Home
Class 10
MATHS
The hand borewell driller charges Rs. 20...

The hand borewell driller charges Rs. 200 for the first one meter only and raises drilling charges @ 30/- for every subsequent meter. Write a progression for the above data.

Promotional Banner

Topper's Solved these Questions

  • PROGRESSIONS

    BEYOND PUBLICATION|Exercise EXERCISE|259 Videos
  • PROBABILITY

    BEYOND PUBLICATION|Exercise EXERCISE|136 Videos
  • QUADRATIC EQUATIONS

    BEYOND PUBLICATION|Exercise EXAMPLE|135 Videos

Similar Questions

Explore conceptually related problems

A lending library has fixed charge for the first three days and an additional charges for each day thereafter. John paid 27 for a book kept for seven days. If the fixed charges be x and subsequent per day charges be y, then write the linear equation representing the above information and draw the graph of the same. From the graph, find fixed charges for the first three if additional charges for each day thereafter is 4. Find additional charges for each day thereafter if the fixed charges for the first three days of 7.

It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of sqrt2 (h)/(2pi) corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum : Orbital angular momentum, L=sqrt(l(l+1))h/(2pi) and spin angular momentum, arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition mush compensate for the angular momentum carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as m((h)/(2pi)) the permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively The spin-only magnetic moment of free ion is sqrt(8) B.M. The spin angular momentum of electron will be

It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of sqrt2 (h)/(2pi) corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum : Orbital angular momentum, L=sqrt(l(l+1))h/(2pi) and spin angular momentum, arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition must compensate for the angular momentum carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as m((h)/(2pi)) the permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively The orbital angular momentum of an electron in p-orbital makes an angle of 45^@ from Z-axis. Hence Z-component of orbital angular momentum of election is :

It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of sqrt2 (h)/(2pi) corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum : Orbital angular momentum, L=sqrt(l(l+1))h/(2pi) and spin angular momentum, arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition must compensate for the angular momentum carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as m((h)/(2pi)) he permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively The maximum orbital angular momentum of an electron with n= 5 is

BEYOND PUBLICATION-PROGRESSIONS-EXERCISE
  1. The hand borewell driller charges Rs. 200 for the first one meter only...

    Text Solution

    |

  2. For the following A.Ps write the first term and the common difference....

    Text Solution

    |

  3. For the following A.Ps write the first term and the common difference....

    Text Solution

    |

  4. For the following A.Ps write the first term and the common difference....

    Text Solution

    |

  5. For the following A.Ps write the first term and the common difference....

    Text Solution

    |

  6. For the following A.Ps write the first term and the common difference....

    Text Solution

    |

  7. Write the first four terms of the A.P. when the first term 'a' and the...

    Text Solution

    |

  8. Write the first four terms of the A.P. when the first term 'a' and the...

    Text Solution

    |

  9. Write the first four terms of the A.P. when the first term 'a' and the...

    Text Solution

    |

  10. Write the first four terms of the A.P. when the first term 'a' and the...

    Text Solution

    |

  11. Write the first four terms of the A.P. when the first term 'a' and the...

    Text Solution

    |

  12. Which of the following are A.Ps? If they form A.P. find the common dif...

    Text Solution

    |

  13. Which of the following are A.Ps? If they form A.P. find the common dif...

    Text Solution

    |

  14. Which of the following are A.Ps? If they form A.P. find the common dif...

    Text Solution

    |

  15. Which of the following are A.Ps? If they form A.P. find the common dif...

    Text Solution

    |

  16. Which of the following are A.Ps? If they form A.P. find the common dif...

    Text Solution

    |

  17. Which of the following are A.Ps? If they form A.P. find the common dif...

    Text Solution

    |

  18. Find the 30^(th) term of the A.P. 7,11,15,19,….

    Text Solution

    |

  19. Find the 10^(th) term of the A.P: -11,-9,-7,….

    Text Solution

    |

  20. Find the 18^(th) term and 23^(rd) term of the A.P 16,11,6,….

    Text Solution

    |

  21. The first term of an A.P. is -2 and the 10^(th) term is 16. Determine ...

    Text Solution

    |