MCQ questions on the Physics chapter errors in measurement for class 11, NEET, JELET, JEE and diploma and degree first year Engineering physics (polytechnic and B.Tech)
1. A certain body weights 22.42 gm and has a measured volume of 4.7 cc. The possible error in the measurement of mass and volume are 0.01 gm and 0.1 cc. Then maximum error in the density will be
a. 22%
b. 2%
c. 0.2%
d. 0.02%
option b
Density=\frac{Mass}{Volume}
D=\frac{m}{V}
Taking logarithm in both sides
ln(D) = ln (m) – ln(V)
Differentiating the above equation
\frac{dD}{D}=\frac{dm}{m}-\frac{dV}{V}
So, the maximum error is (neglect negative sign to calculate the maximum error)
\frac{\Delta D}{D} \times 100\%=\frac{\Delta m}{m}\times 100\%+\frac{\Delta V}{V} \times 100\%
Here, \Delta m = 0.01 gm, m = 22.42 gm, \Delta V = 0.1cc, V= 4.7cc
\frac{\Delta D}{D} \times 100 \%=\frac{0.01}{22.42}\times 100\%+\frac{0.1}{4.7} \times 100\%
\frac{\Delta D}{D} \times 100\%= 2\%
2. If the error in the measurement of radius of a sphere is 2 %, then the error in the determination of volume of the sphere will be
a. 4%
b. 6%
c. 8%
d. 2%
option b
Volume of a sphere of radius r is
V=\frac{4}{3} \pi r^{3}
Taking logarithm
ln (V) = ln ( \frac{4}{3} \pi) +3ln (r)
By differentiating this equation
\frac{dV}{V}=3\frac{dr}{r}
So, the error in measurement is
\frac{\Delta V}{V} \times 100\%=3\frac{\Delta r}{r} \times 100\%
The error in radius measurement is
\frac{\Delta r}{r} \times 100\%=2\%
Therefore,
\frac{\Delta V}{V}\times 100\%=3\times 2\%
\frac{\Delta V}{V}\times 100\%=6\%
3. In an experiment four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity P is calculated as follows
P=\frac{a^{3}b^{2}}{cd} , % error in P is
a.10%
b. 7%
c. 4%
d. 14%
option d
P=\frac{a^{3}b^{2}}{cd}
Taking logarithm
ln (P) = 3ln (a)+2ln(b)-lnc-lnd
Differentiating
\frac{dP}{P} = 3\frac{da}{a}+2\frac{db}{b}-\frac{dc}{c}-\frac{dd}{d}
The percentage error in P is (neglect negative sign to calculate the maximum error)
\frac{\Delta P}{P} \times 100\% = 3\frac{\Delta a}{a} \times 100\%+2\frac{\Delta b}{b} \times 100\%+\frac{\Delta c}{c} \times 100\%+\frac{\Delta d}{d} \times 100\%
Here, the given values are
\frac {\Delta a}{a} \times 100\% = 1\%, \frac{\Delta b}{b} \times 100\% = 2\%, \frac{\Delta c}{c} \times 100\%=3\%, \frac{\Delta d}{d} \times 100\% = 4\%
so,
\frac{\Delta P}{P} = (3\times 1\%)+(2\times 2\%)+(3\%)+(4\%) = 14\%
4. The percentage errors in the measurements of mass and speed are 2% and 3% respectively. The error in kinetic energy obtained by measuring mass and speed will be
a. 12%
b. 10%
c. 8%
d. 2%
option c
Kinetic energy of mass m and speed v is given by
K=\frac{1}{2}mv^{2}
Taking logarithm
ln(K)=ln(\frac{1}{2})+ln(m)+2ln(v)
Differentiating this
\frac{dK}{K}=\frac{dm}{m}+2\frac{dv}{v}
The percentage error of kinetic energy is
\frac{\Delta K}{K}\times 100\%=\frac{\Delta m}{m}\times 100\%+2\frac{\Delta v}{v}\times 100\%
The given values are
\frac{\Delta m}{m}\times 100\%=2\% and \frac{\Delta v}{v}\times 100\% = 3\%
By putting these values, we can write
\frac{\Delta K}{K}\times 100\%=2\%+(2\times 3\%) =8\%
5. A physical quantity A is dependent on other four physical quantities p, q, r and s as given by A=\frac{\sqrt{pq}}{r^{2}s^{3}} . The percentage error in the measurement of p, q, r and s are 1%, 3%, 0.5% and 0.33% respectively, then the maximum percentage error in A is
a. 2%
b. 0%
c. 4%
d. 3%
option c
The given relation is
A=\frac{\sqrt{pq}}{r^{2}s^{3}}
Taking log in both sides
ln(A) = \frac{1}{2}ln(p)+\frac{1}{2}ln(q)-2ln(r)-3ln(s)
Differentiation of the above equation gives
\frac{dA}{A}=\frac{1}{2}\frac{dp}{p}+\frac{1}{2}\frac{dq}{q}-2\frac{dr}{r}-3\frac{ds}{s}
So, the maximum percentage error (taking only positive values)
\frac{\Delta A}{A}\times 100\%=\frac{1}{2}\frac{\Delta p}{p}\times 100\%+\frac{1}{2}\frac{\Delta q}{q}\times 100\%+2\frac{\Delta r}{r}\times 100\%+3\frac{\Delta s}{s}\times 100\%
Putting all the values in the above equation
\frac{\Delta A}{A}\times 100\%=\frac{1}{2}(1\%)+\frac{1}{2}(3\%)+2(0.5\%)+3(0.33)\%
\frac{\Delta A}{A}\times 100\%=4\%
6. A physical quantity ρ is related to four variables α, β, γ and η as \rho =\frac{\alpha ^{3}\beta ^{2}}{\eta \sqrt{\gamma }} , the percentage errors of measurements in α, β, γ and η are 1%, 3%, 4% and 2% respectively. Find the percentage error in ρ.
