Question:
If water at \(100^{\circ}\text{C}\) cools in 10 minutes to \(80^{\circ}\text{C}\) and to \(65^{\circ}\text{C}\) in the next 10 minutes, then the room temperature will be ...
If water at \(100^{\circ}\text{C}\) cools in 10 minutes to \(80^{\circ}\text{C}\) and to \(65^{\circ}\text{C}\) in the next 10 minutes, then the room temperature will be ...
Show Hint
Newton law gives a geometric progression of temperature differences over equal times.
Updated On: Oct 1, 2026
- \(30^{\circ}\text{C}\)
- \(15^{\circ}\text{C}\)
- \(25^{\circ}\text{C}\)
- \(20^{\circ}\text{C}\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Set up
Let the room temperature be \(s\). By Newton's law, the excess temperature falls by the same ratio in equal times: \(\frac{80-s}{100-s} = \frac{65-s}{80-s}\).
Step 2: Solve
\((80-s)^2 = (100-s)(65-s)\). Expand: \(6400 - 160s + s^2 = 6500 - 165s + s^2\), so \(5s = 100\) and \(s = 20\).
Step 3: Check
Differences: \(100-20 = 80\), \(80-20 = 60\), \(65-20 = 45\). The ratios are \(\frac{60}{80} = \frac34\) and \(\frac{45}{60} = \frac34\), equal as required. Option (D).
Final Answer:
The room temperature is 20 degrees Celsius. \[ \boxed{\text{(D)}\ 20^{\circ}\text{C}} \]
Let the room temperature be \(s\). By Newton's law, the excess temperature falls by the same ratio in equal times: \(\frac{80-s}{100-s} = \frac{65-s}{80-s}\).
Step 2: Solve
\((80-s)^2 = (100-s)(65-s)\). Expand: \(6400 - 160s + s^2 = 6500 - 165s + s^2\), so \(5s = 100\) and \(s = 20\).
Step 3: Check
Differences: \(100-20 = 80\), \(80-20 = 60\), \(65-20 = 45\). The ratios are \(\frac{60}{80} = \frac34\) and \(\frac{45}{60} = \frac34\), equal as required. Option (D).
Final Answer:
The room temperature is 20 degrees Celsius. \[ \boxed{\text{(D)}\ 20^{\circ}\text{C}} \]
Was this answer helpful?
0
0
Top MHT CET Mathematics Questions
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- If $\int\frac{f\left(x\right)}{log \left(sin\,x\right)}dx = log\left[log\,sin\,x\right]+c$ then $f\left(x\right)=$
- If $\int\limits^{K}_0 \frac{dx}{2 + 18 x^2} = \frac{\pi}{24}$, then the value of K is
- If $\int^{\pi/2}_{0} \log\cos x dx =\frac{\pi}{2} \log\left(\frac{1}{2}\right)$ then $ \int^{\pi/2}_{0} \log\sec x dx = $
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
View More Questions
Top MHT CET Differential equations Questions
- The differential equation of all parabolas whose axis is $y-axis$ is
- The particular solution of the differential equation $xdy + 2ydx = 0$, when $x = 2, y = 1$ is
- If $\log_{10} \left(\frac{x^{3} - y^{3} }{x^{3} + y^3} \right) = 2$ then $ \frac{dy}{dx} = $
- If $y = \left(\tan^{-1} x\right)^{2}$ then $ \left(x^{2} + 1\right)^{2} \frac{d^{2}y}{dx^{2} } + 2x \left(x^{2} + 1 \right) \frac{dy}{dx} = $
- The equation of the locus of a point equidistant from the point A(1, 3) and B(-2, 1) is:
View More Questions
Top MHT CET Questions
- During r- DNA technology, which one of the following enzymes is used for cleaving DNA molecule ?
- In non uniform circular motion, the ratio of tangential to radial acceleration is (r = radius of circle, $v =$ speed of the particle, $\alpha =$ angular acceleration)
- The temperature of $32^{\circ}C$ is equivalent to
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- The heat of formation of water is $ 260\, kJ $ . How much $ H_2O $ is decomposed by $ 130\, kJ $ of heat ?
View More Questions