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

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

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

= $2 + 6$

= $8$% 

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

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. $\left (\dfrac {\Delta a}{a-b}-\dfrac {\Delta b}{a-b}\right )\times 100\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $= \left (\dfrac {\Delta a}{a-b}+\dfrac {\Delta b}{a-b}\right )\times 100$%

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

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%
Reveal answer Fill a bubble to check yourself
C Correct answer
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,
$\dfrac { \Delta \rho  }{ \rho  } =\pm (2\dfrac { 0.02 }{ 0.2 } +\dfrac { 0.1 }{ 150 } +\dfrac { 2 }{ 60 } )=0.234=23.4$%

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

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%
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $13\%$

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 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%
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$= 4+2\times 2= 8$%

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

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. $\pm \, 5\, \%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Acceleration due to gravity   $g = \dfrac{4\pi^2 \ L}{T^2}$
Percentage error in $g$ is   $\dfrac{\Delta g}{g}\times 100 = \dfrac{\Delta L}{L}\times 100 +\times \dfrac{\Delta T}{T}\times 100$
$\implies \ \dfrac{\Delta g}{g}\times 100 = 2+2\times 3 = \pm 8$%
Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

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

Since $V=lbw$

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