Messages posted by varun.tinkle

simple sum unnecessaruy complexed by the presence of kr/16lets find the vel at time t which is equl to -radistance can be found by by finding the speed so the speed is actually the magnitude of vel and here the magnitude is also ra (only sign is changing)now the time at which position is same is t=1/anow distance travelled wiill be v*t (herev refers to speed ) and the answer comes

i think the answer should be c as we can conserve angular momentum abt the bottom most pnt so mvr=mvr+iw here vand w can be found out easily

since no ext force acts on the systme positon of com will rmeian same thereofre m1x1=m2(x-x1)

i misintepreted it as an altitude other than latitude

special theory of relativity

2) answer 64 ms

see it is a simple sum.....

use first

p1v=rt here n=1 for poth cases we know t and also

for part 2

p2v=rt since rt and v is p1=p2

according to daltons law of pressure p=p1+p2 we know p1 and p2 and from this we get the volume of the container

solving it we get c=1800/1 atm

now we get v

for n2

1mole n2=0.7mole n2 +0.3 mole n

for o2

1mole 02=0.5moleo2 +0.5mole0

now again use

pv=nrt so we can know it

by again using datons law of pressure

see the avg speed of a molecule is independent of the mixture this means that the speed of the h2 will remain indepnedent of 02 molecues i hope now u can solve it

u can also solve it by normal calculus

i think the answr should be -rt

simple sum with a twist .... 4=k(l1-l) 5=k(l2-l) and 9=k(l3-l) where k is the natural lenght of the spring and l3 is the extension of the spring when a force of 9n is applied

no i\t quickly undergoes beta decay

area under the graph is equal to the displacement since the slope of the line is 4............and c=-9 at displacement 1.2m y=mx +c so apply it and get h of the trinagle and the base is 1.2ms=area under vt graph =h*b/2 solve it similarly for t=3.6 s

see this is how it should be

dw/dt=0.002t

or w=0.01t^2

or radial acc is w^2r =10^-4t^4

and tangential acc................................

is dw/dt is 0.02t

now proceeding as above

tan 60=radial acc/tang acc

actuallly thers no mistake in the first solution

kk if i am not wrong krishna sir is using ... the screening effect the simpler explaination is this .. after removing one electorn helium becomes a hydrogen like atom so enrgy of that electron =13.6z^2 here z=2 or it is equal to 13.6*4 so the toal enrgy required is 13.6*4+24.6

suresh sir i ahd actually this doubt only the asnwer by calculus comes different

from calculus

d@/dt=-k(@-10)

olving it like this we get @ as log (5/3)

now again using it we get t as 32.... where hv i gone wrong

first use constraint eq/....

ignoring the modulus sign

dv/dt=v^2/r

dv/v^2=dt/r

or -1/v=t/r

now intwgrating beween v(x) and v and from t=0 and t=t we get v

now we got v as a function of t now we use ds/dt to get s as a function of t now substitute the values to get v as a function of s

simple sum dude first since block m moves down 10cm so does block a move up 10cm.... now in the x axis the resultant horizontal force is 0 motioin of com in x qaxis is 0 so............ let the mass 2m mve a distance x in the left dirwection so from the ground frame2mcos45*(10-x)+10cos45*(10-X)=2MX ..... olve it if any mistake then nudge me

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor

@ INR 1,995/-

For Quick Info