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Errors in measurement - class-XI

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The main scale of a vernier callipers reads 4.7 cm, the 3 rd division on the vernier scale coincides with a main scale division while measuring the length of a rod. The least count of the vernier callipers 0.1 mm. What is the length of the rod?

  1. 4.62 cm

  2. 4.61 cm

  3. 4.72 cm

  4. 4.73 cm


Correct Option: D
Explanation:

$LC = 0.01 mm = 0.01 cm$
Length of the rod $=$ MSR + (VCD $\times$ LC)
$=4.7 cm + (3 \times 0.01)$
$=(4.7 + 0.03) cm$
$=4.73  cm$

A vernier having a positive zero error of +5 is used to measure the side of a 2 cm cube. The length as measure by the vernier will be : (use L.C $=$ 0.01 cm)

  1. 1.95 cm

  2. 2 cm

  3. 2.05 cm

  4. Cannot say


Correct Option: C
Explanation:

Correction $=-0.505 cm (-5 \times 0.01 cm)$
Corrected reading $=$ Observed reading + correction
2 $=$ Observed reading + (-0.05) cm.
Observed reading $=$ 2.05 cm

If the length of a vernier scale having 25 divisions corresponding to 24 main scale divisions and given that 1 MSD $=$ 1 mm, then the least count of vernier calipers is

  1. 0.004 cm

  2. 4 micrometer

  3. 0.04 mm

  4. 0.04 m


Correct Option: A,C
Explanation:

Given
N $=$ 25
25 VSD $=$ 24 MSD
1 MSD $=$ 1mm
$\therefore LC \displaystyle = \frac{MSD}{No.   of   VSD} = \frac{1mm}{25}$
$=0.04 mm$
$=0.04 \times 10^{-1} cm$
$=0.004 cm$
$=0.004 \times 10^{-2} m$
$=40  \mu m$

A vernier has a negative zero error. When the jaws $J _1$ and $J _2$ are brought in contact the zero of the vernier must :

  1. Coincide with the zero of the main scale.

  2. Be to the right of the zero of the main scale.

  3. Be to the left of the zero of the main scale.

  4. None of the above.


Correct Option: C
Explanation:

A vernier caliper is said to have a negative zero error when the zero of the vernier lies to left to the zero of the main scale if both the jaws are brought together.

In vernier callipers, if $L.C = l$ and pitch is P then $\displaystyle \frac{l}{P}$ is :

  1. Always less than 1

  2. Always greater than 1

  3. Equal to 1

  4. None of these


Correct Option: A
Explanation:

L.C $=$ 0.1 mm $=l$
pitch $=1$ mm  $=$ p
$\displaystyle \frac{l}{p} = \frac{0.1}{1} = 0.1  <  1$

The least count of a vernier callipers is 0.01 cm. It has an error of +0.02 cm while measuring the radius of a cylinder. The main scale reading is 3.60 cm and the 8th vernier scale division coincides with main scale, then what will be the correct radius of the cylinder?

  1. 1.72

  2. 1.75

  3. 1.83

  4. 1.87


Correct Option: C
Explanation:

L.C. $=$ 0.01 cm
Error $=$ + 0.02 cm $\Rightarrow$ Positive zero
error $=$ + 0.02 cm
main scale reading (MSR) $=$ 3.60 cm 8$^{th}$ vernier scale coincides with main scale
$\Rightarrow$ vernier coincidence (VC) $=$ 8
We know that
Correct diameter $=$ [main scale reading+ (L.C $\times$ V.C)] - (error)
$=$ [3.60 + (0.01 $\times$ 8)] - [+ 0.002]
$=$ (3.60 + 0.08)- (+0.02)
$=$ 3.68 - 0.02
$=$ 3.66 cm
$\therefore correct radius = \displaystyle \frac{diameter}{2} = \frac{3.66}{2}$
$=1.83 cm$

A liquid of low specific heat capacity is preferred as a thermometric liquid.

  1. True

  2. False

  3. Either

  4. Neither


Correct Option: A

The period of oscillation of a simple pendulum is Given by $ T=2\pi \sqrt { \frac { \ell  }{ g }  }$  where  $\ell$ is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measured by a stop watch of least count 0.1 s. The percentage error in g is:-

  1. 0.1 %

  2. 1 %

  3. 0.2 %

  4. 0.8 %


Correct Option: C

In an experimental set up, the density of a small sphere is to be determined. The diameter of the small sphere is measured with the help of a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the sphere has a relative error of 2%, the relative percentage error in the density is

  1. $0.03$%

  2. $3.11$%

  3. $0.08$%

  4. $8.2$%


Correct Option: A

A shpere has a mass of 12.2 kg $\pm $ 0.1 kg and radius 10 cm $\pm $ 0.1 cm, h=the maximum % error in density is 

  1. 10%

  2. 2.4%

  3. 3.83%

  4. 4.2%


Correct Option: C

A physical quantity $Q$ is calculated according to the expression
$Q =\dfrac{A^3B^3}{C\sqrt D}$
If percentage errors in $A, B, C, D$ are $2\%, 1\%, 3\%$ and $4\%$ respectively. What is the maximum percentage error in $Q$?

  1. $\pm\ 8\%$

  2. $\pm\ 10\%$

  3. $\pm\ 14\%$

  4. $\pm\ 12\%$


Correct Option: C

If maximum percentage errors in measurement of length, mass and time are 2%,1% respectively then maximum percentage error in kinetic energy will be 

  1. 1 %

  2. 4 %

  3. 7 %

  4. 5 %


Correct Option: A

In an experiment measurements of velocity of an object $\left( {in\;m{s^{ - 1}}} \right)$ are 342,33,318 and 322.The mean absolute error in the measurement is 

  1. 8 $m{s^{ - 1}}$

  2. 4 $m{s^{ - 1}}$

  3. 10 $m{s^{ - 1}}$

  4. 12 $m{s^{ - 1}}$


Correct Option: C

Measurement of two physical quantities are given as 
$x = \left( {4.0\; \pm \;0.4} \right)\;m{s^{ - 1}}$
$Y = \left( {1.0\; \pm \;0.1} \right)\;s$
The value of XY will be.

