Q. If the quadratic equation x^2 + 2px + p^2 - 4 = 0 has roots that are equal, what is the value of p?
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Solution
Setting the discriminant to zero: (2p)^2 - 4(1)(p^2 - 4) = 0 leads to p = ±2.
Correct Answer: C — -2
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Q. If the quadratic equation x^2 + 2x + k = 0 has equal roots, what is the value of k?
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Solution
For equal roots, the discriminant must be zero: 2^2 - 4*1*k = 0, leading to k = 1.
Correct Answer: C — -1
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Q. If the quadratic equation x^2 + 2x + k = 0 has no real roots, what is the condition for k?
A.
k < 0
B.
k > 0
C.
k >= 0
D.
k <= 0
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Solution
For no real roots, the discriminant must be less than zero: 2^2 - 4*1*k < 0 => 4 - 4k < 0 => k > 1.
Correct Answer: A — k < 0
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Q. If the quadratic equation x^2 + 2x + k = 0 has no real roots, what is the condition on k?
A.
k < 0
B.
k > 0
C.
k >= 0
D.
k <= 0
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Solution
For no real roots, the discriminant must be less than zero: 2^2 - 4*1*k < 0, hence k > 1.
Correct Answer: A — k < 0
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Q. If the quadratic equation x^2 + 2x + k = 0 has roots that are equal, what is the value of k?
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Solution
For equal roots, the discriminant must be zero: 2^2 - 4*1*k = 0 leads to k = -1.
Correct Answer: D — -2
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Q. If the quadratic equation x^2 + 4x + c = 0 has one root equal to -2, what is the value of c?
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Solution
If one root is -2, then substituting x = -2 gives: (-2)^2 + 4(-2) + c = 0 => 4 - 8 + c = 0 => c = 4.
Correct Answer: A — 0
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Q. If the quadratic equation x^2 + 4x + k = 0 has roots -2 and -2, what is the value of k?
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Solution
Using the formula for roots, k = (-2)^2 - 4*(-2) = 4 + 8 = 12.
Correct Answer: B — 4
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Q. If the quadratic equation x^2 + 6x + 9 = 0 is solved, what is the nature of the roots?
A.
Real and distinct
B.
Real and equal
C.
Complex
D.
None of the above
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Solution
The discriminant is 0, indicating that the roots are real and equal.
Correct Answer: B — Real and equal
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Q. If the quadratic equation x^2 + 6x + k = 0 has roots -2 and -4, what is the value of k?
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Solution
Using Vieta's formulas, k = (-2)(-4) = 8.
Correct Answer: B — 12
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Q. If the quadratic equation x^2 + 6x + k = 0 has roots that are both negative, what is the condition for k?
A.
k > 9
B.
k < 9
C.
k = 9
D.
k < 0
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Solution
For both roots to be negative, k must be greater than the square of half the coefficient of x, hence k > 9.
Correct Answer: A — k > 9
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Q. If the quadratic equation x^2 + bx + 9 = 0 has roots 3 and -3, what is the value of b?
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Solution
The sum of the roots is 3 + (-3) = 0, so b = -0.
Correct Answer: C — -6
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Q. If the quadratic equation x^2 + kx + 16 = 0 has equal roots, what is the value of k?
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Solution
For equal roots, the discriminant must be zero: k^2 - 4*1*16 = 0, thus k = -8.
Correct Answer: A — -8
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Q. If the quadratic equation x^2 + kx + 9 = 0 has no real roots, what is the condition on k?
A.
k < 6
B.
k > 6
C.
k < 0
D.
k > 0
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Solution
The discriminant must be less than zero: k^2 - 4*1*9 < 0 => k^2 < 36 => |k| < 6.
Correct Answer: B — k > 6
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Q. If the quadratic equation x^2 + mx + n = 0 has roots 1 and -3, what is the value of n?
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Solution
Using Vieta's formulas, the product of the roots is n = 1 * (-3) = -3.
Correct Answer: A — -3
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Q. If the quadratic equation x^2 + mx + n = 0 has roots 1 and -3, what is the value of m?
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Solution
Using Vieta's formulas, m = -(1 + (-3)) = 2.
Correct Answer: A — 2
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Q. If the quadratic equation x^2 + mx + n = 0 has roots 2 and -3, what is the value of m + n?
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Solution
Using Vieta's formulas, m = -(-1) = 1 and n = 2*(-3) = -6, thus m + n = 1 - 6 = -5.
