Tag: vernier calliper and screw gauge

Questions Related to vernier calliper and screw gauge

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.

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

  1. 0.4%

  2. 0.8%

  3. 1.6 %

  4. 3.2%

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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 :-

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

  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} $

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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 

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

  1. $64^{0}$

  2. $54^{0}$

  3. $60^{0}$

  4. $68^{0}$

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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:

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

  1. 10%

  2. 13%

  3. 30%

  4. 20%

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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.

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

  1. x+y

  2. x-y

  3. 2x+y

  4. 2x-y

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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 :

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

  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

Show answer
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?

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%

  3. 1.2%

  4. 1.8%

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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 :

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

  1. 8%

  2. 6%

  3. 18%

  4. 12%

Show answer
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 :

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

  1. 2%

  2. 0%

  3. 4%

  4. 3%

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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?

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

  1. 1%

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

  3. 3%

  4. 10%

Show answer
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 $ %