Errors in Measurements
137 Questions
Start Allen Test
JEE (Main) 2023
Q1
Allen
5. JEE Main PYQ
MCQ
A cylindrical wire of mass $(0.4\pm0.01)g$ has length $(8\pm0.04)cm$ and radius $(6\pm0.03)mm$ . The maximum error in its density will be
A.
1%
B.
3.5%
C.
4%
D.
5%
JEE (Main) 2023
Q2
Allen
5. JEE Main PYQ
MCQ
In an experiment with Vernier callipers of least count 0.1 mm, when two jaws are joined together the zero of Vernier scale lies right to the zero of the main scale and $6^{th}$ division of Vernier scale coincides with the main scale division. While measuring the diameter of a spherical bob, the zero of vernier scale lies in between 3.2 cm and 3.3 cm marks, and $4^{th}$ division of vernier scale coincides with the main scale division. The diameter of bob is measured as :
A.
3.18 cm
B.
3.25 cm
C.
3.26 cm
D.
3.22 cm
JEE (Main) 2023
Q3
Allen
5. JEE Main PYQ
MCQ
For the logic circuit shown, the output waveform at Y is:
A.
B.
C.
D.
JEE (Main) 2023
Q4
Allen
5. JEE Main PYQ
MCQ
A zener diode of power rating 1.6W is to be used as voltage regulator. If the zener diode has a breakdown of 8V and it has to regulate voltage fluctuating between 3V and 10V. The value of resistance $R_{s}$ for safe operation of diode will be :
A.
$13.3\Omega$
B.
12 $\Omega$
C.
10$\Omega$
D.
$13\Omega$
JEE (Advanced) 2023
Q5
Allen
7. JEE Advanced PYQ
Numerical
In an experiment for determination of the focal length of a thin convex lens, the distance of the object from the lens is $10 \pm 0.1$ cm and the distance of its real image from the lens is $20 \pm 0.2$ cm. The error in the determination of focal length of the lens is n %. The value of n is ____.
Correct Answer: 1
Explanation:
Ans. (1)
u = 10 ± 0.1 cm, v = 20 ± 0.2 cm $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v^{2}} dv + \frac{1}{u^{2}} du = -\frac{1}{f^{2}} df$ $\frac{1}{20} + \frac{1}{10} = \frac{1}{f} \Rightarrow \frac{1}{f} = \frac{3}{20} \Rightarrow f = \frac{20}{3} cm$ $\Rightarrow \frac{1}{(20)^{2}}(0.2) + \frac{1}{(10)^{2}}(0.1) = \frac{9}{400} df$ $df = \frac{1}{9}\left(\frac{400}{400} \times 0.2 + \frac{400}{100} \times 0.1\right)$ $df = \frac{1}{9}(0.2 + 0.4) \Rightarrow df = \frac{0.6}{9}$ $\frac{df}{f} = \frac{0.6}{9} \times \frac{3}{20} = \frac{1}{100}$ % error = 1 %
## JEE (Main) Practice Paper
u = 10 ± 0.1 cm, v = 20 ± 0.2 cm $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v^{2}} dv + \frac{1}{u^{2}} du = -\frac{1}{f^{2}} df$ $\frac{1}{20} + \frac{1}{10} = \frac{1}{f} \Rightarrow \frac{1}{f} = \frac{3}{20} \Rightarrow f = \frac{20}{3} cm$ $\Rightarrow \frac{1}{(20)^{2}}(0.2) + \frac{1}{(10)^{2}}(0.1) = \frac{9}{400} df$ $df = \frac{1}{9}\left(\frac{400}{400} \times 0.2 + \frac{400}{100} \times 0.1\right)$ $df = \frac{1}{9}(0.2 + 0.4) \Rightarrow df = \frac{0.6}{9}$ $\frac{df}{f} = \frac{0.6}{9} \times \frac{3}{20} = \frac{1}{100}$ % error = 1 %
## JEE (Main) Practice Paper
JEE (Advanced) 2022
Q6
Allen
7. JEE Advanced PYQ
Match the Columns
Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is 0.5 mm. The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.
What are the diameter and cross-sectional area of the wire measured using the screw gauge?
| Column-I | Column-II | Column-III |
|---|---|---|
| Measurement condition | Main scale reading | Circular scale reading |
| Two arms of gauge touching each other without wire | 0 division | 4 division |
| Attempt-1: With wire | 4 divisions | 20 divisions |
| Attempt-2: With wire | 4 divisions | 16 divisions |
A.
