JEE Units and Measurements Previous Year Questions (PYQs) Class 11 Physics with solutions are essential for building a strong foundation in Physics. This chapter plays a crucial role in improving accuracy in calculations and understanding concepts like dimensional analysis, significant figures, and error estimation. Practicing PYQs helps students become familiar with exam patterns, avoid common mistakes, and strengthen their problem-solving skills for JEE Main. Follow the link for JEE Main PYQs Units and Measurements Set-1.
JEE Main PYQs Previous Year Questions and Solutions – Units and Measurements
Problem : The period of oscillation of a simple pendulum is $T = 2\pi\sqrt{\dfrac{L}{g}}$. Measured value of $L$ is $1.0 \text{ m}$ from metre scale having a minimum division of $1 \text{ mm}$ and time of one complete oscillation is $1.95 \text{ s}$ measured from stopwatch of $0.01 \text{ s}$ resolution. The percentage error in the determination of $g$ will be:
(a) $1.13\%$
(b) $1.03\%$
(c) $1.33\%$
(d) $1.30\%$
[2021, 24 Feb Shift-II]
Solution: Ans. (a)
Given, $T = 1.95 \text{ s}$ and $L = 1.0 \text{ m}$
$\Delta T = 0.01 \text{ s}$ and $\Delta L = 1 \text{ mm} = 10^{-3} \text{ m}$
From $T = 2\pi\sqrt{\dfrac{L}{g}}$, we have $g = 4\pi^2 \dfrac{L}{T^2}$
$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$
$\dfrac{\Delta g}{g} = \dfrac{10^{-3}}{1} + 2 \times \dfrac{0.01}{1.95}$
$\dfrac{\Delta g}{g} = 0.001 + 0.01025 = 0.01125$
Percentage error $= 0.01125 \times 100 = 1.125\% \approx 1.13\%$
Follow the Units and Measurements chapter to build a strong foundation in basics.
Problem : A physical quantity $z$ depends on four observables $a, b, c$ and $d$, as $z = \dfrac{a^2 b^{2/3}}{\sqrt{c} d^3}$. The percentages of error in the measurement of $a, b, c$ and $d$ are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in $z$ is:
(a) 13.5%
(b) 16.5%
(c) 14.5%
(d) 12.25%
[2020, 5 Sep Shift-I]
Solution: Ans. (c)
Given, $z = \dfrac{a^2 b^{2/3}}{\sqrt{c} d^3}$
Using the concept of relative error:
$\dfrac{\Delta z}{z} = 2\dfrac{\Delta a}{a} + \dfrac{2}{3}\dfrac{\Delta b}{b} + \dfrac{1}{2}\dfrac{\Delta c}{c} + 3\dfrac{\Delta d}{d}$
Percentage error in $z$:
$\% \text{ error in } z = 2(\% \text{ error in } a) + \frac{2}{3}(\% \text{ error in } b) + \frac{1}{2}(\% \text{ error in } c) + 3(\% \text{ error in } d)$
$= 2(2\%) + \frac{2}{3}(1.5\%) + \frac{1}{2}(4\%) + 3(2.5\%)$
$= 4\% + 1\% + 2\% + 7.5\% = 14.5\%$
Problem : The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is $(\dfrac{x}{100})\%$. If the relative errors in measuring the mass and the diameter are 6.0% and 1.5% respectively, the value of $x$ is ……………………….
[2020, 6 Sep Shift-I]
Solution: Ans. (1050)
Density $\rho = \dfrac{\text{Mass}}{\text{Volume}} = \dfrac{M}{\frac{4}{3}\pi (\dfrac{d}{2})^3} = \dfrac{6M}{\pi d^3}$
Percentage error in density:
$\dfrac{\Delta \rho}{\rho} \times 100 = \dfrac{\Delta M}{M} \times 100 + 3 \left( \dfrac{\Delta d}{d} \times 100 \right)$
Given, $\dfrac{\Delta M}{M} \times 100 = 6\%$ and $\dfrac{\Delta d}{d} \times 100 = 1.5\%$
$\% \text{ error in } \rho = 6\% + 3(1.5\%) = 6\% + 4.5\% = 10.5\%$
According to the question:
$\dfrac{x}{100} = 10.5 \Rightarrow x = 1050$
Problem : For the four sets of three (given) measured physical quantities as given below.
