Units and Measurements Test

Result

Units and Measurements Test - 3

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

If the error in measuring the radius of the sphere is 2% and that in measuring its mass is 3%, then the error in measuring the density of material of the sphere is

A
5%

B
7%

C
9%

D
11%

Solution

∴ The percentage error in density is

= 3% + 3(2%) = 3% + 6% = 9%
Question 2
4 / -1

Which of the following is the most precise instrument for measuring length?

A
Metre rod of least count 0.1cm

B
Vernier callipers of least count 0.01cm

C
Screw gauge of least count 0.001cm

D
None of these

Solution

Screw gauge has minimum least count of 0.001cm. Hence, it is most precise instrument.
Question 3
4 / -1

Which of the following time measuring devices is most precise?

A
A wall clock

B
An atomic clock

C
A digital watch

D
A stop watch

Solution

An atomic clock is the most precise time measuring device because atomic oscillations are repeated with a precision of 1s in 10¹³s.
Question 4
4 / -1

Which of the following statements is incorrect?

A
Every measurement by measuring instrument has some error.

B
A measurement can have more accuracy but less precision and vice versa.

C
Every calculated quantity that is based on measured values has some error.

D
The magnitude of the difference between the true value of the quantity and the individual measurement value is called the relative error of the measurement.

Solution

The magnitude of the difference between the true value of the quantity and the individual measurement value is called the absolute error of the measurement. Hence, option (d) is an incorrect statement while all other statements are correct.
Question 5
4 / -1

Which of the following instruments has minimum least count?

A
A vernier callipers with 20 divisions on the vernier scale coinciding with 19 main scale divisions

B
A screw gauge of pitch 1mm and 100 divisions on the circular scale

C
A spherometer of pitch 0.1mm and 100 divisions on the circular scale

D
An optical instrument that can measure length to within a wavelength of light

Solution

(a) Least count of vernier callipers

(∵ 1MSD = 1mm)
(b) Least count of screw gauge

(c) Least count of spherometer

= 0.0001cm
(d) Wavelength of light, λ = 10⁻⁵ cm = 0.00001cm
Clearly the optical instrument is the most precise.
Question 6
4 / -1

The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, then the least count of the microscope is

A
0.02 mm

B
0.05 mm

C
0.01 mm

D
0.1 mm

Solution

1 MSD = 0.5mm
50VSD = 49MSD
1 VSD = 49/50 MSD
Least Count = 1 MSD - 1 VSD
Question 7
4 / -1

In an experiment, the period of oscillation of a simple pendulum was observed to be 2.63s, 2.56s, 2.42s, 2.71s and 2.80s. The mean absolute error is

A
0.11s

B
0.12s

C
0.13s

D
0.14s

Solution

The mean period of oscillation of the pendulum is

(Rounded off to two decimal places)
The absolute errors in the measurement are
ΔT₁ = 2.62s − 2.63s = −0.01s
ΔT₂ = 2.62s − 2.56s = 0.06s
ΔT₃ = 2.62s − 2.42s = 0.20s
ΔT₄ = 2.62s − 2.71s = −0.09s
ΔT₅ = 2.62s − 2.80s = −0.18s
Mean absolute error is
Question 8
4 / -1

If and ΔA, ΔB, ΔC, and ΔD are their absolute errors in A, B, C and D respectively. The relative error in Z is then

A

B

C

D

Solution

The relative error in Z is given by
Question 9
4 / -1

The temperatures of two bodies measured by a thermometer are t₁ = 20 °C ± 0.5 °C and t₂ = 50 °C ± 0.5 °C. The temperature difference and the error there is

A
30 °C ± 1 °C

B
70 °C ± 0.5 °C

C
30 °C ± 0.5 °C

D
70 °C ± 1 °C

Solution

Here, t₁ = 20 °C ± 0.5 °C; t₂ = 50 °C ± 0.5 °C
The temperature difference of two bodies is
t = t₂ - t₁ = 50 °C - 20 °C = 30 °C
The error in temperature difference is given by
Δt = (Δt₁, + Δt₂) = (0.5 °C + 0.5 °C) = 1 °C
∴ The temperature difference is 30 °C ± 1 °C.
Question 10
4 / -1

Two resistors of resistances R₁ = (300 ± 3)Ω and R₂ = (500 ± 4)Ω are connected in series. The equivalent resistance of the series combination is

A
(800 ± 1)Ω

B
(800 ± 7) Ω

C
(200 ± 7) Ω

D
(200 ± 1) Ω

Solution

The equivalent resistance of series combination is
Rs = R₁ + R₂= 300Ω + 500Ω = 800Ω
The error in equivalent resistance is given by
ΔR = (ΔR₁ + ΔR₂) = (3 + 4)Ω = 7Ω
Hence, the equivalent resistance along with error is (800 ± 7)Ω.