Q. If the area of a rectangle is calculated as 50 m² with a length of 10 m and an uncertainty of ±0.1 m in length, what is the uncertainty in the area?
A.
1 m²
B.
0.5 m²
C.
0.2 m²
D.
0.1 m²
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Solution
Uncertainty in area = 2 * length * uncertainty in length = 2 * 10 * 0.1 = 2 m².
Correct Answer: B — 0.5 m²
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Q. If the area of a triangle is 30 square units and the base is 10 units, what is the height?
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Solution
Area = 1/2 * base * height => height = 30/(1/2 * 10) = 6.
Correct Answer: B — 5
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Q. If the area of triangle ABC is 30 cm² and the base BC = 10 cm, what is the height from A to BC?
A.
5 cm
B.
6 cm
C.
7 cm
D.
8 cm
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Solution
Area = 1/2 * base * height. Therefore, 30 = 1/2 * 10 * height, height = 6 cm.
Correct Answer: B — 6 cm
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Q. If the area of triangle ABC is 30 square units and the base BC = 10 units, what is the height from A to BC?
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Solution
Area = 1/2 * base * height => 30 = 1/2 * 10 * height => height = 30 / 5 = 6.
Correct Answer: A — 5
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Q. If the area of triangle ABC is 30 square units and the base BC is 10 units, what is the height from A to BC?
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Solution
Area = 1/2 * base * height. Thus, 30 = 1/2 * 10 * height. Height = 6.
Correct Answer: B — 5
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Q. If the area of triangle ABC is 60 cm² and the base BC = 12 cm, what is the height from A to BC?
A.
5 cm
B.
10 cm
C.
12 cm
D.
15 cm
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Solution
Area = (1/2) * base * height. Therefore, 60 = (1/2) * 12 * height, height = 10 cm.
Correct Answer: B — 10 cm
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Q. If the area of triangle JKL is 30 cm² and the base JK is 10 cm, what is the height from point L?
A.
3 cm
B.
6 cm
C.
5 cm
D.
4 cm
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Solution
Area = 1/2 * base * height. 30 = 1/2 * 10 * height. Therefore, height = 6 cm.
Correct Answer: C — 5 cm
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Q. If the azimuthal quantum number l = 1, what is the shape of the orbital?
A.
Spherical
B.
Dumbbell
C.
Double dumbbell
D.
None of the above
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Solution
The azimuthal quantum number l = 1 corresponds to p orbitals, which have a dumbbell shape.
Correct Answer: B — Dumbbell
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Q. If the balance length of a potentiometer is 50cm for a cell of unknown emf, and the potential gradient is 4 V/m, what is the emf of the cell?
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Solution
The emf is calculated as V = potential gradient × length = 4 V/m × 0.5 m = 2 V.
Correct Answer: B — 4V
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Q. If the balancing length of a potentiometer is found to be 40 cm for a cell of emf 2V, what is the potential gradient if the total length of the wire is 100 cm?
A.
5 V/m
B.
2 V/m
C.
4 V/m
D.
3 V/m
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Solution
The potential gradient is V/L = 2V/0.4m = 5 V/m.
Correct Answer: A — 5 V/m
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Q. If the center of a circle is at (0, 0) and it passes through the point (3, 4), what is the radius of the circle?
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Solution
The radius is the distance from the center to the point, which is √(3² + 4²) = √25 = 5.
Correct Answer: A — 5
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Q. If the center of a circle is at (0, 0) and it passes through the point (3, 4), what is the equation of the circle?
A.
x² + y² = 25
B.
x² + y² = 12
C.
x² + y² = 7
D.
x² + y² = 16
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Solution
The radius is 5 (distance from (0,0) to (3,4)), so the equation is x² + y² = 5² = 25.
Correct Answer: A — x² + y² = 25
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Q. If the charge density of a non-conducting sphere increases linearly with radius, how does the electric field vary inside the sphere?
A.
Linearly with radius
B.
Quadratically with radius
C.
Constant
D.
Inversely with radius
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Solution
The electric field inside a non-conducting sphere with linearly increasing charge density varies linearly with radius.
Correct Answer: A — Linearly with radius
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Q. If the charge density of a non-uniform spherical charge distribution varies as ρ(r) = kr², what is the electric field at the center of the sphere?
A.
0
B.
k/3ε₀
C.
k/4ε₀
D.
k/2ε₀
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Solution
At the center of a non-uniform spherical charge distribution, the electric field is zero due to symmetry.
Correct Answer: A — 0
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Q. If the charge density of a spherical charge distribution increases linearly with radius, how does the electric field vary inside the sphere?
A.
Linearly with radius
B.
Quadratically with radius
C.
Inversely with radius
D.
Constant
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Solution
The electric field inside a non-uniform charge distribution varies quadratically with radius due to the increasing charge density.
Correct Answer: B — Quadratically with radius
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Q. If the charge inside a closed surface is doubled, what happens to the electric flux through the surface?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It becomes zero
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Solution
According to Gauss's law, the electric flux is directly proportional to the enclosed charge, so it doubles.
