Notes : Class 11 Physics in Hindi Chapter Wise भौतिक विज्ञान नोट्स

By -

 Class 11 Physics Notes in Hindi – अध्यायवार भौतिकी नोट्स, आसान भाषा में समझाया गया। NCERT आधारित संक्षिप्त सारांश और महत्वपूर्ण सूत्र, परीक्षा तैयारी के लिए उपयोगी।

  1. इकाइयाँ और मापन
  2. सीधी रेखा में गति
  3. तल में गति
  4. गति के नियम
  5. कार्य, ऊर्जा और शक्ति
  6. कणों का तंत्र और घूर्णन गति
  7. गुरुत्वाकर्षण
  8. ठोसों के यांत्रिक गुण
  9. द्रवों के यांत्रिक गुण
  10. पदार्थ के ऊष्मीय गुण
  11. ऊष्मागतिकी
  12. गतिज सिद्धांत
  13. दोलन
  14. तरंगें

Comments

Post a Comment

Popular posts from this blog

Experiment 8 : To find the refractive index of a liquid using a concave mirror and a plane mirror.

By -
Image
Experiment 8 : To find the refractive index of a liquid using a concave mirror and a plane mirror. AIM :  To find the refractive index of any liquid (water) using a concave mirror. APPARATUS :  A concave spherical mirror, water, an optical needle, a clamp stand, one meter scale, plumb line, etc. THEORY :  If the tip of object needle $O$ is at the centre of curvature $C$, the tip of the image will exactly coincide with it. ( Principle axis is verticle to the plane ). When water is filled in the concave mirror, the object needle is moved to a new position $C'$ to remove the parallax between the tips of the object needle and its image. A ray starting from $C$ reaches $E$ without deviation because it is along the radius of curvature. Due to water in the concave mirror the position of object and image shifts to $C'$.i.e now ray starting from $C'$ after refraction moves along $ED$ and then  $DC$ to make the apparent centre of curvature. $\angle NDC'=i \text{(incidence angl...
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