Explore the Laws That Bend Space and Time

Detailed Solution for Test 1

Solution to Questions

Converstion of Young's Modulus

Problem:

Young's modulus of the material of a wire is \( 18 \times 10^{11}\ \text{dyne cm}^{-2} \). Its value in SI is:

Given

  • Original value: \( Y = 18 \times 10^{11}\ \text{dyne cm}^{-2} \)
  • Base conversions: \( 1\ \text{dyne} = 10^{-5}\ \text{N} \), \( 1\ \text{cm} = 10^{-2}\ \text{m} \)

Step 1 — Convert the unit dyne/cm² to N/m²

\[ 1\ \frac{\text{dyne}}{\text{cm}^{2}} = \frac{10^{-5}\ \text{N}}{(10^{-2}\ \text{m})^{2}} = \frac{10^{-5}}{10^{-4}}\ \frac{\text{N}}{\text{m}^{2}} = 10^{-1}\ \frac{\text{N}}{\text{m}^{2}} = 0.1\ \text{Pa} \]

Rule of thumb: multiply a value in dyne/cm² by 0.1 to get Pascals.

Step 2 — Apply the factor to the given value

\[ Y = 18 \times 10^{11}\ \frac{\text{dyne}}{\text{cm}^{2}} = 18 \times 10^{11} \times 10^{-1}\ \frac{\text{N}}{\text{m}^{2}} = 1.8 \times 10^{11}\ \text{N m}^{-2} = 1.8 \times 10^{11}\ \text{Pa} \]

Final

Young’s modulus in SI: \( \boxed{1.8 \times 10^{11}\ \text{Pa}} \) (i.e., \(18 \times 10^{10}\ \text{N m}^{-2}\)).

Sanity check: Metals typically have \(Y \sim 10^{11}\) Pa, so the result is in a realistic range.



Significant Figures

Problem: The respective number of significant figures for the numbers \( 23.023 \), \( 0.0003 \), and \( 2.1 \times 10^{-3} \) are:

Rules We’ll Use

  • All non-zero digits are significant.
  • Zeros between non-zero digits are significant.
  • Leading zeros (to the left of the first non-zero digit) are not significant.
  • Trailing zeros after a decimal point are significant.
  • In scientific notation, only the digits in the coefficient (mantissa) count; the exponent does not affect the count.

Worked Out

A) 23.023

\[ 23.023 \;=\; \underbrace{2}_{\text{sig}}\,\underbrace{3}_{\text{sig}}\,\underbrace{0}_{\text{between non-zeros ⇒ sig}}\,\underbrace{2}_{\text{sig}}\,\underbrace{3}_{\text{sig}} \;\Rightarrow\; \boxed{5\ \text{significant figures}} \]

Explanation: The zero is sandwiched between non-zero digits, so it is significant.

B) 0.0003

\[ 0.0003 \;=\; \underbrace{0.000}_{\text{leading zeros ⇒ not sig}}\,\underbrace{3}_{\text{sig}} \;\Rightarrow\; \boxed{1\ \text{significant figure}} \]

Explanation: All zeros before the first non-zero digit are leading zeros and are not significant.

C) \(2.1 \times 10^{-3}\)

\[ 2.1 \times 10^{-3} \quad \Rightarrow \quad \text{Coefficient } 2.1 \text{ has } \boxed{2 \text{ significant figures}} \]

Final

  • 23.023 → 5 significant figures
  • 0.0003 → 1 significant figure
  • 2.1 × 10−3 → 2 significant figures


Screw Gauge — Worked Solution

Problem:

student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of−0004 . cm, the correct diameter of the ball is..

Given

  • Least count (LC) = 0.001 cm
  • Main/Sleeve scale reading (MSR/PSR) = 5 mm = 0.5 cm
  • Circular scale reading (CSR) = 25 divisions
  • Zero error = −0.004 cm

1) Observed Reading (before zero-error correction)

Circular contribution:

\[ \text{Circular contribution} = \text{CSR} \times \text{LC} = 25 \times 0.001 = 0.025\ \text{cm} \]

Observed diameter:

\[ D_{\text{obs}} = \text{MSR} + \text{Circular contribution} = 0.5 + 0.025 = 0.525\ \text{cm} \]

2) Apply Zero-Error Correction (sign rule)

Use the standard correction rule:

\[ D_{\text{true}} = D_{\text{obs}} - (\text{Zero error}) \]
  • Positive zero error (over-read) → subtract a positive number → result decreases.
  • Negative zero error (under-read) → subtract a negative number → effectively add its magnitude.

Here, zero error = \(-0.004\ \text{cm}\):

\[ D_{\text{true}} = 0.525 - (-0.004) = 0.525 + 0.004 = 0.529\ \text{cm} \]

Final

Correct diameter = 0.529 cm.

Note: If the zero error were +0.004 cm instead, then \[ D_{\text{true}} = 0.525 - 0.004 = 0.521\ \text{cm}. \] The difference between 0.529 cm and 0.521 cm comes solely from the sign of the zero error.



Percentage Error in Volume

Problem:

If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be

If the percentage error in the radius is 2%, what is the percentage error in the volume of the sphere?

Model & Key Idea

Volume of a sphere depends on the cube of its radius:

\[ V = \frac{4}{3}\,\pi r^{3} \]

For a quantity of the form \( V \propto r^{n} \), the fractional (percentage) error multiplies by \( n \):

\[ \frac{\Delta V}{V} = n\,\frac{\Delta r}{r} \quad\Rightarrow\quad \%\text{error in }V = n \times \%\text{error in }r \]

Work the Numbers

Here, \( n = 3 \) (since \( V \propto r^3 \)) and \( \%\text{error in } r = 2\% \):

\[ \%\text{error in }V = 3 \times 2\% = 6\% \]

Final

Percentage error in volume = 6%.

Note: This result uses the small-error propagation rule for powers: \( X = A^{k} \Rightarrow \frac{\Delta X}{X} = k\,\frac{\Delta A}{A} \).
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