A marble rolls down (from rest) from a cliff of height 5m, and then encounters a vertical drop of 1m. Find the range of the marble on the horizontal planes...
For those of you who don't know already, marbles are solid spheres, and are quite small..
hi amit...i had once tried that question at 100p forum...though i was never able to get the correct answer:
apply conservation of energy:
gain in surface energy: sigma*{2*4pi(2R)^2}-sigma*{2*4pi(R)^2}
for finding the loss in electrostatic potential energy, use the capacitance of a shell: 4(pi)eR, and that the energy is q^2/2C...C increases when R becomes 2R..so EPE decreases.
Also, there is some work done against the atmospheric pressure, given by P*increase in volume.
But combining all this did not give me the correct answer
you're correct till this point...but as this question is labeled "difficult", you cannot ignore the fact that the sphere infact rotates (by SOME angle), about the edge...
@ankush sachdeva , it will not rotate by cos inverse 10/17 bcoz it is no fallling from rest position initially .it is moving with some velocity .that's why the angle would change ...
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ReplyDeletei think x=root(80/7)
ReplyDeleteafter falling there it will go upto infinity of friction is acting on horizontal plane
bhaiya plz solve this ques also
http://forum.100percentile.com/iitjee_aieee_forums/posts/list/Physics-Discussions-electrostatics-29619.htm
hi amit...i had once tried that question at 100p forum...though i was never able to get the correct answer:
ReplyDeleteapply conservation of energy:
gain in surface energy:
sigma*{2*4pi(2R)^2}-sigma*{2*4pi(R)^2}
for finding the loss in electrostatic potential energy, use the capacitance of a shell: 4(pi)eR, and that the energy is q^2/2C...C increases when R becomes 2R..so EPE decreases.
Also, there is some work done against the atmospheric pressure, given by P*increase in volume.
But combining all this did not give me the correct answer
Getting same root 80/7
ReplyDeletesqrt(80/7);
ReplyDeleteyou know...i don't really give out easy questions..so, my sixth sense tells me that root (80/7) can't be the answer...
ReplyDeleteI made that extra effort in "its a sphere"...
you know..i don't really hand out easy problems...
ReplyDeletei have used the fact that it is a sphere
ReplyDeletethe balls KE at the edge=5/7mgh
you're correct till this point...but as this question is labeled "difficult", you cannot ignore the fact that the sphere infact rotates (by SOME angle), about the edge...
ReplyDeleteit will rotate by angle cos inv (10/17)
ReplyDeleteits velocity at the edge will make the same agle with horizontal anticlockwise
then a very huge calculation of solving the quadratic
i get ans
RANGE = 0.65 m (approx)
@ankush sachdeva , it will not rotate by cos inverse 10/17 bcoz it is no fallling from rest position initially .it is moving with some velocity .that's why the angle would change ...
ReplyDelete