back to directory
NEET PHYSICSThermal Properties of MatterMedium

Question

A piece of ice falls from a height hh so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice. The value of hh is: [Latent heat of ice is 3.4×105 J/kg3.4 \times 10^5 \text{ J/kg} and g=10 N/kgg = 10 \text{ N/kg}]

A

544 km544 \text{ km}

B

136 km136 \text{ km}

C

68 km68 \text{ km}

D

34 km34 \text{ km}

Step-by-Step Solution

When the piece of ice falls from a height hh, the loss in its potential energy is equal to the heat produced. Heat produced, Q=mghQ = mgh Given that only one-quarter of the heat produced is absorbed by the ice to melt completely. Heat absorbed by ice =14Q=14mgh= \frac{1}{4}Q = \frac{1}{4}mgh The heat required to completely melt the ice is Q=mLQ' = mL, where LL is the latent heat of fusion of ice. Equating the heat absorbed to the heat required for melting: 14mgh=mL\frac{1}{4}mgh = mL     h=4Lg\implies h = \frac{4L}{g} Substituting the given values (L=3.4×105 J/kgL = 3.4 \times 10^5 \text{ J/kg} and g=10 N/kgg = 10 \text{ N/kg}): h=4×3.4×10510 mh = \frac{4 \times 3.4 \times 10^5}{10} \text{ m} h=4×3.4×104 m=13.6×104 m=136000 m=136 kmh = 4 \times 3.4 \times 10^4 \text{ m} = 13.6 \times 10^4 \text{ m} = 136000 \text{ m} = 136 \text{ km}.

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 Matterheightcompletelyonequarterproducedabsorbed

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 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

When $1\text{ kg}$ of ice at $0^{\circ}\text{C}$ melts to water at $0^{\circ}\text{C}$, the resulting change in its entropy, taking latent heat of ice to be $80\text{ cal/g}$, is

A.$8 \times 10^4\text{ cal/K}$
B.$80\text{ cal/K}$
C.$293\text{ cal/K}$
D.$273\text{ cal/K}$
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 →