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21
The length of a rod as measured in an experiment is 2.56 m, 2.51 m, 2.49 m, 2.58 m, 2.48 m and 2.55 m respectively. The average length is
Correct Option
Option B
Explanation
Sum = 15.17; average = 15.17/6 = 2.528333... which rounded as a direct arithmetic mean gives 2.5283 m.
22
10. The number of significant figures in 0.005005 are:
Correct Option
Option A
Explanation
Leading zeros are not significant; the digits 5,0,0,5 are significant, giving 4 significant figures.
23
The length of a side of a cube is 4.4 cm. The volume of the cube according to idea of significant figures is
Correct Option
Option C
Explanation
Side 4.4 has two significant figures, so volume (4.4³ ≈ 85.184) should be reported to two significant figures as 85 cm³.
24
Significant figures in 2.00 × 10^0 are:
Correct Option
Option C
Explanation
The number 2.00 has three significant figures due to the two trailing zeros after the decimal.
25
A calculator with eight figure capability is used to compute the area of desk top with dimensions: length L = 2.49 m, width W = 1.07 m and finds area A = L × W = 2.664... m². The area of the desk top with regard to significant figures is
Correct Option
Option A
Explanation
Both factors have 3 significant figures, so result should have 3 sig figs: 2.49×1.07 ≈ 2.6643 → 2.66 m².
26
Given: n = 3.14. The value of n2 with due regard for significant figures is
Correct Option
Option C
Explanation
3.14 has three significant figures, so n^2 rounded to three significant figures is 9.86.
27
In which of the following numbers are all zeros significant?
Correct Option
Option A
Explanation
In 20.000 all zeros are significant because the number has a decimal and trailing zeros indicate measured precision.
28
Multiplying 3.233, 2.105 and 1.05, rounded off answer is recorded as:
Correct Option
Option C
Explanation
The OCR marks option A; using significant-figure rules, the product is rounded to three significant figures giving 7.15.
29
Mass of air at NTP in a room about 100 cubic meter is approximately
Correct Option
Option C
Explanation
Air density ≈1.29 kg/m^3, so mass in 100 m^3 ≈129 kg ≈130 kg, making option C correct.
30
With the rise in temperature, the angle of contact is
Correct Option
Option A
Explanation
Increasing temperature generally reduces surface tension, which tends to decrease the contact angle for typical liquid–solid pairs.
