Notes : Class 12 Physics Chapter 8 Electromagnetic Waves - Physicskund

By -

Notes : CBSE Class 12 Physics Chapter 8 Electromagnetic Waves - Physicskund

1. Basic idea of displacement current, 

2. Electromagnetic waves, their characteristics, their transverse nature (qualitative idea only).

3. Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays,gamma rays) including elementary facts about their uses.

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 - Param Himalaya

By -
Image
Ncert Solution CBSE Class 11 Chapter 10 THERMAL PROPERTIES OF MATTER - Param Himalaya  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...
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