back to directory
NEET PHYSICSThermal Properties of MatterEasy

Question

On observing light from three different stars PP, QQ, and RR, it was found that the intensity of the violet colour is maximum in the spectrum of PP, the intensity of the green colour is maximum in the spectrum of RR and the intensity of the red colour is maximum in the spectrum of QQ. If TPT_P, TQT_Q, and TRT_R are the respective absolute temperatures of PP, QQ, and RR, then it can be concluded from the above observations that:

A

TP>TQ>TRT_P > T_Q > T_R

B

TP>TR>TQT_P > T_R > T_Q

C

TP<TR<TQT_P < T_R < T_Q

D

TP<TQ<TRT_P < T_Q < T_R

Step-by-Step Solution

According to Wien's displacement law, the wavelength corresponding to maximum spectral emissive power is inversely proportional to the absolute temperature of the emitting body (λmax1T\lambda_{\text{max}} \propto \frac{1}{T}). The order of wavelengths for the given colours in the visible spectrum is λviolet<λgreen<λred\lambda_{\text{violet}} < \lambda_{\text{green}} < \lambda_{\text{red}}. Given that stars PP, RR, and QQ have maximum intensity at violet, green, and red colours respectively, we have λP<λR<λQ\lambda_P < \lambda_R < \lambda_Q. Since temperature is inversely proportional to the wavelength of maximum intensity, the temperatures of the stars will follow the reverse order: TP>TR>TQT_P > T_R > T_Q.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermal Properties of Matterobservingdifferentintensityvioletcolour

More Thermal Properties of Matter Questions

View all

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

A.$\frac{3}{4}$
B.$\frac{4}{3}$
C.$\frac{256}{81}$
D.$\frac{81}{256}$
MediumSolve

A piece of ice falls from a height $h$ so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice. The value of $h$ is: [Latent heat of ice is $3.4 \times 10^5 \text{ J/kg}$ and $g = 10 \text{ N/kg}$]

A.$544 \text{ km}$
B.$136 \text{ km}$
C.$68 \text{ km}$
D.$34 \text{ km}$
MediumSolve

A black body at $227^\circ\text{C}$ radiates heat at the rate of $7 \text{ cal cm}^{-2}\text{s}^{-1}$. At a temperature of $727^\circ\text{C}$, the rate of heat radiated in the same units will be:

A.$60$
B.$50$
C.$112$
D.$80$
MediumSolve

A copper rod of $88 \text{ cm}$ and an aluminium rod of an unknown length have an equal increase in their lengths independent of an increase in temperature. The length of the aluminium rod is: ($\alpha_{\text{Cu}} = 1.7 \times 10^{-5} \text{ K}^{-1}$ and $\alpha_{\text{Al}} = 2.2 \times 10^{-5} \text{ K}^{-1}$)

A.$68 \text{ cm}$
B.$6.8 \text{ cm}$
C.$113.9 \text{ cm}$
D.$88 \text{ cm}$
MediumSolve

A black body at $227^{\circ}\text{C}$ radiates heat at the rate of $7\text{ cal cm}^{-2}\text{s}^{-1}$. At a temperature of $727^{\circ}\text{C}$, the rate of heat radiated in the same units will be

A.$60$
B.$50$
C.$112$
D.$80$
MediumSolve

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_1$ and $T_2$ ($T_1 > T_2$). The rate of heat transfer $\frac{dQ}{dt}$ through the rod in a steady state is given by:

A.$\frac{dQ}{dt} = \frac{KL(T_1 - T_2)}{A}$
B.$\frac{dQ}{dt} = \frac{K(T_1 - T_2)}{LA}$
C.$\frac{dQ}{dt} = KLA(T_1 - T_2)$
D.$\frac{dQ}{dt} = \frac{KA(T_1 - T_2)}{L}$
EasySolve

Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at $100^{\circ}\text{C}$, while the other one is at $0^{\circ}\text{C}$. If the two bodies are brought into contact, then assuming no heat loss, the final common temperature is -

A.$50^{\circ}\text{C}$
B.more than $50^{\circ}\text{C}$
C.less than $50^{\circ}\text{C}$ but greater than $0^{\circ}\text{C}$
D.$0^{\circ}\text{C}$
MediumSolve

The total radiant energy per unit area per unit time, normal to the direction of incidence, received at a distance $R$ from the centre of a star of radius $r$, whose outer surface radiates as a black body at a temperature $T\text{ K}$ is given by (where $\sigma$ is Stefan's constant):

A.$\frac{\sigma r^2 T^4}{R^2}$
B.$\frac{\sigma r^2 T^4}{4\pi r^2}$
C.$\frac{\sigma r^4 T^4}{r^4}$
D.$\frac{4\pi \sigma r^2 T^4}{R^2}$
MediumSolve

This neet physics practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice physics sets →