Tag: vernier calliper and screw gauge

Questions Related to vernier calliper and screw gauge

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

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

  1. 12%

  2. 10%

  3. 8%

  4. 2%

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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:

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

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

Show answer
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:

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

  1. 0.6 %

  2. 1.2 %

  3. 1.8 %

  4. 0.9 %

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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.

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

  1. 5.85%

  2. 11.7%

  3. 23.4%

  4. 35.1%

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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

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

  1. 9%

  2. 13%

  3. 12%

  4. 7%

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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:

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

  1. 1%

  2. 2%

  3. 6%

  4. 8%

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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 ? 

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

  1. 2%

  2. 8%

  3. 4%

  4. 1%

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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

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

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

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

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

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

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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 ?

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

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

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

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

  4. None of these

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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 :

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

  1. 2%

  2. 8%

  3. 4%

  4. 1%

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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.