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Communication System Quiz Details

Q.
A screw gauge gives the following readings when used to measure the diameter of a wire

Main scale reading : 0 mm

Circular scale reading : 52 divisions

Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is

A
0.52 cm

B
0.026 cm

C
0.26 cm

D
0.052 cm

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
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. 4
c) $X = \frac{A^{2} B^{1 / 2}}{C^{1 / 3} D^{3}}$

$\ln X = 2 \ln A + \frac{1}{2} \ln B - \frac{1}{3} \ln C - 3 \ln D$

${(\frac{d X}{X})}_{m a x} = 2 \frac{d A}{A} + \frac{1}{2} \frac{d B}{B} + \frac{1}{3} \frac{d C}{C} + \frac{3 d D}{D}$

$E r r o r X_{m a x} = 2 \times 1 + \frac{2}{2} + \frac{1}{3} \times 3 + 3 \times 4 = + 16 %$

Q. 5
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. 5
d) $L . C . = \frac{P i t c h}{N u m b e r o f d i v i s i o n s o n c i r c u l a r s c a l e}$

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

$P i t c h = 0.5 m m$

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