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Upthrust = Weight of fluid displayed = 6N

Pressure = \(\frac{\text{Force}}{\text{cross-sectional area}}\) From the equation above Option A is incorrect: increases with an increase in surface area

Upthrust = 96 - 32 = 64N mass(m) = \(\frac{64}{10} = 6.4 kg\) volume = \(\frac{mass}{Density} = \frac{6.4}{1000}\) = 6.4 x 10-3m3

Given that: Weight of the block: 15N Horizontal force applied: 3N Since the applied force (3 N) is less than the maximum static friction (which can be greater than or equal to the weight of the block), the friction force will match the applied force to prevent motion Therefore, The friction force on the block is: 3N

For a body moving at a constant speed, the distance x traveled can be represented by the equation: x = U t (constant speed = zero acceleration)

The amplitude of the motion of a body performing simple harmonic motion (SHM) decreases over time primarily due to energy loss, often caused by friction or other resistive forces. Therefore OPTION A is correct.

amax = \(\omega^2 r = 10^2 \times \frac{5}{100} = 5ms^{-2}\)

The time rate of change of momentum is defined as force.

In an elastic collision, both kinetic energy and momentum are conserved. Therefore, the correct answer is C. Both kinetic energy and momentum are conserved.

Electrical resistance is a property that causes electrical energy to be converted into heat energy Therefore, the correct option is B heat energy.

M.A = \(\frac{load}{effort} \frac{5000N}{500N} = 10\) V.R = \(\frac{M.A \times 100}{V} = \frac{10 \times 100}{75} = 13.3\)

P = \(\frac{25}{50} = 0.5\) V.R = \(\frac{2\pi D}{p} = \frac{2 \times 3.142 \times 12}{0.5}\) = 151

E x 3 = 80 x 0.5 E3 = 40 E = \(\frac{40}{3} = 13.3N\)

85\(\Omega\) ........100oC 52\(\Omega\) ........\(\theta\)oC 20\(\Omega\) ........0oC \(\frac{52 - 20}{85 - 20} = \frac{\theta - 0}{100 - 0}\) \(\theta = \frac{32 \times 100}{65} = 49.2^oC\)

\(P_2 = \frac{p_1 \times T_2}{T_1} = \frac{500 \times 520}{260}\) = 1000N\(m^{-2}\)