1n by d way arshad,wat formula hav u applied.....sum of infinite g.p is =a/(1-r)
here according to u a=t and r=1/2,,,so sum is=t/{1-(1/2)}....it comes as 2t newton and not 2 newton...
............isnt it arshad???
A highly expexted good question for JEE which is made by me with a GREAT MIND.
Find the tension in the thread at the last block placed at infinity.
Each block is of mass m and the masses of the thread and pulleys are negligible.
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18 Answers
1Kaymant sir , plzzzzz check whether ans given by Arshad is right or not ????
1
1yup i realized my mistake...
the acceleration of left side block is coming out to be g/2
1Arshad u are underestimating the question to an alarming degree
also your method of summing all the tensions on one side is physically insignificant
1On solving we get the acceleration of the topmost block as g/2 (actually all the blocks move with this acceleration upwards in their respective frames)
My solution is a bit tedious to write without latex,
if no one else posts the correct solution then I will do so.
1sir actually i missed one brackett..
what i wanted to write was that sum of an infinte gp=
t/(1-(t/2))=2
66No, arshad's answer is not correct.
@arshad, how does t/1-(t/2) becomes 2 N
66I'll add to the problem: Find the acceleration of the topmost block on the left.
(And this question is not going to come in IIT JEE)
1and eureka the feyman lectures one is a bit different....
here the masses are given to be equal....
1i think yes sir....if we consider the tension to be in newton....
1tension in each succesive pulley varies by a factor of t/2
so its forming an infinite gp like t,t/2,t/4........
hence the tension in the infinite pulley will be 0....
now kaymant sirs question
total tension in the right side=t/1-(t/2) =2 N
so acceleration of topmost block on the left a=(mg-2)/m
1Thanx Anant sir, for making the ques more complicated and interesting.
Why this ques is not going to come in jee? These type of questions should come in JEE