In a room where the temperature is $30^{\circ} C$, a body cools from $61^{\circ} C$ to $59^{\circ} C$ in $4$ minutes. The time taken by the body to cool from $51^{\circ} C$ to $49^{\circ} C$ will be:

  • A
    $4$ minutes
  • B
    $6$ minutes
  • C
    $5$ minutes
  • D
    $8$ minutes

Explore More

Similar Questions

$A$ body cools from a temperature $3T$ to $2T$ in $10$ minutes. The room temperature is $T$. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of the next $10$ minutes will be

$A$ body takes $T$ minutes to cool from $62^{\circ}C$ to $61^{\circ}C$ when the surrounding temperature is $30^{\circ}C$. The time taken by the body to cool from $46^{\circ}C$ to $45^{\circ}C$ is:

$A$ solid copper cube of edges $1\;cm$ is suspended in an evacuated enclosure. Its temperature is found to fall from $100^{\circ}C$ to $99^{\circ}C$ in $100\;s$. Another solid copper cube of edges $2\;cm$,with similar surface nature,is suspended in a similar manner. The time required for this cube to cool from $100^{\circ}C$ to $99^{\circ}C$ will be approximately ...... $s$.

$A$ body cools from $50.0^{\circ}C$ to $49.9^{\circ}C$ in $5\;s$. How long will it take to cool from $40.0^{\circ}C$ to $39.9^{\circ}C$? Assume the temperature of surroundings to be $30.0^{\circ}C$ and Newton's law of cooling to be valid. The time taken is ....... $s$.

The circuit below is used to heat water kept in a bucket. Assuming heat loss only by Newton's law of cooling, the variation in the temperature of the water in the bucket as a function of time is depicted by

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo