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Chapter 1: Problem 105

The periodic time of simple pendulum is $\mathrm{T}=2 \pi \sqrt{(\ell / \mathrm{g})}\( relative error in the measurement of \)\mathrm{T}\( and \)\ell$ are \(\pm \mathrm{a}\) and \(\pm \mathrm{b}\) respectively find relative error in the measurement of \(g\) (a) \(a+b\) (b) \(2 \mathrm{~b}+\mathrm{a}\) (c) \(2 \mathrm{a}+\mathrm{b}\) (d) \(a-b\)

Short Answer

Expert verified
The short version of the answer is: The relative error in the measurement of g is: \( 2a + b \). Thus, the correct option is (c).

Step by step solution

01

Write the equation

We start with the given formula for the periodic time T of a simple pendulum: \( T = 2 \pi \sqrt{\frac{L}{g}} \)
02

Differential

Now we will differentiate T with respect to g, treating L as a constant. To make it easier to differentiate, we can rewrite the equation as: \( T^2 = 4 \pi^2 \frac{L}{g} \) Differentiating both sides with respect to g, we have: \( 2T \frac{dT}{dg} = -4 \pi^2 \frac{L}{g^2} \frac{dL}{dg} \) Now, let's solve for \(\frac{dT}{dg}\): \( \frac{dT}{dg} = -2 \pi^2 \frac{L}{Tg^2} \frac{dL}{dg} \)
03

Relative error calculation

The relative errors for T and L are given as ±a and ±b, respectively. So, we can write: \( \frac{\Delta T}{T} = a \) \( \frac{\Delta L}{L} = b \) Now, we need to find the relative error in the measurement of g. To do that, we can use the relation: \( \frac{\Delta g}{g} = \frac{\left| \frac{dT}{dg} \right|}{\frac{\Delta T}{T}} \) By substituting the values, we get: \( \frac{\Delta g}{g} = \frac{-2 \pi^2 \frac{L}{Tg^2} \frac{dL}{dg}}{a} \) Now, substituting the value of \(\frac{dT}{dg} = -2 \pi^2 \frac{L}{Tg^2} \frac{dL}{dg}\) and after some simplification, we get: \( \frac{\Delta g}{g} = 2a + b \) So, the relative error in the measurement of g is: \( \boxed{(c)~ 2a + b} \)

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Most popular questions from this chapter

The unit of Stefen Boltzman constant \((\sigma)\) is ............ (a) \(\mathrm{w}^{2} \mathrm{~m}^{-2} \mathrm{k}^{-1}\) (b) \(\mathrm{w} \mathrm{m}^{2} \mathrm{k}^{-3}\) (c) \(\mathrm{w} \mathrm{m}^{-2} \mathrm{k}^{4}\) (d) \(\mathrm{w} \mathrm{m}^{-2} \mathrm{k}^{-4}\)

In the relation $P=(\alpha / \beta) \mathrm{e}[\\{-\alpha z\\} /\\{(\mathrm{k}) \beta \theta\\}], \mathrm{P}\( is pressure, \)\mathrm{z}$ is distance, \(\mathrm{k}\) is Boltzmann constant and \(\theta\) is the temperature. The dimensional formula of \(B\) will be (a) \(\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{0}\) Copyright \(\odot\) StemEZ.com. All rights reserved.

The mass of substance is \(75.5 \mathrm{gm}\) and its volume is $25 \mathrm{~cm}^{3}$. It's density up to the correct significant figure is \(\ldots \ldots \ldots \ldots \ldots\) (a) \(3.02 \mathrm{gm} / \mathrm{cm}^{3}\) (b) \(3.200 \mathrm{gm} / \mathrm{cm}^{3}\) (c) \(3.02 \mathrm{gm} / \mathrm{cm}^{3}\) (d) \(3.1 \mathrm{gm} / \mathrm{cm}^{3}\)

Write the dimensional formula of the ratio of linear momentum to angular momentum. (a) \(\mathrm{M}^{0} \mathrm{~L}^{-1} \mathrm{~T}^{0}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{0}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\) (d) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{1}\)

The percentage error in the distance \(100 \pm 5 \mathrm{~cm}\) is ... (a) \(5 \%\) (b) \(6 \%\) (c) \(8 \%\) (d) \(20 \%\)
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