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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
A. 2.528 m
B. 2.5283 m
C. m
D. 2.53 m
Answer & Solution
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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:
A. 4
B. 3
C. 7
D. 2
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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
A. 85.184 cm³
B. 85.18 cm³
C. 85 cm³
D. 85.2 cm³
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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:
A. One
B. Two
C. Three
D. Four
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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
A. 2.66 m²
B. 2.67 m²
C. 2.664 m²
D. 2.6843 m²
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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
A. 9.85960
B. 9.8596
C. 9.86
D. 9.859
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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?
A. 20.000
B. 0.00004
C. 0.800
D. 0.0060
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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:
A. 714.573 × 10^-2
B. 7146 × 10^-3
C. 7.15
D. 7.146
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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
A. 13.0 gm
B. 1.30 gm
C. 130 kg
D. 1.30 kg
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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
A. decreases
B. Increases
C. remains constant
D. sometimes increases and sometimes decreases
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Explanation
Increasing temperature generally reduces surface tension, which tends to decrease the contact angle for typical liquid–solid pairs.