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Q. Calculate the perimeter of a square with side length 4 cm. (2015)

A. 16 cm
B. 12 cm
C. 8 cm
D. 20 cm

Solution

Perimeter = 4 × side = 4 × 4 cm = 16 cm.

Correct Answer: A — 16 cm

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Q. Calculate the perimeter of a square with side length 6 cm. (2015)

A. 24 cm
B. 20 cm
C. 18 cm
D. 30 cm

Solution

Perimeter = 4 × side = 4 × 6 = 24 cm.

Correct Answer: A — 24 cm

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Q. Calculate the pH of a 0.01 M solution of NaHCO3. (2023)

A. 8.3
B. 9.0
C. 7.5
D. 8.0

Solution

NaHCO3 is a weak base. The pH can be calculated using the formula pH = 7 + 0.5(pKa - log[C]). pKa of HCO3- is about 10.3, so pH ≈ 8.3.

Correct Answer: A — 8.3

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Q. Calculate the pH of a 0.05 M NH4Cl solution (Kb for NH3 = 1.8 x 10^-5).

A. 4.75
B. 5.25
C. 5.75
D. 6.25

Solution

Using the formula for weak bases, pH = 14 - 0.5(pKb - logC) = 14 - 0.5(4.74 - log(0.05)) = 5.25.

Correct Answer: B — 5.25

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Q. Calculate the pH of a 0.1 M acetic acid solution (Ka = 1.8 x 10^-5).

A. 2.87
B. 3.87
C. 4.87
D. 5.87

Solution

Using the formula for weak acids, pH = 0.5(pKa - log[C]), where pKa = -log(1.8 x 10^-5) ≈ 4.74. Thus, pH = 0.5(4.74 - log(0.1)) = 3.87.

Correct Answer: B — 3.87

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Q. Calculate the pH of a 0.1 M NaOH solution.

A. 12
B. 13
C. 14
D. 11

Solution

pOH = -log[OH-] = -log(0.1) = 1, thus pH = 14 - pOH = 14 - 1 = 13.

Correct Answer: C — 14

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Q. Calculate the pH of a 0.2 M solution of KOH.

A. 12
B. 13
C. 14
D. 11

Solution

pOH = -log(0.2) = 0.7, thus pH = 14 - 0.7 = 13.3.

Correct Answer: B — 13

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Q. Calculate the pH of a buffer solution containing 0.1 M acetic acid and 0.1 M sodium acetate.

A. 4.76
B. 5.76
C. 6.76
D. 7.76

Solution

Using Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]); pKa of acetic acid = 4.76, so pH = 4.76 + log(1) = 4.76

Correct Answer: B — 5.76

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Q. Calculate the pH of a solution that is 0.1 M in acetic acid (Ka = 1.8 x 10^-5).

A. 2.87
B. 3.87
C. 4.87
D. 5.87

Solution

Using the formula for weak acids, pH = 0.5(pKa - logC) = 0.5(4.74 - log(0.1)) = 3.87.

Correct Answer: B — 3.87

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Q. Calculate the range of the data set: 12, 15, 22, 30, 5.

A. 25
B. 27
C. 30
D. 22

Solution

Range = Maximum - Minimum = 30 - 5 = 25.

Correct Answer: A — 25

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Q. Calculate the range of the data set: 4, 8, 15, 16, 23, 42.

A. 38
B. 36
C. 34
D. 32

Solution

Range = Maximum - Minimum = 42 - 4 = 38.

Correct Answer: A — 38

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Q. Calculate the range of the data set: 8, 12, 15, 20, 22.

A. 10
B. 12
C. 14
D. 15

Solution

Range = Max - Min = 22 - 8 = 14.

Correct Answer: A — 10

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Q. Calculate the range of the following data set: 12, 15, 20, 22, 30.

A. 10
B. 12
C. 18
D. 20

Solution

Range = Maximum - Minimum = 30 - 12 = 18.

Correct Answer: C — 18

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Q. Calculate the range of the following data set: 15, 22, 8, 19, 30.

A. 7
B. 15
C. 22
D. 30

Solution

Range = max - min = 30 - 8 = 22.

Correct Answer: D — 30

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Q. Calculate the range of the following data set: 4, 8, 15, 16, 23, 42.

A. 38
B. 36
C. 34
D. 32

Solution

Range = Maximum - Minimum = 42 - 4 = 38.

Correct Answer: A — 38

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Q. Calculate the range of the following data set: 8, 12, 15, 7, 10.

