JEE Advanced PYQ – Errors in Measurement & Experimental Physics

Q 1 :

The dimensions of a cone are measured using a scale with a least count of 2 mm. The diameter of the base and the height are both measured to be 20.0 cm. The maximum percentage error in the determination of the volume is ____. [2024]

(3)

$Volume of a cone = V = \frac{1}{3} π R^{2} H$

$\Rightarrow \frac{d V}{V} = 2 \cdot \frac{d R}{R} + \frac{d H}{H}$

$∴$ $Percentage error = [2 \times \frac{0.2}{20} + \frac{0.2}{20}] \times 100$

$= 3$

Q 2 :

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 ____. [2023]

(1)

Given: Differentiating lens formula, $u = 10 \pm 0.1 cm, v = 20 \pm 0.2 cm$

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v^{2}} d v + \frac{1}{u^{2}} d u = - \frac{1}{f^{2}} d f$

$\frac{1}{20} + \frac{1}{10} = \frac{1}{f} \Rightarrow \frac{1}{f} = \frac{3}{20} \Rightarrow f = \frac{20}{3} cm$

$\Rightarrow \frac{1}{{(20)}^{2}} (0.2) + \frac{1}{{(10)}^{2}} (0.1) = \frac{9}{400} d f$

$d f = \frac{1}{9} (\frac{400}{400} \times 0.2 + \frac{400}{100} \times 0.1)$

$\Rightarrow d f = \frac{1}{9} (0.2 + 0.4) \Rightarrow d f = \frac{0.6}{9}$

$\Rightarrow \frac{d f}{f} = \frac{0.6}{9} \times \frac{3}{20} = \frac{1}{100}$

$∴$ $% error = \frac{d f}{f} \times 100 = \frac{1}{100} \times 100 = 1 %$

Q 3 :

The energy of a system as a function of time $t$ is given as $E (t) = A^{2} \exp (- α t),$ where $α = 0.2 s^{- 1}$ . The measurement of $A$ has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of $E (t)$ at $t = 5 s$ is _____. [2015]

(4)

$E_{(t)} = A^{2} e^{- \propto t} = A^{2} e^{- 0.2 t}$

$∴$ $\log_{e} E = 2 \log_{e} A - 0.2 t$

On differentiating we get $= \frac{d E}{E} = 2 \frac{d A}{A} - 0.2 \frac{d t}{t} \times t$

As errors always add up

$∴$ $\frac{d E}{E} \times 100 = 2 (\frac{d A}{A} \times 100) + 0.2 t (\frac{d t}{t} \times 100)$

$\Rightarrow \frac{d E}{E} \times 100 = 2 \times 1.25 % + 0.2 \times 5 \times 1.5 % = 4 %$

Q 4 :

During Searle’s experiment, zero of the Vernier scale lies between $3.20 \times 10^{- 2} m$ and $3.25 \times 10^{- 2} m$ of the main scale. The $20^{th}$ division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{- 2} m$ and $3.25 \times 10^{- 2} m$ of the main scale but now the $45^{th}$ division of the Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is $8 \times 10^{- 7} m^{2}$ . The least count of the Vernier scale is $1.0 \times 10^{- 5} m$ . The maximum percentage error in Young’s modulus of the wire is _____. [2014]

(4)

$Young's modulus Y = \frac{F L}{A \times l}$

$Here F, A and L are accurately known.$

$∴$ $Percentage error in Young's modulus$

$l = (45 - 20) \times L.C. = 25 \times 10^{- 5} m$

$\frac{Δ Y}{Y} \times 100 = \frac{Δ l}{l} \times 100 = \frac{1.0 \times 10^{- 5}}{25 \times 10^{- 5}} \times 100 = 4 %$

Q 5 :

Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter D of a tube. The measured value of D is : [2025]

0.12 cm
0.11 cm
0.13 cm
0.14 cm

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(3)

From figure 1

$10 MSD = 1 cm, so 1 MSD = 0.1 cm$

$7 MSD = 10 VSD$

$∴ 1 VSD = 0.07 cm$

$Least count (L.C.) = 1 MSD - 1 VSD = 0.1 - 0.07 = 0.03 cm$

Vernier scale 1 marking matches with main scale division

$Measurement of diameter, D = MSR + VSR \times L.C.$

$= 0.1 + 1 \times 0.03 = 0.13 cm$

Q 6 :

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 [2021]

3.07 cm
3.11 cm
3.15 cm
3.17 cm

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(3)

