JEE Mains 2025 Practice Test: Units, Measurement, and Dimensions
Total Marks: 40 (4 marks per question)
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Instructions: Select an option for each question. Correct answers turn green (+4 marks), wrong answers turn red (-1 mark). Click "Show Answer" for detailed solutions.
\( E(t) = \alpha^3 e^{-\beta t} \), where \( t \) is time, and \( \beta = 0.3 \, \text{s}^{-1} \). The errors in the measurement of \( \alpha \) and \( t \) are 1.2% and 1.6%, respectively. At \( t = 5 \, \text{s} \), the maximum percentage error in the energy is:
Solution:
We are given \( E(t) = \alpha^3 e^{-\beta t} \), with \( \beta = 0.3 \, \text{s}^{-1} \), and errors in \( \alpha \) and \( t \) as 1.2% and 1.6%, respectively, at \( t = 5 \, \text{s} \). We need to find the maximum percentage error in \( E \).
In error propagation, for a function \( f(x, y) \), the relative error is approximated as:
\[ \frac{\Delta f}{f} \approx \left| \frac{\partial f}{\partial x} \frac{\Delta x}{f} \right| + \left| \frac{\partial f}{\partial y} \frac{\Delta y}{f} \right| \]
Or, using logarithmic differentiation for \( E = \alpha^3 e^{-\beta t} \):
\[ \ln E = 3 \ln \alpha - \beta t \]
Differentiating:
\[ \frac{dE}{E} = 3 \frac{d\alpha}{\alpha} - \beta dt \]
For maximum error, we take absolute values:
\[ \frac{\Delta E}{E} = 3 \frac{\Delta \alpha}{\alpha} + |\beta \Delta t| \]
Percentage error is:
\[ \% \text{error} = \left( 3 \frac{\Delta \alpha}{\alpha} + \beta \Delta t \right) \times 100\% \]
Given:
- \( \frac{\Delta \alpha}{\alpha} = 1.2\% = 0.012 \)
- \( \frac{\Delta t}{t} = 1.6\% = 0.016 \), so \( \Delta t = 0.016 \times t = 0.016 \times 5 = 0.08 \, \text{s} \)
- \( \beta = 0.3 \, \text{s}^{-1} \)
Calculate each term:
\( 3 \frac{\Delta \alpha}{\alpha} = 3 \times 0.012 = 0.036 \) (or 3.6%)
\( \beta \Delta t = 0.3 \times 0.08 = 0.024 \) (or 2.4%)
Total maximum percentage error:
\[ 0.036 + 0.024 = 0.06 \text{ or } 6\% \]
Thus, the answer is A. 6%.
The position of a particle moving on x-axis is given by \( x(t) = A \sin t + B \cos 2t + C t^2 + D \), where \( t \) is time. The dimension of \( \frac{ABC}{D} \) is:
Solution:
The position is \( x(t) = A \sin t + B \cos 2t + C t^2 + D \), where \( t \) is time. Since \( x(t) \) is position, its dimension is [L]. Each term must have the same dimension.
For \( A \sin t \):
\( \sin t \) is dimensionless (assuming \( t \) is \( \omega t \), but here treated as time with [T]), so if \( \sin t \) is dimensionless, [A] = [L].
For \( B \cos 2t \): Similarly, [B] = [L].
For \( D \): [D] = [L].
For \( C t^2 \): [C] [T]\(^2\) = [L], so [C] = [L T\(^{-2}\)].
Now, \( \frac{ABC}{D} \):
\[ [A] [B] [C] / [D] = [L] [L] [L T^{-2}] / [L] = [L^2 T^{-2}] \]
Thus, the dimension is L\(^2\)T\(^{-2}\), so the answer is C. L\(^2\)T\(^{-2}\).
The maximum percentage error in the measurement of density of a wire is given: mass = \( (0.60 \pm 0.003) \, \text{g} \), radius = \( (0.50 \pm 0.01) \, \text{cm} \), length = \( (10.00 \pm 0.05) \, \text{cm} \). Options are:
Solution:
Density \( \rho = \frac{m}{V} \), where \( V = \pi r^2 l \). We need the maximum percentage error in \( \rho \).
Relative error in \( \rho \):
\[ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V} \]
For volume \( V = \pi r^2 l \):
\[ \frac{\Delta V}{V} = 2 \frac{\Delta r}{r} + \frac{\Delta l}{l} \] (since \( V \propto r^2 \))
Given:
- \( m = 0.60 \, \text{g}, \Delta m = 0.003 \, \text{g} \), so \( \frac{\Delta m}{m} = \frac{0.003}{0.60} = 0.005 \) or 0.5%
- \( r = 0.50 \, \text{cm}, \Delta r = 0.01 \, \text{cm} \), so \( \frac{\Delta r}{r} = \frac{0.01}{0.50} = 0.02 \) or 2%
- \( l = 10.00 \, \text{cm}, \Delta l = 0.05 \, \text{cm} \), so \( \frac{\Delta l}{l} = \frac{0.05}{10.00} = 0.005 \) or 0.5%
\( \frac{\Delta V}{V} = 2 \times 0.02 + 0.005 = 0.04 + 0.005 = 0.045 \) or 4.5%
Total error in \( \rho \):
\[ \frac{\Delta \rho}{\rho} = 0.005 + 0.045 = 0.05 \text{ or } 5\% \]
However, options suggest re-evaluation. Typically, percentage error is rounded or interpreted differently in JEE context. Let’s compute numerically:
\( \rho = \frac{m}{\pi r^2 l} \), but for max error, sum all:
\[ \% \text{error} = 0.5\% + 4\% + 0.5\% = 5\% \]
Correct answer is 5%. A. 5.
