Q. If the gravitational field strength at a point is 10 N/kg, what is the gravitational potential at that point assuming it is at a distance of 2 m from the mass?
A.
-20 J/kg
B.
-10 J/kg
C.
0 J/kg
D.
-5 J/kg
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Solution
Gravitational potential V = -g * r = -10 * 2 = -20 J/kg.
Correct Answer: A — -20 J/kg
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Q. If the gravitational field strength at a point is 10 N/kg, what is the gravitational potential at that point assuming it is zero at infinity?
A.
-10 J/kg
B.
-5 J/kg
C.
0 J/kg
D.
10 J/kg
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Solution
The gravitational potential V is related to the gravitational field strength g by V = -g * r. If we consider r to be 1 meter, V = -10 * 1 = -10 J/kg.
Correct Answer: A — -10 J/kg
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Q. If the gravitational force between two objects is 20 N and the distance between them is doubled, what will be the new force?
A.
5 N
B.
10 N
C.
20 N
D.
40 N
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Solution
F is inversely proportional to r². If r is doubled, F becomes F/4 = 20 N / 4 = 5 N.
Correct Answer: A — 5 N
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Q. If the gravitational potential at a point is -15 J/kg and the gravitational field strength is constant at 3 N/kg, what is the distance from the mass?
A.
5 m
B.
10 m
C.
15 m
D.
20 m
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Solution
Using V = -g * r, we have -15 = -3 * r, thus r = 15/3 = 5 m.
Correct Answer: B — 10 m
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Q. If the gravitational potential at a point is -20 J/kg and the gravitational field strength is 4 N/kg, what is the distance from the mass?
A.
5 m
B.
10 m
C.
15 m
D.
20 m
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Solution
Using V = -g * r, we have -20 = -4 * r, thus r = 20/4 = 5 m.
Correct Answer: B — 10 m
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Q. If the gravitational potential at a point is -30 J/kg and the gravitational field strength is 5 N/kg, what is the distance from the mass?
A.
6 m
B.
3 m
C.
5 m
D.
4 m
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Solution
Using V = -g * r, we have -30 = -5 * r, thus r = 30/5 = 6 m.
Correct Answer: B — 3 m
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Q. If the gravitational potential at a point is -30 J/kg, what is the gravitational field strength at that point?
A.
3 N/kg
B.
30 N/kg
C.
0 N/kg
D.
Cannot be determined
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Solution
The gravitational field strength can be found using the relation E = -dV/dr. If V = -30 J/kg, E = 3 N/kg.
Correct Answer: A — 3 N/kg
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Q. If the HCF of 24 and 36 is 12, what is the LCM of these two numbers? (2023)
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Solution
Using the relation HCF * LCM = Product of the numbers, we have LCM = (24 * 36) / 12 = 72.
Correct Answer: A — 72
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Q. If the HCF of three numbers is 7 and their LCM is 420, what is the product of the three numbers? (2023)
A.
2940
B.
840
C.
2100
D.
1470
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Solution
The product of the three numbers = HCF * LCM = 7 * 420 = 2940.
Correct Answer: A — 2940
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Q. If the HCF of two numbers is 1, what can be inferred about the two numbers? (2023)
A.
They are both prime.
B.
They are coprime.
C.
They are multiples of each other.
D.
They are both even.
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Solution
If the HCF is 1, it means the two numbers are coprime, meaning they have no common factors other than 1.
Correct Answer: B — They are coprime.
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Q. If the HCF of two numbers is 5 and their LCM is 100, what is the sum of the two numbers? (2023)
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Solution
Let the two numbers be 5x and 5y. Then, HCF(5x, 5y) = 5 and LCM(5x, 5y) = 100. This gives xy = 20. The pairs (x, y) that satisfy this are (4, 5) or (5, 4), leading to the sum 5(4 + 5) = 45.
Correct Answer: A — 30
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Q. If the HCF of two numbers is 5 and their product is 100, what is their LCM?
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Solution
Using the relationship between HCF, LCM, and product: HCF * LCM = Product. Thus, LCM = 100 / 5 = 20.
Correct Answer: B — 25
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Q. If the HCF of two numbers is 5 and their product is 1000, what is their LCM? (2023)
A.
200
B.
100
C.
250
D.
150
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Solution
Using the relation HCF * LCM = Product of the numbers, we have LCM = 1000 / 5 = 200.
Correct Answer: A — 200
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Q. If the HCF of two numbers is 8 and their LCM is 48, what could be one of the possible pairs of numbers? (2023)
A.
16 and 24
B.
8 and 32
C.
4 and 12
D.
20 and 16
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Solution
The numbers can be 16 and 24, as their HCF is 8 and LCM is 48.
Correct Answer: A — 16 and 24
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Q. If the HCF of two numbers is 8 and their product is 288, what is their LCM? (2023)
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Solution
Using the relationship between HCF, LCM, and the product of two numbers: LCM = Product / HCF = 288 / 8 = 36.
