E maximum potential = 1 /2 C x₀²

If x = 1 /2 x₀ E pot = 1 / 2 C (1/ 2 x_o)² = 1 /4 x (1 / 2 C x₀²)

E pot = 1/ 4 E maximum potential

Thus, for x = 1/ 2 x₀ KE = 3/ 4 E maximum potential = 3 / 4 x 16 J = 12 J

Answer C is correct.

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An alpha decay can be considered as an explosion.

The momentum before the explosion is zero and due to the law of conservation of momentum the momentum after the explosion is zero as well.

Let u be the velocity of the nucleus after the decay.

0 = m nucleus x u + m alpha x v

An alpha particle is built up by a helium nucleus and its mass is 4 atomic units.

The nucleus after the decay has a mass (210 – 4) = 206

Thus, 0 = 206 x u + 4 x v

u = (-) 4 v ÷ 206

Answer D is correct.

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If the intensity of the unporalized light = I₀ , then after sheet 1 the intensity = 1 /2 I₀.

The polarizing sheets are initially parallel and the intensity after sheet 2 is 1 /2 I₀ as well, because the second sheet does not affect the polarisation of the light coming into the second sheet when it is parallel to sheet 1.

However, by turning sheet 2 by an angle θ the intensity after sheet 2 becomes according to Malus’s Law 1/ 2 I₀ cos² θ.

To get an intensity after sheet 2 equal to 1 /4 I₀ , cos² θ should be equal to 1 / 2.

And cos θ = √1/ 2 = 1/2 √ 2 .

And thus, the angle θ = cos^-1 (1/2 √ 2 ) .

It does not matter whether you turn sheet 2 and keep sheet 1 fixed or you turn sheet 1 and keep sheet 2 fixed.
What matters is their mutual orientation.

Answer C is correct.

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Apply the formula s = 1/ 2 g t²

The fourth second lasts from t = 3 s to t = 4 s.

t = 3 seconds s = 1 /2 x 10 x (3)² = 45 m

t = 4 seconds s = 1 /2 x 10 x (4)² = 80 m

Distance travelled in the fourth second = 80 m - 45 m = 35 m

Answer C is correct.

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