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\(\frac{load}{area}\) = \(\frac{effort}{area}\) \(\frac{load}{5}\) = \(\frac{100}{0.5}\) load = 1000N

A floating body displaces its own weight of the liquid in which it floats. i.e mass of block = mass of water = mass of liquid let the fraction of the volume of the block in liquid be q \(\frac{2}{3}\)V x 1 = Vq X 0.8 q = \(\frac{2V}{2.4V}\) = \(\frac{5}{6}\) Therefore, the fraction of the height above the liquid level is (1–\(\frac{5}{6}\)) = \(\frac{1}{6}\).

pressure = h x g x e 105 = 10 x h x 1 h = 104

Since the resistors are in parallel, the p.d across them is the same, P = VI but I = \(\frac{ V}{R}\) P = \(\frac{ V^2}{R}\) = power dissipated by 5\Omega resistor, 40 = \(\frac{ V^2}{5}\) \(V^2\) = 200V. Power dissipated by the 10\Omega resistor, P = \(\frac{ V^2}{R}\) = \(\frac{200}{10}\) P = 20W.

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

The slope of the graph of distance versus time squared is equal to \(\frac{1}{2}\)a, which means the slope measures half the acceleration. Also from s = ut + \(\frac{1}{2}\)at\(^2\) when u = 0 then, s = \(\frac{1}{2}\)at\(^2\) So, \(\frac{\Delta\text{s}}{\Deltat^2}\) = \(\frac{1}{2}\)a

Moment = force x distance = m x a x d = kg x \(\frac{m}{s^2}\) x \(\frac{m}{1}\) = \(ML^2T^{-2}\)

Mass = Density x Volume 4.2 x 7.2 x 102 = v x 8.4 x 102 = 3.6m3

Heat energy = electrical energy mL = I2Rt 5 x L = 2 x 2 x 5 x 30 L = 120Jg-1

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

V.R = \(\frac{1}{sin 30}\) = 2 Efficiency = \(\frac{m.A}{V.R}\) x 100% 60 = \(\frac{120}{effort}\) x \(\frac{1}{2}\) x 100 Effort = 100N

R = \(\sqrt{12^2 + 5^2}\) R = 13N and R = F = ma 13 = 5 x a = 2.60ms\(^{-2}\)

i. There is tension on the water surface. The surface tension of water supports the needle, allowing it to float against the force of gravity. ii. Kerosene reduces the density of water so that the needle becomes denser than water. Kerosene does not change the density of water. It forms a separate layer on top, while the density of water remains unchanged. iii. Kerosene reduces the surface tension of water. Adding kerosene decreases the surface tension of water, which can eliminate the buoyant force that keeps the needle afloat. Therefore, The accurate statements are i and iii. are correct.

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

W = \(\frac{v}{r}\) = \(\frac{0.88}{0.02}\) = 44 rad/s 2\(\pi\)F = W = 44 F = \(\frac{44}{2 \times 3.142}\) = 7Hz