A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The ti me taken by a similar cup of coffee to cool from 80°C to 60°C at room temperature same at 20°C. (1) 13/10 t (2) 13/5 t (3) 10/13 t (4) 5/13 t

Ruhul Amin Alig

Question is Below this important notice-

Physics

NEET (UG) 2021 PHYSICS PAPER

By- Organised Notes
(free eduction for all)

Mapping with NCERT Class XI & XII Books

IMPORTANT NOTE : With every question below a mapping tag is added which shows the reference of

NCERT Book Page Number and the concerned question from which that question is matching or derived.

If students carefully analyze every question and related NCERT mapping then you can feel that proper

coverage of NCERT Book and PYQs are covered properly with derived level questions then NEET

2021 was a cake walk for aspirants.

Hope this will be an eye-opener for NEET 2022 & upcoming NEET aspirants and boost them for their

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Question: From NCERT NEET 2021

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 a similar cup of coffee to cool from 80°C to 60°C at room temperature same at 20°C.
(1) 13/10 t
(2) 13/5 t
(3) 10/13 t
(4) 5/13 t

Mapping : Concept explained on Page-296 and question matching with solved example-11.8

Physics>Thermal Properties of Matter>Newton's Law of Cooling

Answer ▽

BY LOGIC;

BY FORMULA:

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Concept used NEET 2021 Physics;

Newton’s Law of Cooling;

the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surroundings.
dQ/dt ∝ (q – qs)
dQ/dt = -k[q – qs)]

Methods to Apply Newton’s Law of Cooling

when we need only approximate values from Newton’s law, we can assume a constant rate of cooling, which is equal to the rate of cooling corresponding to the average temperature of the body during the interval.
i.e. dθ\dt = k(<q> – q0)

Derived PYQs; ⬇️

Question: From NCERT NEET Derived PYQ

A body at temperature 40ºC is kept in a surrounding of constant temperature 20ºC. It is observed that its temperature falls to 35ºC in 10 minutes. Find how much more time will it take for the body to attain a temperature of 30ºC.

a) 10 min

b)20 min

c) 8 min

d) 14 min

Answer ▽

d) 14 min

By Logic;

By Formula;

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Question: From NCERT NEET Derived PYQ

The oil is heated to 70oC. It cools to 50oC after 6 minutes. Calculate the time taken by the oil to cool from 50oC to 40oC given the surrounding temperature Ts = 25oC.

a)10 min

b) 5 min

c) 15 min

d) 20 min

Answer ▽

5min

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Question: From NCERT NEET Derived PYQ

Water is heated to 80oC for 10 min. How much would be the temperature if k = 0.56 per min and the surrounding temperature is 25oC?

a) 50

b) 25

c) 56

d) 35

Answer ▽

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Question: From NCERT NEET Derived PYQ

A body cools from $70$ $^{o}$ C to $50$ $^{o}$ C in $5$ minutes. Temperature of surroundings is $20$ $^{o}$ C. Its temperature after next $10$ minutes is

25^{\circ} C

30^{\circ} C

35^{\circ} C

45^{\circ} C

Answer ▽

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Question: From NCERT NEET Derived PYQ

Hot water cools from $6 0^{o} C$ to $5 0^{o} C$ in the first $10$ minutes and to $4 2^{o} C$ in the next $10$ minutes. The temperature of the surroundings is :

$2 5^{o} C$

$1 0^{o} C$

$1 5^{o} C$

$2 0^{o} C$

Medium

JEE Mains

Answer ▽

$1 0^{o} C$

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Question: From NCERT NEET 2014

Certain quantity of water cools from $7 0^{o} C$ to $6 0^{o} C$ in the first $5$ minutes and to $5 4^{o} C$ in the next $5$ minutes. The temperature of the surroundings is ;

$4 5^{o} C$

$2 0^{o} C$

$4 2^{o} C$

$1 0^{o} C$

Medium

NEET

Mapping : Physics>Thermal Properties of Matter>Newton's Law of Cooling

Answer ▽

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Question: From NCERT NEET Derived PYQ

The plots of intensity versus wavelength for three black bodies at temperatures $T_{1}, T_{2}$ and $T_{3}$ respectively are shown in fig. Their temperatures are such that:

$T_{1} > T_{2} > T_{3}$

$T_{1} > T_{3} > T_{2}$

$T_{2} > T_{3} > T_{1}$

$T_{3} > T_{2} > T_{1}$

Medium

Answer ▽

Correct option is B)

