Errors in Measurements
137 Questions
Start Allen Test
Q126
Allen
JEE Advanced Practice Paper
MCQ
In Wheatstone bridge experiment as shown in figure -
A.
Key $K_{1}$ should be pressed first and then $K_{2}$ .
B.
Key $K_{2}$ should be pressed first and then $K_{1}$ .
C.
Any key can be pressed in any order.
D.
Both keys should be pressed simultaneously.
Q127
Allen
JEE Advanced Practice Paper
MCQ
In a resonance column method, resonance occurs at two successive level of $l_{1} = 30.7 \, cm$ and $l_{2} = 63.2 \, cm$ using a tuning fork of f = 512 Hz. What is the maximum error in measuring speed of sound using relations $v = f\lambda$ and $\lambda = 2(l_{2} - l_{1})$
A.
256 cm/sec
B.
92 cm/sec
C.
128 cm/sec
D.
102.4 cm/sec
Q128
Allen
JEE Advanced Practice Paper
MSQ
In the Searle's experiment, after every step of loading, why should wait for two minutes before taking the readings?
A.
So that the wire can have its desired change in length.
B.
So that the wire can attain room temperature.
C.
So that vertical oscillations can get subsided.
D.
So that the wire has no change in its radius.
Q129
Allen
JEE Advanced Practice Paper
MSQ
Two resistance are $R_{1} = 5.0 \pm 0.2 \Omega$ and $R_{2} = 10.0 \pm 0.1 \Omega$ . Equivalent resistance with limit of possible error is $R_{s}$ and $R_{p}$ when both resistances are connected in series and parallel respectively.
A.
$R_{s} = 15.0 \Omega \pm 2\%$
B.
$R_{p} = 3.3 \Omega \pm 3\%$
C.
$R_{s} = 15.0 \Omega \pm 1\%$
D.
$R_{p} = 3.3 \Omega \pm 2\%$
Q130
Allen
JEE Advanced Practice Paper
MSQ
A physical quantity P is related to four observables A, B, C and D as $P = 4\pi^{2}A^{3}B^{2}/(\sqrt{C}D)$ . The percentage error of the measurement in A, B, C and D are 1%, 3% and 2%, 4% respectively. Value of P is calculated 3.763.
A.
Percentage error is 12%
B.
Percentage error is 14%
C.
Absolute error is 0.53
D.
Absolute error is 0.27
Q131
Allen
JEE Advanced Practice Paper
MSQ
The period of oscillation of a simple pendulum in an experiment is recorded as 2.63 sec, 2.56 sec, 2.42 sec, 2.71 sec and 2.80 sec respectively:-
A.
Time period is 2.62 sec
B.
Average absolute error is 0.01 sec
C.
Percentage error is $2.4\%$
D.
Percentage error is $4.2\%$
Q132
Allen
JEE Advanced Practice Paper
Numerical
The time period of oscillation of a simple pendulum is given by $T = 2\pi\sqrt{\ell/g}$ . The length of the pendulum is measured as $\ell = 10.0 \pm 0.1cm$ and the time period as $T = 0.50 \pm 0.02s$ . Determine percentage error in the value of g.
Correct Answer: 9
Explanation:
Ans. (9)
All. (9) $T = 2\pi \sqrt{\frac{\ell}{g}}$ $\Rightarrow g = \frac{2\pi\ell}{T^2}$ $\Rightarrow \frac{\Delta g}{g} = \frac{\Delta\ell}{\ell} + \frac{2\Delta T}{T}$ Here $\ell = 10.0$ , $\Delta\ell = 0.1$ $T = 0.50$ , $\Delta T = 0.02$ $\frac{\Delta g}{g} \times 100 = \left(\frac{10^{-1}}{10} + 2 \times \frac{2 \times 10^{-2}}{\frac{1}{2}}\right) \times 100$ $= (10^{-2} + 8 \times 10^{-2}) \times 100 = 9\%$
All. (9) $T = 2\pi \sqrt{\frac{\ell}{g}}$ $\Rightarrow g = \frac{2\pi\ell}{T^2}$ $\Rightarrow \frac{\Delta g}{g} = \frac{\Delta\ell}{\ell} + \frac{2\Delta T}{T}$ Here $\ell = 10.0$ , $\Delta\ell = 0.1$ $T = 0.50$ , $\Delta T = 0.02$ $\frac{\Delta g}{g} \times 100 = \left(\frac{10^{-1}}{10} + 2 \times \frac{2 \times 10^{-2}}{\frac{1}{2}}\right) \times 100$ $= (10^{-2} + 8 \times 10^{-2}) \times 100 = 9\%$
Q133
Allen
JEE Advanced Practice Paper
Numerical
The edge of a cube is measured using a vernier calliper. (9 division of the main scale is equal to 10 division of Vernier scale and 1 main scale division is 1 mm). The main scale division reading is 10 and 1 division of Vernier scale was found to be coinciding with the main scale. The mass of the cube is 2.736 g. The density is $\rho g/cm^{3}$ upto correct significant figures. Find 100 $\rho$ .
