
A ball is dropped on a floor from a height of 2m. After the collision it rises upto a height of 1m. Assuming that $20\% $ of mechanical energy is lost in the form of thermal energy. If the specific heat capacity of the ball is $800J/K$ then the rise in temperature of the ball during collision is (Take $g = 10m/{s^2}$).
A. $2.5 \times {10^{ - 3}}$ $^oC$
B. $7.5 \times {10^{ - 3}}$ $^oC$
C. $1.5 \times {10^{ - 3}}$ $^oC$
D. $5 \times {10^{ - 3}}$ $^oC$
Answer
138k+ views
Hint: As initially ball is at height of 2m and after collision ball height becomes 1m. So after collision loss in potential energy will increase the thermal energy of the ball i.e. decrease in potential energy will be the cause of increase in temperature. Hence change in potential energy will be equal to thermal energy.
Mathematically $mg({h_1} - {h_2}) = ms\Delta T$.
Complete step by step answer:
As we already know that change is potential energy will be equal to increase in thermal energy
$\therefore $given only 20% 0f mechanical energy will be convert into thermal energy
Hence, we apply the equation: $mg({h_1} - {h_2}) \times \dfrac{{20}}{{100}} = ms\Delta T$.
In the above equation m=mass of balL
g= acceleration due to gravity
${h_1}$=initial height
${h_2}$=final height
s=specific heat capacity
$\Delta T$=change in temperature of body
So after putting all given values in the above equation:
$
mg(2 - 1) \times \dfrac{{20}}{{100}} = m \times 800 \times \Delta T \\
\\
$ here $g = 10m{s^{ - 2}}$
$m \times 10 \times \dfrac{1}{5} = m \times 800 \times \Delta T$
$2 = 800 \times \Delta T$
$\Delta T = \dfrac{2}{{800}}$
$\Delta T = 2.5 \times {10^{ - 3}}$
Hence, increase in temperature of ball be $2.5 \times {10^{ - 3}}$
Option number A will be the correct option.
Note: Potential energy is the energy of the body associated with height of the body i.e. if a body will be more height from the surface of the earth then it will have more potential energy and if the body has more potential energy then it will have more mechanical energy.
Thermal energy is the energy of the body associated with the temperature of the body, thermal energy also depends on the specific heat of the body.
Mathematically $mg({h_1} - {h_2}) = ms\Delta T$.
Complete step by step answer:
As we already know that change is potential energy will be equal to increase in thermal energy
$\therefore $given only 20% 0f mechanical energy will be convert into thermal energy
Hence, we apply the equation: $mg({h_1} - {h_2}) \times \dfrac{{20}}{{100}} = ms\Delta T$.
In the above equation m=mass of balL
g= acceleration due to gravity
${h_1}$=initial height
${h_2}$=final height
s=specific heat capacity
$\Delta T$=change in temperature of body
So after putting all given values in the above equation:
$
mg(2 - 1) \times \dfrac{{20}}{{100}} = m \times 800 \times \Delta T \\
\\
$ here $g = 10m{s^{ - 2}}$
$m \times 10 \times \dfrac{1}{5} = m \times 800 \times \Delta T$
$2 = 800 \times \Delta T$
$\Delta T = \dfrac{2}{{800}}$
$\Delta T = 2.5 \times {10^{ - 3}}$
Hence, increase in temperature of ball be $2.5 \times {10^{ - 3}}$
Option number A will be the correct option.
Note: Potential energy is the energy of the body associated with height of the body i.e. if a body will be more height from the surface of the earth then it will have more potential energy and if the body has more potential energy then it will have more mechanical energy.
Thermal energy is the energy of the body associated with the temperature of the body, thermal energy also depends on the specific heat of the body.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

At which height is gravity zero class 11 physics JEE_Main

Other Pages
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
