A body cool from 90^@C to 70^@C in 10 mi...

A body cool from `90^@C` to `70^@C` in 10 minutes if temperature of surrounding is `20^@C` find the time taken by body to cool from `60^@C` to `30^@C`. Assuming Newton’s law of cooling is valid.

20 minutes

40 minutes

60 minutes

80 minutes

Text Solution

Verified by Experts

The correct Answer is:

Topper's Solved these Questions

DIFFERENTIAL EQUATIONS
TARGET PUBLICATION|Exercise COMPETITIVE THINKING|124 Videos
DIFFERENTIAL EQUATIONS
TARGET PUBLICATION|Exercise EVALUATION TEST|25 Videos
DIFFERENTIAL EQUATIONS
TARGET PUBLICATION|Exercise EVALUATION TEST|25 Videos
DEFINITE INTEGRALS
TARGET PUBLICATION|Exercise EVALUATIO TEST|30 Videos
DIFFERENTIATION
TARGET PUBLICATION|Exercise EVALUATION TEST|30 Videos

Similar Questions

Explore conceptually related problems

A body takes 5 minutes to cool from 90^(@)C to 60^(@)C . If the temperature of the surroundings is 20^(@)C , the time taken by it to cool from 60^(@)C to 30^(@)C will be.

A body cools from 100° C to 70° C in 8 second. If temperature of surrounding is I 5° C, then time required for body to cool from 70° C to 40° C is

A body cools from 80^(@)C to 60^(@)C in 5 minutes. Then its average rate of cooling is

A body takes 4 min. to cool from 61^(@)C to 59^(@)C . If the temperature of the surroundings is 30^(@)C , the time taken by the body to cool from 51^(@)C to 49^(@)C is :

A body cools from 70^(@)C to 50^(@)C in 5minutes Temperature of surroundings is 20^(@)C Its temperature after next 10 minutes is .

A liquid cools from 70 °C to 60 °C in 5 minutes. If the temperature of the surrounding is constant at 30 °C, then the time taken by the liquid to cool from 60 °C to 50 °C is

A body takes T minutes to cool from 62^@C to 61^@C when the surrounding temperature is 30^@C . The time taken by the body to cool from 46^@C to 45^@C is

A body cools from 50^(@)C to 49.9^(@)C in 5 s. the temperature of the surroundings is 30^(@)C . How long will the body, take to cool from 40^(@)C to 39.9^(@)C ?

A liquid contained in a beaker cools from 80^(@)C to 60^(@)C in 10 minutes. If the temperature of the surrounding is 30^(@)C , find the time it will take to cool further to 48^(@)C . Here cooling take place ac cording to Newton's law of cooling.

TARGET PUBLICATION-DIFFERENTIAL EQUATIONS -CRITICAL THINKING

The integrating factor of the differential equation (1-x^(2))(dy)/(d...
Text Solution
|
An integrating factor of the differential equation x(dy)/(dx) +y l...
Text Solution
|
The integrating factor of (1+y^(2)) dx = (tan^(-1)y-x) dy is -
Text Solution
|
The solution of the differential equation (dy)/(dx)+2ycotx=3x^2cosec^2...
Text Solution
|
The solution of "dy"/"dx"+2ytan x =sin x is
Text Solution
|
The solution of the equation dy/dx=1/(x+y+1) is
Text Solution
|
The solution of the differential equation x"dy"/"dx"+y=x^2 +3x +2 is
Text Solution
|
The solution of the differential equation xlogx(dy)/(dx)+y=2logx is
Text Solution
|
The solution of the differential equation (dy)/(dx) +(3x^(2))/(1+x...
Text Solution
|
Solution of the differential equation (dy)/(dx)+ysec^2 x = tan x sec^2...
Text Solution
|
Solution of the equation (x+2y^3)(dy)/(dx)-y=0 , (y > 0) is
Text Solution
|
Solution of the equation (x+log y ) dy + y dx =0 is
Text Solution
|
The integrating factor of the differential equation (dy)/(dx)=y tan x-...
Text Solution
|
The solution of the differential equation x(dy)/(dx) = 2y + x^(3)e^(x...
Text Solution
|
If x dy =y (dx+y dy), y(1) =1and Y(x)gt 0. Then, y (-3) is epual to
Text Solution
|
The slope of the tangent to the curve at any point is equal to y + 2x....
Text Solution
|
The population of a town increases at a rate proportional to the popul...
Text Solution
|
The rate of increases of bacteria in a certain culture is proportional...
Text Solution
|
A body cool from 90^@C to 70^@C in 10 minutes if temperature of surrou...
Text Solution
|
The rate o growth of bacteria is proportional to the number present . ...
Text Solution
|