(i) $A_1 = 24.36, B_1 = 0.0724, C_1 = 256.2$
(ii) $A_2 = 24.44, B_2 = 16.082, C_2 = 240.2$
(iii) $A_3 = 25.2, B_3 = 19.2812, C_3 = 236.183$
(iv) $A_4 = 25, B_4 = 236.191, C_4 = 19.5$
Which of the following options is correct?
(a) $A_1+B_1+C_1 < A_3+B_3+C_3 < A_2+B_2+C_2 < A_4+B_4+C_4$
(b) $A_4+B_4+C_4 < A_1+B_1+C_1 = A_2+B_2+C_2 = A_3+B_3+C_3$
(c) $A_4+B_4+C_4 < A_1+B_1+C_1 = A_3+B_3+C_3 < A_2+B_2+C_2$
(d) $A_1+B_1+C_1 = A_2+B_2+C_2 = A_3+B_3+C_3 = A_4+B_4+C_4$
[2020, 9 Jan Shift-II]
Solution: Ans. (None matching)
In addition, the result should have the same number of decimal places as the quantity with the least number of decimal places.
(i) $A_1+B_1+C_1 = 24.36 + 0.0724 + 256.2 = 280.6324 \rightarrow 280.6$ (least decimal place is 1)
(ii) $A_2+B_2+C_2 = 24.44 + 16.082 + 240.2 = 280.722 \rightarrow 280.7$
(iii) $A_3+B_3+C_3 = 25.2 + 19.2812 + 236.183 = 280.6642 \rightarrow 280.7$
(iv) $A_4+B_4+C_4 = 25 + 236.191 + 19.5 = 280.691 \rightarrow 280.7$
So, $A_1+B_1+C_1 < A_2+B_2+C_2 = A_3+B_3+C_3 = A_4+B_4+C_4$.
None of the options is matching with result.
Problem : In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to:
(a) 0.7%
(b) 6.8%
(c) 3.5%
(d) 0.2%
[2019, 8 April Shift-II]
Solution: Ans. (b)
$g = \dfrac{4\pi^2 L}{T^2}$
$\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta T}{T}$
Given: $\Delta L = 1 \text{ mm} = 0.1 \text{ cm}$, $L = 55.0 \text{ cm}$
$\Delta T = 1 \text{ s}$, $T = 30 \text{ s}$
$\% \text{ error } = \left( \dfrac{0.1}{55} + 2 \times \dfrac{1}{30} \right) \times 100$
$= \left( \dfrac{10}{55} + \dfrac{200}{30} \right) = 0.18 + 6.66 = 6.84\% \approx 6.8\%$
Problem : In the density measurement of a cube, the mass and edge length are measured as $(10.00 \pm 0.10) \text{ kg}$ and $(0.10 \pm 0.01) \text{ m}$, respectively. The error in the measurement of density is:
(a) $0.01 \text{ kg/m}^3$
(b) $0.10 \text{ kg/m}^3$
(c) $0.07 \text{ kg/m}^3$
(d) $0.31 \text{ kg/m}^3$
[2019, 9 April Shift-I]
Solution: Ans. (None matching)
Density $\rho = \dfrac{M}{L^3}$
Relative error $\dfrac{\Delta \rho}{\rho} = \dfrac{\Delta M}{M} + 3\dfrac{\Delta L}{L}$
$\dfrac{\Delta \rho}{\rho} = \dfrac{0.10}{10.00} + 3 \times \frac{0.01}{0.10} = 0.01 + 0.3 = 0.31$
The question asks for the “error” (likely relative error), which is unitless, but options have units. Hence, no option is correct.
Problem : The area of a square is $5.29 \text{ cm}^2$. The area of 7 such squares taking into account the significant figures is:
(a) $37.030 \text{ cm}^2$
(b) $37.0 \text{ cm}^2$
(c) $37.03 \text{ cm}^2$
(d) $37 \text{ cm}^2$
[2019, 9 April Shift-II]
Solution: Ans. (c)
Area of 1 square $= 5.29 \text{ cm}^2$ (3 significant figures)
Total area $= 7 \times 5.29 = 37.03 \text{ cm}^2$
In multiplication with a constant (exact number), the number of significant figures is preserved from the measured value.
Since 5.29 has two decimal places, the result maintains the precision of the sum ($5.29 + 5.29 + 5.29…… 7 times$) which leads to $37.03 \text{ cm}^2$.
As we know that, if in the measured values to be added/subtracted the least number of significant digits after the decimal is n.
Then, in the sum or difference also, the number of significant digits after the decimal should be n.
