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CBSE Class 12-science Answered

1. Magnetic force acting on a particle moving in magnetic field is _________ [ answer using q, B and Vd]
2. An electron rotates on circular path of radius r in normal direction of magnetic field B. The kinetic energy gain by the electron when it completes half rotation is ___________[half mv^2, one fourth mv^2, [pi]rBev or zero?]
3. Two metal spheres of radii 5cm and 4cm are charged to same potential. the surface charge densities of two spheres are in the ratio ___________
4.stopping potential for one polished surface is 2 volt. the maximum speed of electron emitteed from surface is ____________m/s
5. A magnetic needle placed in uniform magnetic field has magnetic moment 2x10^-2Am^2 and moment of inertia 16x10^-6kgm^2. It performs 10 oscillations in 20 seconds. what is the magnitude of magnetic field. (consider pi^2 = 10)
Asked by Dhriti | 24 Oct, 2014, 11:56: AM
answered-by-expert Expert Answer
3. Two metal spheres of radii 5cm and 4cm are charged to same potential. the surface charge densities of two spheres are in the ratio ___________
 
begin mathsize 14px style Let space straight sigma subscript 1 space end subscript space and space straight sigma subscript 2 space be space the space surface space charge space densities space of space the space two space spheres space 1 space and space 2. Let space straight r subscript 1 space and space straight r subscript 2 space end subscript be space the space radii space of space the space two space spheres space 1 space and space 2. We space know space that space the space potential space at space the space surface space of space straight a space charged space sphere space comma space straight V equals fraction numerator 1 over denominator 4 πε subscript 0 end fraction straight Q over straight r space space space space equals space fraction numerator straight k space straight Q over denominator straight r end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets fraction numerator 1 over denominator 4 πε subscript 0 end fraction equals straight k close square brackets So space the space potential space at space the space surface space of space sphere 1 space can space be space writtten space as : straight V subscript 1 space equals fraction numerator straight k space straight Q subscript 1 over denominator straight r subscript 1 end fraction space space space minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus left parenthesis 1 right parenthesis and space that space of space sphere space 2 space is : straight V subscript 1 space equals fraction numerator straight k space straight Q subscript 2 over denominator straight r subscript 2 end fraction space space minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus left parenthesis 2 right parenthesis We space know space that space surface space charge space density space is space defined space as space charge space per space unit space area. space straight i. straight e space straight sigma subscript 1 space end subscript space equals straight Q subscript 1 over straight A subscript 1 space and space straight sigma subscript 2 space end subscript space equals straight Q subscript 2 over straight A subscript 2 From space this space we space can space write space that : open table attributes columnalign right end attributes row cell straight Q subscript 1 equals straight sigma subscript 1 space end subscript cross times 4 πr subscript 1 squared space space and straight Q subscript 2 equals straight sigma subscript 2 space end subscript cross times 4 πr subscript 2 squared end cell row blank end table close curly brackets minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus minus left parenthesis 3 right parenthesis On space substituting space left parenthesis 3 right parenthesis space in space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis space we space get : straight V subscript 1 space equals fraction numerator kσ subscript 1 space end subscript cross times 4 πr subscript 1 squared space over denominator straight r subscript 1 end fraction space straight V subscript 1 space equals fraction numerator kσ subscript 2 space end subscript cross times 4 πr subscript 2 squared space over denominator straight r subscript 2 end fraction space Given space that space both space the space spheres space are space charged space to space the space same space potential space so space straight V subscript 1 equals straight V subscript 2 straight i. straight e. space fraction numerator kσ subscript 1 space end subscript cross times 4 πr subscript 1 squared space over denominator straight r subscript 1 end fraction space equals fraction numerator kσ subscript 2 space end subscript cross times 4 πr subscript 2 squared space over denominator straight r subscript 2 end fraction space straight sigma subscript 1 space end subscript cross times 4 πr subscript 1 equals straight sigma subscript 2 space end subscript cross times 4 πr subscript 2 straight sigma subscript 1 space end subscript over straight sigma subscript 2 space end subscript equals fraction numerator 4 πr subscript 2 over denominator 4 πr subscript 1 end fraction equals straight r subscript 2 over straight r subscript 1 Given space straight r subscript 1 equals 5 space cm space and space straight r subscript 2 equals space 4 space cm rightwards double arrow straight sigma subscript 1 space end subscript over straight sigma subscript 2 space end subscript equals 4 over 5 equals 0.8 end style
The ratio of surface charge densities = 0.8 
 
Note: Please ask each questions as seperate seperate queries
Answered by Jyothi Nair | 25 Oct, 2014, 01:51: PM

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