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Communication System Quiz Details

Q.
In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4%, respectively. The maximum percentage of error in the measurement X, where

$X = \frac{A^{2} B^{1 / 2}}{C^{1 / 3} D^{3}}$ , will be

A
10%

B
$(\frac{3}{13}) %$

C
16%

D
-10%

Related Questions :-

Q. 1
The angle of 1' (minute of arc) in radian is nearly equal to

A. 1
(b) $1^{'} = {(\frac{1}{60})}^{o} = \frac{1}{60} \times \frac{π}{180}$

$= 2.91 \times 10^{- 4} r a d$

Q. 2
In an experiment, four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4%, respectively. Quantity P is calculated as follows:

$P = \frac{a^{3} b^{2}}{c d}, %$ error in P is

A. 2
(a) $P = \frac{a^{3} b^{2}}{c d}$

$\Rightarrow \frac{∆ P}{P} \times 100 = 3 (\frac{∆ a}{a} \times 100) + 2 (\frac{∆ b}{b} \times 100) + (\frac{∆ c}{c} \times 100) + (\frac{∆ d}{d} \times 100)$

$= 3 \times 1 + 2 \times 2 + 3 + 4$

$= 3 + 4 + 3 + 4 = 14 %$

Q. 3
A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level.

If screw gauge has a zero error of -0.004 cm, the correct diameter of the ball is

Q. 4
A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is

A. 4
d) $L . C . = \frac{P i t c h}{N u m b e r o f d i v i s i o n s o n c i r c u l a r s c a l e}$

$0.01 = \frac{P i t c h}{50}$

$P i t c h = 0.5 m m$

Q. 5
Time intervals measured by a clock give the following readings: 1.25 s, 1.24 s, 1.27 s, 1.21 s and 1.28 s

What is the percentage relative error of the observations

A. 5
a) $\bar{t} = \frac{1.25 + 1.24 + 1.27 + 1.21 + 1.28}{5}$

$= 1.25 S$

${(∆ t)}_{1} = 1.25 - 1.25 = 0$

${(∆ t)}_{2} = 1.24 - 1.25 = - 0.01$

${(∆ t)}_{3} = 1.27 - 1.25 = 0.02$

${(∆ t)}_{4} = 1.21 - 1.25 = - 0.04$

${(∆ t)}_{5} = 1.28 - 1.25 = 0.03$

$| ∆ \bar{t} | = \frac{| {(∆ t)}_{1} | + | {(∆ t)}_{2} | + | {(∆ t)}_{3} | + | {(∆ t)}_{4} | + | {(∆ t)}_{5} |}{5}$

$= \frac{0 + 0.01 + 0.02 + 0.04 + 0.03}{5} = 0.02$

$\frac{| ∆ \bar{t} |}{\bar{t}} \times 100 = \frac{0.02}{1.25} \times 100 = 1.6 %$

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