Correct Answer: 266
Explanation:
Ans. (266)
$1 M S D = 1 m m, 9 M S D = 1 0 V S D \Rightarrow 1 V S D = \frac {9}{1 0} M S D$
$\mathrm{Leastcount} (L C) = M S D - 1 V S D = M S D - \frac {9}{1 0} M S D$
$\Rightarrow L C = \frac {1}{1 0} M S D = \frac {1 m m}{1 0} = 0. 1 m m$
$\mathrm{Reading} = 1 0 \mathrm{mm} + 1 \times 0. 1 \mathrm{mm} = 1 0. 1 \mathrm{mm} = 1. 0 1 \mathrm{cm}$
$\mathrm{Density} = \frac {m}{a ^ {3}} = \frac {2 . 7 3 6 g}{(1 . 0 1 c m) ^ {3}} = \frac {2 . 7 3 6 g m}{1 . 0 3 0 3 0 1 c m ^ {3}}$
$\Rightarrow \text { density } = 2. 6 5 5 5 3 4 6 g / c m ^ {3}$
$\mathrm{density} = 2. 6 6 \mathrm{g} / \mathrm{cm} ^ {3}$
$1 M S D = 1 m m, 9 M S D = 1 0 V S D \Rightarrow 1 V S D = \frac {9}{1 0} M S D$
$\mathrm{Leastcount} (L C) = M S D - 1 V S D = M S D - \frac {9}{1 0} M S D$
$\Rightarrow L C = \frac {1}{1 0} M S D = \frac {1 m m}{1 0} = 0. 1 m m$
$\mathrm{Reading} = 1 0 \mathrm{mm} + 1 \times 0. 1 \mathrm{mm} = 1 0. 1 \mathrm{mm} = 1. 0 1 \mathrm{cm}$
$\mathrm{Density} = \frac {m}{a ^ {3}} = \frac {2 . 7 3 6 g}{(1 . 0 1 c m) ^ {3}} = \frac {2 . 7 3 6 g m}{1 . 0 3 0 3 0 1 c m ^ {3}}$
$\Rightarrow \text { density } = 2. 6 5 5 5 3 4 6 g / c m ^ {3}$
$\mathrm{density} = 2. 6 6 \mathrm{g} / \mathrm{cm} ^ {3}$
Q134
Allen
JEE Advanced Practice Paper
Numerical
The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47 $^{th}$ division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. The curved surface area (in $cm^{2}$ ) of the wire in appropriate number of significant figures is A. Find $A \times 10$ .
Correct Answer: 26
Explanation:
Ans. (26)
Pitch = 1 mm
Division on circular scale = 100
L.C. = $\frac{\text{Pitch}}{\text{Circular scale division}}$ $\Rightarrow$ L.C. = $\frac{1mm}{100}$ = 0.01mm
MSR = 1 mm
VSR = 47 × LC = 0.47 mm
Measured diameters = 1.47 mm = 0.147 cm
Length of wire = 5.6 cm
Curved surface area = 2πrℓ
A = 2π × $\frac{0.147}{2}$ × 5.6
A = 2.5816cm²
In this operation, least significant digit is two in 5.6 cm length.
So answer of surface must not be consist more than two significant digit.
i.e. A = 2.6 cm²
Pitch = 1 mm
Division on circular scale = 100
L.C. = $\frac{\text{Pitch}}{\text{Circular scale division}}$ $\Rightarrow$ L.C. = $\frac{1mm}{100}$ = 0.01mm
MSR = 1 mm
VSR = 47 × LC = 0.47 mm
Measured diameters = 1.47 mm = 0.147 cm
Length of wire = 5.6 cm
Curved surface area = 2πrℓ
A = 2π × $\frac{0.147}{2}$ × 5.6
A = 2.5816cm²
In this operation, least significant digit is two in 5.6 cm length.