$Since 10 VSD = 9 MSD$ $∴ 1 VSD = \frac{9}{10} MSD$

$Least count (L.C.) = 1 MSD - 1 VSD = (1 - \frac{9}{10}) MSD$

$= 0.1 MSD = 0.1 \times 0.1 cm = 0.01 cm$

Since 0 of Vernier scale lies before 0 of main scale

$Zero error = - [10 - 6] L.C. = - 4 \times 0.01 cm$

$= - 0.04 cm$ $∴$ Correct diameter of the sphere

$= Reading - Zero error = 3.1 + 1 \times L.C. - (- 0.04)$

$= 3.1 + 1 \times 0.01 + 0.04 = 3.11 + 0.04 cm = 3.15 cm$

Q 7 :

A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is $δ T = 0.01$ seconds and he measures the depth of the well to be $L = 20$ meters. Take the acceleration due to gravity $g = 10 {ms}^{- 2}$ and the velocity of sound is $300 {ms}^{- 1}$ . Then the fractional error in the measurement, $δ L / L$ , is closest to [2017]

0.2 %
1 %
3 %
5 %

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(2)

$Depth of the well = L = 20 m$

$∴$ $Time taken by stone to reach the bottom of well,$

$t_{1} = \sqrt{\frac{2 L}{g}}$ $(using, L = \frac{1}{2} g t^{2})$

$Time taken by impact sound to reach the person,$

$t_{2} = \frac{L}{v}$

$Total time taken in the process is given by,$

$T = t_{1} + t_{2} = \sqrt{\frac{2 L}{g}} + \frac{L}{v}$ $∴$ $δ T = \sqrt{\frac{2}{g}} \cdot \frac{δ L}{2 \sqrt{L}} + \frac{δ L}{v}$

$On substituting the given values we get,$

$δ T = \frac{16}{300} δ L$

$∴$ $Fractional error, \frac{δ L}{L} \times 100 = \frac{300}{16} \cdot \frac{δ T}{L} \times 100$ $= \frac{300}{16} \times \frac{0.01}{20} \times 100 = 1 %$

Q 8 :

There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers $(C_{1})$ has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper $(C_{2})$ has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers $C_{1}$ and $C_{2}$ , respectively, are [2016]

2.85 and 2.82
2.87 and 2.83
2.87 and 2.86
2.87 and 2.87

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(2)

$For C_{1} vernier calliper,$

$L.C. = 1 MSD - 1 VSD = 1 mm - 0.9 mm = 0.1 mm = 0.01 cm$

$[∵ 10 VSD = 9 MSD = 9 mm]$

$Reading = MSR + L.C. \times VSR = 2.8 + (0.01) \times 7 = 2.87 cm$

$For C_{2} vernier calliper,$

$L.C. = 1 mm - 1.1 mm$ $[∵ 10 VSD = 11 MSD = 11 mm]$

$L.C. = - 0.1 mm = - 0.01 cm$

$Reading = 2.8 + (10 - 7) \times 0.01 = 2.83 cm$

Q 9 :

In the determination of Young’s modulus $(Y = \frac{4 M L g}{π l d^{2}})$ by using Searle’s method, a wire of length $L = 2 m$ and diameter $d = 0.5 mm$ is used. For a load $M = 2.5 kg$ , an extension $l = 0.25 mm$ in the length of the wire is observed. Quantities $d$ and $l$ are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement [2012]

due to the errors in the measurements of $d$ and $l$ are the same.
due to the error in the measurement of $d$ is twice that due to the error in the measurement of $l$ .
due to the error in the measurement of $l$ is twice that due to the error in the measurement of $d$ .
due to the error in the measurement of $d$ is four times that due to the error in the measurement of $l$ .

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(1)

$The maximum possible error in Y due to l and d$

$\frac{Δ Y}{Y} = \frac{Δ l}{l} + \frac{2 Δ d}{d}$

$Least count = \frac{Pitch}{No. of divisions on circular scale}$

$= \frac{0.5}{100} mm = 0.005 mm$

$Here, Δ d = Δ l = 0.005 mm$

$Error contribution of l = \frac{Δ l}{l} = \frac{0.005 mm}{0.25 mm} = \frac{1}{50}$

$Error contribution of d = \frac{2 Δ d}{d} = \frac{2 \times 0.005 mm}{0.5 mm} = \frac{1}{50}$

$Hence contribution to the maximum possible error in the measurement of y due to l and d is the same.$

Q 10 :

The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is [2011]

0.9%
2.4%
3.1%
4.2%

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(3)