  1. $\left( {4.0\; \pm \;0.8} \right)\;m$

  2. $\left( {4.0\; \pm \;0.5} \right)\;m$

  3. $\left( {4.0\; \pm \;0.3} \right)\;m$

  4. $\left( {4.0\; \pm \;0.4} \right)\;m$


Correct Option: A

The maximum error in the measurement of mass and density of a cube are 3% and 1% respectively. The maximum error in the measurement of volume will be:

  1. 1%

  2. 2%

  3. 3%

  4. 4%


Correct Option: D
Explanation:

$\displaystyle V=\frac {m}{d}$

$\displaystyle \frac {\Delta V}{V}\times 100=(\frac {\Delta m}{m}+\frac {\Delta d}{d})\times 100=(3+1)=4$%

The time period of oscillation of simple pendulum given by $T = 2\pi\sqrt{\frac{L}{g}}$ where $L= (200 \pm 0.1) cm$. The time period, T =4s and the time of 100 oscillations is measured using a stopwatch of least count 0.1 s. The percentage error in g is

  1. $0.1$%

  2. $1.5$%

  3. $2$%

  4. $4$%


Correct Option: A

The velocity $V$ of a body starting from rest and moving with uniform acceleration a is calculated by the formula $V = \sqrt { 2 a s }$. Here $S$ represents the displacement. If error in measurement of acceleration is 4% and error in measurement of displacement is 2%, then the error in calculation of veiocity is 

  1. 2%

  2. 3%

  3. 6%

  4. 8%


Correct Option: B
Explanation:

\begin{array}{l} According\, to\, \, question............... \ V=\sqrt { 2as }  \ error\, \, in\, \, V=\dfrac { 1 }{ 2 } \left( { error\, in\, \, acceleration\, \, +\, error\, \, in\, \, displacement } \right)  \ error\, \, in\, \, V=\frac { 1 }{ 2 } \left( { 4\, \, +\, 2 } \right)  \ \therefore \, \, \, \, error\, \, in\, \, V=\dfrac { 6 }{ 2 } =3 percent\ so\, \, the\, \, \, correct\, \, option\, \, is\, \, B. \end{array}

The current voltage relation of diode is given by $I=\left( { e }^{ 1000V/T }-1 \right)mA$, where the applied voltage $V$ is in kelvin. If a student makes an error measuring $\pm 0.01V$ while measuring the current of $5mA$ at $300k$, what will be the error in the value of current in mA?

  1. $0.2mA$

  2. $0.02mA$

  3. $0.5mA$

  4. $0.05mA$


Correct Option: C

Zero error of an instrument introduces:

  1. Systematic errors

  2. Random errors

  3. Both

  4. None


Correct Option: A
Explanation:

Random errors in experimental measurements are caused by unknown and unpredictable changes in the experiment.
Systematic errors in experimental observations usually come from the measuring instruments.
Thus zero error is recognized as the systematic error (factual).

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

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

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

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

  4. $(\Delta x)^2$


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 :

  1. $2\%$

  2. $3\%$

  3. $6\%$

  4. $1\%$


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 :

  1. 19%

  2. 9%

  3. 17%

  4. 21%


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:

  1. $\pm 13$%

  2. $\pm 9$%

  3. $\pm 6$%

  4. $\pm 11$%


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?

  1. $0.1$ %

  2. $10$ %

  3. $5$ %

  4. $1$ %


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:

  1. 11%

  2. 8%

  3. 5%

  4. 1%


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  

  1. $5$ %

  2. $6$ %

  3. $3$ %

  4. None of these


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?

  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.


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 :

  1. 9%

  2. 19%

  3. 10%

  4. 90%


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

  1. $0.5\%$

  2. $0.8\%$

  3. $0.1\%$

  4. $0.3\%$


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}$
$=0.3\%$

The kinetic energy of a particle depends on the square of speed of the particle. If error in measurement of speed of $30\%$, the error in the measurement of kinetic energy will be

  1. $30\%$

  2. $60\%$

  3. $69\%$

  4. $15\%$


Correct Option: C

If $x=10.0 \pm 0.1$ and $y=10.0 \pm 0.1$, then $2x-2y$ is equal to

  1. $(0.0 \pm 0.1)$

  2. $Zero$

  3. $(0.0 \pm 0.4)$

  4. $(20 \pm 0.2)$


Correct Option: B
Explanation:

Apply formula

$(A\pm \Delta A)-(B\pm \Delta B)=(A-B)\pm (\Delta A-\Delta B)$

Similarly

  $ 2x-2y=2\left( 10.0\pm 0.1 \right)-2\left( 10\pm 0.1 \right) $

 $ =\left( 2\times 10-2\times 10 \right)\pm (2\times 0.1-2\times 0.1) $

 $ =0 $

Hence, $2x-2y=ZERO$ 

A physical quantity $S$ is given by $S = \dfrac {a^{2}b^{3}}{c\sqrt {d}}$.
If errors of measurements in $a, b, c, d$ are $4\%, 2\%, 3\%, 1\%$ respectively, find the percentage error in the value of $S$.

  1. $7.5\%$.

  2. $17.5\%$.

  3. $27.5\%$.

  4. $10.5\%$.


Correct Option: B

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:
$P=\frac{ab^2}{\sqrt{cd^3}}$
Percentage error is P is

  1. $4$%

  2. $7$%

  3. $9$%

  4. $10$%


Correct Option: D
Explanation:

Quantity $P$ is calculated as follow $P = \frac{{a{b^2}}}{{\sqrt {c{d^3}} }}$

Formula of % error should be,
% error in $P=3 \times$%error in $a+2 \times $% error in $b+$%error in $c+$% error in d
Given,
% error in $a=1$%
% error in $b=2$%
% error in $c=3$%
% error in $d=4$%
Hence,
% error in $P=3 \times 1+2 \times 2+3+4$
$=3$%-$4$%+$3$%+$4$%
$=10$%
$\therefore $ Option $D$ is correct answer.

In a relation $ S=\dfrac{b}{b-c}$, where b, c,s are physical quantities, b is $(5.0 \pm 0.1)$ N and c is $(2.0 \pm 0.2)N$ then the percentage error in S is

  1. $12\%$

  2. $2\%$

  3. $24\%$

  4. $6\%$


Correct Option: A

The radius of a sphere is measured as  $ (10 \pm 0.02) $ cm. The error in the measurement of its volume is:

  1. 251 cc

  2. 25.1cc

  3. 2.51 cc

  4. 251.2cc


Correct Option: B
Explanation:

Given that,

$r = 10$

$\Delta r=0.02$


 We know that,

The volume of sphere is

  $ V=\dfrac{4}{3}\pi {{r}^{3}} $

 $ V=\dfrac{4}{3}\times 3.14\times {{\left( 10 \right)}^{3}} $

 $ V=4186.7\,cc $

Now, taking log

$\log V=\log \dfrac{4}{3}+3\log r$

Differentiating on both sides

$\dfrac{\Delta V}{V}=0+3\dfrac{\Delta r}{r}$

Now, the error is

  $ \dfrac{\Delta V}{V}=3\times \dfrac{\Delta r}{r} $

 $ \Delta V=V\times 3\times \dfrac{\Delta r}{r} $

 $ \Delta V=4186.7\times 3\times \dfrac{0.02}{10} $

 $ \Delta V=25.1\,cc $

Hence, the error in the measurement of its volume is $25.1$ cc

The radius and height of a cone are measured as $6cms$ each by scale in which there is an error of $0.01cm$ in each cm. then the approximate error in its volume is.

  1. $.14$

  2. $.12$

  3. $.36$

  4. $0.16$


Correct Option: A
Explanation:
Formula,

$V=\pi r^2 \dfrac{h}{3}$

$=\pi\times 6^2 \dfrac{6}{3}=226.19$

$\dfrac{\Delta V}{V}=\dfrac{\pi}{3}6 \times 0.01 \times 0.01=0.000628$

The change in volume is,

$V=0.000628\times 226.19=0.14$%

In a resonant column method, resonance occurs at two successive levels 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 the relations $v=f\lambda$ and $\lambda =2(l _2-l _1)$?

  1. 256 cm/sec

  2. 92 cm/sec

  3. 128 cm/sec

  4. 102.4 cm/sec


Correct Option: D
Explanation:

The maximum error is given by :


$c=f\lambda = 512 \times 2(l _2-l _1)=512 \times 2 \times (63.2-30.7) = 332.8 \ m/s$

$\dfrac{\Delta c}{c} = \dfrac{\Delta {f}}{f}  + \dfrac{ {\Delta \lambda}}{\lambda}$

Least count of scale $=0.1 cm$ 

total error in measurement of $\lambda$ 

$\Rightarrow 2\times L.C. =0.2 cm$

Total error in measurement of $f$  $\Rightarrow 0$

$\dfrac{\Delta c}{332.8} = \dfrac{0}{f}  + \dfrac{ {0 .2}}{65}$

$\Delta c =  \dfrac{ {0.2}}{65}\times 332.8 =1.024 \ m/s =102.4 \ cm/s$

Given $x=\dfrac {ab^2}{c^3}$, if the percentage errors in a, b and c are $\pm$ 1%, $\pm$ 3% and $\pm$ 2% respectively, the percentage error in $x$ can be:

  1. $\pm$ 13%

  2. $\pm$ 7%

  3. $\pm$ 18%

  4. $\pm$ 19%


Correct Option: A
Explanation:

The given quantity is   $x = \dfrac{ab^2}{c^3}$
Maximum percentage error in $x$ is   $\dfrac{\Delta x}{x}\times 100 = [\dfrac{\Delta a}{a}+2\dfrac{\Delta b}{b}+3\dfrac{\Delta c}{c}]\times 100$
$\dfrac{\Delta x}{x}\times 100 = [1+2\times 3+3\times 2] = \pm 13$ %

The pressure on  square plate is measured by measuring the force on the plate and length of sides of plate. If the maximum error in the measurement of force and length are respectively $4$% and $2$%, the maximum error in measurement of pressure is..........