A. 5
B. 8
C. 10
D. 12

Solution

Range = Maximum - Minimum = 15 - 7 = 8.

Correct Answer: A — 5

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Q. Calculate the RMS speed of a gas with molar mass 0.028 kg/mol at 300 K. (R = 8.314 J/(mol K))

A. 500 m/s
B. 600 m/s
C. 700 m/s
D. 800 m/s

Solution

Using v_rms = sqrt(3RT/M), we find v_rms = sqrt(3 * 8.314 * 300 / 0.028) = 600 m/s.

Correct Answer: B — 600 m/s

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Q. Calculate the scalar product of A = (1, 1, 1) and B = (2, 2, 2).

A. 3
B. 4
C. 5
D. 6

Solution

A · B = 1*2 + 1*2 + 1*2 = 2 + 2 + 2 = 6.

Correct Answer: D — 6

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Q. Calculate the scalar product of the vectors (1, 0, 0) and (0, 1, 0).

A. 0
B. 1
C. 2
D. 3

Solution

Scalar product = 1*0 + 0*1 + 0*0 = 0.

Correct Answer: A — 0

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Q. Calculate the scalar product of the vectors (1, 2, 3) and (4, 5, 6).

A. 32
B. 33
C. 34
D. 35

Solution

Scalar product = 1*4 + 2*5 + 3*6 = 4 + 10 + 18 = 32.

Correct Answer: A — 32

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Q. Calculate the scalar product of the vectors (2, 3, 4) and (4, 3, 2).

A. 28
B. 29
C. 30
D. 31

Solution

Scalar product = 2*4 + 3*3 + 4*2 = 8 + 9 + 8 = 25.

Correct Answer: A — 28

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Q. Calculate the scalar product of the vectors (3, 0, -3) and (1, 2, 1).

A. 0
B. 1
C. 2
D. 3

Solution

Scalar product = 3*1 + 0*2 + (-3)*1 = 3 + 0 - 3 = 0.

Correct Answer: A — 0

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Q. Calculate the scalar product of the vectors A = (1, 2, 3) and B = (4, 5, 6).

A. 32
B. 30
C. 28
D. 26

Solution

A · B = 1*4 + 2*5 + 3*6 = 4 + 10 + 18 = 32.

Correct Answer: B — 30

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Q. Calculate the scalar product of the vectors A = (4, -1, 2) and B = (2, 3, 1).

A. 10
B. 8
C. 6
D. 12

Solution

A · B = 4*2 + (-1)*3 + 2*1 = 8 - 3 + 2 = 7.

Correct Answer: A — 10

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Q. Calculate the scalar product of the vectors K = (0, 1, 2) and L = (3, 4, 5).

A. 10
B. 11
C. 12
D. 13

Solution

K · L = 0*3 + 1*4 + 2*5 = 0 + 4 + 10 = 14.

Correct Answer: A — 10

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Q. Calculate the term containing x^3 in the expansion of (2x + 5)^6. (2000)

A. 1500
B. 1800
C. 2000
D. 2500

Solution

The term containing x^3 is C(6,3) * (2)^3 * (5)^(6-3) = 20 * 8 * 125 = 20000.

Correct Answer: B — 1800

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Q. Calculate the term containing x^3 in the expansion of (x + 2)^7.

A. 56
B. 84
C. 112
D. 128

Solution

The term containing x^3 is C(7,3) * (2)^4 = 35 * 16 = 560.

Correct Answer: B — 84

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Q. Calculate the term independent of x in the expansion of (2x - 3)^5.

A. -243
B. 0
C. 243
D. 81

Solution

The term independent of x is C(5,5) * (2x)^0 * (-3)^5 = 1 * 1 * (-243) = -243.

Correct Answer: A — -243

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Q. Calculate the term independent of x in the expansion of (2x^2 - 3x + 4)^3.

A. 12
B. 24
C. 36
D. 48

Solution

The term independent of x occurs when the powers of x cancel out. The term is C(3,0)(4)^3 + C(3,1)(-3x)(4)^2 + C(3,2)(-3x)^2(4) + C(3,3)(-3x)^3 = 64 - 36 + 0 + 0 = 28.

Correct Answer: B — 24

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Q. Calculate the term independent of x in the expansion of (2x^2 - 3x + 4)^5.

A. 80
B. 120
C. 160
D. 200

Solution

The term independent of x occurs when the powers of x cancel out. This can be calculated using the binomial expansion.

Correct Answer: C — 160

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Major Competitive Exams

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