As we know that from emissive power $E \propto T^{4}$ and from wein's displacement law $λ_{m a x} \times T = c o n s t a n t$ . So if emissivity will be maximum them simultaneously it's temperature will be maximum, but from the wein's law we can say that the particle which has maximum temperature will have lower wavelength.
So, $E \propto T \propto \frac{1}{λ}$
So, we can say:
$T_{1} > T_{3} > T_{2}$
$E_{1} > E_{3} > E_{2}$
But, $λ_{1} < λ_{3} < λ_{2}$

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Question: From NCERT NEET Derived PYQ

A body cools down from $80 ° C to 70 ° C$ in 5 minutes. The temperature of the body will fall from $70 ° C to 60 ° C$ in a time

1. Less than 5 minutes

2. equal to 5 minutes

3. More than 5 minutes

4. Can't say anything like the temperature of surrounding is not known

Answer ▽

3. More than 5 minutes

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Question: From NCERT NEET 2016

A body cools from a temperature 3T to 2T in10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be -

(a) $\frac{7}{4} T$ (b) $\frac{3}{2} T$

Mapping : Physics>Thermal Properties of Matter>Newton's Law of Cooling

Answer ▽

b) $\frac{3}{2} T$

From newton's law of cooling

$\\ \frac{3T-2T}{10}=K.(\frac{3T+2T}{2}-T)=\frac{3}{2}KT$

$\frac{T}{10}=\frac{3KT}{2}\ \: \: \: \:or\: \: \: \: K=\frac{1}{15}\ \: \: \: ..........(1)$

Let us say temperature after 10 minutes become T1

Then $\frac{2T-T_{1}}{10}=K\left [ \frac{2T+T_{1}}{2}-T \right ]$

$\\ \frac{2T-T_{1}}{10}=\frac{1}{15}(\frac{T_{1}}{2})\\ or\ \: \: 6T-3T_{1}=T_{1}\ \: \: \: or\ \: \: T_{1}=\frac{3}{2}T$

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Question: From NCERT NEET 2019

An object kept in a large room having an air temperature of 25° C takes 12 minutes to cool from 80° C to 70° C. The time taken to cool for the same object from 70° C to 60° C would be nearly -

1. 10 min

2. 12 min

3. 20 min

4. 15 min

Mapping : Physics>Thermal Properties of Matter>Newton's Law of Cooling

Answer ▽

By Formula;

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Question: From NCERT NEET Derived PYQ

A bucket full of hot water cools from 75 $° C$ $° C$ $° C$ to 70 $° C$ $° C$ $° C$ in time $T_{1}$ $T_{1}$ $T_{1}$ , from 70 $° C$ $° C$ $° C$ to 65 $° C$ $° C$ $° C$ in time $T_{2}$ $T_{2}$ $T_{2}$ and from 65 $° C$ $° C$ $° C$ to 60 $° C$ $° C$ $° C$ in time $T_{3}$ $T_{3}$ $T_{3}$ , then
(1) $T_{1} = T_{2} = T_{3}$ $T_{1} = T_{2} = T_{3}$ $T_{1} = T_{2} = T_{3}$ $T_{1} = T_{2} = T_{3}$ $T_{1} = T_{2} = T_{3}$ $T_{1} = T_{2} = T_{3}$ $T_{1} = T_{2} = T_{3}$ $T_{1} = T_{2} = T_{3}$ $T_{1} = T_{2} = T_{3}$
(2) $T_{1} > T_{2} > T_{3}$ $T_{1} > T_{2} > T_{3}$ $T_{1} > T_{2} > T_{3}$ $T_{1} > T_{2} > T_{3}$ $T_{1} > T_{2} > T_{3}$ $T_{1} > T_{2} > T_{3}$ $T_{1} > T_{2} > T_{3}$ $T_{1} > T_{2} > T_{3}$ $T_{1} > T_{2} > T_{3}$

(3)

T_{1} < T_{2} < T_{3}

T_{1} < T_{2} < T_{3}

T_{1} < T_{2} < T_{3}

T_{1} < T_{2} < T_{3}

T_{1} < T_{2} < T_{3}

T_{1} < T_{2} < T_{3}

T_{1} < T_{2} < T_{3}

T_{1} < T_{2} < T_{3}

T_{1} < T_{2} < T_{3}

(4)

T_{1} > T_{2} < T_{3}

T_{1} > T_{2} < T_{3}

T_{1} > T_{2} < T_{3}

T_{1} > T_{2} < T_{3}

T_{1} > T_{2} < T_{3}

T_{1} > T_{2} < T_{3}

T_{1} > T_{2} < T_{3}

T_{1} > T_{2} < T_{3}

Answer ▽

(3) $T_{1} < T_{2} < T_{3}$

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Question: From NCERT NEET Derived PYQ