$Least count of screw gauge$

$= \frac{Pitch}{divisions on circular scale} = \frac{0.5}{50} = 0.01 mm = Δ r$

$Diameter, r = M.S.R. + (C.S.R.) \times (L.C.)$

$Diameter, r = 2.5 mm + 20 \times \frac{0.5}{50} = 2.70 mm$

$\frac{Δ r}{r} = \frac{0.01}{2.70} or \frac{Δ r}{r} \times 100 = \frac{1}{2.7}$

$Now, density, d = \frac{m}{V} = \frac{m}{\frac{4}{3} π {(\frac{r}{2})}^{3}}$

$∴$ $Percentage error in density, \frac{Δ d}{d} \times 100$

$= {\frac{Δ m}{m} + 3 (\frac{Δ r}{r})} \times 100$ $= \frac{Δ m}{m} \times 100 + 3 \times \frac{Δ r}{r} \times 100$

$= 2 % + 3 \times \frac{1}{2.7} = 3.11 %$

Q 11 :

A student performs an experiment to determine the Young's modulus of a wire, exactly 2 m long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of $\pm 0.05 mm$ at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of $\pm 0.01 mm$ . Take $g = 9.8 {m/s}^{2}$ (exact). The Young's modulus obtained from the reading is [2007]

$(2.0 \pm 0.3) \times 10^{11} {N/m}^{2}$
$(2.0 \pm 0.2) \times 10^{11} {N/m}^{2}$
$(2.0 \pm 0.1) \times 10^{11} {N/m}^{2}$
$(2.0 \pm 0.05) \times 10^{11} {N/m}^{2}$

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(2)

$Y = \frac{F / A}{ℓ / L} = \frac{4 m g L}{π D^{2} ℓ} = \frac{4 \times 1 \times 9.8 \times 2}{π {(0.4 \times 10^{- 3})}^{2} \times (0.8 \times 10^{- 3})}$

$= 2.0 \times 10^{11} {N/m}^{2}$

$Now \frac{Δ Y}{Y} = \frac{2 Δ D}{D} + \frac{Δ ℓ}{ℓ}$ $[∵ the value of m, g and L are exact]$

$= 2 \times \frac{0.01}{0.4} + \frac{0.05}{0.8} = 2 \times 0.025 + 0.0625$

$= 0.05 + 0.0625 = 0.1125$

$\Rightarrow Δ Y = Y \times 0.1125 = 2 \times 10^{11} \times 0.1125 = 0.225 \times 10^{11}$

$= 0.2 \times 10^{11} {N/m}^{2}$

$∴$ $Reading of Young's modulus = (Y \pm Δ Y)$

$= (2 \pm 0.2) \times 10^{11} {N/m}^{2}$

Q 12 :

A student performs an experiment for determination of $(g = \frac{4 π^{2} ℓ}{T^{2}} .)$ The error in length $ℓ$ is $Δ ℓ$ and in time $T$ is $Δ T$ and $n$ is the number of times the reading is taken. The measurement of $g$ is most accurate for [2006]

$Δ ℓ = 5 mm, Δ T = 0.2 sec, n = 10$
$Δ ℓ = 5 mm, Δ T = 0.2 sec, n = 20$
$Δ ℓ = 5 mm, Δ T = 0.1 sec, n = 10$
$Δ ℓ = 1 mm, Δ T = 0.1 sec, n = 50$

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(4)

$Relative error in g, \frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + 2 \frac{Δ T}{T}$

$Δ ℓ and Δ T are least and number of readings taken are maximum in option (d),$

$∴ the measurement of g is most accurate.$

Q 13 :

In a screw gauge, the zero of main scale coincides with fifth division of circular scale in figure (i). The circular divisions of screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball in figure (ii) is: [2006]

2.25 mm
2.20 mm
1.20 mm
1.25 mm

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(3)

$Least count = \frac{Pitch}{no. of divisions on circular scale} = \frac{0.5}{50} = 0.01 mm$

$Zero error = 5 \times L.C. = 5 \times 0.01 mm = 0.05 mm$

$Diameter of ball = [Reading on main scale] + [Reading on circular scale \times L.C.] - Zero error$

$= 0.5 \times 2 + 25 \times 0.01 - 0.05 = 1.20 mm$

Q 14 :

A wire of length $ℓ = 6 \pm 0.06 cm$ and radius $r = 0.5 \pm 0.005 cm$ and mass $m = 0.3 \pm 0.003 gm.$ Maximum percentage error in density is [2004]

4
2
1
6.8

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(1)

