Home
Class 11
PHYSICS
Show dimensionally that the relation t =...

Show dimensionally that the relation `t = 2pi((I)/(g))`
is incorrect, where I is length and t is time period of a simple pendulum , g is acc. Due to gravity. Find the correct form of the relation, dimensionally

Text Solution

Verified by Experts

`RHS = 2pi((I)/(g)) = (L)/(LT^(-2))`
`=T^2 != t(LHS)`
:. Formula is incorrect
Let `t =k l^a g^b …..(i)`
`[M^0L^0T^1] = L^a (LT^(-2))^b = L^(a +b) T^(-2b)`
Using principle of homogeneity of dimensions,
` a +b =0, -2b =1, b = -(1)/(2)`
`:. a =-b =- (-(1)/(2)) = (1)/(2)`
From (i), `t = kl^(1//2) g^(-1//2) = k sqrt((I)/(g))`
Promotional Banner

Topper's Solved these Questions

  • PHYSICAL WORLD AND MEASUREMENT

    PRADEEP|Exercise Conceptual Problem|65 Videos
  • PHYSICAL WORLD AND MEASUREMENT

    PRADEEP|Exercise NCERT Exercises|40 Videos
  • OSCILLATIONS AND WAVES

    PRADEEP|Exercise multiple choice Questions|13 Videos
  • PROPERTIES OF BULK MATTER

    PRADEEP|Exercise Multiple choice questions|7 Videos

Similar Questions

Explore conceptually related problems

Derive dimensionally the relation s = ut +(1)/(2)f t^(2) .

The time period T of a simple pendulum depends on length L and acceleration due to gravity g Establish a relation using dimensions.

The graph of time period (T) of a simple pendulum versus its length (I) is

Test dimensionally if the formula t= 2 pi sqrt(m/(F/x)) may be corect where t is time period, F is force and x is distance.

If the time period of a simple pendulum is T = 2pi sqrt(l//g) , then the fractional error in acceleration due to gravity is

The time period of simple pendulum is T. If its length is increased by 2%, the new time period becomes

Find the dimensions of K in the relation T = 2pi sqrt((KI^2g)/(mG)) where T is time period, I is length, m is mass, g is acceleration due to gravity and G is gravitational constant.

If 'L' is length of simple pendulum and 'g' is acceleration due to gravity then the dimensional formula for (l/g)^(1/2) is same as that for

The graph between time period ( T ) and length ( l ) of a simple pendulum is

PRADEEP-PHYSICAL WORLD AND MEASUREMENT-Competiton Focus Jee Medical Entrance
  1. Show dimensionally that the relation t = 2pi((I)/(g)) is incorrect, ...

    Text Solution

    |

  2. Assertion : Number of significant figure in 0.005 is one and that is 0...

    Text Solution

    |

  3. Assersion : Out of three meansurements l = 0.7 m, l = 0.70m and l = 0....

    Text Solution

    |

  4. Statement-1 : nm is not same as m N Statement -2 : 1 nm = 10^(-9) m ...

    Text Solution

    |

  5. Assertion: The dimensional formula of surface energy is [M^(0)L^(2)T^(...

    Text Solution

    |

  6. Statement -1 : Distance travelled in nth second has the dimensions of ...

    Text Solution

    |

  7. Statement-1 : Velocity gradient has the dimensions of frequency. Sta...

    Text Solution

    |

  8. Statement-1 If error in measurement of distance and time are 3% and 2%...

    Text Solution

    |

  9. Statement-1 : The dimensional formula of electric potential is [ML^2 T...

    Text Solution

    |

  10. Consider a Vernier callipers in which each 1cm on the main scale is di...

    Text Solution

    |

  11. The vernier constant of a vernier callipers is 0.1 mm and it has a pos...

    Text Solution

    |

  12. What is the use of thin strip at the back of vernier calliper ?

    Text Solution

    |

  13. When the two jaws of a vernier callipers are in touch, zero of vernier...

    Text Solution

    |

  14. The circular scale of a screw gauge has 200 divisions. When it is give...

    Text Solution

    |

  15. While measuring diameter of a wire using a screw gauge the main scale ...

    Text Solution

    |

  16. When a screw gauge is completely closed, zero of circular scale is 4 ...

    Text Solution

    |

  17. Two spherometers A and B have the same pitch. A has 100 division on pe...

    Text Solution

    |

  18. If h is the height or depth (sagitta) of a spherical surfacce and I is...

    Text Solution

    |

  19. A student measured the length of a rod and wrote it as 3.50 cm. Which ...

    Text Solution

    |