Units and Measurement

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New answer posted

4 weeks ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

As the factors controlling temperature and voltage supply are beyond prediction and control so the error occurred due to unpredictable fluctuations of temperature and voltage would be random errors.

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

V = (4/3)πr³

⇒ ΔV/V = 3 (Δr/r) ⇒ Relative error

% error in volume = (ΔV/V) * 100 = 3 (Δr/r) * 100 = 3 * (0.85/7.50) * 100 = 33.999 ≈ 34

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

l = 10.0 ± 0.1cm.
T = (100 ± 1) / 200 = 0.5 ± 0.005

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Positive zero error = 0.02 cm

Reading = 8.5 + 6 * 0.01 = 8.56 cm

Actual reading = Reading - Positive zero error

Actual reading = 8.56 - 0.02 = 8.54 cm

New question posted

a month ago

0 Follower 1 View

New answer posted

a month ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Since significant figures show the degree of correctness of any measurement, so in any mathematical calculation we cannot increase the number of significant digits. Because the four reading has 3 significant digits so the answer should also have 3 significant digits only.

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

x = (IFV²)/ (WL? )
I = [ML²]
F = [MLT? ²]
V² = [L² T? ²]
W = [ML² T? ²]
Q? = [L? ]
X = [ML? ¹ T? ²]

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

 K = (Q?)Δx / AΔT
⇒ (ML²T?²)(L) / (L²)(θ)(T)
⇒ M¹L¹T?³θ?¹

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