$Density, ρ = \frac{m}{v} = \frac{m}{A ℓ} = \frac{m}{ℓ π r^{2}}$

Taking log and differentiating for errors,

$\frac{Δ ρ}{ρ} = \frac{Δ m}{m} + 2 \frac{Δ r}{r} + \frac{Δ ℓ}{ℓ} \dots (i)$

From question, putting the values of

$Δ ℓ = 0.06 cm, ℓ = 6 cm; Δ r = 0.005 cm, r = 0.5 cm,$

$m = 0.3 g; Δ m = 0.003 g$ in eqn. (i) we get $\frac{Δ ρ}{ρ} = \frac{4}{100}$

$∴$ $Percentage error in density,$

$\frac{Δ ρ}{ρ} \times 100 = \frac{4}{100} \times 100 = 4 % .$

Q 15 :

A cube has a side of length $1.2 \times 10^{- 2} m$ . Calculate its volume. [2003]

$1.7 \times 10^{- 6} m^{3}$
$1.73 \times 10^{- 6} m^{3}$
$1.70 \times 10^{- 6} m^{3}$
$1.732 \times 10^{- 6} m^{3}$

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(1)

$Volume of cube, V = ℓ^{3} = {(1.2 \times 10^{- 2} m)}^{3}$

$= 1.728 \times 10^{- 6} m^{3} \Rightarrow V = 1.7 \times 10^{- 6} m^{3}$

As length has two significant figures, so volume also has two significant figures.

Q 16 :

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 _______ . [2019]

(1.39)

$u \pm Δ u = (75 - 45) \pm (\frac{1}{4} + \frac{1}{4}) = (30 \pm 0.5) cm$

$v \pm Δ v = (135 - 75) \pm (\frac{1}{4} + \frac{1}{4}) = (60 \pm 0.5) cm$

We know that

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} ∴ \frac{Δ v}{v^{2}} + \frac{Δ u}{u^{2}} = \frac{Δ f}{f^{2}} \dots (i)$

Now $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{60} - \frac{1}{- 30} = \frac{1}{f}$

$\Rightarrow \frac{1}{f} = \frac{1}{20} ∴ f = 20 cm$

Substituting the values in eqn. (i)

$\frac{0.5}{{(60)}^{2}} + \frac{0.5}{{(30)}^{2}} = \frac{Δ f}{{(20)}^{2}}$ $\Rightarrow Δ f = 0.277$

Hence percentage error in the measurement of focal length

$\frac{Δ f}{f} \times 100 = \frac{0.277}{20} \times 100 = 1.388 % \approx 1.39 % .$

Q 17 :

A steel wire of diameter $0.5 mm$ and Young's modulus $2 \times 10^{11} {N m}^{- 2}$ carries a load of mass M. The length of the wire with the load is $1.0 m$ . A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count $1.0 mm$ , is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is _________. Take $g = 10 {m s}^{- 2}$ and $π = 3.2$ . [2018]

(3)

We know that $Δ l = \frac{F l}{A Y}$

$= \frac{1.2 \times 10 \times 1}{π {(\frac{5 \times 10^{- 4}}{2})}^{2} \times 2 \times 10^{11}} \approx 0.3 mm$

$L.C. of vernier calliper = 1 MSD - 1 VSD$

$= (1 - \frac{9}{10}) = 0.1 mm$

$As 9 MSD = 10 VSD$

The third marking of vernier scale will coincide with the main scale because least count is 0.1 mm.

Q 18 :

The side of a cube is measured by vernier callipers (10 divisions of a vernier scale coincide with 9 divisions of main scale, where 1 division of main scale is 1 mm). The main scale reads 10 mm and first division of vernier scale coincides with the main scale. Mass of the cube is 2.736 g. Find the density of the cube in appropriate significant figures. [2005]

(2.66)

$L.C. = 1 MSD - 1 VSD$

$= 1 MSD - \frac{9}{10} MSD (∵ 9 MSD = 10 VSD)$

$= (1 - \frac{9}{10}) MSD = \frac{1}{10} MSD = (\frac{1}{10} \times 1) mm = 0.1 mm (∵ 1 MSD = 1 mm)$

$Reading of side, l = MSR + VSR (LC) = 10 mm + 1 \times 0.1 mm = 10.1 mm = 1.01 cm$

$Now, density = \frac{mass}{volume} = \frac{m}{l^{3}} = \frac{2.736 g}{{(1.01)}^{3}} = 2.66 {g/cm}^{3}$

$(Rounding off to 3 significant figures)$

Q 19 :