$2.22 \pm 0.02 \, mm$ , $\pi(1.23 \pm 0.02) \, mm^{2}$
B.
$2.22 \pm 0.01 \, mm$ , $\pi(1.23 \pm 0.01) \, mm^{2}$
C.
$2.14 \pm 0.02 \, mm$ , $\pi(1.14 \pm 0.02) \, mm^{2}$
D.
$2.14 \pm 0.01 \, mm$ , $\pi(1.14 \pm 0.01) \, mm^{2}$
JEE (Main) 2021
Q7
Allen
5. JEE Main PYQ
MCQ
If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is:
A.
86.4 s
B.
4.32 s
C.
43.2 s
D.
8.64 s
JEE (Main) 2021
Q8
Allen
5. JEE Main PYQ
MCQ
In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00 mm. The measured current in the conductor is 2.00 A. The maximum permissible percentage error in the resistivity of the conductor is :-
A.
3.9
B.
8.4
C.
7.5
D.
3.0
JEE (Main) 2021
Q9
Allen
5. JEE Main PYQ
Match the Columns
A student determined Young's Modulus of elasticity using the formula $Y = \frac{MgL^{3}}{4bd^{3}\delta}$ . The value of g is taken to be $9.8\ m/s^{2}$ , without any significant error, his observation are as following.
Then the fractional error in the measurement of Y is:
| Column-I | Column-II | Column-III |
|---|---|---|
| Physical Quantity | Least Count of the Equipment used for measurement | Observed value |
| Mass (M) | 1 g | 2 kg |
| Length of bar (L) | 1 mm | 1 m |
| Breadth of bar (b) | 0.1 mm | 4 cm |
| Thickness of bar (d) | 0.01 mm | 0.4 cm |
| Depression (Ī“) | 0.01 mm | 5 mm |
A.
0.0083
B.
0.0155
C.
0.155
D.
0.083
JEE (Main) 2021
Q10
Allen
5. JEE Main PYQ
Numerical
In a semiconductor, the number density of intrinsic charge carriers at $27^{\circ}C$ is $1.5 \times 10^{16}/m^{3}$ . If the semiconductor is doped with impurity atom, the hole density increases to $4.5 \times 10^{22}/m^{3}$ . The electron density in the doped semiconductor is ____ $\times 10^{9}/m^{3}$ .
Correct Answer: 5
Explanation:
Ans. (5)
$n _ {\mathrm{e}} n _ {\mathrm{h}} = n _ {\mathrm{i}} ^ {2}$
$n _ {e} = \frac {n _ {i} ^ {2}}{n _ {h}} = \frac {(1 . 5 \times 1 0 ^ {1 6}) ^ {2}}{4 . 5 \times 1 0 ^ {2 2}} = \frac {1 . 5 \times 1 . 5 \times 1 0 ^ {3 2}}{4 . 5 \times 1 0 ^ {2 2}}$
$5 \times 1 0 ^ {9} / m ^ {3}$
$n _ {\mathrm{e}} n _ {\mathrm{h}} = n _ {\mathrm{i}} ^ {2}$
$n _ {e} = \frac {n _ {i} ^ {2}}{n _ {h}} = \frac {(1 . 5 \times 1 0 ^ {1 6}) ^ {2}}{4 . 5 \times 1 0 ^ {2 2}} = \frac {1 . 5 \times 1 . 5 \times 1 0 ^ {3 2}}{4 . 5 \times 1 0 ^ {2 2}}$
$5 \times 1 0 ^ {9} / m ^ {3}$
JEE (Main) 2021
Q11
Allen
5. JEE Main PYQ
MCQ
Choose the correct waveform that can represent the voltage across R of the following circuit, assuming the diode is ideal one:
A.
B.
C.
D.
JEE (Main) 2021
Q12
Allen
5. JEE Main PYQ
MCQ
Statement-I:
To get a steady dc output from the pulsating voltage received from a full wave rectifier can connect a capacitor across the output parallel to the load $R_{L}$ .
Statement-II :
To get a steady dc output from the pulsating voltage received from a full wave rectifier can connect an inductor in series with $R_{L}$ .
In the light of the above statements,
choose the most appropriate answer from the options given below:
A.
Statement I is true, but Statement II is false.
B.
Statement I is false, but Statement II is true.
C.
Both Statement I and Statement II are false.
D.
Both Statement I and Statement II are true.
JEE (Main) 2021
Q13
Allen
5. JEE Main PYQ
MCQ
The logic circuit shown above is equivalent to:
A.
B.
C.
D.