a. 10%
b. 11%
c. 12%
d. 13%
option d
The given relation is
\rho =\frac{\alpha ^{3}\beta ^{2}}{\eta \sqrt{\gamma }}
Taking logarithm
ln(\rho ) =3 ln(\alpha )+2ln(\beta )-ln(\eta)-\frac{1}{2}ln(\gamma)
Differentiating above equation
\frac{d\rho }{\rho } == 3\frac{d\alpha }{\alpha }+2\frac{d\beta }{\beta }-\frac{d\eta }{\eta }-\frac{1}{2}\frac{d\gamma }{\gamma }
The percentage error in \rho is
\frac{\Delta \rho }{\rho }\times 100\% = (3\frac{\Delta \alpha }{\alpha }\times 100\%)+(2\frac{\Delta \beta }{\beta }\times 100\%)+(\frac{\Delta \eta }{\eta }\times 100\%)+(\frac{1}{2}\frac{\Delta \gamma }{\gamma })\times 100\%)
The percentage errors of measurements in α, β, γ and η are 1%, 3%, 4% and 2% respectively. So,
\frac{\Delta \rho }{\rho }\times 100\% = (3\times 1\%)+(2\times 3\%)+(2\%)+(\frac{1}{2}\times 4\%)=13 \%
7. The least count of a stop watch is 0.2 s. The time of 20 oscillations of a pendulum is measured to be 25 s. The percentage error in the time period is
a. 1.2%
b. 0.8%
c. 1.8%
d. None of these
option b
Time of 20 oscillation (n) is 25 s (t) and least count is 0.2s (\Delta t) .
T=\frac{t}{n}
Where T = time period
The percentage error in time period (T) will be
\frac{\Delta T}{T}\times 100\% = \frac{\Delta t}{t}\times 100\%
\frac{\Delta T}{T}\times 100\% = \frac{0.2 s}{25 s}\times 100\% = 0.8\%
8. The length of a simple pendulum is about 100 cm known to have an accuracy of 1 mm. Its period of oscillation is 2 s determined by measuring the time for 100 oscillations using a clock of 0.1 s resolution. What is the accuracy in the determined value of g (acceleration due to gravity)?
a. 0.2%
b. 0.5 %
c. 0.1 %
d. 2 %
option a
Time period, T
T=2\pi \sqrt{\frac{l}{g}}
g=\frac{4\pi^{2}l}{T^{2}}
ln (g) = ln ( 4\pi^{2} )+ln(l)-2ln(T)
differentiating this
\frac{dg}{g}=\frac{dl}{l}-2\frac{dT}{T}
The maximum percentage error in g (acceleration due to gravity) is
\frac{\Delta g}{g}\times 100\%=\frac{\Delta l}{l}\times 100\%+2\frac{\Delta T}{T}\times 100\% ———————– (1)
The value of \frac{\Delta l}{l}\times 100\% = \frac{1mm}{100cm}\times 100\% = \frac{0.1cm}{100cm}\times 100\% =0.1\%
Time to complete 100 oscillation (n) is 2 s (t) and error is 0.1s (\Delta t) .
T=\frac{t}{n}
The percentage error in T will be
\frac{\Delta T}{T}\times 100\% = \frac{\Delta t}{n}\frac{1}{T}\times 100\%
\frac{\Delta T}{T}\times 100\% = \frac{1}{100}\frac{0.1 s}{2s}\times 100\% = 0.05\%
So, by putting all values in equation (1)
\frac{\Delta g}{g}\times 100\%=0.1\%+(2\times 0.05\%) = 0.2 \%
9. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in measuring clock is 1 s, then the reported mean time should be
a. 92±1.8 s
b. 92±3s
c. 92±2 s
d. 92±5.0 s
option c
Sum of all observations = 90+91+95+92 = 368 s
Mean value = 368/4 = 92 s
The errors in each measurement are (92-90 = 2s), (92-91=1s), (92-92=0s) and (95-92=3s).
Sum of all errors = 2+1+0+3 = 6s
Average error = 6/4 = 1.5 s
The minimum division in the clock is 1s, so the average error in measurement should be rounded of to 2s.
The mean time is 92±2 s
Rules for Finding Significant Figure
Rule 1: All non-zero digits are significant. Example 12.3545 has 6 significant figures.
Rule 2: Zeros appearing in between two non-zero digits are significant. Example: 23.056 has 5 significant figures.
Rule 3: All trailing zeros after decimal are significant.
Example: 8.9300 has 5 significant figures
Rule 4: Powers of 10 are not counted in significant figures.
Example: 3.3× 10-8 has only two significant figure
Rule 5: All zeros appearing before a non zero digit are not significant for numbers less than 1.
Example: 0.000123 has 3 significant figure
Rule 6: Changing unit does not change significant figure.
Example: If a length is measured 4.3 mm then it has 2 significant figure. Changing unit does not change significant figure.
Rule 7: Trailing zeros without decimal point are not significant.
Example: 43600 has only 3 significant figures.
Some Examples on determination of significant figures
| Measurement | Number of significant figures | Rule number |
| 25478 cm | 5 | 1 |
| 365.035 m | 6 | 2 |
| 25470 cm | 4 | 7 |
| 0.00345 m | 3 | 5 |
| 3.1250 cm | 5 | 3 |
| 1.50✕107 kg | 3 | 4 |
| 3.0120 | 5 | 2 & 3 |