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Sat October 25, 2014  

Ans: Ethene on  hydrogenation in the presence of catalyst like platinum gives ethane.   CH 2 = CH 2   → CH 3 - CH 3

Sat October 25, 2014  

Ans: Diatomic molecules are molecules composed of two atoms, of either the same or different chemical elements. Non-metallic elements have 4, 5, 6 or 7 electrons in their outermost shell. Chlorine is a non-metal with atomic number 17 and electronic configuration 2 , 8, 7. It requires one electron to complete its octet configuration. So two chlorine atoms approach each other and share their one electron each and an electron pair is shared between the two chlorine atoms.  This results in the formation of a chlorine molecule. A bond is formed by sharing of electrons. This type of a bond is called Covalent bond.

Sat October 25, 2014  

Ans: Different groups show different trends in boiling and melting points.  The boiling and melting points decrease as you move down the group in group 1 and group 2. The boiling and melting points mostly increase as you move down the group for transition metals, but they increase for the zinc family.  In Groups 13 and 14, there is a decrease in boiling and melting points as you move down the group. In Groups 15, 16, and 17, the melting and boiling points tend to increase in both.  Group 18 decrease in their boiling and melting points down the group.                               

Sat October 25, 2014  

Ans: Dear kumar.ashlesha@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
We have not understood the query that you have posted. We would request you to clarify / provide additional details so that we may answer this to the best of the ability. In order to compare the equations, you need another equation, therefore we would like you to repost the proper equations.

Regards

Topperlearning Team.

Sat October 25, 2014  

Ans: L e n g t h space o f space t h e space t r i a n g l e comma space l space equals 12 H e i g h t space o f space t h e space t r i a n g l e comma space h equals 4 A r e a space o f space t h e space t r i a n g l e equals 1 half cross times l cross times h equals 1 half cross times 12 cross times 4 equals 24 space s q. space u n i t s A r e a space o f space t r a i n g l e space equals 24 space s q. space u n i t s

Sat October 25, 2014  

Ans: Gravitational force is actually the force of attraction between two bodies due to their masses. The force of attraction between the earth and any other body lying on or near its surface is called gravitational force of earth or gravity. By Newtons second law we know that a body acted upon by an unbalanced force gets accelerated in the direction of force. The acceleration produced on a freely falling body by the gravitataional  pull of the earth is called the acceleration due to gravity and is denoted as 'g' .  The value of g near the surface of earth is taken as 9.8 m/s 2  . Sometimes 10 m/ s 2  is also used in certain cases which is just an appoximation of the value 9.8 m/s 2 . The correct value of g is 9.8 m/s 2  .

Sat October 25, 2014  

Ans: A f t e r space c o m p l e t i n g space t h e space t a s k space o f space i n d e f i n i t e space i n t e g r a t i o n comma space s i m p l y space s u b s t i t u t e space t h e space v a l u e s space o f space t h e space l i m i t s space i n space t h e space f u n c t i o n. space S u b s t i t u t e space t h e space u p p e r space l i m i t s space f i r s t space s u b t r a c t e d space b y space t h e space l o w e r space l i m i t. F o r space E x a m p l e : integral subscript 1 superscript 4 x cubed. d x equals open square brackets x to the power of 4 over 4 close square brackets subscript 1 superscript 4 equals open square brackets 4 to the power of 4 over 4 minus 1 to the power of 4 over 4 close square brackets equals open square brackets 64 minus 1 fourth close square brackets equals 255 over 4

Sat October 25, 2014  

Ans: Dear NAKSHATRA123@GMAIL.COM

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
A similar question is already been answered, please refer the already answered questions.

Regards

Topperlearning Team.

Sat October 25, 2014  

Ans: 5 cross times 7 cross times 11 cross times 13 plus 11 equals 11 cross times left parenthesis 5 cross times 7 cross times 13 plus 1 right parenthesis space left square bracket t a k i n g space 11 space c o m m o n right square bracket equals 11 cross times 456 S i n c e space i t space c a n space b e space w r i t t e n space a s space f a c t o r s space o t h e r space t h a n space i t s e l f comma space t h e r e f o r e space t h e space n u m b e r space i s space a space c o m p o s i t e space n u m b e r.

Sat October 25, 2014  

Ans: square root of 1200 equals square root of 100 cross times 12 end root equals square root of 100 cross times square root of 12 equals 10 cross times square root of 4 cross times 3 end root equals 10 cross times 2 cross times square root of 3 equals 20 square root of 3 T h e r e f o r e space 3 space i s space t h e space s m a l l e s t space n u m b e r space t h a t space n e e d s space t o space b e space m u l i t p l i e d space w i t h space t h e space a b o v e space n u m b e r comma space s o space a s space t o space m a k e space t h e space p r o d u c t space a space r a t i o n a l space n u m b e r.

Fri October 24, 2014  

Ans: Dear jaiswalr550@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
The scope of this helpdesk is to answer subject related queries which are relevant to understanding concepts and solving problems. Hence we regret that we would not be able to answer your query.

Regards

Topperlearning Team.

Fri October 24, 2014  

Ans: Dear jaiswalr550@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
The scope of this helpdesk is to answer subject related queries which are relevant to understanding concepts and solving problems. Hence we regret that we would not be able to answer your query.

Regards

Topperlearning Team.

Fri October 24, 2014  

Ans: Dear jaiswalr550@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
The scope of this helpdesk is to answer subject related queries which are relevant to understanding concepts and solving problems. Hence we regret that we would not be able to answer your query.

Regards

Topperlearning Team.

Fri October 24, 2014  

Ans: Dear annaya4ever@outlook.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
We cannot entertain  5 questions in a single query. In case of multiple questions within a query, please post each question individually and let us know where you are getting stuck so that we would be able to explain things better.   Answer to your first query is given below: 1.What defect creates when some P atom is added to Si? In the processing of Si crystals for microelectronics, a very small concentration of phosphorus atoms is added to silicon. Such additions create defects in the arrangement of atoms in silicon that impart special electrical properties to different parts of the silicon crystal. This in turn allows us to make useful devices such as transistors which are the building blocks that have enabled computers and the information technology revolution.         Regards

Topperlearning Team.

Fri October 24, 2014  

1. The focal length of an equiconvex lens in air is equal to either of its radii of curvature. the refractive index of the material of lens is __________
2. A charge of 4x10^-8C is uniformly distributed over the surface of sphere radius 1cm. another hollow sphere of radius 5cm is concentric with the small sphere. find intensity of electric field at a distance 2cm from centre. [K=9x10^9 MKS] [ans in NC^-1]
3. A light of 4560angstrom and 1mW is made incident on photosensitive surface of Cs. If efficiency of surface is 0.1%, then photoelectric current produces is___________
4.A spherical drop of water has 3x10^-10C charge on its surface. 300 volt electric potential exists on its surface. if 27 such identical drops are combined to form a single drop, calculate electric potential on the surface of new drop._____________
Ans: 2.    Given charge Q = 4 × 10 -8 C.  begin mathsize 14px style Let space straight P space be space straight a space point space at space straight a space distance space of space straight r space left parenthesis 2 space cm right parenthesis space away space from space the space centre. To space find space the space field space at space straight P space let space us space consider space straight a space gaussian space surface space through space the space point space straight P space as space shown space in space the space figure. We space can space write space that space the space flux space through space this space surface space equals space contour integral space straight E with rightwards arrow on top. dS with rightwards arrow on top space space space space space space equals space contour integral straight E. space dS space space space space space space equals straight E contour integral dS space space space space space space equals space 4 cross times straight pi cross times straight r squared cross times straight E Here space straight r space equals space 2 space cm space equals space 2 space cross times 10 to the power of minus 2 space end exponent straight m According space t o space Gauss space Law comma space this space flux space is space equal space to space the space charge space straight Q space enclosed space by space the space surface space divided space by space straight epsilon subscript 0. straight i. straight e. space space contour integral space straight E with rightwards arrow on top. dS with rightwards arrow on top space equals space straight Q over straight epsilon subscript 0 4 cross times straight pi cross times straight r squared cross times straight E space equals straight Q over straight epsilon subscript 0 4 cross times straight pi cross times left parenthesis 2 cross times 10 to the power of minus 2 end exponent right parenthesis space squared cross times straight E space equals straight Q over straight epsilon subscript 0 On space rearranging space we space get space : straight E space equals space fraction numerator straight Q over denominator straight epsilon subscript 0 cross times 4 cross times straight pi cross times left parenthesis 2 cross times 10 to the power of minus 2 end exponent right parenthesis squared space end fraction open parentheses we space know space that space fraction numerator 1 over denominator 4 πε subscript 0 end fraction equals k equals 9 cross times 10 to the power of 9 space Nm squared straight C to the power of minus 2 end exponent space space close parentheses space space space space equals 9 cross times 10 to the power of 9 space Nm squared straight C to the power of minus 2 end exponent cross times fraction numerator 4 cross times 10 to the power of minus 8 end exponent space straight C over denominator 4 cross times 10 to the power of minus 4 end exponent space straight m squared end fraction straight E space equals 9 space cross times space 10 to the power of 5 space straight N space straight C to the power of minus 1 end exponent end style The Electric field at a distance 2cm from centre = 9 × 10 5 N C -1   Note: Please ask other questions (each question) as seperate separate queries.

Fri October 24, 2014  

1. The focal length of an equiconvex lens in air is equal to either of its radii of curvature. the refractive index of the material of lens is __________
2. A charge of 4x10^-8C is uniformly distributed over the surface of sphere radius 1cm. another hollow sphere of radius 5cm is concentric with the small sphere. find intensity of electric field at a distance 2cm from centre. [K=9x10^9 MKS] [ans in NC^-1]
3. A light of 4560angstrom and 1mW is made incident on photosensitive surface of Cs. If efficiency of surface is 0.1%, then photoelectric current produces is___________
4.A spherical drop of water has 3x10^-10C charge on its surface. 300 volt electric potential exists on its surface. if 27 such identical drops are combined to form a single drop, calculate electric potential on the surface of new drop._____________
Ans: 1.   Given Focal length, f = R Being an equiconvex lens, let R 1  = R ,  R 2  = - R  begin mathsize 14px style According space to space lens space makers space formula : space 1 over straight f equals open parentheses straight mu minus 1 close parentheses open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses Substituting space the space values space we space get : 1 over straight R equals open parentheses straight mu minus 1 close parentheses open parentheses 1 over straight R plus 1 over straight R close parentheses 1 over straight R equals open parentheses straight mu minus 1 close parentheses 2 over straight R 1 half plus 1 equals straight mu rightwards double arrow space straight mu space equals space 3 over 2 equals 1.5 end style  The refractive index of the material of lens is 1.5   Note: Please ask other questions (each question) as seperate separate queries.

Fri October 24, 2014  

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)
Ans: 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

Fri October 24, 2014  

Ans: NH 3   + H 2 O → NH 4 + + OH -   K b   = [NH 4 + ] [OH - ]    =  1.77 x 10 -5             [NH 3 ] Before neutralization, [NH 4 + ]  = [OH - ] = x Thus K b  =    x 2     = 1.77x 10 -5          0.10 Hence x = 1.33 x 10 -5 = [OH - ]     Now K w = [H + ][OH - ] So [H + ]  = K w / [OH - ] = 10 -14 / 1.33x 10 -12 = 7.51 x 10 -12          p H = - log(7.51 x 10 -12 ) = 11.12 On addition of 25 ml of 0.1 M HCl solution to 50 ml of 0.1 M ammonia solution , 2.5 mol of ammonia are neutralized. The resulting 75 ml solution contains the remaining unneutralised 2.5 mol of NH 3 molecules and 2.5 mol of NH 4 + .                                            NH 3     + HCl  → NH 4 +    + Cl -                                         2.5         2.5          0             0   At equilibrium                     0            0            2.5        2.5 The resulting 75 ml of solution contains 2.5 mol of NH 4 + ions and 2.5 mol of uneutralised NH 3 molecules. This NH 3 exists in the following equilibrium. NH 4 OH     ⇌      NH 4 +      + OH - 0.033 M- y      y                  y Where y = [OH - ] = [NH 4 + ] The final 75ml solution after neutralization already contains 2.5 mol NH 4 + ions, thus total concentration of NH 4 + ions is given as: [NH 4 + ]  = 0.033 + y As y is small, [NH 4 OH] = 0.033 M and [NH 4 + ] = 0.033M Kb =  [NH 4 + ]  [OH - ]   = y (0.033)  = 1.77x 10 -5 M              [NH 4 OH]          (0.033)  Thus , y = 1.77 x 10 -5 = [OH - ] [H + ] = 10 -14              = 0.56 x 10 -9          1.77 x 10 -5   Hence p H = 9.24

Fri October 24, 2014  

Ans: Dear M.s.srinidhi21@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
We have not understood the query that you have posted. We would request you to clarify / provide additional details so that we may answer this to the best of the ability. Please check the question and repost again.
Regards

Topperlearning Team.

Fri October 24, 2014  

Ans: 5 to the power of 2 x minus 1 end exponent minus 25 to the power of x minus 1 end exponent equals 25000 rightwards double arrow 25 to the power of x over 5 minus 25 to the power of x over 25 equals 25000 rightwards double arrow 25 to the power of x open parentheses 1 fifth minus 1 over 25 close parentheses equals 25000 rightwards double arrow 25 to the power of x open parentheses fraction numerator 5 minus 1 over denominator 25 end fraction close parentheses equals 25000 rightwards double arrow 25 to the power of x cross times 4 over 25 equals 25000 rightwards double arrow 25 to the power of x equals 25000 cross times 25 over 4 rightwards double arrow 25 to the power of x equals 25 cross times 1000 over 4 cross times 25 rightwards double arrow 25 to the power of x equals 25 cross times 250 cross times 25 rightwards double arrow 25 to the power of x equals open parentheses 25 close parentheses cubed cross times 10 I f space i n space t h i s space q u e s t i o n comma space t h e space R H S space w o u l d space h a v e space b e e n space 2500 comma space t h e n space t h e space a n s w e r space w o u l d space h a v e space b e e n space rightwards double arrow 25 to the power of x equals 25 cubed rightwards double arrow x equals 3 space left square bracket I f space b a s e s space a r e space s a m e comma space p o w e r s space a r e space e q u a l right square bracket I f space t h e space R H S space i s space n o t space 2500 comma space o r space t h e r e space a r e space a n y space o t h e r space c h a n g e s comma space t h e n space log a r i t h m space h a v e space t o space b e space u s e d. P l e a s e space i n f o r m space a b o u t space t h e space c o r r e c t space q u e s t i o n.
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