Consider two hot bodies $B_{1}$ and $B_{2}$ which have temperatures 100 $° C$ and 80 $° C$ respectively at t=0. The temperature of the surroundings is 40 $° C$ . The ratio of the respective rates of cooling $R_{1}$ and $R_{2}$ of these two bodies at t=0 will be
(1) $R_{1} : R_{2} = 3 : 2$

(2) $R_{1} : R_{2} = 5 : 4$

(3) $R_{1} : R_{2} = 2 : 3$

(4) $R_{1} : R_{2} = 4 : 5$

Answer ▽

(1)

R_{1} : R_{2} = 3 : 2

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Question: From NCERT NEET Derived PYQ

Newton's law of cooling is a special case of
(1) Stefan's law

(2) Kirchhoff's law

(3) Wien's law

(4) Planck's law

Answer ▽

(1) Stefan's law

Newton's law cooling is a special case of Stefan-Boltzmann’s law where the temperature difference of the body and the surrounding is very small.

Let us understand what the other laws given in the options tell us.
(i) Kirchoff’s law: It says that the ratio of emissive power to absorptive power is the same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature.
(ii) Wien’s law: According to this law, the product of wavelength for which a black body emits radiation of maximum intensity and the temperature of the body is constant.

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Question: From NCERT NEET Derived PYQ

In Newton's experiment of cooling, the water equivalent of two similar calorimeters is 10 gm each. They are filled with 350 gm of water and 300 gm of a liquid (equal volumes) separately. The time taken by water and liquid to cool from 70°C to 60°C is 3 min and 95 sec respectively. The specific heat of the liquid will be
(1) 0.3 Cal/gm $\times$ °C

(2) 0.5 Cal/gm $\times$ °C

(3) 0.6 Cal/gm $\times$ °C

(4) 0.8 Cal/gm $\times$ °C

Answer ▽

The correct option is C

0.6 Cal/gm ´°C

$S_{l} = \frac{1}{m_{l}} [\frac{t_{l}}{t_{w}} (M_{w} C_{w} + W) - W]$

= $\frac{1}{300} [\frac{95}{3 \times 60} (350 \times 1 + 10) - 10] = 0.6 C a l / g m \times^{\circ} C$

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Question: From NCERT NEET Derived PYQ

The rates of cooling of two different liquids put in exactly similar calorimeters and kept in identical surroundings are the same if

(1) The masses of the liquids are equal

(2) Equal masses of the liquids at the same temperature are taken

(3) Different volumes of the liquids at the same temperature are taken

(4) Equal volumes of the liquids at the same temperature are taken

Answer ▽

(4) Equal volumes of the liquids at the same temperature are taken

Rate of loss of heat will depend on the surface area
So, when volume is same, the surface area will also be same

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Question: From NCERT NEET Derived PYQ

The temperature of a liquid drops from 365 K to 361 K in 2 minutes. Find the time during which temperature of the liquid drops from 344 K to 342 K . Temperature of room is 293 K
(a) 84 sec (b) 72 sec
(c) 66 sec (d) 60 sec

Answer ▽

(a) 84 sec

The correct option is A $84 sec$
Newton's law of cooling says that,
$\frac{θ_{1} - θ_{2}}{t} \propto (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
From the data given in the question,
$\frac{365 - 361}{2} = K [\frac{365 + 361}{2} - 293]$
$\Rightarrow K = \frac{1}{35} {min}^{- 1}$
Again, for change in temperature from $344 K$ to $342 K$ ,
$\frac{344 - 342}{t} = \frac{1}{35} [\frac{344 + 342}{2} - 293]$
$\Rightarrow \frac{2}{t} = \frac{10}{7}$
$\Rightarrow t = \frac{7}{5} min = \frac{7}{5} \times 60 = 84 sec$
Thus, option (a) is the correct answer.

