
If the unit of work is \[100J\], the unit of power is \[1KW\], the unit of time in second is:
A) \[{10^{ - 1}}\]
B) \[{10^{}}\]
C) \[{10^{ - 2}}\]
D) \[{10^{ - 3}}\]
Answer
232.8k+ views
Hint: As we know that, we have to find the unit of time. According to the question we have energy (unit of work) and power. So as per given data we can use the following formula to find the unit of time. It can also be solved using dimensional formulas.
Complete step by step answer:
The data given in the question are
Energy = \[100J\]
Power = \[10KW\] = \[1000W\]
We have to find the unit of time in second,
\[{{Energy }} = {{ Power }} \times {{ Time}}\]
\[{{Time = }}\dfrac{{{{Energy}}}}{{{{Power}}}} = \dfrac{{100}}{{1000}} = \dfrac{1}{{10}} = {10^{ - 1}}\].
Additional information:
Energy: is the capacity when we can do the work. The unit of energy is joule.
Power: it is the rate of doing the work. The unit of power is watt.
Work: the amount of energy transfer that occurs when a particle is moved to some distance by an external force. The unit of work is also joule.
We can also solve the question using dimension formulas as shown below.
Energy = \[100J\] \[ = \left[ {{{M }}{{{L}}^2}{{ }}{{{T}}^{ - 2}}} \right]\]
Power = \[10KW\] = \[1000W\] = \[ = \left[ {{{M }}{{{L}}^2}{{ }}{{{T}}^{ - 3}}} \right]\]
\[{{Time = }}\dfrac{{{{Energy}}}}{{{{Power}}}} = \dfrac{{100}}{{1000}} = \dfrac{{{{M}}{{{L}}^2}{{{T}}^{ - 2}}}}{{{{M}}{{{L}}^2}{{{T}}^{ - 3}}}} = {{{T}}^1} = \dfrac{1}{{10}} = {10^{ - 1}}\].
Note: The base formula used here is Energy = Power × Time.
If data of the time and power is given in the question then we can find energy by multiplying both power and time. Similarly, if we have the data of time and energy then we can find the power by dividing both time and energy. If we have the data of power and energy then we can get the value of time by dividing both power and energy.
The dimension formula for some physical quantities:
Force \[ = {{mass }} \times {{ acceleration = }}\left[ {{M}} \right] \times \left[ {{{L }}{{{T}}^{ - 2}}} \right] = \left[ {{{M L }}{{{T}}^{ - 2}}} \right]\]
Work \[ = {{force }} \times {{ distance = }}\left[ {{{M L }}{{{T}}^{ - 2}}} \right] \times \left[ {{L}} \right] = \left[ {{{M }}{{{L}}^2}{{ }}{{{T}}^{ - 2}}} \right]\]
Energy = work \[ = \left[ {{{M }}{{{L}}^2}{{ }}{{{T}}^{ - 2}}} \right]\]
Power \[ = \dfrac{{{{work}}}}{{{{time}}}} = \dfrac{{\left[ {{{M }}{{{L}}^2}{{ }}{{{T}}^{ - 2}}} \right]}}{{\left[ {{T}} \right]}} = \left[ {{{M }}{{{L}}^2}{{ }}{{{T}}^{ - 3}}} \right]\].
Complete step by step answer:
The data given in the question are
Energy = \[100J\]
Power = \[10KW\] = \[1000W\]
We have to find the unit of time in second,
\[{{Energy }} = {{ Power }} \times {{ Time}}\]
\[{{Time = }}\dfrac{{{{Energy}}}}{{{{Power}}}} = \dfrac{{100}}{{1000}} = \dfrac{1}{{10}} = {10^{ - 1}}\].
Additional information:
Energy: is the capacity when we can do the work. The unit of energy is joule.
Power: it is the rate of doing the work. The unit of power is watt.
Work: the amount of energy transfer that occurs when a particle is moved to some distance by an external force. The unit of work is also joule.
We can also solve the question using dimension formulas as shown below.
Energy = \[100J\] \[ = \left[ {{{M }}{{{L}}^2}{{ }}{{{T}}^{ - 2}}} \right]\]
Power = \[10KW\] = \[1000W\] = \[ = \left[ {{{M }}{{{L}}^2}{{ }}{{{T}}^{ - 3}}} \right]\]
\[{{Time = }}\dfrac{{{{Energy}}}}{{{{Power}}}} = \dfrac{{100}}{{1000}} = \dfrac{{{{M}}{{{L}}^2}{{{T}}^{ - 2}}}}{{{{M}}{{{L}}^2}{{{T}}^{ - 3}}}} = {{{T}}^1} = \dfrac{1}{{10}} = {10^{ - 1}}\].
Note: The base formula used here is Energy = Power × Time.
If data of the time and power is given in the question then we can find energy by multiplying both power and time. Similarly, if we have the data of time and energy then we can find the power by dividing both time and energy. If we have the data of power and energy then we can get the value of time by dividing both power and energy.
The dimension formula for some physical quantities:
Force \[ = {{mass }} \times {{ acceleration = }}\left[ {{M}} \right] \times \left[ {{{L }}{{{T}}^{ - 2}}} \right] = \left[ {{{M L }}{{{T}}^{ - 2}}} \right]\]
Work \[ = {{force }} \times {{ distance = }}\left[ {{{M L }}{{{T}}^{ - 2}}} \right] \times \left[ {{L}} \right] = \left[ {{{M }}{{{L}}^2}{{ }}{{{T}}^{ - 2}}} \right]\]
Energy = work \[ = \left[ {{{M }}{{{L}}^2}{{ }}{{{T}}^{ - 2}}} \right]\]
Power \[ = \dfrac{{{{work}}}}{{{{time}}}} = \dfrac{{\left[ {{{M }}{{{L}}^2}{{ }}{{{T}}^{ - 2}}} \right]}}{{\left[ {{T}} \right]}} = \left[ {{{M }}{{{L}}^2}{{ }}{{{T}}^{ - 3}}} \right]\].
Recently Updated Pages
Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

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

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