So answer of surface must not be consist more than two significant digit.
i.e. A = 2.6 cm²
Q135
Allen
JEE Advanced Practice Paper
Numerical
In a Searle's experiment, the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Maximum error in the measurement of Young's modulus of the material of the wire from these data is $P \times 10^{9}$ N/m $^{2}$ , then value of P is
Correct Answer: 11
Explanation:
Ans. (11)
Diameter D = 0.050 cm, ΔD = 0.001 cm
Length L = 110.0 cm
ΔL = 0.1 cm
Weight W = 50 N
Extension ℓ = 0.125 cm = Δℓ = 0.001 cm
Y = $\frac{\frac{W}{A}}{\frac{\ell}{L}}$ ⇒ Y = $\frac{4WL}{\pi D^2\ell}$ $\Rightarrow$ Y = $\frac{4\times 50\times 1.1}{\pi\times 25\times 10^{-8}\times 125\times 10^{-5}}\frac{N}{m^2}$ $\Rightarrow$ Y = 0.02241 × $10^{13}N/m^2$ Y = 2.241 × $10^{11}N/m^2$ $\frac{\Delta Y}{Y}$ = $\frac{\Delta L}{L}$ + $\frac{2\Delta D}{D}$ + $\frac{\Delta \ell}{\ell}$ $\Rightarrow$ $\frac{\Delta Y}{Y}$ = $\frac{0.1}{110}$ + 2 × $\frac{0.001}{0.050}$ + $\frac{0.001}{0.125}$ $\Rightarrow$ $\frac{\Delta Y}{Y}$ = 0.04891 $\Rightarrow$ $\Delta Y$ = Y × 0.04891 $\Rightarrow$ $\Delta Y$ = 2.241 × $10^{11}$ × 0.04891 $\Rightarrow$ $\Delta Y$ = 0.1096 × $10^{11}N/m^2$ $\Rightarrow$ $\Delta Y$ = 1.096 × $10^{10}N/m^2$ Least significant digit in this operation is two $\Delta Y$ = 1.1 × $10^{10}N/m^2$ = 11 × $10^9 N/m^2$
Diameter D = 0.050 cm, ΔD = 0.001 cm
Length L = 110.0 cm
ΔL = 0.1 cm
Weight W = 50 N
Extension ℓ = 0.125 cm = Δℓ = 0.001 cm
Y = $\frac{\frac{W}{A}}{\frac{\ell}{L}}$ ⇒ Y = $\frac{4WL}{\pi D^2\ell}$ $\Rightarrow$ Y = $\frac{4\times 50\times 1.1}{\pi\times 25\times 10^{-8}\times 125\times 10^{-5}}\frac{N}{m^2}$ $\Rightarrow$ Y = 0.02241 × $10^{13}N/m^2$ Y = 2.241 × $10^{11}N/m^2$ $\frac{\Delta Y}{Y}$ = $\frac{\Delta L}{L}$ + $\frac{2\Delta D}{D}$ + $\frac{\Delta \ell}{\ell}$ $\Rightarrow$ $\frac{\Delta Y}{Y}$ = $\frac{0.1}{110}$ + 2 × $\frac{0.001}{0.050}$ + $\frac{0.001}{0.125}$ $\Rightarrow$ $\frac{\Delta Y}{Y}$ = 0.04891 $\Rightarrow$ $\Delta Y$ = Y × 0.04891 $\Rightarrow$ $\Delta Y$ = 2.241 × $10^{11}$ × 0.04891 $\Rightarrow$ $\Delta Y$ = 0.1096 × $10^{11}N/m^2$ $\Rightarrow$ $\Delta Y$ = 1.096 × $10^{10}N/m^2$ Least significant digit in this operation is two $\Delta Y$ = 1.1 × $10^{10}N/m^2$ = 11 × $10^9 N/m^2$
Q136
Allen
JEE Advanced Practice Paper
Numerical
A glass prism of angle $A = 60^{\circ}$ gives minimum angle of deviation $\theta = 30^{\circ}$ with the max. error of $1^{\circ}$ when a beam of parallel light passed through the prism during an experiment. The permissible error in the measurement of refractive index $\mu$ of the material of the prism is $\frac{x\pi}{y}\%$ then value of $x + y$ is ____.