Here, number of digits after decimal in 5.29 is 2, so our answer also contains only two digits after decimal point.
∴ Area required = $37.03 \text{ cm}^2$.
Problem : The pitch and the number of divisions on the circular scale for a given screw gauge are 0.5 mm and 100, respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale for a thin sheet are 5.5 mm and 48 respectively. The thickness of this sheet is:
(a) 5.950 mm
(b) 5.725 mm
(c) 5.755 mm
(d) 5.740 mm
[2019, 9 Jan Shift-II]
Solution: Ans. (c)
For a measuring device, the least count is the smallest value that can be measured by measuring instrument.
$LC = \dfrac{\text{Pitch}}{\text{No. of divisions}} = \dfrac{0.5}{100} = 0.005 \text{ mm}$
According to question, the zero line of its circular scale lies 3 division below the mean line and the readings of main scale = 5 5 mm
The reading of circular scale = 48
then the actual value is given by
actual value of thickness (t) = (main scale reading) + (circular scale reading + number of division below mean line) × LC
Zero error $= -3 \times 0.005 = -0.015 \text{ mm}$ (below the mean line implies negative error)
Observed reading $= MSR + (CSR \times LC) = 5.5 + (48 \times 0.005) = 5.5 + 0.24 = 5.74 \text{ mm}$
True thickness $= \text{Observed reading} – \text{Zero error}$
True thickness $= 5.74 – (-0.015) = 5.755 \text{ mm}$
Problem : The diameter and height of a cylinder are measured by a meter scale to be $12.6 \pm 0.1 \text{ cm}$ and $34.2 \pm 0.1 \text{ cm}$, respectively. What will be the value of its volume in appropriate significant figures?
(a) $4300 \pm 80 \text{ cm}^3$
(b) $4260 \pm 80 \text{ cm}^3$
(c) $4264.4 \pm 81.0 \text{ cm}^3$
(d) $4264 \pm 81 \text{ cm}^3$
[2019, 10 Jan Shift-II]
Solution: Ans. (b)
Volume $V = \dfrac{\pi D^2 h}{4} = \dfrac{\pi (12.6)^2 (34.2)}{4} \approx 4262.22 \text{ cm}^3$
Rounding to 3 significant figures (as 12.6 and 34.2 have 3),
$V = 4260 \text{ cm}^3$
$\dfrac{\Delta V}{V} = 2\dfrac{\Delta D}{D} + \dfrac{\Delta h}{h} = 2 \times \dfrac{0.1}{12.6} + \dfrac{0.1}{34.2} = 0.0158 + 0.0029 = 0.0187$
$\Delta V = 0.0187 \times 4262.22 \approx 79.7 \approx 80 \text{ cm}^3$
∴For proper significant numbers, volume reading will be
$V = (4260 \pm 80) \text{ cm}^3$
Problem : The least count of the main scale of a screw gauge is $1\text{ mm}$. The minimum number of divisions on its circular scale required to measure $5\text{ µm}$ diameter of a wire is:
(a) 50
(b) 200
(c) 500
(d) 100
[2019, 12 Jan Shift-I]
Solution: Ans. (b)
In a screw gauge,
Least Count ($LC$) = $\dfrac{\text{Measure of 1 main scale division (MSD)}}{\text{Number of divisions on circular scale (N)}}$.
Here, minimum value to be measured/least count is 5μm.
Given, $LC = 5\text{ µm} = 5 \times 10^{-6}\text{ m}$
and $1\text{ MSD} = 1\text{ mm} = 1 \times 10^{-3}\text{ m}$.
∴According to the given values,
$5 \times 10^{-6} = \dfrac{1 \times 10^{-3}}{N}$
$N = \dfrac{10^{-3}}{5 \times 10^{-6}} = \dfrac{1000}{5} = 200$ divisions.
Problem : The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively $1.5\%$ and $1\%$, the maximum error in determining the density is:
(a) $2.5\%$
(b) $3.5\%$
(c) $4.5\%$
(d) $6\%$
[JEE Main 2018]
Solution: Ans. (c)
Density $\rho = \dfrac{\text{Mass}}{\text{Volume}} = \dfrac{M}{L^3}$.
Relative error in density is $\dfrac{\Delta \rho}{\rho} = \dfrac{\Delta M}{M} + 3\dfrac{\Delta L}{L}$.