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Question: From NCERT NEET Derived PYQ

Newton’s law of cooling, holds good only if the temperature difference between the body and the surroundings is
(1) Less than 10 $° C$

(2) More than 10 $° C$

(3) Less than 100 $° C$

(4) More than 100 $° C$

Answer ▽

(1) Less than 10 $° C$

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Question: From NCERT NEET Derived PYQ

The temperature of a body falls from 50 $° C$ to 40 $° C$ in 10 minutes. If the temperature of the surroundings is 20 $° C$ , then temperature of the body after another 10 minutes will be-
(1) 36.6 $° C$

(2) 33.3 $° C$

(3) 35 $° C$

(4) 30 $° C$

Answer ▽

(2) 33.3 $° C$

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Question: From NCERT NEET Derived PYQ

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.
(1) 5 min

(2) 8 min

(3) 11 min

(4) 12 min

Answer ▽

(3) 11 min

$\frac{d T}{d t}$ = $\frac{9 0 - 6 0}{5} = K [\frac{9 0 + 6 0}{2} - 20] \Rightarrow K = \frac{6}{5 5}$

And again $\frac{6 0 - 3 0}{t} = \frac{6}{5 5} [\frac{6 0 + 3 0}{2} - 20]$ $\Rightarrow t = 11$ minute

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Question: From NCERT NEET Derived PYQ

An object is cooled from 75°C to 65°C in 2 minutes in a room at 30°C. The time taken to cool another object from 55°C to 45°C in the same room in minutes is -

(1) 4

(2) 5

(3) 6

(4) 7

Answer ▽

According to Newton's law of cooling,
$- \frac{d θ}{d t} \propto (θ - θ_{0})$
$- \frac{d θ}{d t} = K (θ - θ_{0})$
where, $K$ is a constant of proportionality.
Now, $\frac{75 - 65}{2} = K (\frac{75 + 65}{2} - 30) = K \times 40 \Rightarrow K = \frac{1}{8}$
In case of second identical object
$\frac{55 - 45}{t} = K (\frac{55 + 45}{2} - 30)$
$\frac{10}{t} = \frac{1}{8} \times 20$
$\Rightarrow t = \frac{80}{20} = 4 min$

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Question: From NCERT NEET Derived PYQ

A cane is taken out from a refrigerator at 0°C. The atmospheric temperature is 25°C. If $t_{1}$ is the time taken to heat from 0°C to 5°C and $t_{2}$ is the time taken from 10°C to 15°C, then
(1) $t_{1} > t_{2}$

(2) $t_{1} < t_{2}$

(3) $t_{1} = t_{2}$

(4) There is no relation

Answer ▽

2) $t_{1} < t_{2}$

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Question: From NCERT NEET Derived PYQ

A block of metal is heated to a temperature much higher than the room temperature and allowed to cool in a room free from air currents. Which of the following curves correctly represents the rate of cooling ?

(1)

(2)

(3)

(4)

Answer ▽

(2)

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Question: From NCERT NEET Derived PYQ

A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?

(a) The rate of cooling is constant till milk attains the temperature of the surrounding.

(b) The temperature of milk falls off exponentially with time.

(c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools.

(d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings.

(1) (a, b, c)

(2) (a, d)

(3) (b, c, d)

(4) (a, c, d)

Answer ▽

(3) (b, c, d)

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Question: From NCERT NEET Derived PYQ

A pan filled with hot food cools from 94 °C to 86 °C in 2 minutes when the room temperature is at 20 °C. How long will it take to cool from 71 °C to 69 °C?

1. 50 sec

2. 52 sec

3. 42 sec

4. 48 sec

Answer ▽

3. 42 sec

⬆️Prev____@organised notes_____End⬇️

Subtopic;Newton's Law of Cooling

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A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The ti me taken by a similar cup of coffee to cool from 80°C to 60°C at room temperature same at 20°C. (1) 13/10 t (2) 13/5 t (3) 10/13 t (4) 5/13 t

Question is Below this important notice-

By- Organised Notes
(free eduction for all)

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 a similar cup of coffee to cool from 80°C to 60°C at room temperature same at 20°C.
(1) 13/10 t
(2) 13/5 t
(3) 10/13 t
(4) 5/13 t

Concept used NEET 2021 Physics;

Newton’s Law of Cooling;

Methods to Apply Newton’s Law of Cooling

Derived PYQs; ⬇️

A body cools from $70$ $^{o}$ C to $50$ $^{o}$ C in $5$ minutes. Temperature of surroundings is $20$ $^{o}$ C. Its temperature after next $10$ minutes is

Hot water cools from $6 0^{o} C$ to $5 0^{o} C$ in the first $10$ minutes and to $4 2^{o} C$ in the next $10$ minutes. The temperature of the surroundings is :

$2 5^{o} C$

$1 0^{o} C$

$1 5^{o} C$

$2 0^{o} C$

$1 0^{o} C$

Certain quantity of water cools from $7 0^{o} C$ to $6 0^{o} C$ in the first $5$ minutes and to $5 4^{o} C$ in the next $5$ minutes. The temperature of the surroundings is ;

