AITS doubt- work,power,energy

16 Answers

1
rishabh ·

aditya wats ur aits PT1 total?

1
AVISIKTA UPADHYAY ·

is the answer 2?????????????????

71
Vivek @ Born this Way ·

10?

1
Aritra Chakrabarti ·

calculate work done by friction. (Here only friction of the second surface should be cosidered) from work energy theorem equate this work done to the change in kinetic energy.

262
Aditya Bhutra ·

i am getting the answer as 10 but the answer given is 2 .
@avistika - plz post the soln for us .
@rishab - about 200

21
Shubhodip ·

yup..i too was getting 10

i think the ans may not be 2...there r some mistakes in the solution..wht is the logic behind giving solutions to some problems and not to some others!!

are those so easy that can be solved w/o any workings !! fiitjee = crajee(crazy)

106
Asish Mahapatra ·

to all those Fiitjee aits(ers), Fiitjee is renowned for mistakes in its solutions. So if you are finding mistakes, then they are mistakes (mostly) (iff quite a few people also find so).

:D

21
Shubhodip ·

Ha ha

71
Vivek @ Born this Way ·

Most of the times, just in order to showcase their strength than they can make your mind scratch, overlook elementary ideas or raise many questions in their way of solution (just to bring the answer that they want).
:(
Heavily Discouraging!

1
Aritra Chakrabarti ·

what should be a good score in this first AITS. what do u people think?

71
Vivek @ Born this Way ·

Don't know, I didn't give!

1
s_feynman mukherjee ·

hello guys b4 criticizing fiitjee u need 2 realise dat u r gettin stuck in a question set by it!!! ..nd needless 2 say mistakes do occur in solution but wats important is raising ming boggling Q nd dats wat fiitjee does...!!!
nd dont 4get d famous AITS

21
Shubhodip ·

Dude ,the fact that this problem is wrong was accepted in the errata ...

71
Vivek @ Born this Way ·

Pinks even for this :) ?

Subho! really rocks.. Yaar hamlog toh lik likh kar thak gaye.. kabhi milta hi nahi

21
Shubhodip ·

Now see ;)

1
Vivek Singhal ·

could someone plz point out my mistake......

dE=2(0.1*m*g*r* theta)

0.1=co-efficient of friction; theta is the angle difference

thus,
dEd(theta)=20

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