Ask Questions, Get Answers
Menu
X
JEEMAIN Crash Practice
15 Test Series
NEET Crash Practice
5 Test Series
CBSE XII
Math
JEEMAIN
Math
Physics
Chemistry
Practice Test Series
CBSE XI
Math
NEET
Physics
Chemistry
Biology - XII
Biology - XI
Olympiad class V
Math - 5 Test Series
Olympiad class VI
Math - 5 Test Series
studyplans
JEEMAIN Crash Practice
15 Test Series
NEET Crash Practice
5 Test Series
CBSE XII
Math
JEEMAIN
Math
Physics
Chemistry
Practice Test Series
CBSE XI
Math
NEET
Physics
Chemistry
Biology - XII
Biology - XI
Olympiad class V
Math - 5 Test Series
Olympiad class VI
Math - 5 Test Series
mobile
exams
ask
sample papers
tutors
pricing
sign-in
Download our FREE mobile app with 1000+ tests for CBSE, JEE MAIN, NEET
X
Search
Topics
Want to ask us a question?
Click here
Browse Questions
Student Questions
Ad
Home
>>
CBSE XII
>>
Math
>>
Differential Equations
0
votes
A population grows at the rate of 8% per year. How long does it take for the population to double?
cbse
class12
additionalproblem
kvquestionbank2012
ch9
q21
p34
sec-b
math
Share
asked
Feb 4, 2013
by
meena.p
retagged
Dec 19, 2013
Please
log in
or
register
to add a comment.
Can you answer this question?
Do not ask me again to answer questions
Please
log in
or
register
to answer this question.
Related questions
0
votes
0
answers
When the interest is compounded continuously, the amount of money invested increases at a rate proportional to its size. If Rs.1000 is invested at 10% compounded continuously, in how many years will the original investment double itself?
asked
Feb 4, 2013
by
meena.p
cbse
class12
additionalproblem
kvquestionbank2012
ch9
q20
p34
sec-c
math
0
votes
0
answers
A wet porous substance in the open air losses its moisture at a rate proportional to the moisture content. If a sheet hung in the wind loses half of its moisture during the first hour, when will it have lost 95% moisture, weather conditions remaining the same.
asked
Feb 4, 2013
by
meena.p
cbse
class12
additionalproblem
kvquestionbank2012
ch9
q22
p34
sec-b
math
0
votes
0
answers
Newtons law of cooling states that the rate of change of the temperature T of an object is proportional to the difference between T and the (constant) temperature t of the surrounding medium. We can write it as \[\frac{dT}{dt}=-K(T-t),K>0\; constant.\]A cup of coffee is served at $ 185^\circ F$ in a room where the temperature is $65^\circ F .$ Two minutes later the temperature of the coffee has dropped to $155^\circ F.$(log 3/4=0.144, log3 = 1.09872). Find the time required for coffee to have $105^\circ F$ temperature.
asked
Feb 4, 2013
by
meena.p
cbse
class12
additionalproblem
kvquestionbank2012
ch9
q25
p34
sec-c
math
0
votes
0
answers
The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time 't'.
asked
Feb 4, 2013
by
meena.p
cbse
class12
additionalproblem
kvquestionbank2012
ch9
q23
p34
sec-a
math
–1
vote
0
answers
Solve: \[\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}\] (Hint : take $ x=r\cos \theta,y=r \sin \theta,$ so that $x^2+y^2=r^2)$
asked
Feb 4, 2013
by
meena.p
cbse
class12
additionalproblem
kvquestionbank2012
ch9
q19
p34
sec-c
math
0
votes
0
answers
A curve passing through the point (1,1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of p from the x-axis. Determine the equation of the curve.
asked
Feb 4, 2013
by
meena.p
cbse
class12
additionalproblem
kvquestionbank2012
ch9
q24
p34
sec-b
math
0
votes
0
answers
Solve:\[(x^3-3xy^2)dx=(y^3-3x^2y)dy\]
asked
Feb 4, 2013
by
meena.p
cbse
class12
additionalproblem
kvquestionbank2012
ch9
q17
p34
sec-b
math
Ask Question
Tag:
Math
Phy
Chem
Bio
Other
SUBMIT QUESTION
►
Please Wait
Take Test
JEEMAIN Crash Practice
15 Test Series
NEET Crash Practice
5 Test Series
JEEMAIN
350+ TESTS
NEET
320+ TESTS
CBSE XI MATH
50+ TESTS
CBSE XII MATH
80+ TESTS
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...