$4 5^{o} C$

$2 0^{o} C$

$4 2^{o} C$

$1 0^{o} C$

The plots of intensity versus wavelength for three black bodies at temperatures $T_{1}, T_{2}$ and $T_{3}$ respectively are shown in fig. Their temperatures are such that:

$T_{1} > T_{2} > T_{3}$

$T_{1} > T_{3} > T_{2}$

$T_{2} > T_{3} > T_{1}$

$T_{3} > T_{2} > T_{1}$

$\frac{d T}{d t}$ = $\frac{9 0 - 6 0}{5} = K [\frac{9 0 + 6 0}{2} - 20] \Rightarrow K = \frac{6}{5 5}$

And again $\frac{6 0 - 3 0}{t} = \frac{6}{5 5} [\frac{6 0 + 3 0}{2} - 20]$ $\Rightarrow t = 11$ minute

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A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The ti me taken by a similar cup of coffee to cool from 80°C to 60°C at room temperature same at 20°C. (1) 13/10 t (2) 13/5 t (3) 10/13 t (4) 5/13 t

Question is Below this important notice-

By- Organised Notes (free eduction for all)

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 a similar cup of coffee to cool from 80°C to 60°C at room temperature same at 20°C. (1) 13/10 t (2) 13/5 t (3) 10/13 t (4) 5/13 t

Concept used NEET 2021 Physics;

Newton’s Law of Cooling;

Methods to Apply Newton’s Law of Cooling

Derived PYQs; ⬇️

A body cools from 70oC to 50oC in 5 minutes. Temperature of surroundings is 20oC. Its temperature after next 10 minutes is

Hot water cools from 60oC to 50oC in the first 10 minutes and to 42oC in the next 10 minutes. The temperature of the surroundings is :

25oC

10oC

15oC

20oC

10oC

Certain quantity of water cools from 70oC to 60oC in the first 5 minutes and to 54oC in the next 5 minutes. The temperature of the surroundings is ;

45oC

20oC

42oC

10oC

The plots of intensity versus wavelength for three black bodies at temperatures T1​,T2​ and T3​ respectively are shown in fig. Their temperatures are such that:

T1​>T2​>T3​

T1​>T3​>T2​

T2​>T3​>T1​

T3​>T2​>T1​

A bucket full of hot water cools from 75°C°C to 70°C°C in time T1T1, from 70°C°C to 65°C°C in time T2T2 and from 65°C°C to 60°C°C in time T3T3, then (1) T1=T2=T3T1=T2=T3 (2) T1>T2>T3T1>T2>T3

dtdT​ = 590−60​=K[290+60​−20]⇒K=556​ And again t60−30​=556​[260+30​−20] ⇒t=11 minute

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By- Organised Notes
(free eduction for all)

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 a similar cup of coffee to cool from 80°C to 60°C at room temperature same at 20°C.
(1) 13/10 t
(2) 13/5 t
(3) 10/13 t
(4) 5/13 t

A body cools from $70$ $^{o}$ C to $50$ $^{o}$ C in $5$ minutes. Temperature of surroundings is $20$ $^{o}$ C. Its temperature after next $10$ minutes is

Hot water cools from $6 0^{o} C$ to $5 0^{o} C$ in the first $10$ minutes and to $4 2^{o} C$ in the next $10$ minutes. The temperature of the surroundings is :

$2 5^{o} C$

$1 0^{o} C$

$1 5^{o} C$

$2 0^{o} C$

$1 0^{o} C$

Certain quantity of water cools from $7 0^{o} C$ to $6 0^{o} C$ in the first $5$ minutes and to $5 4^{o} C$ in the next $5$ minutes. The temperature of the surroundings is ;

$4 5^{o} C$

$2 0^{o} C$

$4 2^{o} C$

$1 0^{o} C$

The plots of intensity versus wavelength for three black bodies at temperatures $T_{1}, T_{2}$ and $T_{3}$ respectively are shown in fig. Their temperatures are such that:

$T_{1} > T_{2} > T_{3}$

$T_{1} > T_{3} > T_{2}$

$T_{2} > T_{3} > T_{1}$

$T_{3} > T_{2} > T_{1}$

$\frac{d T}{d t}$ = $\frac{9 0 - 6 0}{5} = K [\frac{9 0 + 6 0}{2} - 20] \Rightarrow K = \frac{6}{5 5}$

And again $\frac{6 0 - 3 0}{t} = \frac{6}{5 5} [\frac{6 0 + 3 0}{2} - 20]$ $\Rightarrow t = 11$ minute