Notes : Class 12 Physics Chapter 2 Electrostatic Potential And Capacitance - Physicskund

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Notes : Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance - Physics Kund

Chapter 2: Electrostatic Potential and Capacitance

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

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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...
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Ncert Solution CBSE Class 11 Chapter 10 THERMAL PROPERTIES OF MATTER

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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^{–...
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