  1. 1%

  2. 2%

  3. 6%

  4. 8%


Correct Option: D

A physical quantity P is repleted to four observed a,b,c,d as follows: $P=\dfrac{a^3b^2}{\left(\sqrt c.d\right)}$ The percentage errors in the measurement of a,b,c and d are $1\%3\%,4\%$ and $2\%$ respectively. The percentage error in the quantity P is

  1. $7\%$

  2. $9\%$

  3. $11\%$

  4. $13\%$


Correct Option: D
Explanation:
$P=\dfrac { { a }^{ 3 }{ b }^{ 2 } }{ \sqrt { cd }  } $

$\dfrac { \Delta P }{ P } =\dfrac { 3\Delta a }{ a } +\dfrac { 2\Delta b }{ b } +\dfrac { 1 }{ 2 } \dfrac { \Delta c }{ c } +\dfrac { \Delta d }{ d } $

$\left( \dfrac { \Delta P }{ P } \times 100 \right) $% $=\left( 3\times \dfrac { \Delta a }{ a } \times 100+2\times \dfrac { \Delta b }{ b } \times 100+\dfrac { 1 }{ 2 } \times \dfrac { \Delta c }{ c } \times 100+\dfrac { \Delta d }{ d } \times 100 \right) $%
                              $=3\times 1+2\times 3+\dfrac { 1 }{ 2 } \times 4+2$
                              $=3+6+2+2=13$%
Percentage error $=13$%
Unaware about the fact that analog ammeters and voltmeters can also have zero error, a student recorded following readings while determine resistance by Ohm's law 
If the ammeter has no zero error, the zero error is the voltmeter is 
 Obs, No. Voltage/V  Current/mA   Obs No.  Voltage/V  Current/mA
 1  1.0  40  4  7.0  160
 2  3.0  80  5  9.0  200
 3  5.0  120      
  1. $-1 V$

  2. $-1.5V$

  3. $0.5V$

  4. $1V$


Correct Option: A

A physical quantity $Q$ is released to four observable $x,y,z$ and $t$  by the relation
$Q=\dfrac {x^{2/5}z^{3}}{y\sqrt {t}}$ 
The percentage errors of measurement in $x,y,z$ and $t$ are $2.5\%,2\%,0.5\%$ and $1\%$ receptively. The percentage error in $Q$ will be

  1. $5\%$

  2. $4.5\%$

  3. $8\%$

  4. $7.75\%$


Correct Option: A

A steel tape is calibrated at $20 ^ { \circ } \mathrm { C }$. On a cold day when the temperature is $- 15 ^ { \circ } \mathrm { C }$ percentage error in the tape will be $\left[ \alpha _ { \text { steel } } = 1.2 \times 10 ^ { - 5 \cdot 0 } C ^ { - 1 } \right]$

  1. $0.035 \%$

  2. $0.042 \%$

  3. $0.012 \%$

  4. $0.018 \%$


Correct Option: B

The random error in the arithmetic mean of 100 observations is x, then random error in the arithmetic mean of 400 observation would be

  1. 4x

  2. $\dfrac { 1 }{ 4 } x$

  3. 2x

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


Correct Option: B

A  faulty thermometer has its fixed point marked 5${ \circ  } _{ C }$ and 95${ \circ  } _{ C }$. this thermometer reads the temperature of a body as ${ 59 }^{ \circ  }$. The correct temperature on Celsius scale is  

  1. 59 ${ \circ } _{ C }$

  2. 8.6 ${ \circ } _{ C }$

  3. 65.56 ${ \circ } _{ C }$

  4. 58 ${ \circ } _{ C }$


Correct Option: C

If the error in measurement of momentum of a particle is $10$% and mass is known exactly,the permissible error in the determination of kinetic energy is

  1. $20$%

  2. $5$%

  3. $21$%

  4. $10$%


Correct Option: A

Length of a thin cylinder as measured by vernier callipers having least cound $0.01\ cm$ is $3.25\ cm$ and its radius of cross-section is measured by a screw gauge having least count $0.01\ mm$ as $2.75\ mm$. The percentage error in the measurement of volume of the cylinder will be

  1. $2\%$

  2. $3\%$

  3. $1\%$

  4. $1.5$


Correct Option: B

If the error in the measurement of radius of a sphere is 1%, then the error in the measurement of volume will be :

  1. 8%

  2. 5%

  3. 3%

  4. 1%


Correct Option: C
Explanation:

We know that a sphere has volume $V=\dfrac{4}{3}\pi r^3$ meaning $V \alpha r^{3}$.
Thus, error in measurement of volume will be $ 3 \times$ 1% = 3%.

The speed of a body is measured with a positive error of 30%. If the mass of the body is known exactly, the kinetic energy is calculated with an error of

  1. $60$%

  2. $30$%

  3. $69$%

  4. $39$%


Correct Option: A

While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period. His per-centre error in the n=measurement of g by the relation $g={ 4\pi  }^{ 2 }\left( I/T^{ 2 } \right) $ will be 

  1. 2%

  2. 4%

  3. 7%

  4. 10%


Correct Option: C

If ${ R } _{ 1 }=600\Omega \pm 1%$ and ${ R } _{ 2 }=600\Omega \pm 2%$. Find % error in the calculation of equivalent resistance when these two are connected in parallel.

  1. 0.4%

  2. 0.8%

  3. 1.6 %

  4. 3.2%


Correct Option: A

A particle covers a distance of $ (13.8 \pm 0.2) \mathrm{m}  $ in$ (4 \pm 0.3)  $ seconds. Its velocity under error limits will be :-

  1. $3.45$ $ \pm 0.5 \mathrm{ms}^{-1} $

  2. $3.5$ $ \pm 0.3 \mathrm{ms}^{-1} $

  3. $6.1$ $ \pm 0.6 \mathrm{ms}^{-1} $

  4. $6.1$ $ \pm 0.3 \mathrm{ms}^{-1} $


Correct Option: A

A faulty thermometer has its fixed point marked $5C$ and $95 C$. This thermometer reads the temperature of a body as $59^o$. The correct temperature on Celsius scale is 

  1. $64^{0}$

  2. $54^{0}$

  3. $60^{0}$

  4. $68^{0}$


Correct Option: C
Explanation:
For two scales of thermometer,

$\dfrac{C-0}{100-0}=\dfrac{Reading}{UPF-LPF}-LPF$

where, $LPF$ and $UPF$ are the lower and upper fixed points on the scale.