If \( B \) is magnetic field and \( \mu_0 \) is permeability of free space, then the dimensions of \( \frac{B}{\mu_0} \) is:
Solution:
\( B \) has dimensions [MT\(^{-2}\)A\(^{-1}\)] (from \( F = ILB \)).
\( \mu_0 \) has dimensions [MLT\(^{-2}\)A\(^{-2}\)] (from \( F = \frac{\mu_0 I_1 I_2 L}{2\pi d} \)).
\[ \frac{B}{\mu_0} = \frac{[MT^{-2}A^{-1}]}{[MLT^{-2}A^{-2}]} = [M^0 L^{-1} T^0 A^{1}] = [L^{-1} A] \]
Answer is B. L\(^{-1}\)A.
Given: Statement I: In a vernier callipers, one vernier scale division is always smaller than one main scale division. Statement II: The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions. Options:
Solution:
Statement I: In a vernier calliper, 1 VSD < 1 MSD is false. Typically, \( n \) VSD = (n-1) MSD, so 1 VSD = \( \frac{n-1}{n} \) MSD, which is less than 1 MSD, making I true.
Statement II: Vernier constant (LC) = 1 MSD - 1 VSD, not MSD × number of VSDs, so II is false.
Answer is B. Statement I is true, II is false.
The de-Broglie wavelength is \( \lambda = \frac{h}{\sqrt{2mE}} \). The dimensional formula for Planck’s constant is:
Solution:
\( \lambda = [L] \), \( m = [M] \), \( E = [ML^2 T^{-2}] \).
\[ \sqrt{2mE} = \sqrt{[M] [ML^2 T^{-2}]} = [M^{1/2} L T^{-1}] \]
\[ h = \lambda \sqrt{2mE} = [L] [M^{1/2} L T^{-1}] = [ML^2 T^{-1}] \]
Answer is C. ML\(^2\)T\(^{-1}\).
The dimensional formula of latent heat is:
Solution:
Latent heat \( L = \frac{Q}{m} \), where \( Q = [ML^2 T^{-2}] \), \( m = [M] \).
\[ [L] = \frac{[ML^2 T^{-2}]}{[M]} = [M^0 L^2 T^{-2}] \]
Answer is C. M\(^0\)L\(^2\)T\(^{-2}\).
If \( \epsilon_0 \) is permittivity of free space and \( E \) is electric field, then \( \epsilon_0 E^2 \) has dimensions:
Solution:
\( \epsilon_0 = [M^{-1} L^{-3} T^4 A^2] \) (from \( F = \frac{1}{4\pi \epsilon_0} \frac{q^2}{r^2} \)).
\( E = [MLT^{-3} A^{-1}] \) (from \( F = qE \)).
\[ E^2 = [MLT^{-3} A^{-1}]^2 = [M^2 L^2 T^{-6} A^{-2}] \]
\[ \epsilon_0 E^2 = [M^{-1} L^{-3} T^4 A^2] [M^2 L^2 T^{-6} A^{-2}] = [ML^{-1} T^{-2}] \]
Answer is A. ML\(^{-1}\)T\(^{-2}\).
For a screw gauge with pitch 1 mm, 100 circular scale divisions, zero error at 5 divisions below reference, reading 4 MSD and 60 CSD, the diameter is:
Solution:
Pitch = 1 mm, LC = \( \frac{1}{100} = 0.01 \, \text{mm} \).
Zero error = 5 divisions below, so +5 × 0.01 = +0.05 mm.
Observed reading = 4 MSD + 60 CSD = 4 mm + 60 × 0.01 = 4.60 mm.
Corrected diameter = 4.60 - 0.05 = 4.55 mm.
Answer is C. 4.55 mm.
Vernier caliper: 9 MSD = 10 VSD, MSD = 1 mm, reading 2 cm + 2 VSD, mass = 8.635 g. Density of sphere is:
Solution:
LC = \( \frac{1}{10} = 0.1 \, \text{mm} \) (since 10 VSD = 9 MSD).
Diameter = 20 mm + 2 × 0.1 = 20.2 mm = 2.02 cm.
Radius = 1.01 cm.
Volume = \( \frac{4}{3} \pi (1.01)^3 \approx 4.31 \, \text{cm}^3 \).
Density = \( \frac{8.635}{4.31} \approx 2.0 \, \text{g/cm}^3 \).
Answer is B. 2.0 g/cm\(^3\).
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