Q.Time for 20 oscillations of a pendulum is measured as $t_1 = 39.6$ s; $t_2 = 39.9$ s; $t_3 = 39.5$ s. What is the precision in the measurements? What is the accuracy of the measurement?

The key concept here is Fraunhofer diffraction through a circular aperture. When a plane wave passes through a circular opening (here, the lens itself acts as the aperture), the diffraction pattern on a screen at the focal plane consists of a central bright disk (the Airy disk) surrounded by concentric dark and bright rings. The radius of the first dark ring defines the size of the central bright spot.

Why this approach works:

The lens focuses the parallel beam to a point in its focal plane, but diffraction at the lens aperture spreads the light into a pattern. The angular radius of the first minimum is $\theta = 1.22 \lambda / D$, where $D$ is the aperture diameter. Since the focal plane is at distance $f$ from the lens, the linear radius is $r = f \theta = 1.22 \lambda f / D$. This is the standard result for the Airy disk radius.

Step-by-step solution:

1. Identify the given quantities

   • Wavelength: $\lambda = 500\ \text{nm} = 500 \times 10^{-9}\ \text{m}$
   • Focal length: $f = 20\ \text{cm} = 0.20\ \text{m}$
   • Aperture diameter: $D = 5\ \text{cm} = 0.05\ \text{m}$

2. Recall the formula for the radius of the central bright spot

   For a circular aperture, the first minimum occurs at an angle $\theta$ satisfying

(small-angle approximation is valid here because $\lambda \ll D$).

The linear radius on the focal plane is

3. Plug in the numbers

Simplify step by step:

First, $0.20 / 0.05 = 4$, so

The key idea is simple: a telescope makes distant objects appear larger by bending light through two lenses. The objective lens (big, long focal length) collects light and forms a real, inverted image at its focal plane. The eyepiece (short focal length) then acts like a magnifying glass to view that image. The angular magnification tells you how many times larger the angle subtended at the eye becomes compared to looking directly.

For a telescope in normal adjustment — where the final image is at infinity (relaxed eye viewing) — the angular magnification $M$ is given by the ratio of the focal lengths:

This formula is derived from geometry: the objective creates an image that subtends an angle $\theta_i \approx h / f_o$ (where $h$ is the image height), and the eyepiece, when the image is placed at its focal point, presents that same height at an angle $\theta_e \approx h / f_e$ to the eye. The ratio $\theta_e / \theta_i$ gives $f_o / f_e$.

Now let’s apply it step by step.

1. Identify the focal lengths. The objective focal length is $f_o = 20\ \text{m}$. The eyepiece focal length is $f_e = 1\ \text{cm}$.

2. Convert to consistent units. Since the formula is a ratio, units must match. Convert everything to metres or centimetres — either works. Let’s use metres: $f_e = 1\ \text{cm} = 0.01\ \text{m}$.

3. Plug into the magnification formula.

The units cancel, leaving a pure number. …

In scientific measurements, significant figures represent the precision of a measurement. They include all the digits that are known with certainty plus one estimated digit. Understanding the rules for significant figures is crucial for correctly reporting and interpreting experimental data, ensuring that the precision of a calculation reflects the precision of the input measurements.

Let's analyze each statement regarding significant figures:

1. Analyze statement (A): "All the zeroes between two non-zero digits are significant, no matter where the decimal point is"

   This statement describes "sandwich zeros" or "captive zeros". These zeros are always considered significant because they are part of the measured value and contribute to its precision.

   • Example 1: In $2005 \text{ m}$, the two zeros between $2$ and $5$ are significant. The number has $4$ significant figures.
   • Example 2: In $1.08 \text{ g}$, the zero between $1$ and $8$ is significant. The number has $3$ significant figures. This statement is correct.

2. Analyze statement (B): "The terminal zero(s) in a number without a decimal point are significant"

   This statement refers to trailing zeros in whole numbers. This is a common point of ambiguity in significant figures. When a number like $100 \text{ kg}$ is written, it's unclear whether the zeros are significant (meaning the measurement is precise to the nearest unit) or if they are just placeholders for the magnitude (meaning the measurement is only precise to the nearest hundred).

   • If $100 \text{ kg}$ was measured to the nearest kilogram, it would have $3$ significant figures.
   • If $100 \text{ kg}$ was measured to the nearest ten kilograms, it would have $2$ significant figures ($1.0 \times 10^2 \text{ kg}$).
   • If $100 \text{ kg}$ was measured to the nearest hundred kilograms, it would have $1$ significant figure ($1 \times 10^2 \text{ kg}$). Because these zeros may or may not be significant depending on the precision of the measurement, stating that they "are significant" (implying always significant) is incorrect. This ambiguity is precisely why scientific notation is preferred for such numbers. This statement is not correct.

3. Analyze statement (C): "The terminal zero(s) in a number with a decimal point are significant"

   This statement refers to trailing zeros when a decimal point is present. When a decimal point is explicitly written, any trailing zeros are considered significant. This is because they indicate that the measurement was made to that level of precision. …

The concept here is significant figures — the digits in a number that carry meaningful information about its precision. The trick is that zeros can be tricky: leading zeros (like the ones before the 7) are never significant because they only set the decimal place, but trailing zeros after a decimal point are significant because they indicate the measurement was made to that level of precision. In $0.079000$, the zeros after the 9 tell us the length was measured to the nearest millionth of a meter, not just to the nearest thousandth.

Let’s work through it step by step:

1. Identify all non-zero digits.

   The digits 7 and 9 are non-zero, so they are always significant. That gives us 2 significant figures so far.

2. Check the zeros between non-zero digits.

   There are no zeros between 7 and 9, so nothing to add here.

3. Examine the leading zeros.

   The zeros before the 7 (the one immediately after the decimal and the one before it? Actually, $0.079000$ has one leading zero: the first zero after the decimal point. But wait — the number is $0.079000$, so the digits are: 0 . 0 7 9 0 0 0. The first zero (the one before the decimal) is not a significant figure; it’s just a placeholder. The zero right after the decimal is also a leading zero — it only tells us the number is less than one-tenth. So these leading zeros are not significant.

4. Examine the trailing zeros after the decimal. …

The concept first

Hooke's law in the language of materials is defined by Young's modulus:

Read the formula physically:

• Longer wire $\Rightarrow$ more elongation, because the same fractional stretch acts on a greater original length.
• Thicker wire $\Rightarrow$ less elongation, because the load is shared over a bigger cross-section (lower stress).
• $Y$ is a property of the material only, so with all four wires of the same material $Y$ cancels out of the comparison. The load $F$ is the same for all (that is implicit in the comparison).

For a circular wire, $A=\dfrac{\pi d^{2}}{4}$, so

Notice the square on the diameter — that is what makes thinness so powerful. Halving the diameter quadruples the stretch, whereas doubling the length only doubles it. Many students rank by length alone and pick (C); the $d^2$ is the trap.

Step-by-step

1. Set up the comparison quantity. Use consistent units within the ratio (cm and mm are fine, since we only compare the same combination across the four wires):

2. Evaluate for each wire.

| Wire | $L$ (cm) | $d$ (mm) | $d^2$ | $k=L/d^2$ |

|---|---|---|---|---| …

The key idea is that when a physical quantity is expressed as a product of powers of measured variables, the relative (or percentage) error in the result is the sum of the relative errors of each variable, each weighted by the magnitude of its exponent. This comes directly from the rules of error propagation using differentials.

For a relation $S = k \cdot a^p b^q c^r d^s$ (where $k$ is a constant), the fractional error is:

The absolute values are crucial — errors always add, never cancel, because each measurement's uncertainty contributes positively to the total uncertainty.

Let's apply this step by step.

1. Rewrite the given expression in power form.

   $S = \dfrac{\sqrt{a b}}{c^3 d^4} = \dfrac{a^{1/2} b^{1/2}}{c^3 d^4} = a^{1/2} b^{1/2} c^{-3} d^{-4}$

   The exponents are: $a$ has $+\frac12$, $b$ has $+\frac12$, $c$ has $-3$, $d$ has $-4$.

2. Identify the percentage errors given.

   $\frac{\Delta a}{a} \times 100 = 2\%$, $\frac{\Delta b}{b} \times 100 = 1\%$, $\frac{\Delta c}{c} \times 100 = 1\%$, $\frac{\Delta d}{d} \times 100 = 1\%$.

3. Apply the error propagation formula.

   The percentage error in $S$ is:

4. Compute each term.

   • From $a$: $\frac12 \times 2\% = 1\%$
   • From $b$: $\frac12 \times 1\% = 0.5\%$
   • From $c$: $3 \times 1\% = 3\%$
   • From $d$: $4 \times 1\% = 4\%$

5. Sum them up.

   $1\% + 0.5\% + 3\% + 4\% = 8.5\%$

Now, 8.5% is not among the given options. This means we need to check the standard convention used in such problems. In many exam contexts (especially in error analysis), the exponent on $d$ is taken as $4$ (not $-4$), and the percentage error contribution is $4 \times 1\% = 4\%$. But wait — we already did that. The issue is that the problem likely expects the exponents to be treated as positive magnitudes only, and the sum $1 + 0.5 + 3 + 4 = 8.5$ still doesn't match any option.

Let's re-examine the expression: $S = \dfrac{\sqrt{a b}}{c^3 d^4}$. The square root applies to the product $ab$, so $\sqrt{ab} = (ab)^{1/2} = a^{1/2} b^{1/2}$. That's correct.

Perhaps the intended interpretation is that the percentage errors are given as 2%, 1%, 1%, 1% and the formula for error uses the absolute values of the powers as coefficients. Then:

• $a$: exponent $\frac12$, contribution $\frac12 \times 2 = 1$
• $b$: exponent $\frac12$, contribution $\frac12 \times 1 = 0.5$ …