575 - Physics Questions Answers
A uniform electric field is established by connecting the plates of a parallel plate capacitor to a 12v battery. A charge particle of +6.24x10 to the power -6c moves from rest from the positive plate to the negative plate. when the charge reaches the negative plate,it acquires a speed of 3.4m/s.what is{a} the mass of the charge particle and{b} its final kinectic energy?
energy gained by the particle between the plates of the capacitor = qV
this will provide K.E. to the particle so qV = 1/2 mv2
solve for mass
then for K.E.
Calculate potential difference across a and b ,
since net potential = 0, answer will be 0 for the given diagram. By interchanging the poles of any one battery this question will become meannigfull so change it and apply Kirchhoff law.
WHEN TWO SUBSTANCES OF DENSITYρ1 & ρ2 MIXED IN EQUAL VOLUME THEN RELATIVE DENSITY OF MIXTURE IS 4 , WHEN THEY R MIXED IN EQUAL MASSES THEN RELATIVE DENSITY =3. THEN ρ1 X ρ 2 =?
ANS 12
in first case 4 = [ρ1V + ρ2V]/2V
and 3 = [m+m]/ [m/ρ1 + m/ρ2]
solve now
A BALL FLOATS ON D SURFACE OF H2O IN A CONTAINER EXPOSED TO ATMOSPHERE . IF D CONTAINER IS COVERED & AIR IS COMPRESSED , D BALL WILL RISE . WHY?
beacause due to compression density of air will increase and upthrust by air will make some contribution in the net upthrust so some extra part of the body will come outside water.
a double star consist of 2 stars of mass m & .2m. distance b/w them is r. d 2 stars revolve under their mutual gravitatn force then d 2 stars have equal periods of revolution . HOW?
according to newtons third law both star will give equal force to each other
so for first star of mass m, m(2r/3)ω2 = F or ω = [3F/2mr]1/2
similarly for second 2m(r/3)ω2 = F or ω = [3F/2mr]1/2
here 2r/3 and r/3 are taken according to centre of mass criteria.
a coin placed on a gramophone record rotating at 33rpm flies off d record , if it is placed at a dist. of more than 16cmfrom d axis of rotation . If d record is revolving at 66rpm , the coin will fly off if it is placed a distance not less than ?
ans=4cm
coin will fly off, if it will gain the required centrifugal
so apply mrω2 = mRW2
a 1 kg block is attached (& held at rest with outside support) to free end of vertically hanging spring of force const. 10Ncm-1. when block is released wat max. extensn does it cause when it comes to rest instantaneously?
apply mgx = 1/2 kx2
for max. extension.
a block of mass m is suspended by a light thread from an elevator . the elevator starts moving up with a uniform acc . the work done during first t sec by tensn in thread ?
although i know how to calculate this ques . i.e w.d =Fds =m(g+a)Xds where ds=ut+1/2(at2) because acc is const a/c to ques BUT I want to know for ds why acc is a ,why not (a+g)
tension in the thread depends on g an a both.
but displacement covered by elevator depends only on the acceleration of the elevator, it does not depend upon the acceleration g thats why we do not use g in that case for finding displacement.
remember that in the problem we have to calculate work done by tension in the thread
a 24.6 mev alpha particle moves towards gold nucleus[atomic no 79] calculate the distance of closest approach of alpha particle
upto closest approach total KE of α particle will become PE
so use the formula KE = 9*109 * Ze2e/r
here KE = 24.6 mev convert it into joule first
solve equation for r
FROM REST A BODY ROTATE 120 rotation/min FOR 20 sec THEN IT ROTATE FOR 5sec WITH SAME ROTATION . FIND HOW MANY ROTATn IT WILL MAKE?
ω = ω0+ αt
so 4π = 0 + α20 so α = π/5
so total angular displacement = 1/2 π/5 202 = 40 π = 20 rotation
and with angular velocity 2 rotation/sec total rotations in 5 sec. = 10
so answer will be 20+10 = 30