51 - Mechanics Part 2 Questions Answers
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
can the linear speed of two objects be different if they have the same angular speed?
angular speed depends on the distance from pivot also so v may be same or different
formula affilated is v = r ω
NAME THE OUTERMOST LAYER OF SUN ????
i think its corona
A STONE WEIGHS 9N AT A PLACE ON THE SURFACE OF EARTH HAVING LATITUDE π/2 . ITS MASS AT CENTRE OF EARTH WILL BE (kg)
1) 10/ 37√2
2 ) 10√2
3) 5
4) 15
π/2 means it is pole so mg = 9 implies that m = 9/9.8
same will be the mass at centre of earth
THE TERMINAL SPEED ATTAINED BY AN Al SPERE OF RADIUS 1mm FALLING THROUGH H2O AT 200C WILL BE CLOSE TO ??
ANS 4.6 m/s
(assume laminar flow , specific gravity of Al=2.7 & η H2O =8X 10-4 PL.)
v = 2r2g(D-d)/9η
so v = 2(.001)2*9.8(2700-1000)/9*8*10-5
solve it
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.
my previous ques from fluid topic is not solved yet so pls reply that ques & that of chem of some other one
very soon you will get the answers