576 - Physics Questions Answers
A SUBMERGED BLOCK RISES BY 2cm WHEN MASS OF 10kg IS REMOVED ( BE4 THAT 10kg MASS WAS KEPT ON IT)
WHAT WILL BE THE VOLUME OF THE BLOCK
According to the given conditions
in first case block is completely inside water by putting 10 kg in it means density of block will be less than that of water
according to the law of floatation we know that
wt. of solid = upthrust
so 10g + AL(0.01)Dg = AL(0.01)dg
in second case
AL(0.01)Dg = A[L-2](0.01)dg
substracting 2nd eq. by 1st eq. we will get A = 1/2
now initial lenght in start is not provided by you, if it is given then calculate V = AL(0.01)
In thid question d and D are densities of liquid and solid, A and L are area of bottom of block and its length in cm. 0.01 is for converting cm. into m.
the density of two substances are in ratio 5:6 & specific heat are in ratio 3:5. their thermal capacity per unit volume will be in ratio?
ans 1:2
thermal capacity = ms = Vds
so thermal capacity per unit volume = ds
now calculate answer
it takes 6 hours for ice layer thickness to grow from 1cm to 2cm on a lake in Siberia, ideally the time taken for thickness to grow from 3cm to 4cm?
ans 14 hrs
for this type of question relation between time and thickness will be
[x2] α t
so 22 - 12 = k6
and 42 - 32 = kt
on dividing we will get 3/7 = 6/t
or t = 14
a cylinderical tube open at both ends has a fundamental frequency f in air the tube is dipped vertically in H2O so that half of it is in H2O the fundamental frequency of the air column is
1)f/2
2)3f/4
3)2f
4) f
aieee 2012
for first case l = λ/2 so l = v/2f
for second case l/2 = λ/4 so l/2 = v/4f
so f in both case are same.
two cars of masses m &M are moving in circles of radii r & R their speeds are such that they make complete circle in same time t. ratio of centripetal acceleration
1) m:M
2)R:r
3)1:1
4) MR:mr
aieee 2012 is the ans is (4)
since both make complete circle in same time t so angular velocities of the two will be same
now use F = mrω2
a cyclist is moving in a circle of radius 20m with velocity 5ms-1mass =60kg of both boy & cycle find static friction on tyre
v2/rg = µ
so µ = 5*5/20*10 = 1/8
Fs = µ * N = 1/8 * 60*10 = 600/8 = 75 N
µ is the coefficent of static friction, Fs is static friction and N is normal reaction
This answer is given by SHUBS and is correct answer
A POINT CHARGE q IS ROTATED ALONG A CIRCLE IN D ELECTRIC FIELD GENERATED BY ANOTHER POINT CHARGE Q D WORK DONE BY ELECTRIC FIELD ON ROTATING CHARGE IN ONE COMPLETE REVOLUTn IS ZERO
WHY???
Because this particle is revolving in an equipotential surface generated by other and work done = potential diff * q = 0
TWO EQ. +VE CHARGES R KEPT AT PT A & B D ELECTRIC POTENTIAL AT D PT B/W A & B (EXCLUDING THESE POINTS ) IS STUDIED WHILE ,MOVING FROM A TO B D POTENTIAL DECREASE THEN INCREASE WHY???
at start when the point will be nearer to first charge, potential due to the nearer charge will be very high, at the middle point of the two charges it will be minimum and after that it will increase again due to its approach towards another charge.
There are two fixed spheres of charges Q and 4Q and the distance between their centres is 6R. The Radius of first sphere is R and radius of the secind one is 2R. A charge -q is to be projected from the along the line joining their centres ( it is on the surface of the sphere with radius R), so that it can reach the surface of other solid sphere of charge 4Q. What will be the minimum velocity to be given so that it reaches there?
sir i have used energy conservation in this problem :
-kqQ/R - kQ4q/5R +.5 mv^2 = -kQq/4r - kqQ4/2R
but im not getting the answer .....is anything left in ths ..please telll
first find a point midway between the two spheres where net E will be 0
Then consider v = 0 at that point and compare the potential energy + kinetic energy of the starting point and that obtained point.
Define Potential Energy?
Potential energy is the energy of a particle due to its position relative to the reference point taken.