Tag: least count of 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

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. $\pm\ 12\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Q = (A^3 * B^3) / (C * D^0.5). Error dQ/Q = 3*dA/A + 3*dB/B + dC/C + 0.5*dD/D. dQ/Q = 3(2%) + 3(1%) + 3% + 0.5(4%) = 6% + 3% + 3% + 2% = 14%.

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

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. $\left( {4.0\; \pm \;0.4} \right)\;m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

X = 4.0 +/- 0.4, Y = 1.0 +/- 0.1. Product XY = 4.0 * 1.0 = 4.0. Relative error in XY = (dX/X) + (dY/Y) = (0.4/4.0) + (0.1/1.0) = 0.1 + 0.1 = 0.2. Absolute error d(XY) = 0.2 * (XY) = 0.2 * 4.0 = 0.8. Result = 4.0 +/- 0.8.

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

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%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

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. $4$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

g = 4*pi^2*L/T^2. dg/g = dL/L + 2*dT/T. dL = 0.1, L = 200, dL/L = 0.0005. dT = 0.1/100 = 0.001, T = 4, dT/T = 0.00025. dg/g = 0.0005 + 2(0.00025) = 0.001 or 0.1%.

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

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%
Reveal answer Fill a bubble to check yourself
B Correct answer
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}

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

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. $0.05mA$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

Zero error of an instrument introduces:

  1. Systematic errors

  2. Random errors

  3. Both

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
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).

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

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. $(\Delta x)^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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: 
$\dfrac{\Delta f}{f} = \dfrac{2x \Delta x}{x^2}$

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