Tag: vernier calliper and screw gauge

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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%
Reveal answer Fill a bubble to check yourself
C Correct answer
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\% $

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

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 %
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The given quantity is   $t = \dfrac{xy^2}{z^3}$
Percentage error in $t$ is given by    $\dfrac{\Delta t}{t}\times 100  =1\times(\dfrac{\Delta x}{x}\times 100)+2(\dfrac{\Delta y}{y}\times 100)+3(\dfrac{\Delta z}{z}\times 100)$
$\implies \ \dfrac{\Delta t}{t}\times 100 = 1\times 1+2\times 3+3\times 2 = 13$%
Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

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 %
Reveal answer Fill a bubble to check yourself
A Correct answer
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\%$

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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  $3$%.

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

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. $15$
Reveal answer Fill a bubble to check yourself
C Correct answer
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: 
$g=\dfrac{\Delta g}{g}\times 100=(\dfrac{\Delta l}{l}+2\dfrac{\Delta T}{T})\times 100$
   $=(\dfrac{0.01}{1.01}+2\dfrac{3}{63})\times 100$% $=10$%

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

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. $3.6 \%$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $=\dfrac{\Delta g}{g}\times 100=\dfrac{\Delta l}{l}\times 100+2\dfrac{\Delta t}{t}\times 100=\dfrac{0.1}{60}\times 100+2\dfrac{0.1}{100}\times 100=0.36 \%$

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

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. $1.5 \%$
Reveal answer Fill a bubble to check yourself
B Correct answer
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  $=\dfrac{\Delta V}{V}\times 100= \left(2\dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}\right)\times 100=\left(2\dfrac{0.005}{0.5}+\dfrac{0.05}{5}\right)100=3\%$

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

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. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
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$

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

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

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

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. $0.6$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 , $\dfrac{\Delta R}{R}=2\dfrac{\Delta l}{l}+\dfrac{\Delta h}{h}+\dfrac{\Delta h}{h}=2\dfrac{0.1}{3}+2\dfrac{0.005}{0.045}=0.3$