Uncategorized - NEET Physics Chapterwise MCQs & PYQs

NEET Uncategorized MCQs & PYQs

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Question 21:

moderate

A cup of coffee cools from $90^\circ\text{C}$ to $80^\circ\text{C}$ in $t$ minutes, when the room temperature is $20^\circ\text{C}$. The time taken by a similar cup of coffee to cool from $80^\circ\text{C}$ to $60^\circ\text{C}$ at a room temperature same at $20^\circ\text{C}$ is: (2021)

Using average form of Newton's law of cooling: $\frac{90-80}{t} = K(\frac{90+80}{2}-20) \Rightarrow \frac{10}{t} = K(65)$. For second case: $\frac{80-60}{t'} = K(\frac{80+60}{2}-20) \Rightarrow \frac{20}{t'} = K(50)$. Dividing the two equations yields $t' = \frac{13}{5}t$.

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Question 22:

moderate

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 next $10$ minutes will be: (2016 – II)

By Newton's law of cooling: $\frac{3T-2T}{10} = K(\frac{3T+2T}{2}-T) \Rightarrow \frac{T}{10} = K(1.5T)$. For next 10 mins: $\frac{2T-T'}{10} = K(\frac{2T+T'}{2}-T)$. Substituting $K = \frac{1}{15}$, we get $2T - T' = \frac{1}{15}(0.5T' + T) \times 10$. Solving gives $T' = \frac{3}{2}T$.

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Question 23:

moderate

Certain quantity of water cools from $70^\circ\text{C}$ to $60^\circ\text{C}$ in the first $5$ minutes and to $54^\circ\text{C}$ in the next $5$ minutes. The temperature of the surroundings is: (2014)

Using Newton's law of cooling: $\frac{70-60}{5} = K(65-T_s) \Rightarrow 2 = K(65-T_s)$ and $\frac{60-54}{5} = K(57-T_s) \Rightarrow 1.2 = K(57-T_s)$. Dividing gives $\frac{2}{1.2} = \frac{65-T_s}{57-T_s} \Rightarrow 5(57-T_s) = 3(65-T_s)$. Solving for $T_s$, we get $T_s = 45^\circ\text{C}$.

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Question 24:

moderate

A beaker full of hot water is kept in a room. If it cools from $80^\circ\text{C}$ to $75^\circ\text{C}$ in $t_1$ minutes, from $75^\circ\text{C}$ to $70^\circ\text{C}$ in $t_2$ minutes and from $70^\circ\text{C}$ to $65^\circ\text{C}$ in $t_3$ minutes, then: (1995)

According to Newton's law of cooling, the rate of cooling is directly proportional to the temperature difference between the body and surroundings. As the water cools, the temperature difference decreases, slowing the rate of cooling. Thus, successive intervals take more time, giving $t_1 < t_2 < t_3$.

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Question 25:

moderate

Three stars A, B, C have surface temperatures $T_A$, $T_B$, $T_C$ respectively. Star A appears bluish, star B appears reddish and star C yellowish. Hence, (2020-Covid)

According to Wien's displacement law, $\lambda_{max} \propto \frac{1}{T}$. The wavelength of red is greater than yellow, which is greater than blue ($\lambda_B > \lambda_C > \lambda_A$). Therefore, the temperatures will be in the reverse order: $T_A > T_C > T_B$.

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Question 26:

moderate

The power radiated by a black body is $P$ and it radiates maximum energy at wavelength, $\lambda_0$. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength $\frac{3}{4}\lambda_0$, the power radiated by it becomes $nP$. The value of $n$ is: (2018)

From Wien's displacement law, $T \propto 1/\lambda_{max}$. Thus $T'/T = \lambda_0 / (3\lambda_0/4) = 4/3$. According to Stefan's law, Power $P \propto T^4$. So, $P'/P = (T'/T)^4 = (4/3)^4 = 256/81$.

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Question 27:

moderate

A spherical black body with a radius of $12\text{ cm}$ radiates $450\text{ watt}$ power at $500\text{ K}$. If the radius were halved and the temperature doubled, the power radiated in watt would be: (2017-Delhi)

Power radiated $P = \sigma A T^4 = \sigma (4\pi r^2) T^4 \Rightarrow P \propto r^2 T^4$. Given $r' = r/2$ and $T' = 2T$. Thus, $P' = P(1/2)^2 (2)^4 = P(1/4)(16) = 4P = 4 \times 450 = 1800\text{ W}$.

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Question 28:

moderate

A black body is at a temperature of $5760\text{ K}$. The energy of radiation emitted by the body at wavelength $250\text{ nm}$ is $U_1$, at wavelength $500\text{ nm}$ is $U_2$ and that at $1000\text{ nm}$ is $U_3$. Wien’s constant, $b = 2.88 \times 10^6\text{ nmK}$. Which of the following is correct? (2016 – I)

From Wien's displacement law, $\lambda_{max} = \frac{b}{T} = \frac{2.88 \times 10^6}{5760} = 500\text{ nm}$. This means maximum energy is radiated at $500\text{ nm}$. Therefore, the energy $U_2$ is maximum, so $U_2 > U_1$ and $U_2 > U_3$.

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Question 29:

moderate

On observing light from three different stars P, Q and R, it was found that intensity of violet color is maximum in the spectrum of P, the intensity of green color is maximum in the spectrum of R and the intensity of red color is maximum in the spectrum of Q. If $T_P$, $T_Q$ and $T_R$ are the respective absolute temperatures of P, Q and R then it can be concluded from the above observations that: (2015)

The wavelengths corresponding to maximum intensity are $\lambda_P < \lambda_R < \lambda_Q$ because violet has the shortest wavelength and red has the longest. From Wien's law, $T \propto 1/\lambda_{max}$. Thus, the temperatures are in the reverse order: $T_P > T_R > T_Q$.

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Question 30:

moderate

A piece of iron is heated in a flame. It first becomes dull red then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using: (2013)

As the temperature of the iron increases, the wavelength at which it emits maximum energy decreases, according to Wien's displacement law ($lambda_{max} T = text{constant}$). This causes the color to shift from longer wavelengths (red) to shorter wavelengths (yellow, then all visible mixing to white).

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