Tag: measurements and experimentation

Questions Related to measurements and experimentation

Two resistors of resistances $R _1$ = (100 $\pm$ 3) $\Omega$ and $R _2$ = (200 $\pm$ 4) $\Omega$ are connected in parallel. The equivalent resistance of the parallel combination is:

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

  1. (66.7 $\pm$ 1.8) $\Omega$

  2. (66.7 $\pm$ 4.0) $\Omega$

  3. (66.7 $\pm$ 3.0) $\Omega$

  4. (66.7 $\pm$ 7.0) $\Omega$

Show answer
Correct Option: A
Explanation:

Here, $R _1 (100 \pm 3) \Omega; R _2 = (200 \pm 4) \Omega$ The equivalent resistance in parallel combination is

$\displaystyle \frac{1}{R _1} = \frac{1}{R _1} + \frac{1}{R _2}, \frac{1}{R _p} = \frac{1}{100} + \frac{1}{200} = \frac{3}{200}, R _p = \frac{200}{3} = 66.7 \Omega$

The error in equivalent resistance is given by
$\displaystyle \frac {\Delta R _p}{R _p^2} = \frac {\Delta R _1}{R _1^2} + \frac {\Delta R _2}{R _2^2}; \Delta R _p = \Delta R _1 (\frac{R _p}{R _1})^2 + \Delta R _2 (\frac {R _p}{R _2})^2 = 3 (\frac {66.7}{100})^2 + 4 (\frac {66.7}{200})^2 = 1.8 \Omega $ 
Hence, the equivalent resistance along with error in parallel combination is (66.7 $\pm$ 1.8)$\Omega$.

The percentage errors in quantities $P, Q, R$ and $S$ are $0.5$%, $1$%, $3$% and $1.5$% respectively in the measurement of a physical quantity $A = \dfrac {P^{3}Q^{2}}{\sqrt {R}S}$.
The maximum percentage error in the value of $A$ will be

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

  1. $8.5\%$

  2. $6.0\%$

  3. $7.5\%$

  4. $6.5\%$

Show answer
Correct Option: D
Explanation:

Percentage error $\dfrac{\Delta A}{A} \times 100 = 3 \dfrac{\Delta P}{P} \times 100 + 2 \dfrac{\Delta Q}{Q} + \dfrac{1}{2} \dfrac{\Delta R}{R} \times 100 + \dfrac{\Delta S}{S} \times 100$ 

$ = 3 \times 0.5 + 2 \times 1+ 0.5 \times 3 + 1.5 = 6.5 %$

The percentage errors in the measurement of length and time period of a simple pendulum are 1% and 2% respectively. Then, the maximum error in the measurement of acceleration due to gravity is:

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

  1. $3%$

  2. $4%$

  3. $6%$

  4. $5%$

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Correct Option: D

The dimensions of a rectangular block measured with callipers having least count of 0.01 cm are 5mm x 10mm x 5 mm. The maximum percentage error in the measurement of the volume of the block is

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

  1. 5%

  2. 10%

  3. 15%

  4. 20%

Show answer
Correct Option: A
Explanation:
Given :  $l = 10 \ mm = 1 \ cm$     $b = 5 \ mm = 0.5 \ cm$        $h = 5 \ mm  = 0.5 \ cm$
Error in the measurements   $\Delta l = \Delta b = \Delta h = 0.01 \ cm$
Maximum percentage error in volume   $\dfrac{\Delta V}{V}\times 100 = \dfrac{\Delta l}{l}\times 100 +\dfrac{\Delta b}{b}\times 100+\dfrac{\Delta h}{h}\times 100$
$\implies \ \dfrac{\Delta V}{V}\times 100 = \dfrac{0.01}{1}\times 100+\dfrac{0.01}{0.5}\times 100+\dfrac{0.01}{0.5}\times 100 = 5$ %
Correct answer is option A.

A force $\vec { F } $ is applied on a square plate of length $L$. If the percentage error in the determination of $L$ is $3$% and in $F$ is $4$%, the permissible error in the calculation of pressure is

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

  1. $13$%

  2. $10$%

  3. $7$%

  4. $12$%

Show answer
Correct Option: B
Explanation:

Pressure  $P=\cfrac { F }{ A } =\cfrac { F }{ { L }^{ 2 } } $
Percentage error in pressure  $\cfrac { \Delta P }{ P } \times 100=\cfrac { \Delta F }{ F } \times 100+2\cfrac { \Delta L }{ L } \times 100$
$ = 4+2\times 3 = 10$  %