Correct Answer: 23
Explanation:
$\begin{array}{r l} & {\mathrm{Ans.} (2 3)} \\ & {A = 6 0 ^ {\circ}} \\ & {\delta_ {m i n} \quad = 3 0 ^ {\circ}} \\ & {\varDelta \delta = 1 ^ {\circ}} \\ & {\mu = \frac {s i n \left(\frac {A + \delta}{2}\right)}{s i n \left(\frac {A}{2}\right)}} \\ & {\qquad \mathrm{Differencing} \mu \mathrm{w.r.t.} \delta} \\ & {\Rightarrow \varDelta \mu = \left[ c o s e c \left(\frac {A}{2}\right) \right] \left[ (c o s \left(\frac {A + \delta}{2}\right)) \times \frac {\varDelta \delta}{2} \right]} \\ & {\Rightarrow \varDelta \mu = [ \operatorname{cosec} (3 0 ^ {\circ}) ] \left[ \left(c o s \left(\frac {6 0 ^ {\circ} + 3 0 ^ {\circ}}{2}\right)\right) \times \frac {1 ^ {\circ}}{2} \right]} \\ & {\Rightarrow \varDelta \mu = 2 \times \frac {1}{\sqrt {2}} \times \frac {1 ^ {\circ}}{2} = \left(\frac {1}{\sqrt {2}}\right) ^ {\circ}} \\ & {\Rightarrow \varDelta \mu = \frac {1}{\sqrt {2}} \times \frac {\pi}{1 8 0} \qquad ... (1)} \\ & {\qquad \mu = \frac {\frac {s i n (A + \delta)}{2}}{s i n \left(\frac {A}{2}\right)} = \frac {s i n \left(\frac {6 0 ^ {\circ} + 3 0 ^ {\circ}}{2}\right)}{s i n 3 0 ^ {\circ}}} \\ & {\qquad \mu = \frac {\frac {1}{\sqrt {2}}}{\frac {1}{2}} \Rightarrow \sqrt {2}} \\ & {\qquad \frac {\varDelta \mu}{\mu} \times 1 0 0 = \frac {\pi}{\sqrt {2} \times 1 8 0} \times \frac {1}{\sqrt {2}} \times 1 0 0 \%} \\ & {\qquad \frac {\varDelta \mu}{\mu} \times 1 0 0 = \frac {1 0 0 \pi}{3 6 0} = \frac {5 \pi}{1 8} \%} \end{array}\tag{...(1}$
Q137
Allen
JEE Advanced Practice Paper
MCQ
In a vernier calipers the main scale and the vernier scale are made up of different materials. When the room temperature increases by $\Delta T^{\circ}C$ , it is found the reading of the instrument remains the same. Earlier it was observed that the front edge of the wooden rod placed for measurement crossed the $N^{th}$ main scale division and $(N + 2)$ MSD coincided with the $2^{\text{nd}}$ VSD. Initially, 10 VSD coincided with 9 MSD. If coefficient of linear expansion of the main scale is $\alpha_{1}$ and that of the vernier scale is $\alpha_{2}$ . If $\alpha_{1}/\alpha_{2}$ is $\frac{x + 0.8}{(N + 2)}$ then value of 'x' is (Ignore the expansion of the rod on heating)
Correct Answer: A
Explanation:
Ans. (1)
$10VSD = 9MSD$
$\Rightarrow 1VSD = \frac{9}{10}MSD$
$\ell_{MSR} = (N+2)(MSD)$
$\Delta \ell_{MSR} = (N+2)(MSD)\times\alpha_1\Delta T \qquad ...(1)$
$\ell_{VSR} = (2)(VSD)$
$\Delta \ell_{VSR} = (2)(VSD)\times\alpha_2\Delta T \qquad ...(2)$
$\text{Given Reading of instrument remain same}$
$\Rightarrow \Delta \ell_{MSR} = \Delta \ell_{VSR}$
$\Rightarrow (N+2)(MSD)\alpha_1\Delta T = 2(VSD)\alpha_2\Delta T$
$\Rightarrow (N+2)(MSD)\alpha_1 = 2\times\left(\frac{9}{10}\right)(MSD)\alpha_2$
$\Rightarrow \frac{\alpha_1}{\alpha_2} = \frac{18}{10(N+2)} = \frac{1.8}{N+2}$
$10VSD = 9MSD$
$\Rightarrow 1VSD = \frac{9}{10}MSD$
$\ell_{MSR} = (N+2)(MSD)$
$\Delta \ell_{MSR} = (N+2)(MSD)\times\alpha_1\Delta T \qquad ...(1)$
$\ell_{VSR} = (2)(VSD)$
$\Delta \ell_{VSR} = (2)(VSD)\times\alpha_2\Delta T \qquad ...(2)$
$\text{Given Reading of instrument remain same}$
$\Rightarrow \Delta \ell_{MSR} = \Delta \ell_{VSR}$
$\Rightarrow (N+2)(MSD)\alpha_1\Delta T = 2(VSD)\alpha_2\Delta T$
$\Rightarrow (N+2)(MSD)\alpha_1 = 2\times\left(\frac{9}{10}\right)(MSD)\alpha_2$
$\Rightarrow \frac{\alpha_1}{\alpha_2} = \frac{18}{10(N+2)} = \frac{1.8}{N+2}$