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Class 11 Physics Laboratory Mannual Practical and Experiment Experiments 1(a) To measure the diameter of a small spherical/cylindrical body using vernier calliper  1(b)  To measure the internal diameter and depth of a given beaker/calorimeter using vernier calliper and hence, find its volume. 2(a)  To measure the diameter of a given wire using screw gauge 2(b) To measure thickness of a given sheet using screw gauge. 3. To determine volume of an irregular lamina using screw gauge. 4 To determine radius of curvature of a given spherical surface by a spherometer. 5. To determine the mass of two different objects using a beam balance  6.To find the weight of a given body using parallelogram law of vectors 7. Using a simple pendulum, plot its $LT^{2}$ graph and use it to find the effective length of second's pendulum 8. To study the variation of time period of a simple pendulum of a given length by taking bobs of same sue but different masses and interpret the result. 9. ...

NCERT Solutions for class 9 Science chapter 8 Forces and Laws of Motion is prepared and uploaded for reference by academic team of expert members of Param Himalaya . Get solutions of all chapters of NCERT class 9 Science from Param himalaya. use as a reference of the following NCERT solutions of chapter 8 prepared by paramhimalaya. Read the theory of chapter-8 Forces and Laws of Motion while before going to exam. Page no :  91 Question 1 to 4 :  Question 1. Which of the following has more inertia : (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five rupees coin and a one-rupee coin? Solution : (a) a stone of the same size will have more inertia than a rubber ball. (b) A train will have more inertia than a bicycle. (c) A five rupees coin will have more inertia than a one-rupee coin. Question 2. In the following example, try to identify the number of times the velocity of the ball changes: “A football player kicks a football to another player o...

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...

Lyrics - Wo Purane Din, Wo Suhane Din , वो पुराने दिन वो सुहाने दिन Poem by Piyush Mishra and sing  वो पुराने दिन वो सुहाने दिन आशिकाने दिन ओस की नमी में भीगे वो पुराने दिन दिन गुजर गए हम किधर गए पीछे मुड़ के देखा पाया सब ठहर गए अकेले है खड़े कदम नही बढ़े चल पड़ेंगे जब भी कोई राह चल पड़े जायेंगे कहां है कुछ पता नही कह रहे है वो हम की है खता नही वो पुराने दिन आशिकाने दिन — पीयूष मिश्रा

NCERT Solutions for Class 11 Physics Chapter 11 Thermodynamics - Param Himalaya  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 s...

12.1 Choose the correct alternative from the clues given at the end of the each statement: (a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than.) Solution: No different From  (b) In the ground state of ..........electrons are in stable equilibrium,while in .......... electrons always experience a net force.(Thomson’s model/ Rutherford’s model.) Solution: (i) Thomson's model ,(ii) Rutherford’s model (c) A classical atom based on .......... is doomed to collapse.(Thomson’s model/ Rutherford’s model.) Solution: Rutherford's model (d) An atom has a nearly continuous mass distribution in a ..........but has a highly non-uniform mass distribution in ..........(Thomson’s model/ Rutherford’s model.) Solution : Thomson's model, Rutherford's model (e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford’s model/both the models.) Solution:   Bot...