Tag: measurement of length

Questions Related to measurement of length

In a measurement, the uncertainty of length $L$ is $\pm a$ and the uncertainty of width $W$ is $\pm b$. Assuming a and b very small, find the the uncertainty in measurement of area. 

physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. $a/L+b/W$

  2. $aL+bW$

  3. $aW+bL$

  4. $a^2/W+b^2/L$

Show answer
Correct Option: C
Explanation:

Here given, $\Delta L=a$ and $\Delta W=b$
Area, $A=LW$
Take ln and then differentiate, $\dfrac{\Delta A}{A}=\dfrac{\Delta L}{L}+\dfrac{\Delta W}{W}=a/L+b/W$
So,$\Delta A=(a/L+b/W)(LW)=aW+bL$

In an experiment, the value of refractive index of a plastic has been found 1.33,1.30, 1.34 and 1.29 in successive measurements. Find the mean absolute error for refractive index.  

physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. $0.12$

  2. $0.02$

  3. $0.10$

  4. $0.01$

Show answer
Correct Option: B
Explanation:

Here the mean value, $\bar{n}=\dfrac{1.33+1.30+1.34+1.29}{4}=1.31$
The absolute error in each measurements are, 
$\Delta n _1=\bar{n}-n _1=1.31-1.33=-0.02$;
$\Delta n _2=\bar{n}-n _2=1.31-1.30=0.01$;
$\Delta n _3=\bar{n}-n _3=1.31-1.34=-0.03$;
$\Delta n _4=\bar{n}-n _4=1.31-1.29=0.02$;
The mean absolute error, $\Delta \bar{n}=\dfrac{|\Delta n _1|+|\Delta n _2|+|\Delta n _3|+|\Delta n _4|}{4}=\dfrac{0.02+0.01+0.03+0.02}{4}=0.02$

The capacitance of two capacitors are $C _1=(5 \pm 0.1)\mu F$ and $C _2=(10 \pm 0.1)\mu F$, If they are connected in series then the percentage error is 

physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. $3.33 $%

  2. $4.03 $%

  3. $3.0 $%

  4. $4.33 $%

Show answer
Correct Option: D
Explanation:

When two capacitors are in series, the equivalent capacitance is $C=\dfrac{C _1C _2}{C _1+C _2}$
Thus, $\dfrac{\Delta C}{C}=\dfrac{\Delta C _1}{C _1}+\dfrac{\Delta C _2}{C _2}+\dfrac{\Delta C _1+\Delta C _2}{C _1+C _2}$

The % error, $\dfrac{\Delta C}{C}\times 100=\left(\dfrac{\Delta C _1}{C _1}+\dfrac{\Delta C _2}{C _2}+\dfrac{\Delta C _1+\Delta C _2}{C _1+C _2}\right)\times 100$
                                         $=(\dfrac{0.1}{5}+ \dfrac{0.1}{10}+\dfrac{0.1+0.1}{5+10})\times 100=4.33$%

In an experiment, mass of an object is measured by applying a known force on it, and then measuring its acceleration. If, in the experiment, the measured values of applied force and the measured acceleration are $F=10.0\pm 0.2N$ and $a=1.00\pm 0.01m/{s}^{2}$, respectively, the mass of the object is

physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. $10.0Kg$

  2. $10.0\pm 0.1Kg$

  3. $10.0\pm 0.3Kg$

  4. $10.0\pm 0.4Kg$

Show answer
Correct Option: C
Explanation:

$F=Ma$
$M=\cfrac { f }{ a } $
$\cfrac { \Delta M\times 100 }{ M } =\cfrac { \Delta f }{ f } \times 100+\cfrac { \Delta a }{ a } \times 100$
$=\cfrac { 0.2 }{ 10 } +\cfrac { 0.01 }{ 1 } $
$\Delta M=0.03\times 10$
$\therefore$ $10.0\pm 0.3Kg$

A physical quantity $X$ is represented by $X = [M^{\eta}L^{-\theta} T^{-\phi}]$. The maximum percentage errors in the measurement of $M, L$ and $T$, respectively are $\alpha$%, $\beta$% and $\gamma$%. The maximum percentage error in the measurement of $X$ will be

physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. $(\eta \alpha - \theta \beta - \phi \gamma)$%

  2. $(\theta \beta + \phi \gamma - \eta \alpha)$%

  3. $\left (\dfrac {\alpha}{\eta} - \dfrac {\beta}{\theta} - \dfrac {\gamma}{\phi}\right )$%

  4. $(\eta \alpha + \theta \beta + \phi \gamma)$%

Show answer
Correct Option: D
Explanation:

Given, $X=M^{\eta}L^{-\theta}T^{-\phi}$

Differentiating, $\Delta X=(\eta M^{\eta-1}\Delta M) L^{-\theta}T^{-\phi}-(\theta L^{-\theta-1}\Delta L)M^{\eta}T^{-\phi}-(\phi T^{-\phi-1}\Delta T)M^{\eta}L^{-\phi}$
So, $\dfrac{\Delta X}{X}=\eta\dfrac{\Delta M}{M}-\theta\dfrac{\Delta L}{L}-\phi\dfrac{\Delta T}{T}$ 

As the error should be both positive and negative , so we can take mod
The % error in X $=|\dfrac{\Delta X}{X}|\times 100$
                            $=\left[\eta\dfrac{\Delta M}{M}+\theta\dfrac{\Delta L}{L}+\phi\dfrac{\Delta T}{T}\right]\times 100$   
                            $=(\eta\alpha+\theta\beta+\phi\gamma) $ %                         

