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No expanded working has been attached for this question yet. The verified answer is Option A.

distance = area under the graph. = \(\frac{1}{2}\times 1 \times 4 ) + \frac{1}{2} \times (4+6)\times 1) + (6 \times 1.5) + (\frac{1}{2}\times(2 + 6)\times 0.5)\) = 18m

P.E\(_{mgh}\) = \(\frac{1}{2}\) * k * x\(^2\) mgh = \(\frac{1}{2}\) * k * x\(^2\) 50 * 10 * 10 = \(\frac{1}{2}\) * 4 * 10\(^4\) * x\(^2\) → x\(^2\) = \(\frac{5000 * 2}{4 * 10^4}\) → x = \sqrt[0.25] : x = 0.50m

\(efficiency = \frac{workout}{workinput}\) = \(\frac{E}{Q_1}\)

Force is a derived quantity, it's unit is a derived unit

Using V2 = U2 - 2gS V = 0ms-1 at greatest height S = \(\frac{u^2}{2g}\)

F = ma and V2 = U2 + 2aX F = m\(\frac{(V^2 - U^2)}{2X}\) = 0.04\(\frac{(500^2 - 0^2)}{2 \times 50}\) F = 100N

Volume = 2 x 4 x 3 = 24m3 mass = V x D = 24 x 103 F = mg = 24 x 103 x 10 = 24 x 104

T = , m = 1000kg, a = 1\(ms^{-1}\), g = 10\(ms^{-2}\) W - T = ma T = W - ma T = mg - ma T = m( g - a ) T = 1000( 10 - 1) = 1000 x 9 = 9000N.

Specific latent heat of fusion of lead is the amount of heat energy required to change the state of 1 kg (unit mass) of solid lead into liquid lead at its melting point, without any change in temperature.

Mercury is employed in the barometer because its density is sufficiently high for a relative short column to be obtained.

\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) \(\frac{300}{27 + 273}\) = \(\frac{V_2}{54 + 273}\) V2 = 327cm3

\(\lambda\) = \(\frac{v}{f}\) = \(\frac{20}{10}\) = 2cm ; distance between adjacent nodes = 2cm \ 2 = 1cm

Beta particles have a negative charge (for electrons) or a positive charge (for positrons). When beta particles enter a magnetic field, they experience a force due to their charge, which causes them to curve in a direction determined by the right-hand rule (for negative charges) or left-hand rule (for positive charges). This results in significant deflection.

No expanded working has been attached for this question yet. The verified answer is Option C.