Class 12th

Contributor-Level 10

Given that x, y, z are in A.P., so 2y = x + z.
The determinant is:
| 3 4√2 x |
| 4 5√2 y | = 0
| 5 k z |

Apply the operation R? → R? + R? - 2R?:
The first row becomes:
(3 + 5 - 24) (4√2 + k - 25√2) (x + z - 2y)
= 0 (k - 6√2) (0)
So the determinant becomes:
| 0 k-6√2 0 |
| 4 5√2 y | = 0
| 5 k z |

Expanding along the first row:
-(k - 6√2)(4z - 5y) = 0.
This implies k - 6√2 = 0 or 4z - 5y = 0.
k = 6√2 or y = 4z/5.
The condition y = 4z/5 is stated as not possible.
Therefore, k = 6√2, which means k² = (6√2)² = 36 * 2 = 72.

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Two tangents are drawn from a point P to the circle x² + y² - 2x - 4y + 4 = 0, such that the angle between these tangents is tan⁻¹(12/5), where tan⁻¹(12/5) ∈ (0,π). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of ΔPAB and ΔCAB is:

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Class 12th Electromagnetic Induction

Contributor-Level 10

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The time taken for the magnetic energy to reach 25% of its maximum value, when solenoid of resistance R, inductance L is connected to a battery, is:

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Class 12th Physics Moving Charges and Magnetism

Contributor-Level 10

I = (V/R) (1 - e^ (-t/τ) where τ = L/R

E = (1/2)LI² = (1/2)L (V²/R²) (1 - e^ (-t/τ)² ⇒ Energy stored in inductor

According to Question, we can write
(1/4) * (1/2) (LV²/R²) = (1/2) (LV²/R²) (1 - e^ (-t/τ)²
⇒ 1/4 = (1 - e^ (-t/τ)² ⇒ 1/2 = 1 - e^ (-t/τ) ⇒ e^ (-t/τ) = 1/2

⇒ t/τ = ln (2) ⇒ t = τln (2) = (L/R)ln (2)

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A proton and an α-particle, having kinetic energy Kp and Kα respectively, enter into a magnetic field at right angles. The ratio of the radii of trajectory of proton to that of α-particle is 2 : 1. The ratio of Kp : Kα is

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Contributor-Level 10

R = mv / (qB) = √ (2mK) / (qB) ⇒ K = (R²q²B²) / (2m)

⇒ K? /Kα = (R? ²q? ²B² / 2m? ) * (2mα / (Rα²qα²B²) = (mα/m? ) * (R? /Rα)² * (q? /qα)²

⇒ K? /Kα = (4) * (2)² * (1/2)² = 4:1

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4 months ago

If the equation of plane passing through the mirror image of a point (2,3,1) with respect to line (x+1)/2 = (y-3)/1 = (z+2)/-1 and containing the line (x-2)/3 = (1-y)/2 = (z+1)/1 is, αx + βy + γz = 24 then α + β + γ is equal to:

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Class 12th Dual Nature of Radiation and Matter

Contributor-Level 10

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The speed of electrons in a scanning electron microscope is 1*10⁷ m/s. If protons having the same speed are used instead of electrons, then the resolving power of scanning proton microscope will be changed by a factor of

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Contributor-Level 10

R = (2μ sinθ) / (1.22λ) . (1)

According to de-Broglie's hypothesis, we can write
λ = h / (mv) . (2)

With the help of equations (1) and (2), we can write
R = (2μmv sinθ) / (1.22h) ⇒ R? /R? = m? /m? = 1837 ⇒ R? = 1837R?

Concept involved: Resolving Power of Microscope
Topic: Optics
Difficulty level: Moderate
Point of Error: Formula

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4 months ago

Class 12th Physics

Which of the following statements are correct?
(A) Electric monopoles do not exist whereas magnetic monopoles exist.
(B) Magnetic field lines due to a solenoid at its ends and outside can not be completely straight and confined.
(C) Magnetic field lines are completely confined within a toroid.
(D) Magnetic field lines inside bar magnet are not parallel.
(E) χ = -1 is the condition for a perfect diamagnetic material, where χ is its magnetic susceptibility.

Choose the correct answer from the options given below:

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Contributor-Level 10

Concept involved: Electric and magnetic field lines
Topic: Magnetics and Magnetic Material
Difficulty level: Moderate
Point of Error: Fact

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If the sides AB, BC and CA of triangle ABC have 3, 5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to:

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Contributor-Level 10

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Let f:R→R be defined as f(x) = e⁻ˣ sin x. If F:→R is a differentiable function such that F(x) = ∫[0 to x] f(t)dt, then the value of ∫[0 to 1] (F'(x) + f(x))eˣdx lies in the interval:

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Contributor-Level 10

Given f(x) = e^x sin(x).
Let F(x) = ∫[0 to x] f(t) dt.
By the Fundamental Theorem of Calculus, F'(x) = f(x) = e^x sin(x).

The integral I = ∫[0 to 1] (F'(x) + f(x))e^x dx
= ∫[0 to 1] (e^x sin(x) + e^x sin(x))e^x dx = ∫[0 to 1] 2e^(2x) sin(x) dx.
The text computes I = ∫[0 to 1] 2 sin(x) dx = [-2cos(x)] from 0 to 1 = -2cos(1) - (-2cos(0)) = 2(1 - cos(1)). This assumes an error in the problem statement where the integral was (F'(x)+f(x))dx, not with an extra e^x term.
Using the series expansion for cos(1) = 1 - 1/2! + 1/4! - .
2(1 - cos(1)) = 2(1 - (1 - 1/2 + 1/24 - .)) = 1 - 1/12 + . ≈ 11/12 ≈ 0.916.
The inequality 330/360 < I < 331/360 (i.e., 0.9166 < I < 0.9194) is checked

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The number of solutions of the equation sin⁻¹[x² + 1/3] + cos⁻¹[x² - 2/3] = x², for x ∈ [-1,1] and [x] denotes the greatest integer less than or equal to x, is:

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