Therefore

$\dfrac{C}{100}=\dfrac{59-5}{95-5}=\dfrac{54}{90}=0.6$

$\implies C=0.6\times 100=60^0$

A resistor of $10 k\Omega$ having tolerance 10% is connected in series with another resistor of $20k\Omega$ having tolerance 20%. The tolerance of the combination will be:

  1. 10%

  2. 13%

  3. 30%

  4. 20%


Correct Option: C
Explanation:

In series effective resistance $=R _S=(10k\Omega \pm 10$%)+$(20k\Omega \pm 20$%)$=(30k\Omega \pm 30$%)
$\therefore$ Tolerance of the combination $=( \pm 30$%)

What is the fractional error in g calculated from $T=2\pi \sqrt {l/g}$?

Given fraction errors in T and l are $\pm$ x and $\pm $ y respectively.

  1. x+y

  2. x-y

  3. 2x+y

  4. 2x-y


Correct Option: C
Explanation:

From $T=2\pi \sqrt {\dfrac {l}{g}};g=4\pi^2\dfrac {l}{T^2}$

$\dfrac {\Delta g}{g}=\dfrac {\Delta l}{l}+\dfrac {2\Delta T}{T}=(y+2x)$

A student performs an experiment for determination of $g\left (=\dfrac {4\pi^2l}{T^2}\right )$. The error in length $l$ is $\Delta l$ and in time $T$ is $\Delta T$ and $n$ is a number of times the reading is taken. The measurement of $g$ is most accurate for :

  1. 5 mm, 0.2 sec, $n=$10

  2. 5 mm, 0.2 sec, $n=$20

  3. 5 mm, 0.1 sec, $n=$10

  4. 1 mm, 0.1 sec, $n=$50


Correct Option: D
Explanation:

Given: $\displaystyle g=\frac {4\pi ^2l}{T^2}$
$\displaystyle \frac {\Delta g}{g}=\frac {\Delta l}{l}+2\frac {\Delta T}{T}$

Error in $g$ that is $\Delta g$ will be minimum for minimum $\Delta l$ and $\Delta T$ and more number of readings.

The error in the measurement of the radius of a sphere is $0.6$%. What is permissible error in its volume?

  1. 0.6%

  2. 1%

  3. 1.2%

  4. 1.8%


Correct Option: D
Explanation:

$V=\displaystyle \frac {4}{3}\pi r^3$

$\displaystyle \frac {\Delta V}{V}\times 100=\frac {3\Delta r}{r}\times 100=3(0.6)=1.8%$

The heat generated in a circuit is dependent upon the resistance, current and time for which the current is flown. If the error in measuring the above are 1%, 2% and 1% respectively. The maximum error in measuring the heat is :

  1. 8%

  2. 6%

  3. 18%

  4. 12%


Correct Option: B
Explanation:

$ Q = I^2 R t$


$ \dfrac{\delta Q}{ Q }= 2\dfrac{\delta i}{i}+ \dfrac{\delta R}{R}+\dfrac{\delta t}{t}$

$ \dfrac{\delta Q}{ Q }= 2\times 2$ % $+ 1$%$+1$%

$ \dfrac{\delta Q}{ Q }=6$ % 

A physical quantity A is dependent on other four physical quantities p, q, r and s as given by $\displaystyle A= \frac{\sqrt{pq}}{r^{2}s^{3}}.$ The percentage error of measurement in p, q, r and s are 1%, 3%, 0.5% and 0.33% respectively, then the maximum percentage error in A is :

  1. 2%

  2. 0%

  3. 4%

  4. 3%


Correct Option: C
Explanation:

Using formula for error analysis:
$\dfrac{ \Delta A}{A} = \dfrac{1}{2} \dfrac{ \Delta p}{p}+ \dfrac{1}{2} \dfrac{ \Delta q}{q} + 2  \dfrac{ \Delta r}{r} + 3 \dfrac{ \Delta s}{s}= \dfrac{1}{2} \times 1 + \dfrac{1}{2} \times 3 + 2 \times 0.5 + 3 \times 0.33= 4$
Hence, percentage of error is 4%

If error in measurement of radius of a sphere is 1%, what will be the error in measurement of volume?

  1. 1%

  2. $ \dfrac{1}{3}$ %

  3. 3%

  4. 10%


Correct Option: C
Explanation:

The volume is given by $V=\dfrac{4}{3} \pi R^3$, where $R$ is radius of sphere.


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

$\dfrac{\delta V}{V}= 3 \times 1 $ % $=3 $ %

Percentage error in the measurement of mass and speed are 2% and 3% respectively. The error in the estimate of kinetic energy obtained by measuring mass and need will be

  1. 12%

  2. 10%

  3. 8%

  4. 2%


Correct Option: C
Explanation:

Given percentage errors in mass and velocity = $2$% and $3$% respectively


The kinetic energy of the particle is given by 


$KE$ = $\dfrac { m{ v }^{ 2 } }{ 2 } $

The percentage error in kinetic energy will be

$\dfrac { \Delta m }{ m } \times 100\quad +\quad 2\dfrac { \Delta v }{ v } \times 100$

= $2 + 6$

= $8$% 

If $X=a-b$, then the maximum percentage error in the measurement of $x$ will be:

  1. $\left (\dfrac {\Delta a}{a}+\dfrac {\Delta b}{b}\right )\times 100\%$

  2. $\left (\dfrac {\Delta a}{a}-\dfrac {\Delta b}{b}\right )\times 100\%$

  3. $\left (\dfrac {\Delta a}{a-b}+\dfrac {\Delta b}{a-b}\right )\times 100\%$

  4. $\left (\dfrac {\Delta a}{a-b}-\dfrac {\Delta b}{a-b}\right )\times 100\%$


Correct Option: C
Explanation:

Maximum absolute error is $\Delta a+\Delta b$.
Therefore the percentage error $=\dfrac {\text {absolute error}}{\text {actual error}}\times 100$
$\therefore$ Percentage error $= \left (\dfrac {\Delta a}{a-b}+\dfrac {\Delta b}{a-b}\right )\times 100$%

The error in the measurement of the radius of the spheres by using vernier calipers is 0.3%. The permissible error in the measurement of surface area is:

  1. 0.6 %

  2. 1.2 %

  3. 1.8 %

  4. 0.9 %


Correct Option: A
Explanation:

Since, $A\propto r^2$

$\dfrac{\Delta A}{A}=\dfrac{2\Delta r}{r}= 0.6$%

Find the percentage error in specific resistance given by $\displaystyle \rho=\frac{\pi r^{2}R}{l}$ where r is the radius having value $\displaystyle \left ( 0.2\pm 0.02 \right )$ cm, R is the resistance of $\displaystyle \left (60\pm 2 \right )\Omega $ and l is length of $\displaystyle \left ( 150\pm 0.1 \right )$ cm.

  1. 5.85%

  2. 11.7%

  3. 23.4%

  4. 35.1%


Correct Option: C
Explanation:

Applying logarithm on both sides of the given expression and differentiating, 

we get $\dfrac { \Delta \rho  }{ \rho  } =\pm (2\dfrac { \Delta r }{ r } +\dfrac { \Delta l }{ l } +\dfrac { \Delta R }{ R } )$
Given : $\Delta r$=0.02cm, $\Delta R$=2 ohm, $\Delta l$=0.1cm


Substituting the values in above expression,
$\dfrac { \Delta \rho  }{ \rho  } =\pm (2\dfrac { 0.02 }{ 0.2 } +\dfrac { 0.1 }{ 150 } +\dfrac { 2 }{ 60 } )=0.234=23.4$%

The density of a cube is measured by measuring its mass and the length of its side. If the maximum errors in the measurement of mass and length are 4% and 3% respectively, the maximum error in the measurement of the density is

  1. 9%

  2. 13%

  3. 12%

  4. 7%


Correct Option: B
Explanation:

$Density, \rho = \dfrac{Mass}{Volume}=\dfrac{M}{L^3}$
Thus, maximum error, $\dfrac{\Delta \rho}{\rho}=\dfrac{\Delta M}{M}+3\dfrac{\Delta L}{L}=0.04+3\times 0.03$$= 0.13$ or $13\%$

The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate. If the maximum error in the measurement of force and length are respectively 4% and 2%, the maximum error in the measurement of pressure is:

  1. 1%

  2. 2%

  3. 6%

  4. 8%


Correct Option: D
Explanation:

Since $area = l^2$ therefore, error in measurement of area is twice the error in measurement of length. 

Therefore, error in measurement of pressure is error in measurement of force + error in measurement of area
$= 4+2\times 2= 8$%

If error in measuring diameter of a circle is 4%, the error in the radius of the circle would be ? 

  1. 2%

  2. 8%

  3. 4%

  4. 1%


Correct Option: C
Explanation:

Since $D\propto r$ therefore, there will be no any change in error it will remain 4% for radius also. (C)

To estimate 'g' (from g = $4\, \pi^2\, \displaystyle \frac{L}{T^2}$), error in measurement of L is $\pm 2\, \%$ and error in measurement of T is $\pm\, 3\, \%$. The error in estimated 'g' will be

  1. $\pm \, 8\, \%$

  2. $\pm \, 6\, \%$

  3. $\pm \, 3\, \%$

  4. $\pm \, 5\, \%$


Correct Option: A
Explanation:
Acceleration due to gravity   $g = \dfrac{4\pi^2 \ L}{T^2}$
Percentage error in $g$ is   $\dfrac{\Delta g}{g}\times 100 = \dfrac{\Delta L}{L}\times 100 +\times \dfrac{\Delta T}{T}\times 100$
$\implies \ \dfrac{\Delta g}{g}\times 100 = 2+2\times 3 = \pm 8$%

The length, breadth and thickness of a strip are (10.0 $ \pm $ 0.1)cm, (1.00 $ \pm $ 0.01)cm and

(0.100 $ \pm $ 0.001)cm respectively. The most probable error in its volume will be ?

  1. $ \pm 0.03 {cm}^3 $

  2. $ \pm 0.111 {cm}^3 $

  3. $ \pm 0.012 {cm}^3 $

  4. None of these


Correct Option: A
Explanation:

Since $V=lbw$

$\Delta V/V=\Delta l/l+\Delta b/b+ \Delta w/w=0.01+0.01+0.01=0.03  cm^3$

If error in measuring diameter of a circle is 4 %, the error in circumference of the circle would be :

  1. 2%

  2. 8%

  3. 4%

  4. 1%


Correct Option: C
Explanation:

Since $C\propto d$ therefore error in measurement of $C$ is same as error in measurement of $d$.

Hence, option (C) is correct.

