Tag: measurement of length

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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 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 :-

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

Velocity v = d/t = 13.8 / 4 = 3.45 m/s. Relative error dv/v = dd/d + dt/t = 0.2/13.8 + 0.3/4 = 0.0145 + 0.075 = 0.0895. Absolute error dv = v * 0.0895 = 3.45 * 0.0895 = 0.308. Rounding gives 3.45 +/- 0.3 m/s. Option A is 0.5, which is likely an overestimate.

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

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 

  1. $64^{0}$

  2. $54^{0}$

  3. $60^{0}$

  4. $68^{0}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

$\dfrac{C}{100}=\dfrac{59-5}{95-5}=\dfrac{54}{90}=0.6$

$\implies C=0.6\times 100=60^0$
Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

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:

  1. 10%

  2. 13%

  3. 30%

  4. 20%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In series effective resistance $=R _S=(10k\Omega \pm 10$%)+$(20k\Omega \pm 20$%)$=(30k\Omega \pm 30$%)
$\therefore$ Tolerance of the combination $=( \pm 30$%)

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

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.

  1. x+y

  2. x-y

  3. 2x+y

  4. 2x-y
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From $T=2\pi \sqrt {\dfrac {l}{g}};g=4\pi^2\dfrac {l}{T^2}$

$\dfrac {\Delta g}{g}=\dfrac {\Delta l}{l}+\dfrac {2\Delta T}{T}=(y+2x)$

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

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 :

  1. 5 mm, 0.2 sec, $n=$10

  2. 5 mm, 0.2 sec,&nbsp;<span>$n=$</span>20

  3. 5 mm, 0.1 sec,&nbsp;<span>$n=$</span>10

  4. 1 mm, 0.1 sec,&nbsp;<span>$n=$</span>50
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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 :

  1. 8%

  2. 6%

  3. 18%

  4. 12%
Reveal answer Fill a bubble to check yourself
B Correct answer
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$ % 

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 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 :

  1. 2%

  2. 0%

  3. 4%

  4. 3%
Reveal answer Fill a bubble to check yourself
C Correct answer
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%

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

If error in measurement of radius of a sphere is 1%, what will be the error in measurement of volume?

  1. 1%

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

  3. 3%

  4. 10%
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $ %