Units and Measurements Previous Year Questions — JEE Main Physics
50 past-year Units and Measurements questions from JEE Main (Physics) — page 8.
Q41 (2024)
A force is represented by $F = ax^2 + bt^{1/2}$
Where $x =$ distance and $t =$ time. The dimensions of $b^2 / a$ are:
- $[ML^3T^{-3}]$
- $[MLT^{-2}]$
- $[ML^{-1}T^{-1}]$
- $[ML^2T^{-3}]$
### Related Formula
$$\text{Principle of Homogeneity: } [F] = [ax^2] = [bt^{1/2}]$$
### Core Logic
By the principle of dimensional homogeneity, each additive term must have the same dimension as the left hand side.
Dimension of force $F = [M L T^{-2}]$.
For the term $ax^2$:
$$[a] = \frac{[F]}{[x^2]} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}]$$
For the term $bt^{1/2}$:
$$[b] = \frac{[F]}{[t^{1/2}]} = \frac{[M L T^{-2}]}{[T^{1/2}]} = [M L T^{-5/2}]$$
### Step 2: Computing Required Ratio
We need the dimension of $\frac{b^2}{a}$:
$$\left[ \frac{b^2}{a} \right] = \frac{[M L T^{-5/2}]^2}{[M L^{-1} T^{-2}]}$$
$$\left[ \frac{b^2}{a} \right] = \frac{[M^2 L^2 T^{-5}]}{[M L^{-1} T^{-2}]}$$
$$\left[ \frac{b^2}{a} \right] = [M^{2-1} L^{2 - (-1)} T^{-5 - (-2)}]$$
$$\left[ \frac{b^2}{a} \right] = [M L^3 T^{-3}]$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units And Measurements
Q30 (2026)
In an experiment the values of two spring constants were measured as $k_{1}=(10\pm0.2)\text{ N/m}$ and $k_{2}=(20\pm0.3)\text{ N/m}$. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :
- 2.67%
- 2.33%
- 1.33%
- 1.67%
### Related Formula
$$K_{\text{eq}} = K_1 + K_2 \quad \text{(Parallel Combination)}$$
$$\Delta K_{\text{eq}} = \Delta K_1 + \Delta K_2$$
$$\text{Percentage Error} = \frac{\Delta K_{\text{eq}}}{K_{\text{eq}}} \times 100$$
### Core Logic
For a parallel combination of springs, the equivalent spring constant is simply the sum of individual constants.
$$K_{\text{eq}} = K_{1} + K_{2} = 10 + 20 = 30\text{ N/m}$$
When quantities are added, their absolute errors are also added:
$$\Delta K_{\text{eq}} = \Delta K_{1} + \Delta K_{2} = 0.2 + 0.3 = 0.5\text{ N/m}$$
### Step 1: Calculating Percentage Error
The percentage error in $K$ is:
$$\text{% Error in } K = \frac{0.5}{30} \times 100 = \frac{5}{3}\% \approx 1.67\%$$
### Pattern Recognition
Addition operation = add absolute errors. For parallel springs, it's just plain addition. Then to get percentage, divide the absolute error sum by the nominal value sum and multiply by 100.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 11 Physics: Oscillations
Q32 (2026)
Consider a modified Bernoulli equation. $\left(\mathrm{P}+\frac{\mathrm{A}}{\mathrm{Bt}^{2}}\right)+\rho\mathrm{g}(\mathrm{h}+\mathrm{Bt})+\frac{1}{2}\rho\mathrm{V}^{2}=\text{constant}$
If t has the dimension of time then the dimensions of A and B are ____, ____ respectively.
- $[ML^{0}T^{-1}]\text{ and }[M^{0}LT]$
- $[ML^{0}T^{-1}]\text{ and }[M^{0}LT^{-1}]$
- $[ML^{0}T^{-2}]\text{ and }[M^{0}LT^{-2}]$
- $[ML^{0}T^{-2}]\text{ and }[M^{0}LT^{-1}]$
### Related Formula
$$\text{Principle of Homogeneity: Only quantities with same dimensions can be added/subtracted.}$$
### Core Logic
From the term $(h + Bt)$, $h$ (height/length) and $Bt$ must have the same dimension.
$[h] = [Bt]$
$$[B] = \frac{[h]}{[t]} = \frac{[L]}{[T]} = [L T^{-1}]$$
So, $B = [M^0 L T^{-1}]$.
### Step 1: Finding Dimension of A
From the term $\left(P + \frac{A}{Bt^2}\right)$, pressure $P$ and $\frac{A}{Bt^2}$ must have the same dimension.
Dimension of Pressure $P = \frac{\text{Force}}{\text{Area}} = [M L^{-1} T^{-2}]$.
$$[P] = \left[\frac{A}{Bt^2}\right]$$
$$[M L^{-1} T^{-2}] = \frac{[A]}{[L T^{-1}] \times [T^2]}$$
$$[M L^{-1} T^{-2}] = \frac{[A]}{[L T]}$$
$$[A] = [M L^{-1} T^{-2}] \times [L T] = [M L^0 T^{-1}]$$
So, $A = [M L^0 T^{-1}]$.
### Pattern Recognition
Apply the Principle of Homogeneity across any $(X + Y)$ terms inside brackets. Extract one unknown, plug it into the next additive bracket group.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 11 Physics: Mechanical Properties of Fluids
Q36 (2026)
Keeping the significant figures in view, the sum of the physical quantities $52.01 \text{ m}$, $153.2 \text{ m}$ and $0.123 \text{ m}$ is :
- $205 \text{ m}$
- $205.333 \text{ m}$
- $205.33 \text{ m}$
- $205.3 \text{ m}$
### Related Formula
Addition Rule for Significant Figures: The result cannot have more digits to the right of the decimal point than any of the original numbers.
### Core Logic
Let's align the numbers and perform exact addition:
$52.01$ (2 decimal places)
$153.2$ (1 decimal place - least precise)
$+ 0.123$ (3 decimal places)
--------
$205.333$
### Step 1: Final Conclusion
According to the rules of significant figures for addition, the sum must be rounded to the least number of decimal places present in the given measurements.
Here, $153.2$ has only $1$ decimal place.
Rounding $205.333$ to one decimal place yields $205.3 \text{ m}$.
### Pattern Recognition
Addition/Subtraction is bounded by decimal places. Multiplication/Division is bounded by total significant figures. In this case, 1 decimal place rules.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Q37 (2026)
A spherical body of radius r and density $\sigma$ falls freely through a viscous liquid having density $\rho$ and viscosity $\eta$ and attains a terminal velocity $v_{0}$. Estimated maximum error in the quantity $\eta$ is :
(Ignore errors associated with $\sigma$, $\rho$ and g, gravitational acceleration)
- $2\frac{\Delta r}{r}-\frac{\Delta v_0}{v_0}$
- $\frac{2\Delta r}{r}+\frac{\Delta v_0}{v_0}$
- $2\left[\frac{\Delta r}{r}+\frac{\Delta v_0}{v_0}\right]$
- $2\left[\frac{\Delta r}{r}-\frac{\Delta v_0}{v_0}\right]$
### Related Formula
$$v_0 = \frac{2}{9}\frac{r^2 g}{\eta} (\sigma - \rho)$$
### Core Logic
Rearranging the formula for viscosity $\eta$:
$$\eta = \frac{2}{9} \frac{r^2 g}{v_0} (\sigma - \rho)$$
Since $\frac{2}{9}$, $g$, $\sigma$, and $\rho$ are constants or have negligible error, the relative error depends only on $r$ and $v_0$.
### Step 1: Error Propagation
Taking natural logarithm on both sides:
$$\ln(\eta) = \ln(\text{constant}) + 2\ln(r) - \ln(v_0)$$
Differentiating for maximum possible error (adding absolute errors):
$$\frac{\Delta \eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$$
### Step 2: Final Conclusion
The maximum fractional error is $\frac{2\Delta r}{r} + \frac{\Delta v_0}{v_0}$.
### Pattern Recognition
For maximum fractional error of a derived quantity $Z = X^a Y^b$, the formula is $\Delta Z/Z = |a|\Delta X/X + |b|\Delta Y/Y$. Always sum the absolute relative errors.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 11 Physics: Mechanical Properties of Fluids