The energy of a system as a function of time t is given as $E(t) = A^2 exp(- \alpha t)$, where $\alpha = 0.2 s^{-1}$. 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:

physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. 2%

  2. 4%

  3. 3%

  4. 5%

Show answer
Correct Option: B
Explanation:

$E(t) = A^2 e^{-\alpha t}$
Taking natural logarithm on both sides,
$ln(E) = 2ln(A) + (- \alpha t)$
Differentiating both sides
$\displaystyle \frac{dE}{E} = 2\frac{dA}{A} + (\alpha dt)$
Errors always add up for maximum error.
$\displaystyle \therefore \frac{dE}{E} = 2\frac{dA}{A} + \alpha \left( \frac{dt}{t} \right) \times t$
Here, $\displaystyle \frac{dA}{A} = 1.25$ %, $\displaystyle \frac{dt}{t} = 1.5$%, $t = 5s$, $\displaystyle \alpha = 0.2 s^{-1}$
$\therefore \displaystyle \frac{dE}{E} = (2 \times 1.25$%$\displaystyle ) + (0.2) \times (1.5$%$) \times 5 = 4$%

The relative error in the determination of the surface area of a sphere is $\alpha$. Then the relative error in the determination of its volume is :

physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. $\cfrac { 2 }{ 3 } \alpha $

  2. $\cfrac { 5 }{ 2 } \alpha $

  3. $\cfrac { 3 }{ 2 } \alpha $

  4. $\alpha $

Show answer
Correct Option: C
Explanation:

S=$4{\pi}R^2$

$\ln { S }$= $\ln ({ 4 {\pi} }) + \ln( { R^2 })$
$\ln { S } = 2\ln { R }$
$\dfrac{\Delta S}{S} = 2 \dfrac{\Delta R}{R} = \alpha$
$\dfrac{\Delta R}{R} = \dfrac {\alpha}{2}$ ------------(1)

V= $\dfrac {4}{3} \pi R^3$
$\ln {V}$ = $\ln ({\dfrac {4}{3} \pi}) + \ln {R^3}$
$\ln {V} = 3 \ln {R}$

$\dfrac{\Delta V}{V} = 3 \dfrac{\Delta R}{R}$

$\dfrac {\Delta V}{V} =3 (\dfrac {\alpha}{2})$

The length  and width of a rectangular room are measured to be $3.95 \pm  0.05 m$ and $3.05 \pm  0.05m$, respectively. The area of the floor is 

physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. $12.05\pm 0.01m^2$

  2. $12.05\pm 0.005m^2$

  3. $12.05\pm 0.34 m^2$

  4. $12.05\pm 0.40m^2$

Show answer
Correct Option: C
Explanation:

Length of the room  $l = 3.95 \pm 0.05 \ m$
Width of the room  $b = 3.05\pm 0.05 \ m$
Absolute area of the room  $A = lb = 3.95\times 3.05 = 12.05 \ m^2$
Error in area of room   $\dfrac{\Delta A}{A} = \dfrac{\Delta l}{l}+\dfrac{\Delta b}{b}$
$\therefore$   $\dfrac{\Delta A}{12.05} = \dfrac{0.05}{3.95}+\dfrac{0.05}{3.05}$
$\implies \ \Delta A = 0.34 \ m^2$
Thus area of the room is written as  $12.05\pm 0.34 \ m^2$ 

If the error in measuring the radius of a sphere is 2%, then the error in the measurement of volume is:

physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. 8%

  2. 6%

  3. 2%

  4. 9%.

Show answer
Correct Option: B
Explanation:
Percentage error in radius is given as $2$% i.e.  $\dfrac{\Delta r}{r}\times 100 = 2$ %
Volume of sphere   $V = \dfrac{4\pi}{3}r^3$
Percentage error in volume   $\dfrac{\Delta V}{V}\times 100 = 3\times \dfrac{\Delta r}{r}\times 100 = 3\times 2 = 6$ %

Two resistors of resistances $R _1$ = (100 $\pm$ 3) $\Omega$ and $R _2$ = (200 $\pm$ 4) $\Omega$ are connected in parallel. The equivalent resistance of the parallel combination is:

physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. (66.7 $\pm$ 1.8) $\Omega$

  2. (66.7 $\pm$ 4.0) $\Omega$

  3. (66.7 $\pm$ 3.0) $\Omega$

  4. (66.7 $\pm$ 7.0) $\Omega$

Show answer
Correct Option: A
Explanation:

Here, $R _1 (100 \pm 3) \Omega; R _2 = (200 \pm 4) \Omega$ The equivalent resistance in parallel combination is

$\displaystyle \frac{1}{R _1} = \frac{1}{R _1} + \frac{1}{R _2}, \frac{1}{R _p} = \frac{1}{100} + \frac{1}{200} = \frac{3}{200}, R _p = \frac{200}{3} = 66.7 \Omega$

The error in equivalent resistance is given by
$\displaystyle \frac {\Delta R _p}{R _p^2} = \frac {\Delta R _1}{R _1^2} + \frac {\Delta R _2}{R _2^2}; \Delta R _p = \Delta R _1 (\frac{R _p}{R _1})^2 + \Delta R _2 (\frac {R _p}{R _2})^2 = 3 (\frac {66.7}{100})^2 + 4 (\frac {66.7}{200})^2 = 1.8 \Omega $ 
Hence, the equivalent resistance along with error in parallel combination is (66.7 $\pm$ 1.8)$\Omega$.