Param Himalaya - परम हिमालय

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...

11.1 Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons. Solution :  Here, $$V = 30 kV = 30 \times 10^{3}volts$$ (a) Maximum frequency of X-rays produced = maximum energy of an accelerated electron. $$E = q.V$$ For electron  $$E = eV$$ $$eV = h f_{max}$$  $$f_{max} = \frac{eV} {h}$$ $$f_{max} = \frac{1.6 \times 10^{-19} \times 30 \times 10^{3}} {6.62 \times 10^{-34}}$$ $$f_{max} = 7.24 \times 10^{18}Hz$$ (b) Minimum wavelength of X-rays produced is  $$\lambda_{min}= \frac{c}{f_{max}}$$ $$\lambda_{min}= \frac{3 \times 10^{8}}{7.24 \times 10^{18}}$$ $$\lambda_{min}= 0.414 \times 10^{-10}m$$ $$\lambda_{min}= 0.414 A^{o}$$ 11.2 The work function of caesium metal is 2.14 eV. When Light of frequency $6 \times 10^{14}Hz$ is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons? ...

10.1 Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33. solution : $$\lambda_{air} = 589 nm = 589 \times 10^{-9}m$$ $$_{a}{\mu}_{w} = 1.33$$ $$c = 3 \times 10^{8}m/s$$ (a) For reflected light . On reflection, no change occurs in wavelength, frequency and speed of the incident light. Wavelength of reflected light , $$\lambda_{air} = 589 nm = 589 \times 10^{-9}m$$ $$f = \frac{c}{\lambda} = \frac{3 \times 10^{8}}{589 \times 10^{-9}}$$ $$f = 5.09 \times 10^{14}Hz$$ Speed of reflected light ,  $$v = c=  3 \times 10^{8}m/s$$ (b) For refracted light ,  On refraction, the frequency f remains the same but both wavelength and speed get changed. Wavelength of refracted light, $$_{a}\mu_{w} = \frac{\lambda_{air}}{\lambda_{water}}$$ $${\lambda_{water}}= \frac{\lambda_{air}}{_{a}\mu_{w}}$$ $$\lambda_{water} = \frac{589 \times 10^{-9...

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments - Param Himalaya  9.1 A small candle. 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirroг should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?  Solution:  Here , h = 2.5 cm , u = - 27 cm , R = - 36 cm  Since $f = \frac{R}{2}$ . Therefore, $f = \frac{-36}{2}$ = -18 cm (i) using mirror formula  $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$ , we get $\frac{1}{v}  = \frac{1}{f} - \frac{1}{u}$ $\frac{1}{v} = \frac{1}{(-18)} - \frac{1}{(-27)}$  $\frac{1}{v} = -\frac{1}{18} + \frac{1}{27}$  $\frac{1}{v} = \frac{-3+2}{54}$ $\frac{1}{v} = -\frac{1}{54}$ $v = - 54 cm$ ( Position of image )  Thus , the screen should be placed at a distance of 54 cm in front of the con...