In Searle’s experiment, which is used to find Young’s Modulus of elasticity, the diameter of experimental wire is $D = 0.05 cm$ (measured by a scale of least count 0.001 cm) and length is $L = 110 cm$ (measured by a scale of least count 0.1 cm). A weight of $50 N$ causes an extension of $X = 0.125 cm$ (measured by a micrometer of least count $0.001 cm$ ). Find maximum possible error in the values of Young’s modulus. Screw gauge and meter scale are free from error. [2004]

(1.09)

$Young's modulus, Y = \frac{\frac{F}{A}}{\frac{X}{L}} = \frac{F}{\frac{π D^{2}}{4}} \times \frac{L}{X}$

$Maximum error in Y$

${(\frac{Δ Y}{Y})}_{\max} = 2 (\frac{Δ D}{D}) + \frac{Δ X}{X} + \frac{Δ L}{L}$

$= 2 (\frac{0.001}{0.05}) + (\frac{0.001}{0.125}) + (\frac{0.1}{110}) = 0.0489$

$Since, W = 50 N; D = 0.05 cm = 0.05 \times 10^{- 2} m;$

$X = 0.125 cm = 0.125 \times 10^{- 2} m;$

$L = 110 cm = 110 \times 10^{- 2} m$

$∴$ $Y = \frac{50 \times 4 \times 110 \times 10^{- 2}}{3.14 (0.05 \times 10^{- 2}) \times (0.125 \times 10^{- 2})} = 2.24 \times 10^{11} {N/m}^{2}$

$∴$ $Maximum possible error in the measurement of Y$

$Δ Y = (0.0489) Y = 0.0489 \times 2.24 \times 10^{11} = 1.09 \times 10^{10} {N/m}^{2}$

Q 20 :

A screw gauge having 100 equal divisions and a pitch of length 1 mm is used to measure the diameter of a wire of length 5.6 cm. The main scale reading is 1 mm and 47th circular division coincides with the main scale. Find the curved surface area of wire in ${cm}^{2}$ to appropriate significant figure. (use $π = \frac{22}{7}$ ). [2004]

(2.6)

$Least count, L.C. = \frac{1 mm}{100} = 0.01 mm$

$Diameter = MSR + CSR \times (L.C.)$

$= 1 mm + 47 \times (0.01) mm = 1.47 mm$

$Curved surface area = 2 π r l = 2 π \frac{D}{2} l = π D l$

$= \frac{22}{7} \times 1.47 \times 56 {mm}^{2} = 2.58724 {cm}^{2}$

$= 2.6 {cm}^{2} (Rounding off to two significant figures)$

Q 21 :

Length, breadth and thickness of a strip having a uniform cross section are measured to be 10.5 cm, 0.05 mm, and 6.0 $μm$ , respectively. Which of the following option(s) give(s) the volume of the strip in ${cm}^{3}$ with correct significant figures: [2025]

$3.2 \times 10^{- 5}$
$32.0 \times 10^{- 6}$
$3.0 \times 10^{- 5}$
$3 \times 10^{- 5}$

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(4)

$Given length L = 10.5 cm \to 3 significant digits$

$Breadth, b = 0.05 mm = 5 \times 10^{- 3} cm \to 1 significant digit$

$Thickness t = 6.0 μ m \to 2 significant digits$

$Therefore volume V = L b t must have only 1 significant digit$

$∴$ $V = 10.5 \times 0.05 \times 10^{- 1} \times 6.0 \times 10^{- 4} {cm}^{3} = 3 \times 10^{- 5} cc$

Q 22 :

In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is $T = 2 π \sqrt{\frac{7 (R - r)}{5 g}} .$ The values of R and r are measured to be $(60 \pm 1) mm$ and $(10 \pm 1) mm$ , respectively. In five successive measurements, the time period is found to be 0.52s, 0.56s, 0.57s, 0.54s and 0.59s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is (are) true? [2016]

The error in the measurement of $r$ is 10%
The error in the measurement of $T$ is 3.75%
The error in the measurement of $T$ is 2%
The error in the determined value of $g$ is 11%

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Select one or more options

(1, 2, 4)

$T_{mean} = \frac{0.52 + 0.56 + 0.57 + 0.54 + 0.59}{5} = 0.556 \approx 0.56 s$

$Error in reading = T_{mean} - T_{1}; T_{mean} - T_{2}; T_{mean} - T_{3}; T_{mean} - T_{4}; T_{mean} - T_{5}$

$Mean error = \frac{0.04 + 0 + 0.01 + 0.02 + 0.03}{5} = 0.016 \approx 0.02 s$

$∴$ $% error in the measurement of T$

$\frac{Δ T}{T} \times 100 = \frac{0.02}{0.56} \times 100 = 3.57 %$

$% error in the measurement of g$

$\frac{Δ g}{g} \times 100 = 2 \frac{Δ T}{T} \times 100 + (\frac{Δ R + Δ r}{R - r}) \times 100$

$= 2 (3.57) + (\frac{1 + 1}{60 - 10}) \times 100 \approx 11 %$

$% error in the measurement of r$

$\frac{Δ r}{r} \times 100 = \frac{1}{10} \times 100 = 10 %$

Q 23 :

Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then: [2015]

If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm
If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm
If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm
If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm

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Select one or more options

(2, 3)

Vernier callipers

$1 MSD = \frac{1 cm}{8} = 0.125 cm$

$5 VSD = 4 MSD$ $∴ 1 VSD = \frac{4}{5} \times 0.125 = 0.1 cm$

$L.C. = 1 MSD - 1 VSD = 0.125 cm - 0.1 cm = 0.025 cm$

Screw gauge

If the pitch of screw gauge is twice the L.C. of vernier callipers, then pitch = $2 \times$ L.C. of vernier calliper $= 2 \times 0.025 = 0.05 cm$

$L.C. of screw gauge = \frac{Pitch}{Total no. of divisions of circular scale}$

$= \frac{0.05}{100} cm = 0.0005 cm = 0.005 mm$

Now if the least count of the linear scale of the screw gauge is twice the least count of vernier callipers,

then L.C. of linear scale of screw gauge $= 2 \times 0.025 = 0.05 cm$

$Then pitch = 2 \times 0.05 = 0.1 cm$

$∴$ $L.C. of screw gauge = \frac{0.1}{100} cm = 0.001 cm = 0.01 mm$

Q 24 :

Using the expression $2 d \sin θ = λ$ , one calculates the values of $d$ by measuring the corresponding angles $θ$ in the range 0 to $90 °$ . The wavelength $λ$ is exactly known and the error in $θ$ is constant for all values of $θ$ . As $θ$ increases from $0 °$ [2013]

The absolute error in $d$ remains constant
The absolute error in $d$ increases
The fractional error in $d$ remains constant
The fractional error in $d$ decreases

1

2

3

4

(4)

$∵$ $2 d \sin θ = λ$ $∴$ $d = \frac{λ}{2} cosec θ . . . (i)$

$∴ \frac{d (d)}{d θ} = \frac{λ}{2} [- cosec θ \cot θ]$

$∴$ $d (d) = - \frac{λ}{2} cosec θ \cot θ d θ . . . (ii)$

$Dividing (ii) by (i), we get$

$| \frac{d (d)}{d} | = \cot θ d θ$

$As θ increases from 0 ° to 90 °, \cot θ decreases$

$∴$ $| \frac{d (d)}{d} | decreases$

$i.e., the fractional error in d decreases.$

Q 25 :

A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true? [2010]

Error $Δ T$ in measuring $T$ , the time period, is 0.05 seconds
Error $Δ T$ in measuring $T$ , the time period, is 1 second
Percentage error in the determination of $g$ is 5%
Percentage error in the determination of $g$ is 2.5%

1

2

3

4

Select one or more options

(1, 3)

The length of the string of simple pendulum is exactly 1 m (given), therefore the error in length $Δ l = 0$ .

Further the possibility of error in measuring time is 1s in 40s as the least count of stop watch is 1s.

$∴$ $\frac{Δ t}{t} = \frac{Δ T}{T} = \frac{1}{40}$

$Time period T = \frac{total time}{no. of oscillations} = \frac{40}{20} = 2 s$

$∴$ $\frac{Δ T}{T} = \frac{1}{40} \Rightarrow \frac{Δ T}{2} = \frac{1}{40} \Rightarrow Δ T = 0.05 s$

$Now for measuring error in g$

$using, T = 2 π \sqrt{\frac{l}{g}} \Rightarrow T^{2} = 4 π^{2} \frac{l}{g} \Rightarrow g = 4 π^{2} \frac{l}{T^{2}}$

$∴$ $Percentage error in g$

$\frac{Δ g}{g} \times 100 = \frac{Δ l}{l} \times 100 + 2 \frac{Δ T}{T} \times 100$

$∴$ $\frac{Δ g}{g} \times 100 = 0 + 2 (\frac{1}{40}) \times 100 = 5 %$

Q 26 :

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 $Δ x$ , $Δ y$ and $Δ z$ , respectively, then

$z \pm Δ z = \frac{x \pm Δ x}{y \pm Δ y} = \frac{x}{y} (1 \pm \frac{Δ x}{x}) {(1 \pm \frac{Δ y}{y})}^{- 1} .$

The series expansion for ${(1 \pm \frac{Δ y}{y})}^{- 1}$ , to first power in $Δ y / y$ , is $1 \mp \frac{Δ y}{y} .$

The relative errors in independent variables are always added. So the error in $z$ will be $Δ z = z (\frac{Δ x}{x} + \frac{Δ y}{y}) .$

The above derivation makes the assumption that $Δ x / x ≪ 1$ , $Δ y / y ≪ 1$ . Therefore, the higher powers of these quantities are neglected. [2018]

Q. 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 $Δ a$ $(Δ a / a ≪ 1)$ , then what is the error $Δ r$ in determining $r$ ?

$\frac{Δ a}{{(1 + a)}^{2}}$
$\frac{2 Δ a}{{(1 + a)}^{2}}$
$\frac{2 Δ a}{(1 - a^{2})}$
$\frac{2 a Δ a}{(1 - a^{2})}$

1

2

3

4

(2)

$r = (\frac{1 - a}{1 + a}) \Rightarrow \frac{Δ r}{r} = \frac{Δ (1 - a)}{1 - a} + \frac{Δ (1 + a)}{1 + a}$

$\Rightarrow \frac{Δ r}{r} = \frac{Δ a}{1 - a} + \frac{Δ a}{1 + a} = \frac{Δ a (1 + a + 1 - a)}{(1 - a) (1 + a)}$

$∴$ $Δ r = \frac{2 Δ a}{(1 - a) (1 + a)} \cdot \frac{(1 - a)}{(1 + a)} = \frac{2 Δ a}{{(1 + a)}^{2}}$

Q 27 :

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 $Δ x$ , $Δ y$ and $Δ z$ , respectively, then

$z \pm Δ z = \frac{x \pm Δ x}{y \pm Δ y} = \frac{x}{y} (1 \pm \frac{Δ x}{x}) {(1 \pm \frac{Δ y}{y})}^{- 1} .$

The series expansion for ${(1 \pm \frac{Δ y}{y})}^{- 1}$ , to first power in $Δ y / y$ , is $1 \mp \frac{Δ y}{y} .$

The relative errors in independent variables are always added. So the error in $z$ will be $Δ z = z (\frac{Δ x}{x} + \frac{Δ y}{y}) .$

The above derivation makes the assumption that $Δ x / x ≪ 1$ , $Δ y / y ≪ 1$ . Therefore, the higher powers of these quantities are neglected. [2018]

Q. In an experiment the initial number of radioactive nuclei is 3000. It is found that $1000 \pm 40$ nuclei decay in the first 1.0 s. For $| x | ≪ 1$ , $l n (1 + x) = x$ up to first power in $x$ . The error $Δ λ$ , in the determination of the decay constant $λ$ , in $s^{- 1}$ , is

0.04
0.03
0.02
0.01

1

2

3

4

(3)

$Using, N = N_{0} e^{- λ t} \Rightarrow \ln N = \ln N_{0} - λ t$

$Differentiation with respect to λ$

$\Rightarrow \frac{1}{N} \frac{d N}{d λ} = 0 - t \Rightarrow Δ λ = \frac{Δ N}{N t} = \frac{40}{2000 \times 1} = 0.02$

Topic Question Set

Q 1 :

The dimensions of a cone are measured using a scale with a least count of 2 mm. The diameter of the base and the height are both measured to be 20.0 cm. The maximum percentage error in the determination of the volume is ____. [2024]

Q 3 :

The energy of a system as a function of time $t$ is given as $E (t) = A^{2} \exp (- α t),$ where $α = 0.2 s^{- 1}$ . The measurement of $A$ has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of $E (t)$ at $t = 5 s$ is _____. [2015]

Q 5 :

Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter D of a tube. The measured value of D is : [2025]

Q 12 :

A student performs an experiment for determination of $(g = \frac{4 π^{2} ℓ}{T^{2}} .)$ The error in length $ℓ$ is $Δ ℓ$ and in time $T$ is $Δ T$ and $n$ is the number of times the reading is taken. The measurement of $g$ is most accurate for [2006]

Q 13 :

In a screw gauge, the zero of main scale coincides with fifth division of circular scale in figure (i). The circular divisions of screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball in figure (ii) is: [2006]

Q 14 :

A wire of length $ℓ = 6 \pm 0.06 cm$ and radius $r = 0.5 \pm 0.005 cm$ and mass $m = 0.3 \pm 0.003 gm.$ Maximum percentage error in density is [2004]

Q 15 :

A cube has a side of length $1.2 \times 10^{- 2} m$ . Calculate its volume. [2003]