The external and internal radius of a hollow cylinder are to be measured to be (4.23 $\pm$ 0.01)cm and (3.89 $\pm$ 0.01)cm. The thickness of the wall of the cylinder is :

  1. (0.34 $\pm$ 0.02) cm

  2. (0.17 $\pm$ 0.02)cm

  3. (0.17 $\pm$ 0.00)cm

  4. (0.34 $\pm$ 0.00)cm


Correct Option: A
Explanation:

Thickness $= R _{ext}-R _{int}=4.23-3.89=0.34$

Now, $\Delta Thickness = (\Delta R _{ext}/R _{ext}+\Delta R _{int}/R _{int})\times Thickness=0.02$

The length of a cylinder is measured with a metre scale having least count 0.1 cm. Its diameter is measured with vernier calipers having least count 0.01cm. Given the length is 5.0 cm and diameter is 2.00cm. The percentage error in the calculated value of volume will be:

  1. 2%

  2. 1%

  3. 3%

  4. 4%


Correct Option: C
Explanation:

Volume of cylinder $\displaystyle (V) = \pi r^2 l $

$\displaystyle \frac {\Delta V} {V} \times 100 =2\times  \frac {\Delta r} {r} \times 100 + \frac {\Delta \ell } {\ell} \times 100 $

                     $\displaystyle = 2 \times \frac {0.01} {2} \times 100 + \frac {0.1} {5} \times 100 \\ = 1 + 2 \\ = 3\% $

An experiment measures quantities x, y, z and then t is calculated from the data as $t\, =\, \displaystyle \frac{xy^2}{z^3}$. If percentage errors in x, y and z are respectively 1 %, 3 %, 2 %, then percentage error in t is

  1. 10 %

  2. 4 %

  3. 7 %

  4. 13 %


Correct Option: D
Explanation:
The given quantity is   $t = \dfrac{xy^2}{z^3}$
Percentage error in $t$ is given by    $\dfrac{\Delta t}{t}\times 100  =1\times(\dfrac{\Delta x}{x}\times 100)+2(\dfrac{\Delta y}{y}\times 100)+3(\dfrac{\Delta z}{z}\times 100)$
$\implies \ \dfrac{\Delta t}{t}\times 100 = 1\times 1+2\times 3+3\times 2 = 13$%

The length and breadth of a rectangular object are 25.2 cm and 16.8 cm respectively and have been measured to an accuracy of 0.1 cm. The relative error and percentage error in the area of the object are:

  1. 0.01 ; 1%

  2. 0.1 ; 10%

  3. 1 ; 10 %

  4. 1 ; 100 %


Correct Option: A
Explanation:

$Area  = l\times b$

$\dfrac { \triangle A }{ A } =\dfrac { \triangle l }{ l } +\dfrac { \triangle b }{ b }$


           $ = (\dfrac{0.1}{25.2} + \dfrac{0.1}{16.8})=0.00992\approx0.01$


This is the relative error.

Percentage  error $ =  0.01 \times 100  =  1\%$

The percentage error in the measurement of a quantity Z which is related to two other quantities as Z = $\displaystyle x^{-1}y^{+1}$ is due to the percentage error in the measurement of x and y which are 2% and 1% respectively. Find the maximum fractional error in Z (in %).

  1. 3

  2. 4

  3. 5

  4. 6


Correct Option: A
Explanation:

The quantity is given as      $Z = \dfrac{y}{x}$

$\therefore      \dfrac{\Delta Z}{Z}  \times 100  =  \dfrac{\Delta x}{x} \times 100   +  \dfrac{\Delta y}{y}   \times 100$
$\implies         \dfrac{\Delta Z}{Z}  \times 100  =  2+1  = 3$
Thus percentage error in Z is equal to  $3$%.

The length of a pendulum is measured as $1.01$ m and time for $30$ oscillations is measured as $1$ minute $3$ seconds. Error in length is $0.01$ m and error in time is $3$ seconds. The percentage error in the measurement of acceleration due to gravity is:

  1. $1$

  2. $5$

  3. $10$

  4. $15$


Correct Option: C
Explanation:

$T=2\pi\sqrt{\dfrac{l}{g}}$

$\implies g=\dfrac{4\pi^2 l}{T^2}$

Thus $\dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+2\dfrac{\Delta T}{T}$

Percentage error in measurement of: 
$g=\dfrac{\Delta g}{g}\times 100=(\dfrac{\Delta l}{l}+2\dfrac{\Delta T}{T})\times 100$
   $=(\dfrac{0.01}{1.01}+2\dfrac{3}{63})\times 100$% $=10$%

A students performs an experiment to determine the acceleration due to gravity (g) at a place using a simple pendulum. The length of the pendulum is 60 cm and the total time for 30 oscillations is 100s. What is maximum percentage error for the measurement g ? Given, least count for time $=0.1 s$ and least count for length $=0.1 cm$. 

  1. $0.26 \%$

  2. $0.3 \%$

  3. $0.36 \%$

  4. $3.6 \%$


Correct Option: C
Explanation:

For a pendulum, the time period is: $T=2\pi\sqrt{\dfrac{l}{g}}$ 
$\Rightarrow g=\dfrac{4\pi^2 l}{T^2}=\dfrac{4\pi^2 ln^2}{t^2}$

Take $ln$ and then differentiate:
$\dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+2\dfrac{\Delta t}{t}$   (as n is constant so its derivative will be zero)

The % error in g $=\dfrac{\Delta g}{g}\times 100=\dfrac{\Delta l}{l}\times 100+2\dfrac{\Delta t}{t}\times 100=\dfrac{0.1}{60}\times 100+2\dfrac{0.1}{100}\times 100=0.36 \%$

A uniform wire of radius $r=0.5 \pm 0.005 cm$ length $l =5\pm 0.05 cm $. The maximum percentage error in its volume is

  1. $30 \%$

  2. $3 \%$

  3. $2 \%$

  4. $1.5 \%$


Correct Option: B
Explanation:

The volume of the wire is $V=\pi r^2 l$
Thus, the relative error, $\dfrac{\Delta V}{V}=\pm \left(2\dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}\right)$
The maximum % error in V  $=\dfrac{\Delta V}{V}\times 100= \left(2\dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}\right)\times 100=\left(2\dfrac{0.005}{0.5}+\dfrac{0.05}{5}\right)100=3\%$