Correct Answer: A — 36
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Q. If the height of a cylinder is doubled while keeping the radius constant, how does the volume change?
A.
It remains the same
B.
It doubles
C.
It triples
D.
It quadruples
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Solution
Volume of a cylinder = πr²h. If height is doubled, volume becomes 2πr²h, which is double the original volume.
Correct Answer: B — It doubles
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Q. If the height of an isosceles triangle is 12 cm and the base is 10 cm, what is the area of the triangle?
A.
60 cm²
B.
70 cm²
C.
80 cm²
D.
90 cm²
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Solution
The area of a triangle is given by (1/2) * base * height = (1/2) * 10 * 12 = 60 cm².
Correct Answer: A — 60 cm²
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Q. If the hexadecimal number A3 is converted to decimal, what is its value? (2023)
A.
163
B.
1630
C.
103
D.
83
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Solution
A3 in hexadecimal is calculated as 10*16^1 + 3*16^0 = 160 + 3 = 163.
Correct Answer: A — 163
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Q. If the image distance is equal to the object distance for a lens, what is the magnification?
A.
0
B.
1
C.
2
D.
Infinity
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Solution
Magnification (m) = v/u = 1 when image distance equals object distance.
Correct Answer: B — 1
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Q. If the image distance of a convex lens is 40 cm and the object distance is 30 cm, what is the focal length of the lens?
A.
10 cm
B.
20 cm
C.
30 cm
D.
15 cm
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Solution
Using the lens formula 1/f = 1/v - 1/u, we have 1/f = 1/40 - 1/(-30) = 1/40 + 1/30. Solving gives f = 20 cm.
Correct Answer: B — 20 cm
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Q. If the incidence rate of a disease is 3 cases per 1000 people per year, how many new cases would you expect in a population of 50,000? (2022)
A.
150
B.
200
C.
300
D.
500
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Solution
Expected new cases = (Incidence rate × Population) / 1000 = (3 × 50000) / 1000 = 150.
Correct Answer: C — 300
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Q. If the inductance of a coil is 0.5H and the frequency is 60Hz, what is the inductive reactance? (2023)
A.
31.42 ohms
B.
18.85 ohms
C.
37.70 ohms
D.
25.13 ohms
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Solution
Xl = 2πfL = 2π × 60 × 0.5 = 31.42 ohms.
Correct Answer: A — 31.42 ohms
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Q. If the inductance of a coil is 2 H and the frequency is 50 Hz, what is the inductive reactance?
A.
100 π
B.
100
C.
314
D.
628
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Solution
Inductive reactance X_L = 2πfL = 2π(50)(2) = 628 Ω.
Correct Answer: C — 314
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Q. If the inductance of a coil is doubled, what happens to its inductive reactance at a constant frequency? (2022)
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
Inductive reactance (X_L) is given by X_L = 2πfL. If the inductance (L) is doubled, X_L will also double.
Correct Answer: A — It doubles
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Q. If the inductive reactance of a coil is 30Ω and the resistance is 10Ω, what is the total impedance? (2018)
A.
10Ω
B.
30Ω
C.
33.33Ω
D.
40Ω
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Solution
Total impedance Z = √(R² + (XL)²) = √(10² + 30²) = √(100 + 900) = √1000 = 31.62Ω.
Correct Answer: C — 33.33Ω
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Q. If the inflation rate is 5% and the current price of a commodity is $200, what will be the price after one year? (2023)
A.
$210
B.
$205
C.
$215
D.
$220
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Solution
Price after one year = Current Price * (1 + Inflation Rate) = 200 * (1 + 0.05) = 200 * 1.05 = $210.
Correct Answer: A — $210
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Q. If the inflation rate is 5%, how much will a product that costs $100 cost after one year? (2023)
A.
$105
B.
$95
C.
$100
D.
$110
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Solution
With a 5% inflation rate, a $100 product will cost $100 + ($100 * 0.05) = $105 after one year.
Correct Answer: A — $105
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Q. If the intensity of light at a point of constructive interference is I, what is the intensity at a point of destructive interference?
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Solution
At destructive interference, the intensity is zero because the waves cancel each other out.
Correct Answer: B — 0
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Q. If the intensity of light is doubled while keeping the frequency constant, what happens to the number of emitted electrons?
A.
It doubles
B.
It remains the same
C.
It is halved
D.
It becomes zero
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Solution
Doubling the intensity of light increases the number of photons incident on the surface, which in turn increases the number of emitted electrons, assuming the frequency is above the threshold frequency.
Correct Answer: A — It doubles
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Q. If the intensity of light is doubled while keeping the frequency constant, what happens to the number of emitted electrons in the photoelectric effect?
A.
It doubles
B.
It remains the same
C.
It is halved
D.
It becomes zero
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Solution
Doubling the intensity of light increases the number of photons incident on the surface, which in turn increases the number of emitted electrons, assuming the frequency is above the threshold frequency.
Correct Answer: A — It doubles
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