Q 21 :

Length, breadth and thickness of a strip having a uniform cross section are measured to be 10.5 cm, 0.05 mm, and 6.0 $μm$ , respectively. Which of the following option(s) give(s) the volume of the strip in ${cm}^{3}$ with correct significant figures: [2025]

Q 24 :

Using the expression $2 d \sin θ = λ$ , one calculates the values of $d$ by measuring the corresponding angles $θ$ in the range 0 to $90 °$ . The wavelength $λ$ is exactly known and the error in $θ$ is constant for all values of $θ$ . As $θ$ increases from $0 °$ [2013]

Q 25 :

A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true? [2010]

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Topic Question Set

Q 1 : The dimensions of a cone are measured using a scale with a least count of 2 mm. The diameter of the base and the height are both measured to be 20.0 cm. The maximum percentage error in the determination of the volume is ____. [2024]

Q 2 : In an experiment for determination of the focal length of a thin convex lens, the distance of the object from the lens is 10±0.1 cm and the distance of its real image from the lens is 20±0.2 cm. The error in the determination of focal length of the lens is n%. The value of n is ____. [2023]

Q 3 : The energy of a system as a function of time t is given as E(t)=A2exp(-αt), where α=0.2 s-1. The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E(t) at t=5 s is _____. [2015]

Q 5 : Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter D of a tube. The measured value of D is : [2025]

Q 12 : A student performs an experiment for determination of (g=4π2ℓT2.) The error in length ℓ is Δℓ and in time T is ΔT and n is the number of times the reading is taken. The measurement of g is most accurate for [2006]

Q 13 : In a screw gauge, the zero of main scale coincides with fifth division of circular scale in figure (i). The circular divisions of screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball in figure (ii) is: [2006]

Q 14 : A wire of length ℓ=6±0.06 cm and radius r=0.5±0.005 cm and mass m=0.3±0.003 gm. Maximum percentage error in density is [2004]

Q 15 : A cube has a side of length 1.2×10-2 m. Calculate its volume. [2003]

Q 21 : Length, breadth and thickness of a strip having a uniform cross section are measured to be 10.5 cm, 0.05 mm, and 6.0 μm, respectively. Which of the following option(s) give(s) the volume of the strip in cm3 with correct significant figures: [2025]

Q 24 : Using the expression 2dsinθ=λ, one calculates the values of d by measuring the corresponding angles θ in the range 0 to 90°. The wavelength λ is exactly known and the error in θ is constant for all values of θ. As θ increases from 0° [2013]

Q 25 : A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true? [2010]

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Q 1 :

The dimensions of a cone are measured using a scale with a least count of 2 mm. The diameter of the base and the height are both measured to be 20.0 cm. The maximum percentage error in the determination of the volume is ____. [2024]

Q 3 :

The energy of a system as a function of time $t$ is given as $E (t) = A^{2} \exp (- α t),$ where $α = 0.2 s^{- 1}$ . The measurement of $A$ has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of $E (t)$ at $t = 5 s$ is _____. [2015]

Q 5 :

Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter D of a tube. The measured value of D is : [2025]

Q 12 :

A student performs an experiment for determination of $(g = \frac{4 π^{2} ℓ}{T^{2}} .)$ The error in length $ℓ$ is $Δ ℓ$ and in time $T$ is $Δ T$ and $n$ is the number of times the reading is taken. The measurement of $g$ is most accurate for [2006]

Q 13 :

In a screw gauge, the zero of main scale coincides with fifth division of circular scale in figure (i). The circular divisions of screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball in figure (ii) is: [2006]

Q 14 :

A wire of length $ℓ = 6 \pm 0.06 cm$ and radius $r = 0.5 \pm 0.005 cm$ and mass $m = 0.3 \pm 0.003 gm.$ Maximum percentage error in density is [2004]

Q 15 :

A cube has a side of length $1.2 \times 10^{- 2} m$ . Calculate its volume. [2003]

Q 21 :

Length, breadth and thickness of a strip having a uniform cross section are measured to be 10.5 cm, 0.05 mm, and 6.0 $μm$ , respectively. Which of the following option(s) give(s) the volume of the strip in ${cm}^{3}$ with correct significant figures: [2025]

Q 24 :

Using the expression $2 d \sin θ = λ$ , one calculates the values of $d$ by measuring the corresponding angles $θ$ in the range 0 to $90 °$ . The wavelength $λ$ is exactly known and the error in $θ$ is constant for all values of $θ$ . As $θ$ increases from $0 °$ [2013]

Q 25 :

A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true? [2010]