Correct Answer: B — 5
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Q. If the quadratic equation x^2 + px + q = 0 has roots 2 and 3, what is the value of p + q?
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Solution
Using Vieta's formulas, p = -(2 + 3) = -5 and q = 2*3 = 6. Thus, p + q = -5 + 6 = 1.
Correct Answer: C — 7
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Q. If the quadratic equation x^2 + px + q = 0 has roots 2 and 3, what is the value of p?
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Solution
The sum of the roots is -p = 2 + 3 = 5, so p = -5.
Correct Answer: A — -5
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Q. If the quadratic equation x^2 - kx + 9 = 0 has equal roots, what is the value of k?
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Solution
For equal roots, the discriminant must be zero: k^2 - 36 = 0, hence k = 6.
Correct Answer: A — 6
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Q. If the radius of a charged sphere is halved while keeping the charge constant, what happens to the electric field at the surface?
A.
It remains the same
B.
It doubles
C.
It halves
D.
It quadruples
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Solution
The electric field at the surface of a sphere is given by E = Q/(4πε₀R²). If R is halved, E increases by a factor of 4.
Correct Answer: B — It doubles
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Q. If the radius of a circular loop carrying current is doubled, how does the magnetic field at the center change?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
The magnetic field at the center of a circular loop is inversely proportional to the radius; thus, doubling the radius halves the magnetic field.
Correct Answer: B — It halves
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Q. If the radius of a circular loop carrying current is doubled, what happens to the magnetic field at the center of the loop?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
The magnetic field at the center of a circular loop is given by B = (μ₀I)/(2r). If the radius is doubled, the magnetic field strength is halved.
Correct Answer: B — It halves
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Q. If the radius of a circular loop carrying current is halved, how does the magnetic field at the center change?
A.
Remains the same
B.
Doubles
C.
Halves
D.
Quadruples
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Solution
The magnetic field at the center is inversely proportional to the radius, so it quadruples.
Correct Answer: D — Quadruples
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Q. If the radius of a disc is doubled while keeping its mass constant, how does its moment of inertia change?
A.
It remains the same
B.
It doubles
C.
It quadruples
D.
It halves
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Solution
The moment of inertia of a disc is I = 1/2 MR^2. If R is doubled, I becomes 1/2 M(2R)^2 = 2MR^2, which is quadrupled.
Correct Answer: C — It quadruples
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Q. If the radius of a disk is doubled while keeping its mass constant, how does its moment of inertia change?
A.
Increases by a factor of 2
B.
Increases by a factor of 4
C.
Remains the same
D.
Decreases by a factor of 4
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Solution
The moment of inertia of a disk is I = 1/2 MR^2. If R is doubled, I becomes 1/2 M(2R)^2 = 2MR^2, which is 4 times the original.
Correct Answer: B — Increases by a factor of 4
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Q. If the radius of a planet is halved while keeping its mass constant, how does the gravitational acceleration at its surface change?
A.
It becomes four times stronger
B.
It becomes twice stronger
C.
It remains the same
D.
It becomes half as strong
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Solution
If the radius is halved, the gravitational acceleration becomes four times stronger, as g is inversely proportional to the square of the radius.
Correct Answer: A — It becomes four times stronger
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Q. If the radius of a planet is halved, what happens to the gravitational acceleration on its surface?
A.
It doubles
B.
It halves
C.
It becomes one-fourth
D.
It remains the same
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Solution
g ∝ 1/R². If R is halved, g becomes 4 times greater.
Correct Answer: A — It doubles
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Q. If the radius of a rotating disc is doubled while keeping the mass constant, how does the angular momentum change if the angular velocity remains the same?
A.
It doubles
B.
It remains the same
C.
It quadruples
D.
It halves
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Solution
Angular momentum L = Iω; if radius is doubled, moment of inertia I increases by a factor of 4, hence L quadruples.
Correct Answer: C — It quadruples
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Q. If the radius of a rotating object is halved while keeping the angular velocity constant, what happens to the linear velocity at the edge?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It becomes zero
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Solution
Linear velocity v = rω. If r is halved and ω remains constant, v also halves.
Correct Answer: B — It halves
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Q. If the radius of a rotating object is halved while keeping the mass constant, how does its moment of inertia change?
A.
It remains the same
B.
It doubles
C.
It halves
D.
It reduces to one-fourth
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Solution
Moment of inertia I is proportional to the square of the radius, so halving the radius reduces I to one-fourth.
Correct Answer: D — It reduces to one-fourth
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