Q. What is the frequency of a wave with a period of 0.5 seconds? (2019)
-
A.
2 Hz
-
B.
1 Hz
-
C.
0.5 Hz
-
D.
4 Hz
Solution
Frequency (f) = 1/Period (T) = 1/0.5 s = 2 Hz.
Correct Answer:
A
— 2 Hz
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Q. What is the frequency of a wave with a speed of 300 m/s and a wavelength of 3 m? (2023)
-
A.
100 Hz
-
B.
150 Hz
-
C.
200 Hz
-
D.
250 Hz
Solution
Frequency (f) = speed (v) / wavelength (λ) = 300 m/s / 3 m = 100 Hz.
Correct Answer:
B
— 150 Hz
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Q. What is the frequency of a wave with a wavelength of 2 m and a speed of 340 m/s?
-
A.
170 Hz
-
B.
340 Hz
-
C.
85 Hz
-
D.
420 Hz
Solution
Frequency (f) is given by the formula f = speed / wavelength. Here, f = 340 m/s / 2 m = 170 Hz.
Correct Answer:
A
— 170 Hz
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Q. What is the functional group of alcohols? (2020)
-
A.
-OH
-
B.
-COOH
-
C.
-CHO
-
D.
-NH2
Solution
The functional group of alcohols is the hydroxyl group (-OH).
Correct Answer:
A
— -OH
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Q. What is the functional group of aldehydes?
-
A.
-OH
-
B.
-CHO
-
C.
-C=O
-
D.
-COOH
Solution
The functional group of aldehydes is -CHO.
Correct Answer:
B
— -CHO
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Q. What is the functional group of phenols? (2019) 2019
-
A.
-OH
-
B.
-COOH
-
C.
-CHO
-
D.
-NH2
Solution
Phenols contain a hydroxyl group (-OH) directly attached to an aromatic ring.
Correct Answer:
A
— -OH
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Q. What is the functional group of phenols? (2021)
-
A.
-OH
-
B.
-COOH
-
C.
-CHO
-
D.
-NH2
Solution
Phenols contain a hydroxyl group (-OH) directly attached to an aromatic hydrocarbon group.
Correct Answer:
A
— -OH
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Q. What is the functional group present in alcohols? (2023) 2023
-
A.
Aldehyde
-
B.
Ketone
-
C.
Hydroxyl
-
D.
Carboxylic acid
Solution
Alcohols contain a hydroxyl (-OH) functional group.
Correct Answer:
C
— Hydroxyl
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Q. What is the functional group present in phenols? (2021)
-
A.
-OH
-
B.
-COOH
-
C.
-CHO
-
D.
-NH2
Solution
Phenols contain a hydroxyl group (-OH) directly attached to an aromatic hydrocarbon group.
Correct Answer:
A
— -OH
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Q. What is the functional group present in phenols? (2021) 2021
-
A.
-OH
-
B.
-COOH
-
C.
-CHO
-
D.
-NH2
Solution
Phenols contain a hydroxyl group (-OH) directly attached to an aromatic hydrocarbon group.
Correct Answer:
A
— -OH
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Q. What is the general formula for secondary amines? (2021) 2021
-
A.
RNH2
-
B.
R2NH
-
C.
R3N
-
D.
RNH
Solution
The general formula for secondary amines is R2NH, where R represents an alkyl or aryl group.
Correct Answer:
B
— R2NH
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Q. What is the general formula for secondary amines? (2023)
-
A.
RNH2
-
B.
R2NH
-
C.
R3N
-
D.
RNH
Solution
The general formula for secondary amines is R2NH, where R represents an alkyl group.
Correct Answer:
B
— R2NH
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Q. What is the general solution of the equation y' + 4y = 0?
-
A.
y = Ce^(-4x)
-
B.
y = Ce^(4x)
-
C.
y = 4x + C
-
D.
y = Cx^4
Solution
This is a separable equation, and integrating gives y = Ce^(-4x).
Correct Answer:
A
— y = Ce^(-4x)
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Q. What is the general solution of the equation y' = 4y + 3?
-
A.
y = Ce^(4x) - 3/4
-
B.
y = Ce^(4x) + 3/4
-
C.
y = 3e^(4x)
-
D.
y = Ce^(3x) + 4
Solution
The integrating factor is e^(-4x). The solution is y = Ce^(4x) + 3/4.
Correct Answer:
B
— y = Ce^(4x) + 3/4
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Q. What is the general solution of the equation y'' - 3y' + 2y = 0?
-
A.
y = C1 e^(x) + C2 e^(2x)
-
B.
y = C1 e^(2x) + C2 e^(x)
-
C.
y = C1 e^(3x) + C2 e^(0)
-
D.
y = C1 e^(0) + C2 e^(3x)
Solution
The characteristic equation is r^2 - 3r + 2 = 0, which factors to (r - 1)(r - 2) = 0. Thus, the general solution is y = C1 e^(2x) + C2 e^(x).
Correct Answer:
B
— y = C1 e^(2x) + C2 e^(x)
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Q. What is the general solution of the equation y'' - 4y' + 4y = 0?
-
A.
y = (C1 + C2x)e^(2x)
-
B.
y = C1 e^(2x) + C2 e^(-2x)
-
C.
y = C1 e^(4x) + C2 e^(-4x)
-
D.
y = C1 cos(2x) + C2 sin(2x)
Solution
The characteristic equation has a repeated root r = 2. The general solution is y = (C1 + C2x)e^(2x).