Correct Answer: A — It doubles
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Q. If the charge on a capacitor is doubled while the voltage remains constant, what happens to the capacitance?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
Capacitance C = Q/V; if Q is doubled and V remains constant, C must halve.
Correct Answer: B — It halves
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Q. If the charge on a capacitor is doubled, what happens to the energy stored in it?
A.
It doubles
B.
It quadruples
C.
It remains the same
D.
It halves
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Solution
The energy stored in a capacitor is given by U = 1/2 * C * V^2. If the charge is doubled, the energy stored quadruples.
Correct Answer: B — It quadruples
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Q. If the circumference of a circle is measured as 31.4 cm with an error of 0.2 cm, what is the percentage error?
A.
0.64%
B.
0.5%
C.
1%
D.
0.2%
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Solution
Percentage error = (Absolute error / Measured value) * 100 = (0.2 / 31.4) * 100 ≈ 0.64%.
Correct Answer: A — 0.64%
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Q. If the circumference of a circle is measured as 31.4 cm with an error of 0.2 cm, what is the radius?
A.
5 cm
B.
4.5 cm
C.
5.0 cm
D.
6 cm
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Solution
Circumference = 2πr. Measured circumference = 31.4 cm. Thus, r = 31.4 / (2π) ≈ 5.0 cm.
Correct Answer: C — 5.0 cm
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Q. If the circumradius of triangle ABC is 10 cm and the area is 48 cm², what is the length of the side opposite to angle A?
A.
12 cm
B.
14 cm
C.
16 cm
D.
18 cm
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Solution
Using the formula R = (abc)/(4 * Area), we can find the side opposite to angle A. Let a = side opposite to A. Then, a = (4 * Area * R) / (bc) = (4 * 48 * 10) / (b * c).
Correct Answer: B — 14 cm
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Q. If the circumradius R of triangle ABC is 5 cm, what is the maximum area of the triangle?
A.
12.5 cm²
B.
15 cm²
C.
20 cm²
D.
25 cm²
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Solution
The maximum area of a triangle with circumradius R is given by the formula Area = (abc)/(4R). For maximum area, the triangle should be equilateral, thus Area = (3√3/4) * (R^2) = (3√3/4) * (5^2) = 25√3/4 cm².
Correct Answer: C — 20 cm²
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Q. If the concentration of reactants is increased in a system at equilibrium, what will happen to the position of equilibrium?
A.
Shift to the right
B.
Shift to the left
C.
No change
D.
Depends on temperature
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Solution
According to Le Chatelier's principle, increasing the concentration of reactants will shift the equilibrium position to the right, favoring the formation of products.
Correct Answer: A — Shift to the right
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Q. If the coordinates of points A and B are (1, 2) and (3, 4) respectively, what is the distance AB?
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Solution
Distance AB = √((3-1)² + (4-2)²) = √(4) = 2.
Correct Answer: A — 2√2
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Q. If the coordinates of points A and B are (1, 2) and (4, 6) respectively, what is the distance AB?
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Solution
Distance AB = √((4-1)² + (6-2)²) = √(9 + 16) = √25 = 5.
Correct Answer: A — 5
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Q. If the coordinates of the vertices of a triangle are (1, 1), (4, 5), and (7, 2), what is the perimeter of the triangle?
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Solution
Calculate distances AB, BC, CA and sum them: AB = 5, BC = 5, CA = 7. Perimeter = 5 + 5 + 7 = 17.
Correct Answer: B — 14
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Q. If the coordinates of the vertices of a triangle are (1, 2), (3, 4), and (5, 2), what is the perimeter of the triangle?
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Solution
Perimeter = AB + BC + CA = √[(3-1)² + (4-2)²] + √[(5-3)² + (2-4)²] + √[(1-5)² + (2-2)²] = 2.83 + 2.83 + 4 = 10.
Correct Answer: B — 10
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Q. If the coordinates of the vertices of a triangle are (1, 2), (4, 6), and (7, 2), what is the perimeter of the triangle?
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Solution
Perimeter = AB + BC + CA = √[(4-1)² + (6-2)²] + √[(7-4)² + (2-6)²] + √[(1-7)² + (2-2)²] = 3 + √(9 + 16) + 6 = 3 + 5 + 6 = 14.
Correct Answer: B — 14
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Q. If the coordinates of the vertices of a triangle are A(1, 1), B(4, 5), and C(7, 2), what is the area of the triangle?
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Solution
Area = 1/2 | x1(y2-y3) + x2(y3-y1) + x3(y1-y2) | = 1/2 | 1(5-2) + 4(2-1) + 7(1-5) | = 1/2 | 3 + 4 - 28 | = 1/2 * 21 = 10.5.
Correct Answer: B — 12
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Q. If the critical angle for a certain material is 30°, what is the refractive index of that material?
A.
1.00
B.
1.15
C.
1.73
D.
2.00
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Solution
The refractive index n can be calculated using n = 1/sin(θc). For θc = 30°, n = 1/sin(30°) = 1/0.5 = 2.00.
Correct Answer: C — 1.73
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