Notes NCERT Class 11 Physics CHAPTER 1 UNITS AND MEASUREMENTS

By -

Notes NCERT Class 11 Physics CHAPTER 1 UNITS  AND MEASUREMENTS 

  1. Need for measurement.
  2. Units of measurement.
  3. Systems of units, SI units, fundamental and derived units.
  4. Definition of Significant Figures Rules for Finding Significant Figures.
  5. Significant Fiqures in Algebraic Operations
  6. Rules of Rounding Off Significant Figures.
  7. Principle of Homogeneity. 
  8. Dimensions of physical quantities 
  9. Dimensional analysis and its applications.



Comments

Post a Comment

Popular posts from this blog

NCERT Solution Class 10 Science Chapter 11 Electricity -

By -
Image
NCERT Exercise Solution Class 10 Science Chapter 11 Electricity  Example 11.1 A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit. Solution: Given  I = 0.5 A ,   t = 10 min = 600 s. We have Q = It = 0.5 A × 600 s = 300 C Page Number 172 – Question with solution 1 to 3 : 1. What does an electric circuit mean? Solution : It is a continuous and closed path through which electric current flows. It contains different components like • Cell or Battery Plug Key • Wires • Electric Components (like Ammeter,Voltmeter) • Bulb etc  2. Define the unit of current. Solution :  The unit of electric current is ampere (A). I  = $\frac{Q}{t}$ When Q = 1 C and t = 1 sec then I = $\frac{1C}{1S}$ I = 1A 1 A is defined as the flow of 1 C of charge through a wire in 1 s. 3. Calculate the number of electrons constituting one coulomb of charge. Solution :  We know  Q...
Read more

Ncert Solution CBSE Class 11 Chapter 10 THERMAL PROPERTIES OF MATTER

By -
Image
Ncert Solution CBSE Class 11 Chapter 10 THERMAL PROPERTIES OF MATTER - 10.1 The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales. Solution :  On Celsius scale  t(°C) = T(K) - 273.15 Triple point of neon = 24.57 - 273.15 = - 248.58°C Triple point of carbon dioxide = 216.55 - 273.15 = -56.6°C On Fahrenheit scale  $t(^{o}F) = \frac{9} {5} (t^{o}C)+32$ Triple point of neon = $\frac{9} {5} (-248.58)+32$  Triple point of neon= -447.444+32=-415.44°F Triple point of carbon dioxide = $\frac{9} {5} (-56.6)+32$  Triple point of carbon dioxide = - 101.88 +32 = -69.88°F 10.2 Two absolute scales A and B have triple points of water defined to be 200 A and 350B. What is the relation between $T_{A}$ and $T_{B}$? Solution :  Triple point of  water in  absolute scale A = 200A absolute scale B = 350 B Temperature on Kelvin scale = 273.16K 200A = 273.16K $1A = \frac{273.16}{...
Read more

NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics

By -
Image
NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics 11.1 A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is $4.0 × 10^{4} J/g$ ?  Solution:   Given  Mass of flowing water , m = 3.0 litre/ min = 3000 g/min The geyser heats the water , raising the temperature from 27 °C to 77 °C. Initial temperature, $T_{1}$ = 27°C Final temperature $T_{2}$ = 77°C Rise in temperature ,  $T = T_{2} -  T_{1}$ = 77-27= 50°C Heat of combustion =$ 4.0 × 10^{4} J/g$ Specific heat of water , $c= 4.2 J/g^{o}C$ Total heat used ,  $$Q = mcT$$ $$Q = 3000 \frac{g}{min} \times 4.2 \frac{J}{g^{o}C}\times 50^{o}C$$ $$Q = 6.3 × 10^{5}J/ min$$ $$Rate \ of \ consumption = \frac{6.3 \times 10^{5}}{(4 \times 10^{4})}$$ $$= 15.75 g/min$$ Therefore, rate of consumption is 15.75 g/min 11.2 What amount of heat must be supplied to $2.0 × 10^{–...
Read more