575 - Physics Questions Answers
Two point masses having mass m and 2m are placed at distance d. The point on the line joining point masses, where gravitational field intensity is zero will be at distance
let the point will be at a distance of x from m then compare the gravitational field of the two masses at that point
If earth suddenly stop rotating, then the weight of an object of mass m at equator will [ ω is angular speed of earth and radius R is its radius]
In rotating condition mass at equator is mg-mRω2
if earth stops rotation then it will become mg so increases by mRω2
Three particles A,B and C each of mass m are lying at the corners of an equilateral triangle of side L. If the particle A is released keeping the particles B and C fixed, the magnitude of instantaneous acceleration of A is
force by the other two on A = Gmm/L2 each but the angle between the two forces is 60 degree so resultant force = √3Gmm/L2
now calculate acceleration?
A large number of identical point masses m are placed along x-axis, at x= 0,1,2,4.........The magnitude of gravitational force on mass at origin (x=0), will be
The magnitude of gravitational force on mass at origin (x=0), will be
Gmm/12 + Gmm/22 + Gmm/42 + Gmm/82 + .........................
= [Gmm/12][1+(1/4)+(1/16)+....................]
= Gmm[1/(1-(1/4)]
solve?
A satellite of mass 200 kg revolves around a planet of mass 5 x 1030kg in a circular orbit of radius 6.6 x 106 m. Binding energy of the satellite.
Binding energy = GMm/2R
What is the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination θ?
friction f = mgsinθ/[1+(mr2/I)]
solve for sphere and compare to μmgcosθ
A particle moving with constant acceleration covers a distance of 30m in the 3rd second. it covers a distance of 50m in the 5th second. What is the acceleration of the particle?
use formula for the distance covered in a particular second
st = u+at-(a/2)
A particle is moving with a velocity of 10m/s due east. In One second its velocity changes to 10m/s due west. If the particle is uniformly accelerated, the change in velocity will be directed as
let east is i
then velocity change = -10i-10i = -20i
A ball is dropped from a height h reaches the ground in time T. What is its height at time T/2
h = 1/2 gT2
in T/2 time h' = 1/2 gT2/4 = 1/8 gT2
so height from ground = h-h'
solve?
see you tube video of iitbrain by the name I E Irodov problem 1.3