Tag: measurements and experimentation

Questions Related to measurements and experimentation

If $f=x^2$, then the relative error in $f$ is :

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

  1. $\dfrac {2\Delta x}{x}$

  2. $\dfrac {(\Delta x)^2}{x}$

  3. $\dfrac {\Delta x}{x}$

  4. $(\Delta x)^2$

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Correct Option: A
Explanation:

Given that: $f = x^2$
Hence, $\dfrac{df}{dx} = 2x$

Therefore: $\Delta f = \dfrac{df}{dx} \Delta x = 2x \Delta x$

The relative error in $f$ is: 
$\dfrac{\Delta f}{f} = \dfrac{2x \Delta x}{x^2}$

$\dfrac{\Delta f}{f} = \dfrac{2 \Delta x}{x}$

The heat generated in a circuit is dependent on the resistance, current and time of flow of electric current. If the percentage errors measured in the above physical quantities are 1%, 2% and 1% respectively, the maximum error in measuring the heat is :

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

  1. $2\%$

  2. $3\%$

  3. $6\%$

  4. $1\%$

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Correct Option: C
Explanation:


$ H={ I }^{ 2 }RT$ (by dimensional analysis)
$\displaystyle \frac { \triangle n }{ n } =(2\frac { \triangle I }{ I } +\frac { \triangle R }{ R } +\frac { \triangle T }{ T } )$%
          $=2(2)+1+1$
          $= 6 $ %

The dimensional formula for a physical quantity $X$ is ${ M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 }. $The errors in measuring the quantities $M, L$ and $T$ respectively are 2%, 3%, and 4%. The maximum percentage error that occurs in measuring the quantity $X$ is :

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

  1. 19%

  2. 9%

  3. 17%

  4. 21%

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Correct Option: A
Explanation:

$ [X] = [M^{-1} L^{3} T^{-2}]$

$ \Rightarrow 
Error  = (\dfrac{\triangle M}{M} +
3 \dfrac{\triangle L}{L} + 2 \dfrac{\triangle T}{T}) %  $

$ \Rightarrow 2+3\times (3) +2\times (4)$

$\Rightarrow   $ 19  %

The percentage errors in a, b, c are $\pm 1$%, $\pm 3$% and $\pm 2$% respectively. The percentage error in $\dfrac{a^2}{bc^3}$ is:

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

  1. $\pm 13$%

  2. $\pm 9$%

  3. $\pm 6$%

  4. $\pm 11$%

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Correct Option: D
Explanation:

Let $Y=\dfrac { { a }^{ 2 } }{ b{ c }^{ 3 } } $
$\displaystyle \frac { \Delta Y }{ Y } \times 100=\pm \left( 2\times \frac { \Delta a }{ a } \times 100+\frac { \Delta b }{ b } \times 100+3\times \frac { \Delta c }{ c } \times 100 \right) $
                    $=\pm \left( 2\times 1+3+3\times 2 \right)$

                    $ =\pm 11$

The error in the measurement of the radius of a sphere is $0.5$ %. Find the permissible error in the measurement of surface area?

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

  1. $0.1$ %

  2. $10$ %

  3. $5$ %

  4. $1$ %

Show answer
Correct Option: D
Explanation:

The percentage error in measurement of radius is given, $\dfrac{\Delta r}{r}\times 100 =0.5$%
Thus, $\dfrac{\Delta r}{r}=0.5/100=0.005$
The surface area of sphere is $S=4\pi r^2$
Take ln and differentiate, $\dfrac{\Delta S}{S}=2\dfrac{\Delta r}{r}=2\times 0.005=0.01 $
The permissible or % error in the measurement of surface area $=\dfrac{\Delta S}{S}\times 100=0.01\times 100=1$%

The percentage error in the measurement of mass and speed are 2% and 3% respectively. The maximum percentage error in the estimation of kinetic energy of a body will be:

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

  1. 11%

  2. 8%

  3. 5%

  4. 1%

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Correct Option: B
Explanation:

$K.E = \dfrac{1}{2}mv^{2}$


$ \dfrac{\triangle K.E}{K.E} =

\dfrac{\triangle m}{m} + 2 \dfrac{\triangle v}{v}$

              $ = 2+2(3) = 8$%

Given $A=\dfrac{x^p}{y^qz^r}$. The percentage errors in measurement of $x, y$ and $z$ are 1 %, 0.5 % and 2% respectively. If $p=3, q=2$ and $r=1$ then the maximum percentage error in A is  

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

  1. $5$ %

  2. $6$ %

  3. $3$ %

  4. None of these

Show answer
Correct Option: B
Explanation:

Given, $A=\dfrac{x^p}{y^qz^r}$
Take ln and then differentiate , $\dfrac{\Delta A}{A}=\pm\left(p\dfrac{\Delta x}{x}+q\dfrac{\Delta y}{y}+r\dfrac{\Delta z}{z}\right)$
Thus the maximum percentage error in A:
$\dfrac{\Delta A}{A}\times 100=p\dfrac{\Delta x}{x}\times 100+q\dfrac{\Delta y}{y}\times 100+r\dfrac{\Delta z}{z}\times 100=(3\times 1)+(2\times 0.5)+(1\times 2)=6$ %

The diameter of an iron rod is given by $(44.42\pm 0.03)$ mm. What does it mean?

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

  1. True value of diameter can be only $44.42$ mm

  2. The relative error in diameter can be $0.03 $

  3. The percentage error in diameter can be $0.03$

  4. True value of diameter can not more than $44.45$ mm and less than $44.39$ mm.

Show answer
Correct Option: D
Explanation:

Here, diameter $d=44.42 $mm and $\Delta d=\pm 0.03 $mm
Thus, true value of diameter will be in between $(44.42+0.03)=44.45$ mm and$(44.42-0.03)=44.39$ mm.
The relative error $=|\dfrac{\Delta d}{d}|=\dfrac{0.03}{44.42}=6.75\times 10^{-4}$ 
The percentage error $=|\dfrac{\Delta d}{d}|\times 100=0.07 $%

The density of a cube can be measured by measuring its mass and the length of its side. If the maximum errors in the measurement of mass and length are 3% and 2% respectively, the maximum error in the measurement of the density of the cube is :

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

  1. 9%

  2. 19%

  3. 10%

  4. 90%

Show answer
Correct Option: A
Explanation:

$ d = ML^{-3}\rightarrow  (By \   dimensional \  analysis)$


$\displaystyle \frac{\triangle d}{d} = \frac{\triangle M}{M}+ 3\frac{\triangle L}{L}$
          $ = 3+3(2)$
          $ = 9$%   

The error in the measurement of radius of a sphere is $0.1\%$ 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. $0.5\%$

  2. $0.8\%$

  3. $0.1\%$

  4. $0.3\%$

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Correct Option: D
Explanation:

$V=\cfrac{4}{3}\pi R^3$

$dV=3\cfrac{4}{3}\pi R^2 dR$
$ \cfrac{dV}{V}=3\cfrac{dR}{R}$
$=0.3\%$