES BY 10% MEANS NOW e' = 9V ( 10/100 X 10 = 1 CHANGE , THUS FINAL = 10-1=9)
I= E-e'/R = ( 12-9)/ 1 = 3A (RESISTANCE IS KEPT CONST.) RESISTANCE WILL CHANGE OR NOT??
yes this method is correct.
if a particle is moving on a circular path with constant speed then the angle between the direction of acceleration and its position vector with respect to centre of circle will be?
since speed is constant so only centripetal acc. will be present and it will be parallel to position vector starting from centre of the circle.
the reduced mass of two particle having masses m and 2m is 2m/m.how????.....
formula for finding reduced mass will be [m1m2/m1+m2]
the moment of inertia of a thin uniform rod about an axis through its end and perpendicular to its length is ML²/3 . The moment of inertia of the rod about an axis parallel to given axis and at a distance L from the center of the rod is
for an axis parellel to the axis of rod and at distance l, rod will act as a set of equidistance particles. so answer will be Ml2
answer given by sarika is not completely correct
velocity of e- in hydrogen atom is 2X106 m/s . the radius of orbit is 5X10-11 m . the mag induction at centre of orbit in T will be??
B = μ0/2 * i/r
and i = e/T
and T = 2πr/v
solve
potential diff b/w dees of cyclotron is V , if charge q comes out of it after completing n revolution then gain in KE of charge is 2nqV HOW?
when charge will move from first dee to second dee it will gain qV and on returning to first dee it will gain qV again so in one revolution it will gain 2qV so in n revolutions it will gain 2qVn
distance b/w plate of capacitor C &q charge is increased to double . work done=??
as distance d is double thus C= Ae/2d thus capacitance will be halved so wd shld be (1/2 )q2/(C/2) ie 1/4 q2/C
BUT Sir the ans given is=( 1/2)q2/C Plzz do me a little favour by solving this query
(1/2 )q2/(C/2) = q2/C not 1/4 q2/C
so work done = new energy - previous energy
An isolated conducting sphere of radius r has given charge q then PE??
ans is q2/8πe0r
But Sir ,why the ans q2/2πe0r is incorrect??
let we have given x charge and want to give dx also to the sphere
so dE = [1/4πε0]xdx/r
now integrate and take limits of x as 0 to q
2 Equal -ve charges -q r placed at pt(0,a) and (o,-a) on y axis , one +ve charge +q at rest is left from pt(2a,0). this charge will not execute SHM . WHY?? even when F α -x condition is satisfying as force is due to both charges is acting & motion is also opposing during oscillating
Solve the question for force, you will get a condition according to which yours statement "even when F α -x condition is satisfying"
is incorrect
A wheel starting from rest is uniformly accelerated at 2 radsec^2 for 20s.It is allowed to rotate uniformly for next 10s and finaly brought to rest in next 20s.Find total angle rotated by wheel ( in radian).
ω = 0 + 2(20) = 40 (according to rotational equation ω = ω0 + αt)
and θ = 0 + 1/2(2)(20)2 = 400 radian
in middle uniform part angle rotated = 40*10 = 400 radian
and since acceleration in first part and retardation in last part are equal so angle rotated in last part = 400 radian
total = 1200 radian