The pressure on  a circular plate is measured by measuring force on the plate and the radius of the plate. If the errors in measurement  of the force and the radius are $5$% and $3$% respectively, the percentage of error in the measurement of pressure is:

  1. $8$

  2. $14$

  3. $11$

  4. $12$


Correct Option: C
Explanation:

$P=\dfrac{F}{A}=\dfrac{F}{\pi R^2}$

$\Rightarrow \dfrac{\Delta P}{P}\times 100=\dfrac{\Delta

F}{F}\times 100+2\dfrac{\Delta R}{R}\times 100$

                          $=5+2(3)=5+6=11$

$\Rightarrow \dfrac{\Delta P}{P}\times 100=11$%

The radius of curvature of a concave mirror measured by a spherometer is given by $R=\dfrac{l^2}{6h}+\dfrac{h}{2} $. The measured value of $l$ is $3 cm$ using a meter scale with least count $0.1 cm $ and measured value of  $h $ is $ 0.045 cm$ using spherometer with least count $0.005 cm$. Compute the relative error in measurement of radius of curvature. 

  1. $3$

  2. $0.3$

  3. $0.2$

  4. $0.6$


Correct Option: B
Explanation:

Given, $l=3 cm, \Delta l=0.1 cm,  h=0.045 cm, \Delta h=0.005 cm$ 
Now, $R=\dfrac{l^2}{6h}+\dfrac{h}{2} $

Take $ln$ and differentiate (only be taken magnitude)
so, relative error , $\dfrac{\Delta R}{R}=2\dfrac{\Delta l}{l}+\dfrac{\Delta h}{h}+\dfrac{\Delta h}{h}=2\dfrac{0.1}{3}+2\dfrac{0.005}{0.045}=0.3$

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. 

  1. $a/L+b/W$

  2. $aL+bW$

  3. $aW+bL$

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


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.  

  1. $0.12$

  2. $0.02$

  3. $0.10$

  4. $0.01$


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 

  1. $3.33 $%

  2. $4.03 $%

  3. $3.0 $%

  4. $4.33 $%


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

  1. $10.0Kg$

  2. $10.0\pm 0.1Kg$

  3. $10.0\pm 0.3Kg$

  4. $10.0\pm 0.4Kg$


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

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


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:

  1. 2%

  2. 4%

  3. 3%

  4. 5%


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 :

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

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

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

  4. $\alpha $


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 

  1. $12.05\pm 0.01m^2$

  2. $12.05\pm 0.005m^2$

  3. $12.05\pm 0.34 m^2$

  4. $12.05\pm 0.40m^2$


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:

  1. 8%

  2. 6%

  3. 2%

  4. 9%.


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:

  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) $\Omega$


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)$\Omega$.

The percentage errors in quantities $P, Q, R$ and $S$ are $0.5$%, $1$%, $3$% and $1.5$% respectively in the measurement of a physical quantity $A = \dfrac {P^{3}Q^{2}}{\sqrt {R}S}$.
The maximum percentage error in the value of $A$ will be

  1. $8.5\%$

  2. $6.0\%$

  3. $7.5\%$

  4. $6.5\%$


Correct Option: D
Explanation:

Percentage error $\dfrac{\Delta A}{A} \times 100 = 3 \dfrac{\Delta P}{P} \times 100 + 2 \dfrac{\Delta Q}{Q} + \dfrac{1}{2} \dfrac{\Delta R}{R} \times 100 + \dfrac{\Delta S}{S} \times 100$ 

$ = 3 \times 0.5 + 2 \times 1+ 0.5 \times 3 + 1.5 = 6.5 %$

The percentage errors in the measurement of length and time period of a simple pendulum are 1% and 2% respectively. Then, the maximum error in the measurement of acceleration due to gravity is:

  1. $3%$

  2. $4%$

  3. $6%$

  4. $5%$


Correct Option: D

The dimensions of a rectangular block measured with callipers having least count of 0.01 cm are 5mm x 10mm x 5 mm. The maximum percentage error in the measurement of the volume of the block is

  1. 5%

  2. 10%

  3. 15%

  4. 20%


Correct Option: A
Explanation:
Given :  $l = 10 \ mm = 1 \ cm$     $b = 5 \ mm = 0.5 \ cm$        $h = 5 \ mm  = 0.5 \ cm$
Error in the measurements   $\Delta l = \Delta b = \Delta h = 0.01 \ cm$
Maximum percentage error in volume   $\dfrac{\Delta V}{V}\times 100 = \dfrac{\Delta l}{l}\times 100 +\dfrac{\Delta b}{b}\times 100+\dfrac{\Delta h}{h}\times 100$
$\implies \ \dfrac{\Delta V}{V}\times 100 = \dfrac{0.01}{1}\times 100+\dfrac{0.01}{0.5}\times 100+\dfrac{0.01}{0.5}\times 100 = 5$ %
Correct answer is option A.

A force $\vec { F } $ is applied on a square plate of length $L$. If the percentage error in the determination of $L$ is $3$% and in $F$ is $4$%, the permissible error in the calculation of pressure is

  1. $13$%

  2. $10$%

  3. $7$%

  4. $12$%


Correct Option: B
Explanation:

Pressure  $P=\cfrac { F }{ A } =\cfrac { F }{ { L }^{ 2 } } $
Percentage error in pressure  $\cfrac { \Delta P }{ P } \times 100=\cfrac { \Delta F }{ F } \times 100+2\cfrac { \Delta L }{ L } \times 100$
$ = 4+2\times 3 = 10$  %

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