Messages posted by indzee

Dude,

Get me the page no.

Dude,

Ans. 1

Energy in both the test tubes will remain same, as both the springs are of same length and same spring constants. There is nothing to explain so elaborately. Here, the compression of spring cannot be taken into picture because the energy of the acid only depends upon the volume and the type of substance dissolved. The only change you will see is that the compressed spring will take more time to dissolve.

It is like dissolving one crystal of sugar in one test tube and one crushed crystal of sugar in another.

Dude,

See I tried my best in solving the problems ?. But I was not able to solve all of them,

I have solved,

1.118

1.119

1.121

1.122

1.125

1.129

1.134

1.135

1.137

1.140

1.162

1.165

1.184

I was incapable to solve the rest of the problems n I am sorry 4 that? I asked my miss for help? she said that most of the problems from Irodov are in IIT syllabus?. probably from 1.118-1.142 and then 1.151-1.185 ? Not sure about the thing? but you must try the problems that I have solved?. I am still in XI (Want to solve H C Verma B4 Irodov). At this elementary level I am completely helpless to help you? but I tried my best!!! Hope you crack IIT with a good rank and best of luck!!

Ill do as you say. I hav cultural day at my college for 3 days. So, it will take time for me to start solving. Ill nudge you the questions as I go on solving. Chalo then Good Night. I have rangoli, n drawing competition tomorrow, so I have to prepare 4 it.

OK

Well Tarin,

I havent yet started solving Irodov (Im still in 11^th).

But dont you worry, Ill see that Ill solve 1.3 and 1.7 as fast as I can and let you know the minimum number of problems you need to solve (till next week), to make things go perfect in mechanics. Ill nudge you when I reply in this forum. Till then, you dont waste time, get perfect with other topics. And one more thing, if you are through with H.C Verma, than you have already cracked AIEEE and 75% of IIT.

Girlfriend of course...

Carbon or Most wanted

Case I (max speed = 60km/hr)

a) Acceleration to reach the max speed:

u = 0m/s; v = 60km/hr = 16.67m/s; t = ?;

s = 1km = 1000m; a = ?

v² = u² + 2as

16.6² = 0² + 2 (a) (1000)

Therefore,

a = 0.13m/s

v = u + at

16.6 = 0 + 0.13 (t)

Therefore,

t = 127.6 sec

b) Declaration at the end of the station:

u = 60km/hr = 16.67m/s; v = 0m/s; t = ?;

s = 0.5km = 500m; a = ?

v² = u² + 2as

0² = 16.67² + 2 (a) (500)

Therefore,

a = - 0.27m/s

v = u + at

0 = 16.67 + (-0.27) (t)

Therefore,

t = 60.18 sec

c) Total time taken

Let the distance between the villages be x meters

Therefore, total time taken

T₁ = 127.6 + 60.18 + (x - 1000 ? 500) / 16.67

T₁ = 187.78 + (x - 1000 ? 500) /16.67

Case 2 (max speed = 20km/hr)

Follow the same procedure,

a) Acceleration to reach the max speed:

b) Declaration at the end of the station:

c) Total time taken

Then when you get T₂

Just subtract T₁ from T₂, to get the final time elapsed.

Now I don?t know how you got (2 min 40 sec) without (x). Some information is also lacking in Case 2. (ie. the distance covered to reach 20km/hr and declarate)

Dude... If I were you, I could have never done this... In such a situation, if you are in lack of time, then try to complete the kinematics, optics, thermodynamics.. n all your favorite chapters perfectly & as fast as possible ... And see that you have time left to complete your mechanics too...

Bcoz IITs hav all the rights to set the whole of the paper from mechanics only (though less probability) ... we cannot say anything about the importance of the chapters!!!

A better advice would be:

Dont waste your time asking such questions in forums, instead study mechanics !!!

Thanks to aatish

So we have finally got the graph of a double derivative!!!

Look at the graphs below (graphs of the example mentioned above)... and say me whether they are right .

Hey dude, that was the perfect answer w.r.t. "me", rather, I could have been too idiot to ask such a stupid question.... I understood what you mean!!

What I understood???

Suppose if,

y = x³+ x²+ x + 1

Then,

dy/dx = 3x²+ 2x + 1

So,

d²x/dy²= 6x + 2 (no need to consider this in graph)

Now, look at these graphs... he he... Yeh! I got it at last ... I cannot resist my xitement!! But say me if I have mistaken somewhere... and "Rate me" coz I am the person who has answered my question perfectly (I mean with graphs) but still, the credit goes to Aatish (special thanx to him)???

@ tarin

Sorry for the abve thing ... posted by mistake^^^

You mean a tangent drawn on a tangent, ie (if x=m as u say) dy/dx should be = dy2/dx2 , ... but thats not possible! coz we get two different values when we differentiate a function twice!! .. well I cannot claim that you are wrong (coz im still in 11th, n new in goiit)... but if I m wrong, then wats da actual answer???

Is there no one in goiit.com who can answer my question!! Plz help.. urgent

Then wats the actual answer!!

... plz plz .. I cannot proceed through my studies!!.. help... If we cannot represent it graphically, then plz gimmie some... reason.!!

Please gimme some satisfactory reply

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