Correct Answer:
A
— y = (C1 + C2x)e^(2x)
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Q. What is the geometry of a molecule with four bonding pairs and no lone pairs around the central atom? (2023)
-
A.
Linear
-
B.
Trigonal planar
-
C.
Tetrahedral
-
D.
Octahedral
Solution
The geometry is tetrahedral when there are four bonding pairs and no lone pairs around the central atom.
Correct Answer:
C
— Tetrahedral
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Q. What is the geometry of a molecule with four bonding pairs and no lone pairs? (2023)
-
A.
Linear
-
B.
Trigonal planar
-
C.
Tetrahedral
-
D.
Octahedral
Solution
A molecule with four bonding pairs and no lone pairs has a tetrahedral geometry.
Correct Answer:
C
— Tetrahedral
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Q. What is the geometry of a molecule with two bonding pairs and no lone pairs on the central atom? (2023)
-
A.
Linear
-
B.
Trigonal planar
-
C.
Tetrahedral
-
D.
Bent
Solution
The geometry is linear when there are two bonding pairs and no lone pairs on the central atom.
Correct Answer:
A
— Linear
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Q. What is the geometry of the complex ion [Ag(NH3)2]+? (2019) 2019
-
A.
Tetrahedral
-
B.
Square planar
-
C.
Octahedral
-
D.
Linear
Solution
The complex [Ag(NH3)2]+ has a linear geometry due to the presence of two ligands around the silver ion.
Correct Answer:
D
— Linear
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Q. What is the gravitational field strength at a distance of 10 m from a mass of 100 kg? (G = 6.67 × 10^-11 N m²/kg²)
-
A.
6.67 × 10^-10 N/kg
-
B.
6.67 × 10^-9 N/kg
-
C.
6.67 × 10^-8 N/kg
-
D.
6.67 × 10^-7 N/kg
Solution
g = G * M / r² = (6.67 × 10^-11) * (100) / (10²) = 6.67 × 10^-10 N/kg
Correct Answer:
A
— 6.67 × 10^-10 N/kg
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Q. What is the gravitational field strength at a distance of 3R from the center of the Earth? (R = radius of the Earth) (2022)
-
A.
g/3
-
B.
g/9
-
C.
g/6
-
D.
g/12
Solution
Gravitational field strength decreases with the square of the distance. At 3R, it is g/9.
Correct Answer:
B
— g/9
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Q. What is the gravitational force acting on a 50 kg object at the surface of the Earth?
-
A.
490 N
-
B.
500 N
-
C.
510 N
-
D.
520 N
Solution
Weight = mass * g = 50 kg * 9.8 m/s² = 490 N
Correct Answer:
A
— 490 N
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Q. What is the gravitational force acting on a 50 kg object on the surface of the Earth?
-
A.
490 N
-
B.
50 N
-
C.
5 N
-
D.
500 N
Solution
Weight W = m * g = 50 kg * 9.8 m/s² = 490 N.
Correct Answer:
A
— 490 N
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Q. What is the gravitational force between two 1 kg masses placed 1 meter apart? (G = 6.67 × 10^-11 N m²/kg²)
-
A.
6.67 × 10^-11 N
-
B.
1.67 × 10^-10 N
-
C.
6.67 × 10^-10 N
-
D.
0 N
Solution
F = G * (m1 * m2) / r² = (6.67 × 10^-11) * (1 * 1) / (1²) = 6.67 × 10^-11 N
Correct Answer:
A
— 6.67 × 10^-11 N
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Q. What is the gravitational force between two 10 kg masses separated by a distance of 2 m?
-
A.
0.25 N
-
B.
5 N
-
C.
10 N
-
D.
20 N
Solution
Using Newton's law of gravitation: F = G(m1*m2)/r² = (6.674 × 10⁻¹¹)(10*10)/(2²) = 0.25 N.
Correct Answer:
A
— 0.25 N
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Q. What is the gravitational force between two objects of mass 5 kg and 10 kg separated by a distance of 2 m? (G = 6.67 × 10^-11 N m²/kg²)
-
A.
1.67 × 10^-10 N
-
B.
1.25 × 10^-10 N
-
C.
1.00 × 10^-10 N
-
D.
2.00 × 10^-10 N
Solution
Using the formula F = G * (m1 * m2) / r², we have F = (6.67 × 10^-11) * (5 * 10) / (2²) = 1.67 × 10^-10 N.
Correct Answer:
A
— 1.67 × 10^-10 N
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Q. What is the gravitational force between two objects of mass m1 and m2 separated by a distance r? (2021)
-
A.
G * (m1 * m2) / r^2
-
B.
G * (m1 + m2) / r^2
-
C.
G * (m1 - m2) / r^2
-
D.
G * (m1 * m2) * r^2
Solution
The gravitational force is given by Newton's law of gravitation: F = G * (m1 * m2) / r^2.
Correct Answer:
A
— G * (m1 * m2) / r^2
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Q. What is the gravitational potential energy of a 10 kg mass at a height of 10 m?
-
A.
100 J
-
B.
200 J
-
C.
500 J
-
D.
1000 J
Solution
Gravitational potential energy (PE) is given by PE = mgh. Here, PE = 10 kg * 9.81 m/s² * 10 m = 981 J.
Correct Answer:
D
— 1000 J
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Q. What is the gravitational potential energy of a 10 kg mass at a height of 5 m? (2019)
-
A.
50 J
-
B.
100 J
-
C.
150 J
-
D.
200 J
Solution
Gravitational potential energy (PE) = m * g * h = 10 kg * 9.81 m/s² * 5 m = 490.5 J.
Correct Answer:
B
— 100 J
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