Question:
A metallic sphere cools from 50\(^\circ\)C to 40\(^\circ\)C in 300 s. If atmospheric temperature around is 20\(^\circ\)C, then the sphere's temperature after the next 5 minutes will be close to :
A metallic sphere cools from 50\(^\circ\)C to 40\(^\circ\)C in 300 s. If atmospheric temperature around is 20\(^\circ\)C, then the sphere's temperature after the next 5 minutes will be close to :
Show Hint
Apply Newton's law of cooling in the form \(\Delta T/\Delta t = K(T_{avg}-T_0)\) for each 300 s interval and eliminate K.
Updated On: Nov 5, 2026
- 35\(^\circ\)C
- 31\(^\circ\)C
- 33\(^\circ\)C
- 28\(^\circ\)C
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Newton's law (average form): \(\dfrac{T_1-T_2}{t} = K\left(\dfrac{T_1+T_2}{2}-T_0\right)\).
First interval: \(\dfrac{50-40}{300} = K(45-20) = 25K\), so \(300K = 0.4\).
Second interval, final temperature T: \(\dfrac{40-T}{300} = K\left(\dfrac{40+T}{2}-20\right)\), so \(40-T = 0.4\cdot\dfrac{T}{2} = 0.2T\).
\(40 = 1.2T\), \(T = 33.3^\circ\mathrm{C} \approx 33^\circ\mathrm{C}\).
Answer: (3) 33\(^\circ\)C
First interval: \(\dfrac{50-40}{300} = K(45-20) = 25K\), so \(300K = 0.4\).
Second interval, final temperature T: \(\dfrac{40-T}{300} = K\left(\dfrac{40+T}{2}-20\right)\), so \(40-T = 0.4\cdot\dfrac{T}{2} = 0.2T\).
\(40 = 1.2T\), \(T = 33.3^\circ\mathrm{C} \approx 33^\circ\mathrm{C}\).
Answer: (3) 33\(^\circ\)C
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
- A particle starts with an initial velocity of 10.0 m/s along x-direction and accelerates uniformly at the rate of 2.0 m/s$^2$. The time taken by the particle to reach the velocity of 60.0 m/s is
View More Questions
Top JEE Main Newton’s Law Of Cooling Questions
- A cup of coffee cools from 90°C to 80°C in \( t \) minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:
- In 5 minutes, a body cools from \(75^\circ\mathrm{C}\) to \(65^\circ\mathrm{C}\) at room temperature of \(25^\circ\mathrm{C}\). The temperature of body at the end of next 5 minutes is ________ \(^\circ\mathrm{C}\).
- A hot body, obeying Newton's law of cooling is cooling down from its peak value \(80^\circ\mathrm{C}\) to an ambient temperature of \(30^\circ\mathrm{C}\). It takes 5 minutes in cooling down from \(80^\circ\mathrm{C}\) to \(40^\circ\mathrm{C}\). How much time will it take to cool down from \(62^\circ\mathrm{C}\) to \(32^\circ\mathrm{C}\) ? (Given \(\ln 2=0.693,\ \ln 5=1.609\) )
- Hot water cools from \(60^\circ\mathrm{C}\) to \(50^\circ\mathrm{C}\) in the first 10 minutes and to \(42^\circ\mathrm{C}\) in the next 10 minutes. The temperature of the surroundings is:
- If a piece of metal is heated to temperature \(\theta\) and then 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 closest to
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions