Consider a modified Bernoulli equation. left(mathrmP+fracmathrmAmathrmBt^2right)+rhomathrmg(mathrmh+mathrmBt)+frac12rhomathrmV^2=textconstant$\left(\mathrm{P}+\frac{\mathrm{A}}{\mathrm{Bt}^{2}}\right)+\rho\mathrm{g}(\mathrm{h}+\mathrm{Bt})+\frac{1}{2}\rho\mathrm{V}^{2}=\text{constant}$
If t has the dimension of time then the dimensions of A and B are ____, ____ respectively.
A.[ML^0T^-1]text and [M^0LT]$[ML^{0}T^{-1}]\text{ and }[M^{0}LT]$
B.[ML^0T^-1]text and [M^0LT^-1]$[ML^{0}T^{-1}]\text{ and }[M^{0}LT^{-1}]$
C.[ML^0T^-2]text and [M^0LT^-2]$[ML^{0}T^{-2}]\text{ and }[M^{0}LT^{-2}]$
D.[ML^0T^-2]text and [M^0LT^-1]$[ML^{0}T^{-2}]\text{ and }[M^{0}LT^{-1}]$
Solution & Explanation
### Related Formula
textPrinciple of Homogeneity: Only quantities with same dimensions can be added/subtracted.$$\text{Principle of Homogeneity: Only quantities with same dimensions can be added/subtracted.}$$
### Core Logic
From the term (h + Bt)$(h + Bt)$, h$h$ (height/length) and Bt$Bt$ must have the same dimension.
[h] = [Bt]$[h] = [Bt]$[B] = frac[h][t] = frac[L][T] = [L T^-1]$$[B] = \frac{[h]}{[t]} = \frac{[L]}{[T]} = [L T^{-1}]$$
So, B = [M^0 L T^-1]$B = [M^0 L T^{-1}]$.
### Step 1: Finding Dimension of A
From the term left(P + fracABt^2right)$\left(P + \frac{A}{Bt^2}\right)$, pressure P$P$ and fracABt^2$\frac{A}{Bt^2}$ must have the same dimension.
Dimension of Pressure P = fractextForcetextArea = [M L^-1 T^-2]$P = \frac{\text{Force}}{\text{Area}} = [M L^{-1} T^{-2}]$.
[P] = left[fracABt^2right]$$[P] = \left[\frac{A}{Bt^2}\right]$$[M L^-1 T^-2] = frac[A][L T^-1] times [T^2]$$[M L^{-1} T^{-2}] = \frac{[A]}{[L T^{-1}] \times [T^2]}$$[M L^-1 T^-2] = frac[A][L T]$$[M L^{-1} T^{-2}] = \frac{[A]}{[L T]}$$[A] = [M L^-1 T^-2] times [L T] = [M L^0 T^-1]$$[A] = [M L^{-1} T^{-2}] \times [L T] = [M L^0 T^{-1}]$$
So, A = [M L^0 T^-1]$A = [M L^0 T^{-1}]$.
### Pattern Recognition
Apply the Principle of Homogeneity across any (X + Y)$(X + Y)$ 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
Keywords:#Consider a modified Bernoulli equation#JEE Main 2026 Morning Q32#Units and Measurements JEE Main 2026#Dimensional Analysis JEE Main 2026
More Units and Measurements Previous-Year Questions — Page 3
Q21jee_main_2025_28_jan_morningErrors in Measurement
A tiny metallic rectangular sheet has length and breadth of 5 mathrm~mm$5 \mathrm{~mm}$ and 2.5 mathrm~mm$2.5 \mathrm{~mm}$ , respectively. Using a specially designed screw gauge which has pitch of 0.75 mathrm~mm$0.75 \mathrm{~mm}$ and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be fracmathrmx100$\frac{\mathrm{x}}{100}$ where mathrmx$\mathrm{x}$ is ________.
Numerical Answer.Answer: 3 to 3
Solution
### Core Logic
First, find the least count of the measurement tool:
textLeast Count = fractextPitchtextNumber of circular scale divisions = frac0.75 mathrm~mm15 = 0.05 mathrm~mm$$\text{Least Count} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.75 \mathrm{~mm}}{15} = 0.05 \mathrm{~mm}$$Least count calculation tracking diagram for Q21
The area of the rectangular metallic sheet is calculated as:
mathrmA = mathrmL cdot mathrmW$$\mathrm{A} = \mathrm{L} \cdot \mathrm{W}$$
Expressing the absolute error via fractional configuration parts:
fracmathrmdAmathrmA = fracmathrmdLmathrmL + fracmathrmdWmathrmW$$\frac{\mathrm{dA}}{\mathrm{A}} = \frac{\mathrm{dL}}{\mathrm{L}} + \frac{\mathrm{dW}}{\mathrm{W}}$$
Substituting the instrument limits (mathrmdL = mathrmdW = 0.05 mathrm~mm$\mathrm{dL} = \mathrm{dW} = 0.05 \mathrm{~mm}$):
fracmathrmdAmathrmA = frac0.055 + frac0.052.5 = frac1100 + frac2100 = frac3100$$\frac{\mathrm{dA}}{\mathrm{A}} = \frac{0.05}{5} + \frac{0.05}{2.5} = \frac{1}{100} + \frac{2}{100} = \frac{3}{100}$$
### Step 1: Final Value Match
Comparing this to the target format fracmathrmx100$\frac{\mathrm{x}}{100}$ gives:
mathrmx = 3$\mathrm{x} = 3$
### Pattern Recognition
The absolute measurement uncertainty matches the instrument's least count value directly. Sum up individual fractional errors to compute the total area uncertainty parameter.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of left[mathrmM^mathrmamathrmL^mathrmbmathrmT^mathrmcright]$\left[\mathrm{M}^{\mathrm{a}}\mathrm{L}^{\mathrm{b}}\mathrm{T}^{\mathrm{c}}\right]$ . If b = 3$b = 3$ , the value of c$c$ is
Numerical Answer.Answer: 0 to 0
Solution
### Core Logic
Let's find the dimensional formula for the ratio of Modulus of Elasticity to Torque:
textTarget Dimensions = frac[textModulus of Elasticity][textTorque]$$\text{Target Dimensions} = \frac{[\text{Modulus of Elasticity}]}{[\text{Torque}]}$$textTarget Dimensions = frac[mathrmM L^-1 mathrmT^-2][mathrmM L^2 mathrmT^-2] = [mathrmM^0 mathrmL^-3 mathrmT^0]$$\text{Target Dimensions} = \frac{[\mathrm{M L}^{-1} \mathrm{T}^{-2}]}{[\mathrm{M L}^2 \mathrm{T}^{-2}]} = [\mathrm{M}^0 \mathrm{L}^{-3} \mathrm{T}^0]$$
### Step 1: Exponent Matching
Comparing this output to the target layout formula [mathrmM^mathrma mathrmL^mathrmb mathrmT^mathrmc]$[\mathrm{M}^{\mathrm{a}} \mathrm{L}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}]$:
mathrmc = 0$\mathrm{c} = 0$
### Pattern Recognition
Both dimensions share identical time dependence factors (mathrmT^-2$\mathrm{T}^{-2}$), meaning they cancel out completely. This leaves the time exponent value as exactly zero.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Choose the correct answer from the options given below:
A. A-IV, B-III, C-II, D-I
B. A-III, B-II, C-I, D-IV
C. A-II, B-IV, C-III, D-I
D. A-I, B-III, C-IV, D-II
Solution
### Related Formula
Newton's Law of Gravitation:
F = Gfracm_1 m_2r^2 implies G = fracFr^2m_1 m_2$$F = G\frac{m_1 m_2}{r^2} \implies G = \frac{Fr^2}{m_1 m_2}$$
Potential Energy:
U = mgh quad [textWork]$$U = mgh \quad [\text{Work}]$$
Potential:
V = fracWm$$V = \frac{W}{m}$$
Acceleration:
g = fractextVelocitytextTime$$g = \frac{\text{Velocity}}{\text{Time}}$$
### Core Logic
Let's perform dimensional analysis for each item:
1. **A. Gravitational constant (G$G$)**:
[G] = frac[F][r^2][M^2] = frac[MLT^-2][L^2][M^2] = [M^-1L^3T^-2]$$[G] = \frac{[F][r^2]}{[M^2]} = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$$
Matches with **IV**.
2. **B. Gravitational potential energy (U$U$)**:
[U] = textDimensions of Work = [ML^2T^-2]$$[U] = \text{Dimensions of Work} = [ML^2T^{-2}]$$
Matches with **III**.
3. **C. Gravitational potential (V$V$)**:
[V] = frac[textEnergy][M] = frac[ML^2T^-2][M] = [L^2T^-2]$$[V] = \frac{[\text{Energy}]}{[M]} = \frac{[ML^2T^{-2}]}{[M]} = [L^2T^{-2}]$$
Matches with **II**.
4. **D. Acceleration due to gravity (g$g$)**:
[g] = [textAcceleration] = [LT^-2]$$[g] = [\text{Acceleration}] = [LT^{-2}]$$
Matches with **I**.
### Step 1: Match and Selection
Let's align our matches:
- A rightarrow$\rightarrow$ IV
- B rightarrow$\rightarrow$ III
- C rightarrow$\rightarrow$ II
- D rightarrow$\rightarrow$ I
This sequence matches option (1).
### Pattern Recognition
To save precious exam time on match-the-column questions, start with the easiest dimensional terms first. You know Acceleration due to gravity is g rightarrow [LT^-2]$g \rightarrow [LT^{-2}]$ (D-I) and energy is [ML^2T^-2]$[ML^2T^{-2}]$ (B-III). Looking at the options, only Option 1 matches this sequence immediately!
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 11 Physics: Gravitation
Q17jee_main_2025_03_april_morningSignificant Figures in Arithmetic
A person measures mass of 3 different particles as 435.42mathrm~g$435.42\mathrm{~g}$, 226.3mathrm~g$226.3\mathrm{~g}$ and 0.125mathrm~g$0.125\mathrm{~g}$. According to the rules for arithmetic operations with significant figures, the additions of the masses of 3 particles will be.
A.661.845mathrm~g$661.845\mathrm{~g}$
B.662mathrm~g$662\mathrm{~g}$
C.661.8mathrm~g$661.8\mathrm{~g}$
D.661.84mathrm~g$661.84\mathrm{~g}$
Solution
### Related Formula
Significant Figures Rule for Addition/Subtraction:
The final result must be rounded off to keep only as many decimal places as there are in the measurement with the **least** number of decimal places.
### Core Logic
Let's look at the decimal places of each measurement:
- 435.42mathrm~g$435.42\mathrm{~g}$ has **2 decimal places**.
- 226.3mathrm~g$226.3\mathrm{~g}$ has **1 decimal place**.
- 0.125mathrm~g$0.125\mathrm{~g}$ has **3 decimal places**.
The minimum number of decimal places is **1 decimal place** (from 226.3mathrm~g$226.3\mathrm{~g}$).
### Step 1: Addition and Rounding
First, perform the standard mathematical addition:
textSum = 435.42 + 226.3 + 0.125 = 661.845mathrm~g$$\text{Sum} = 435.42 + 226.3 + 0.125 = 661.845\mathrm{~g}$$
Now, round this raw sum off to **1 decimal place**:
- The digit after tenths place is 4 (4 < 5$4 < 5$), so we round down.
- Net rounded sum = 661.8mathrm~g$661.8\mathrm{~g}$.
### Pattern Recognition
Bust the common myth: Addition depends on the least number of decimal places, whereas multiplication/division depends on the least number of significant figures. Always distinguish between these two rules during exams!
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements: Error Analysis and Significant Figures
Q4jee_main_2025_04_april_eveningDimensions of Physical Quantities
Given below are two statements:
Statement (I): The dimensions of Planck's constant and angular momentum are same.
Statement (II): In Bohr's model electron revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck's constant.
In the light of the above statements, choose the most appropriate answer from the options given below:
A. Both Statement I and Statement II are correct
B. Statement I is incorrect but Statement II is correct
C. Statement I is correct but Statement II is incorrect
D. Both Statement I and Statement II are incorrect
Solution
### Related Formula
E = hf implies [h] = frac[E][f] = fractextMtextL^2textT^-2textT^-1 = textMtextL^2textT^-1$$E = hf \implies [h] = \frac{[E]}{[f]} = \frac{\text{M}\text{L}^2\text{T}^{-2}}{\text{T}^{-1}} = \text{M}\text{L}^2\text{T}^{-1}$$L = mvr implies [L] = textM cdot (textLtextT^-1) cdot textL = textMtextL^2textT^-1$$L = mvr \implies [L] = \text{M} \cdot (\text{L}\text{T}^{-1}) \cdot \text{L} = \text{M}\text{L}^2\text{T}^{-1}$$L = fracnh2pi$$L = \frac{nh}{2\pi}$$
### Core Logic
Statement I: Comparing the dimensional formula of Planck's constant (h$h$) and angular momentum (L$L$), both are identical [textMtextL^2textT^-1]$[\text{M}\text{L}^2\text{T}^{-1}]$. Hence, Statement I is correct.
Statement II: According to Bohr's second postulate, angular momentum is an integral multiple of frach2pi$\frac{h}{2\pi}$, not an integral multiple of h$h$. Hence, Statement II is incorrect.
### Pattern Recognition
Watch out for exact definitions in standard postulates. Bohr's model requires angular momentum to be quantized in units of hbar = frach2pi$\hbar = \frac{h}{2\pi}$, making statement II a classic trap.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Units and Measurements
Class 12 Physics: Atoms
More Units and Measurements Questions — jee_main_2026_21_jan_morning
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