Home
Class 12
PHYSICS
A body cools in a surrounding which is a...

A body cools in a surrounding which is at a constant temperature of `theta_(0)`. Assume that it obeys Newton's law of cooling. Its temperature `theta` is plotted against time t. Tangent are drawn to the curve at the points `P(theta= theta_(2))` and `Q(theta = theta_(1))`. These tangents meet the time axis at angles of `phi_(2)` and `phi_(1)` as shown , then :-

A

`(tanphi_(2))/(tanphi_(1))=(theta_(1)-theta_(0))/(theta_(2)-theta_(0))`

B

`(tanphi_(2))/(tanphi_(1))=(theta_(2)-theta_(0))/(theta_(1)-theta_(0))`

C

`(tanphi_(1))/(tanphi_(2))=(theta_(1))/(theta_(2))`

D

`(tanphi_(1))/(tanphi_(2))=(theta_(2))/(theta_(1))`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • RACE

    ALLEN |Exercise Basic Maths (Thermal Physics) (Kinetic theory of gasess)|14 Videos
  • RACE

    ALLEN |Exercise Basic Maths (Thermal Physics) (Thermodynamic process)|20 Videos
  • RACE

    ALLEN |Exercise Basic Maths (Thermal Physics) (Calorimetry)|14 Videos
  • NEWTONS LAWS OF MOTION

    ALLEN |Exercise EXERCISE-III|28 Videos
  • SIMPLE HARMONIC MOTION

    ALLEN |Exercise Example|1 Videos

Similar Questions

Explore conceptually related problems

A body cools in a surrounding which is at constant temperature of theta_(0) . Assume that it obeys Newton's law of colling. Its temperature theta is plotted against time t . Tangents are drawn to the curve at the points P(theta=theta_(t)) and Q(theta=theta_(2)) . These tangents meet the time axis at angles of phi_(2) and phi_(1) , as shown

A body with an initial temperature theta_(1) is allowed to cool in a surrounding which is at a constant temperature of theta_(0) (theta lt theta_(1)) Assume that Newton's law of cooling is obeyed Let k = constant The temperature of the body after time t is best experssed by .

In Newton's law of cooling (d theta)/(dt)=-k(theta-theta_(0)) the constant k is proportional to .

If a piece of metal is heated to temperature theta and the allowed to cool in a room which is at temperature theta_0 , the graph between the temperature T of the metal and time t will be closet to

A body cools in a surrounding of constant temperature 30^@C Its heat capacity is 2 J//^@C . Initial temeprature of cooling is valid. The body cools to 38^@C in 10 min The temperature of the body In .^@C denoted by theta . The veriation of theta versus time t is best denoted as

A liquid in a beaker has temperature theta(t) at time t and theta_0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between log_e( theta-theta_0) and t is :

Let theta in (0,pi/4) and t_1=(tan theta)^(tan theta), t_2=(tan theta)^(cot theta), t_3=(cot theta)^(tan theta) and t_4=(cot theta)^(cot theta), then

A body cools in a surrounding of constant temperature 30^(@)C . Its heat capacity is 2J//^(@)C . Initial temperature of the body is 40^(@)C . Assume Newton's law of cooling is valid. The body cools to 36^(@)C in 10 minutes. In further 10 minutes it will cool from 36^(@)C to :

If the points (-2, 0), (-1,(1)/(sqrt(3))) and (cos theta, sin theta) are collinear, then the number of value of theta in [0, 2pi] is

A body cools in a surrounding of constant temperature 30^(@)C . Its heat capacity is 2J//^(@)C . Initial temperature of the body is 40^(@)C . Assume Newton's law of cooling is valid. The body cools to 36^(@)C in 10 minutes. When the body temperature has reached 36^(@)C . it is heated again so that it reaches to 40^(@)C in 10 minutes. Assume that the rate of loss of heat at 38^(@)C is the average rate of loss for the given time . The total heat required from a heater by the body is :