JEE (Main) 2021
Q14
Allen
5. JEE Main PYQ
MCQ
If $V_{A}$ and $V_{B}$ are the input voltages (either 5V or $V_{0}$ ) and $V_{0}$ is the output voltage then the two gates represented in the following circuit (A) and (B) are:-
A.
AND and OR Gate
B.
OR and NOT Gate
C.
NAND and NOR Gate
D.
AND and NOT Gate
JEE (Advanced) 2021
Q15
Allen
7. JEE Advanced PYQ
MCQ
The smallest division on the main scale of a Vernier calipers is 0.1 cm. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is
A.
3.07 cm
B.
3.11 cm
C.
3.15 cm
D.
3.17 cm
JEE (Main) 2020
Q16
Allen
5. JEE Main PYQ
MCQ
A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is:
A.
3.40%
B.
5.40%
C.
4.40%
D.
2.40%
JEE (Main) 2020
Q17
Allen
5. JEE Main PYQ
MCQ
The least count of the main scale of a vernier calipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the $7^{th}$ division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of a cylinder the zero of the vernier scale between 3.1 cm and 3.2 cm and $4^{th}$ VSD coincides with a main scale division. The length of the cylinder is: (VSD is vernier scale division)
A.
3.21 cm
B.
2.99 cm
C.
3.2 cm
D.
3.07 cm
JEE (Main) 2020
Q18
Allen
5. JEE Main PYQ
MCQ
Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as :
A.
2.123 cm
B.
2.125 cm
C.
2.121 cm
D.
2.124 cm
JEE (Main) 2020
Q19
Allen
5. JEE Main PYQ
MCQ
A physical quantity z depends on four observables a, b, c and d, as $z = \frac{a^{2}b^{\frac{2}{3}}}{\sqrt{c}d^{3}}$ . The percentage 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.
12.25%
B.
14.5%
C.
16.5%
D.
13.5%
JEE (Main) 2020
Q20
Allen
5. JEE Main PYQ
MCQ
A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm; 5.65 mm. The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as:
A.
(5.5375 ± 0.0739) mm
B.
(5.538 ± 0.074) mm
C.
(5.54 ± 0.07) mm
D.
(5.5375 ± 0.0740) mm
JEE (Advanced) 2020
Q21
Allen
7. JEE Advanced PYQ
Numerical
Two capacitors with capacitance values $C_{1} = 2000 \pm 10$ pF and $C_{2} = 3000 \pm 15$ pF are connected in series. The voltage applied across this combination is $V = 5.00 \pm 0.02$ V. The percentage error in the calculation of the energy stored in this combination of capacitors is ____.
Correct Answer: 1.30
Explanation:
Ans. (1.30)
$\mathrm{Series} = \left(\frac {C _ {1} C _ {2}}{C _ {1} + C _ {2}}\right) = 1 2 0 0 P F$
$\frac {\varDelta C _ {e q}}{C ^ {2}} = \frac {\varDelta C _ {1}}{C _ {1} ^ {2}} + \frac {\varDelta C _ {2}}{C _ {2} ^ {2}}$
$\frac {\varDelta C _ {e q}}{(1 2 0 0) ^ {2}} = \frac {1 0}{(2 0 0 0) ^ {2}} + \frac {1 5}{(3 0 0 0) ^ {2}}$
$\begin{array}{r} \varDelta C _ {e q} = 1 0 \times \left(\frac {1 2 0 0}{2 0 0 0}\right) ^ {2} + 1 5 \left(\frac {1 2 0 0}{3 0 0 0}\right) ^ {2} \\ = 1 0 \times (0. 3 6) + 1 5 (0. 1 6) \end{array}$
$\Rightarrow \Delta C _ {e q} = 6 P F$
$\Rightarrow U = \frac {1}{2} C V ^ {2}$
$\begin{array}{r l} \Rightarrow \frac {\Delta U}{U} & = \frac {\Delta C}{C} + \left(\frac {2 \Delta V}{V}\right) \\ & = \frac {6}{1 2 0 0} + \frac {2 \times 0 . 0 2}{5} = . 0 0 5 +. 0 0 8 \\ & = 0. 0 1 3 = 1. 3 \% \end{array}$
$\mathrm{Series} = \left(\frac {C _ {1} C _ {2}}{C _ {1} + C _ {2}}\right) = 1 2 0 0 P F$
$\frac {\varDelta C _ {e q}}{C ^ {2}} = \frac {\varDelta C _ {1}}{C _ {1} ^ {2}} + \frac {\varDelta C _ {2}}{C _ {2} ^ {2}}$
$\frac {\varDelta C _ {e q}}{(1 2 0 0) ^ {2}} = \frac {1 0}{(2 0 0 0) ^ {2}} + \frac {1 5}{(3 0 0 0) ^ {2}}$
$\begin{array}{r} \varDelta C _ {e q} = 1 0 \times \left(\frac {1 2 0 0}{2 0 0 0}\right) ^ {2} + 1 5 \left(\frac {1 2 0 0}{3 0 0 0}\right) ^ {2} \\ = 1 0 \times (0. 3 6) + 1 5 (0. 1 6) \end{array}$
$\Rightarrow \Delta C _ {e q} = 6 P F$
$\Rightarrow U = \frac {1}{2} C V ^ {2}$
$\begin{array}{r l} \Rightarrow \frac {\Delta U}{U} & = \frac {\Delta C}{C} + \left(\frac {2 \Delta V}{V}\right) \\ & = \frac {6}{1 2 0 0} + \frac {2 \times 0 . 0 2}{5} = . 0 0 5 +. 0 0 8 \\ & = 0. 0 1 3 = 1. 3 \% \end{array}$
JEE (Main) 2019
Q22
Allen
5. JEE Main PYQ
MCQ
For the circuit shown below, the current through the Zener diode is:
A.
$5mA$
B.
Zero
C.
$14mA$
D.
$9mA$
JEE (Advanced) 2019
Q23
Allen
7. JEE Advanced PYQ
Numerical
An optical bench has 1.5 m long scale having four equal divisions in each cm. While measuring the focal length of a convex lens, the lens is kept at 75 cm mark of the scale and the object pin is kept at 45 cm mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at 135 cm mark. In this experiment, the percentage error in the measurement of the focal length of the lens is.
Correct Answer: 1.35 to 1.45
Explanation:
Ans. (1.35 to 1.45) $1\ MSD = \frac{1}{4} cm \Rightarrow \Delta v = \Delta u = 2\left(\frac{1}{4}cm\right) = \frac{1}{2}cm$ $X_{\mathrm{lens}} = 75\mathrm{cm}, X_{\mathrm{object}} = 45\mathrm{cm}, X_{\mathrm{image}} = 135\mathrm{cm}$ $u = -(75 - 45) = -30\mathrm{cm}$ and $v = (135 - 75) = 60\mathrm{cm}$ $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{60} + \frac{1}{30} \Rightarrow f = 20\mathrm{cm}$ Also $\frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} \Rightarrow \frac{\Delta f}{f}\% = \left(\frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}\right) \times f \times 100\%$ $\frac{\Delta f}{f}\% = \left(\frac{1/2}{(60)^2} + \frac{1/2}{(30)^2}\right) 20 \times 100\% \approx 1.39\%$
JEE (Main) 2018
Q24
Allen
5. JEE Main PYQ
MCQ
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.
3.5 %
B.
4.5 %
C.
6 %
D.
2.5 %
JEE (Advanced) 2018
Q25
Allen
7. JEE Advanced PYQ
MCQ
PARAGRAPH FOR QUESTION NO. 8 and 9
If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z = x/y. If the errors in x, y and z are $\Delta x$ , $\Delta y$ and $\Delta z$ , respectively, then
$z \pm \Delta z = \frac {x \pm \Delta x}{y \pm \Delta y} = \frac {x}{y} \left(1 \pm \frac {\Delta x}{x}\right) \left(1 \pm \frac {\Delta y}{y}\right) ^ {- 1}
$
The series expansion for $\left(1\pm\frac{\Delta y}{y}\right)^{-1}$ , to first power in $\Delta y/y$ , is $1\mp(\Delta y/y)$ . The relative errors in independent variables are always added. So the error in z will be
$
\varDelta z = z \left(\frac {\varDelta x}{x} + \frac {\varDelta y}{y}\right)
$
The above derivation makes the assumption that $\frac{\Delta x}{x} \ll 1$ , $\frac{\Delta y}{y} \ll 1$ . Therefore, the higher powers of these quantities are neglected.
Consider the ratio $r = \frac{(1-a)}{(1+a)}$ to be determined by measuring a dimensionless quantity a. If the error in the measurement of a is $\Delta a(\Delta a/a \ll 1)$ , then what is the error $\Delta r$ in determining r?
A.
$\frac{\Delta a}{(1+a)^{2}}$
B.
$\frac{2\Delta a}{(1+a)^{2}}$
C.
$\frac{2\Delta a}{(1-a^{2})}$
D.
$\frac{2a\Delta a}{(1-a^{2})}$

