
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
233.1k+ 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
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

