Messages posted by HEMANT BHARDWAJ

curved sphere? or hemisphere?

frinstion acts at lowest point which remain at rest in case of pure rolling and hence work done by it is zero

if particle is undergoing uniform circular motion then torque will not be acting on it hence momentum(angular) will remain conserved for all stationary pt

ya surely answer of b is wrong in book if it is what u r saying.

u can read it"s explanation in haliday resnick walker
actually pi was used to just make few calculation easier

great............. but what abt i think it is a joke....... right?

why don"t u try on urself??????????

if we assume that battery has some internal resistance then we can give the some answer otherwise infinite current will flow through tht battery and as there is no resistance anywhere so battery will become dead instantly(in infinitely small time) and sparking will occur

and net rotational Kinetic energy decreases.

Actually net kinetic energy will not decrease but will increase as KE=L²/2I

here L is constant while I is decreasing hence KE shouid increase....

Question : Where does this extra energy come from?

while pulling his arms some work will be done by man and energy will come from his internal energy.

i think answer is B although i m not sure
work will be done coz direction of magnetic field (and this force on north pole ) will be tangential (given by right hand rule) hence some work would be done

if P is replaced by a cell with lower emf than Q then u can not achieve balance point.

now u can think y not............

B-

As no current flows through galvanometre, cell Q , or resistance S when potential metre is at balanced position.

Hence there will be no effect of decreasing\increasing value of S..

A-

if R is increased then the value of current flowing in potential metre shouid decrease which will result in lower value of potential gradient(drop of potential per unit length)

Hence to re-establish balance point u wouid require to shift the point X to right(i.e toward"s point B)

i m giving here some more explanation

initially when the resistance was R then balance point was at X suppose AX=L since potential metre is balanced at this position which mean potential difference b\w A and X is equal to EMF of cell Q(EMF coz no current is flowing through Q hence no potential drop when potential metre is balanced)

Next suppose u increased the resistance which will result decrease in value of current in potential metre so for producing same potential drop=EMF of cell Q u hav to shift the balance point toward"s B..

Any satellite never fall itself !

but when a satellite become old and useless then it is made to fall on earth so that it may not collide with some other space shuttle ot satellite in future

Hence i think abhi has done a small mistake while stating that
(IF GOING UP OR DOWN WITH A CONSTANT ACCn then also no effect)

if lift is going up or down with constant velocity then the tension will be same as it was (when it was at rest) but if it os going upward with some aceeleration (no matter constant or not) then effective value of g will increase and hence tension will also increase. and xactly opposite if going downward with some acceleration.

really tough 1

using very general formula of impulse

impulse=change in momentum=avg Force * time

it is not given that wether he travelled 1Km distance by constant speed or non uniform speed. but since change in linear momentum of man is zero(if man comes to rest after waiking this distanceand was initially at rest) then avg friction force will be zero since friction force is the only external force which can act on the man in horizontan direction.

question is incomplete current or voltage and mass of ice e.t.c must be given

and net rotational Kinetic energy decreases.

if P is replaced by a cell with lower emf than Q then u can not achieve balance point.

now u can think y not............

B-

As no current flows through galvanometre, cell Q , or resistance S when potential metre is at balanced position.

Hence there will be no effect of decreasing\increasing value of S..

A-

if R is increased then the value of current flowing in potential metre shouid decrease which will result in lower value of potential gradient(drop of potential per unit length)

Hence to re-establish balance point u wouid require to shift the point X to right(i.e toward"s point B)

Next suppose u increased the resistance which will result decrease in value of current in potential metre so for producing same potential drop=EMF of cell Q u hav to shift the balance point toward"s B..

Any satellite never fall itself !

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and net rotational Kinetic energy decreases.

if P is replaced by a cell with lower emf than Q then u can not achieve balance point.now u can think y not............

B-As no current flows through galvanometre, cell Q , or resistance S when potential metre is at balanced position.Hence there will be no effect of decreasing\increasing value of S..

A-

if R is increased then the value of current flowing in potential metre shouid decrease which will result in lower value of potential gradient(drop of potential per unit length)Hence to re-establish balance point u wouid require to shift the point X to right(i.e toward"s point B)

Next suppose u increased the resistance which will result decrease in value of current in potential metre so for producing same potential drop=EMF of cell Q u hav to shift the balance point toward"s B..

Any satellite never fall itself !

Preparing for JEE?

Kickstart your preparation with new improved study material - Books & Online Test Series for JEE 2014/ 2015

@ INR 5,443/-

For Quick Info

if P is replaced by a cell with lower emf than Q then u can not achieve balance point.

now u can think y not............

B-

As no current flows through galvanometre, cell Q , or resistance S when potential metre is at balanced position.

Hence there will be no effect of decreasing\increasing value of S..

if R is increased then the value of current flowing in potential metre shouid decrease which will result in lower value of potential gradient(drop of potential per unit length)

Hence to re-establish balance point u wouid require to shift the point X to right(i.e toward"s point B)