Maximum percentage error $= \left( \dfrac{\Delta M}{M} \times 100 \right) + 3 \times \left( \dfrac{\Delta L}{L} \times 100 \right)$
$Maximum percentage error = 1.5\% + 3(1\%) = 1.5\% + 3\% = 4.5\%$.
Problem : A student measured the length of a rod and wrote it as $3.50\text{ cm}$. Which instrument did he use to measure it?
(a) A meter scale
(b) A vernier calliper where the 10 divisions in vernier scale matches with 9 divisions in main scale and main scale has 10 divisions in $1\text{ cm}$
(c) A screw gauge having 100 divisions in the circular scale and pitch as $1\text{mm}$
(d) A screw gauge having 50 divisions in the circular scale and pitch as $1\text{mm}$
[JEE Main 2014]
Solution: Ans. (b)
A measurement of $3.50\text{ cm}$ implies a least count of $0.01\text{ cm}$ (or $0.1\text{ mm}$).
For option (b): $1\text{ MSD} = \frac{1\text{ cm}}{10} = 0.1\text{ cm} = 1\text{ mm}$.
$10\text{ VSD} = 9\text{ MSD} \Rightarrow 1\text{ VSD} = 0.9\text{ MSD} = 0.9\text{ mm}$.
$LC = 1\text{ MSD} – 1\text{ VSD} = 1\text{ mm} – 0.9\text{ mm} = 0.1\text{ mm} = 0.01\text{ cm}$.
This matches the precision of the recorded value.
Problem : The current voltage relation of diode is given by $I = (e^{1000V/T} – 1)\text{ mA}$, where the applied voltage $V$ is in volt and the temperature $T$ is in kelvin. If a student makes an error measuring $\pm 0.01\text{V}$ while measuring the current of $5\text{ mA}$ at $300\text{K}$, what will be the error in the value of current in mA?
(a) $0.2\text{ mA}$
(b) $0.02\text{ mA}$
(c) $0.5\text{ mA}$
(d) $0.05\text{ mA}$
[JEE Main 2013]
Solution: Ans. (a)
$I = e^{1000V/T} – 1$
$\dfrac{dI}{dV} = e^{1000V/T} \cdot \dfrac{1000}{T}$
Since $I + 1 = e^{1000V/T}$, then $dI = (I + 1) \dfrac{1000}{T} dV$.
Given $I = 5\text{ mA}$, $T = 300\text{ K}$, and $dV = 0.01\text{ V}$:
$dI = (5 + 1) \times \dfrac{1000}{300} \times 0.01 = 6 \times \dfrac{10}{3} \times 0.01 = 2 \times 10 \times 0.01 = 0.2\text{ mA}$.
So, error in the value of current is 0.2mA
Problem : Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are $3\%$ each, then error in the value of resistance of the wire is:
(a) $6\%$
(b) zero
(c) $1\%$
(d) $3\%$
[AIEEE 2012]
Solution: Ans. (a)
$R = \dfrac{V}{I}$
$\% \text{ error in } R = \% \text{ error in } V + \% \text{ error in } I = 3\% + 3\% = 6\%$.
Problem : The respective number of significant figures for the numbers $23.023$, $0.0003$ and $2.1 \times 10^{-3}$ are:
(a) 5, 1, 2
(b) 5, 1, 5
(c) 5, 5, 2
(d) 4, 4, 2
[AIEEE 2010]
Solution: Ans. (a)
$23.023$: All non-zero digits and the zero between them are significant. Total = 5.
$0.0003$: Leading zeros are not significant. Only ‘3’ is significant. Total = 1.
$2.1 \times 10^{-3}$: Only digits in the decimal part are significant. Total = 2.
Problem : In an experiment, the angles are required to be measured using an instrument. $29$ divisions of the main scale exactly coincide with the $30$ divisions of the vernier scale. If the smallest division of the main scale is half a degree $(= 0.5^\circ)$, then the least count of the instrument is:
(a) one minute
(b) half minute
(c) one degree
(d) half degree
[AIEEE 2009]
Solution: Ans. (a)
$1\text{ MSD} = 0.5^\circ = 30’$.
$30\text{ VSD} = 29\text{ MSD} \Rightarrow 1\text{ VSD} = \frac{29}{30}\text{ MSD}$.
$LC = 1\text{ MSD} – 1\text{ VSD} = \left( 1 – \frac{29}{30} \right)\text{ MSD} = \frac{1}{30}\text{ MSD}$.
$LC = \frac{1}{30} \times 30′ = 1’$.
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