10 questions · 1 mark each · Correct answer highlighted with explanation
10 questions · 3 marks each · Full model answers included
Absolute error ($\Delta a$) is the magnitude of the difference between the true value ($a_{true}$) and a measured value ($a$): $\Delta a = |a_{true} - a|$.
Relative error is the ratio of the mean absolute error to the true (or mean) value: $\text{Relative error} = \dfrac{\Delta a_{mean}}{a_{mean}}$.
Percentage error is the relative error expressed as a percentage: $\text{Percentage error} = \dfrac{\Delta a_{mean}}{a_{mean}} \times 100\%$.
The principle of homogeneity states that a physical equation is dimensionally correct only if the dimensions of every term on both sides of the equation are the same — you cannot validly add, subtract, or equate quantities with different dimensions.
Example: In $s = ut + \frac{1}{2}at^2$, every term ($s$, $ut$, and $at^2$) must reduce to $[L]$. Checking: $[ut] = [LT^{-1}][T] = [L]$, and $[at^2] = [LT^{-2}][T^2] = [L]$ — both match $[s] = [L]$, confirming the equation is dimensionally homogeneous.
From Newton's law of gravitation: $F = \dfrac{Gm_1m_2}{r^2}$, so $G = \dfrac{Fr^2}{m_1m_2}$.
Substituting dimensions: $[F] = [MLT^{-2}]$, $[r^2] = [L^2]$, $[m_1m_2] = [M^2]$.
$[G] = \dfrac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$
The least count of an instrument is the smallest measurement that can be accurately read using that instrument.
For a vernier caliper, if 1 main scale division (MSD) = $x$, and $n$ vernier scale divisions (VSD) coincide with $(n-1)$ main scale divisions, then 1 VSD = $\dfrac{(n-1)}{n}x$.
Least count = 1 MSD − 1 VSD = $x - \dfrac{(n-1)}{n}x = \dfrac{x}{n}$
Systematic errors occur consistently in the same direction due to a specific identifiable cause (like instrumental defects, zero error, or imperfect experimental technique) — they can often be estimated and corrected. Example: a scale with a worn-out zero mark that consistently reads 0.2 cm too high.
Random errors occur irregularly, with no consistent pattern, due to unpredictable fluctuations (like small variations in observer judgement or environmental conditions) — they are reduced by taking repeated readings and averaging. Example: slightly different stopwatch reaction times across repeated trials.
From $E = h\nu$, we get $h = \dfrac{E}{\nu}$.
$[E] = [ML^2T^{-2}]$ (energy), and $[\nu] = [T^{-1}]$ (frequency).
$[h] = \dfrac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$
Zero error occurs when the zero of the circular scale does not coincide with the reference line on the main scale when the jaws of the screw gauge are fully closed.
If the circular scale's zero lies below the reference line when closed, it is a positive zero error — this makes every measured reading too large by that error amount, so the zero error (found by reading the circular scale division at the reference line) must be subtracted from every observed reading to get the correct value.
Energy has dimensional formula $[M^1L^2T^{-2}]$. Converting 1 erg (CGS) to joule (SI) using $n_1u_1 = n_2u_2$:
$n_2 = n_1 \left(\dfrac{M_1}{M_2}\right)^1\left(\dfrac{L_1}{L_2}\right)^2\left(\dfrac{T_1}{T_2}\right)^{-2}$
With $n_1 = 1$, $M_1=1\text{ g}$, $M_2=1\text{ kg}=1000\text{ g}$, $L_1=1\text{ cm}$, $L_2=1\text{ m}=100\text{ cm}$, $T_1=T_2=1\text{ s}$:
$n_2 = 1 \times \left(\dfrac{1}{1000}\right) \times \left(\dfrac{1}{100}\right)^2 \times 1 = 10^{-3} \times 10^{-4} = 10^{-7}$
So $1\text{ erg} = 10^{-7}\text{ J}$. Therefore: $1\text{ cal} = 4.186\times10^7\text{ erg} = 4.186\times10^7 \times 10^{-7}\text{ J} = \mathbf{4.186\text{ J}}$
10 questions · 5 marks each · Detailed step-by-step solutions
Density $\rho = \dfrac{m}{V}$. For a quotient, percentage errors add:
$\dfrac{\Delta \rho}{\rho} \times 100 = \dfrac{\Delta m}{m}\times 100 + \dfrac{\Delta V}{V}\times 100$
$= \left(\dfrac{0.05}{5.00}\times 100\right) + \left(\dfrac{0.02}{2.00}\times 100\right) = 1\% + 1\% = 2\%$
The percentage error in density is 2%. Calculated density = 5.00/2.00 = 2.50 g/cm³, so the result is reported as (2.50 ± 0.05) g/cm³.
Let $T = k\, l^a m^b g^c$, where $k$ is a dimensionless constant.
Writing dimensions: $[T^1] = [L^a][M^b][LT^{-2}]^c = [M^b L^{a+c} T^{-2c}]$
Comparing powers: M: $b = 0$ · L: $a + c = 0$ · T: $-2c = 1 \Rightarrow c = -\frac{1}{2}$
From L: $a = -c = \frac{1}{2}$
So $T = k\, l^{1/2} g^{-1/2} = k\sqrt{\dfrac{l}{g}}$
This matches the known formula $T = 2\pi\sqrt{\dfrac{l}{g}}$, with $k = 2\pi$ — a value dimensional analysis cannot determine, since it must be found experimentally or derived from the equations of motion.
Total reading = Main scale reading + (Vernier coinciding division number × Least count)
Total reading = $3.4\text{ cm} + (6 \times 0.01\text{ cm})$
Total reading = $3.4\text{ cm} + 0.06\text{ cm} = 3.46\text{ cm}$
The measured length is 3.46 cm.
Step 1 — Least count: LC = Pitch/Number of divisions = 0.5/50 = 0.01 mm
Step 2 — Observed reading: Observed = Main scale reading + (Circular division × LC) = $2\text{ mm} + (25 \times 0.01\text{ mm}) = 2\text{ mm} + 0.25\text{ mm} = 2.25\text{ mm}$
Step 3 — Correct for zero error: True diameter = Observed reading − Zero error = $2.25\text{ mm} - 0.03\text{ mm} = 2.22\text{ mm}$
The true diameter of the wire is 2.22 mm.
LHS: $[v^2] = [LT^{-1}]^2 = [L^2T^{-2}]$
RHS term 1: $[u^2] = [LT^{-1}]^2 = [L^2T^{-2}]$
RHS term 2: $[2as] = [LT^{-2}][L] = [L^2T^{-2}]$ (2 is a dimensionless numerical constant)
Since $[v^2] = [u^2] = [2as] = [L^2T^{-2}]$, all terms have identical dimensions.
The equation is dimensionally correct, satisfying the principle of homogeneity of dimensions.
Area = length × breadth = $16.2 \times 10.1 = 163.62\text{ cm}^2$
Rule for multiplication: The result must have the same number of significant figures as the measurement with the fewest significant figures.
Both 16.2 (3 s.f.) and 10.1 (3 s.f.) have 3 significant figures, so the answer must also be rounded to 3 significant figures.
Area = 164 cm² (rounding 163.62 to 3 significant figures).
Young's modulus $Y = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{F/A}{\Delta l / l}$
Since strain is dimensionless, $[Y] = [\text{Stress}] = \dfrac{[F]}{[A]} = \dfrac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$
Dimensional formula: $[ML^{-1}T^{-2}]$
SI unit: Since $[ML^{-1}T^{-2}]$ corresponds to force/area, the SI unit is newton per square metre (N/m²), which is equivalent to the pascal (Pa).
Pressure has dimensional formula $[ML^{-1}T^{-2}]$. Using $n_1u_1 = n_2u_2$:
$n_2 = n_1\left(\dfrac{M_1}{M_2}\right)^1\left(\dfrac{L_1}{L_2}\right)^{-1}\left(\dfrac{T_1}{T_2}\right)^{-2}$
With $n_1 = 1.013\times10^5$, $M_1=1\text{ kg}=1000\text{ g}$, $M_2=1\text{ g}$, $L_1=1\text{ m}=100\text{ cm}$, $L_2=1\text{ cm}$, $T_1=T_2=1\text{ s}$:
$n_2 = 1.013\times10^5 \times (1000)^1 \times (100)^{-1} \times 1$
$n_2 = 1.013\times10^5 \times 1000 \times \dfrac{1}{100} = 1.013\times10^6$
1 atmosphere = 1.013 × 10⁶ dyne/cm²
From $g = \dfrac{4\pi^2 l}{T^2}$, taking percentage errors (constant $4\pi^2$ contributes no error):
$\dfrac{\Delta g}{g}\times 100 = \dfrac{\Delta l}{l}\times 100 + 2\left(\dfrac{\Delta T}{T}\times 100\right)$
$= \left(\dfrac{0.01}{1.00}\times100\right) + 2\left(\dfrac{0.02}{2.00}\times100\right)$
$= 1\% + 2(1\%) = 1\% + 2\% = 3\%$
The percentage error in the calculated value of $g$ is 3%. Note the factor of 2 for $T$, since $T$ appears squared in the formula.
Let $T = k\, l^a g^b$. Writing dimensions: $[T] = [L^a][LT^{-2}]^b = [L^{a+b}T^{-2b}]$
Comparing powers: T: $-2b = 1 \Rightarrow b = -\frac{1}{2}$ · L: $a + b = 0 \Rightarrow a = \frac{1}{2}$
So $T = k\sqrt{\dfrac{l}{g}}$, matching the standard formula $T = 2\pi\sqrt{l/g}$ with $k = 2\pi$.
Why mass doesn't appear: Since $T$ has dimensions of time only $[T^1]$, and mass $[M]$ cannot combine with $[L]$ and $[T]$ from $l$ and $g$ to cancel out entirely (there is no other mass-dimensioned quantity in the assumed dependency to cancel it), the only dimensionally consistent solution requires the power of $M$ to be zero — confirming that the pendulum's period is genuinely independent of the mass of the bob, a result also derivable from Newton's laws directly.