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F\(_c\) = \(\frac{mv^2}{r}\) mass = 200g = 0.2kg, v = 5 m/s, r = 1.5 m F\(_c\) = \(\frac{0.2 \times 5 \times 5}{1.5}\) ≈ 3.33N At the bottom of the circle, the tension T in the string must counteract the weight of the stone as well as provide the centripetal force: T = F\(_c\) + mg T = 3.33 + 0.2 x 10 = 5.3N

F = mass x acceleration due to gravity F\(_c\) = m x a\(_c\) a\(_c\) = a\(_r\) = 4.3 x 10\(^7\)ms\(^{-2}\), m = 3.6 x 10\(^{-6}\)kg F\(_c\) = m x a\(_c\) = 3.6 x 10\(^{-6}\) x 4.3 x 10\(^7\) = 154.8N ≈ 155N ⇒ 1.55 x 10\(^2\)N

The attraction between molecules of the same substance is referred to as cohesive force. These forces enable molecules to adhere to one another, which affects properties such as surface tension in liquids and the structural integrity of materials.

Density = \(\frac{\text{mass}}{\text{volume}}\) Volume of solid cube = s\(^3\) = 1.5\(^3\) = 3.375cm\(^3\) Density = 2700kgm\(^{-3}\) ⇒ 2.7gcm\(^{-3}\) 2.7 = \(\frac{\text{m}}{3.375}\) mass of the cube = 2.7 x 3.375 = 9.1125g \(\approxeq\) 9.1g

Density = \(\frac{\text{mass}}{\text{volume}}\) mass of hydrometer = y kg 80% hydrometer's volume submerge(20% above the fluid) = \(\frac{4}{5}\) x 2y x 10\(^{-5}\) = \(\frac{8y \times 10^{-5}}{5}\) Density = y x \(\frac{5}{8y\times 10^{-5}}\) = 6.25 x 10\(^4\)kgm\(^3\)

For Hare's apparatus Relative density = \(\frac{h_w}{h_l}\) Given: height of liquid = 0.5cm, height of water = 0.3cm Relative density = \(\frac{0.3}{0.5}\) = 0.6

Upthrust(Force) = volume of object x density of liquid x g = V x \(\rho\) x g U = 4.2 x 10\(^{-4}\) x 1028 x 10 \(\approxeq\) 4.3N

Pressure is defined as force per unit area (P = \(\frac{F}{A}\)). When a person stands on one foot, all their weight is concentrated on a smaller area (the sole of one foot), resulting in higher pressure.

Volume of cube = s\(^3\) = 10\(^3\) = 1000cm\(^3\) Volume of cube submerged = 10 cm x 10 cm x 4.5 cm = 450 cm\(^3\) Volume of water displaced = volume of cube submerged = 450 cm\(^3\) Mass of water displaced = 1 cm\(^3\) x 450 cm\(^3\) = 450g (density of water is 1 cm\(^3\)) Mass of cube = V\(_{cube}\) x \(\rho_{cube}\) But, mass of cube = mass of water displaced 1000 cm\(^3\) x \(\rho_{cube}\) = 450g \(\rho_{cube}\) = \(\frac{450}{1000}\) = 0.45 g/cm\(^3\)

Stress = \(\frac{force}{area}\) Force = weight = y\(^2\), Area = \(\pi\)r\(^2\), Diameter = \(\frac{\text{y}}{\pi}\)m ⇒ r = \(\frac{\text{y}}{2\pi}\)m Stress = \(\frac{y^2}{\pi \times\frac{y^2}{(2\pi)^2}}\) = \(\frac{y^2}{\pi y^2} \times 4(\pi)^2\) = 4\(\pi\)Nm\(^{-2}\)

Young's modulus E is given by the formula: E = \(\frac{\text{STRESS}}{\text{STRAIN}}\) Stress = \(\frac{Force}{Area}\) Dimension of force = MLT\(^{-2}\) Dimension of Area = L\(^2\) Dimension of strain = 1(dimensionless) E = \(\frac{MLT^{-1}}{L^2}\) = ML\(^{-1}\)T\(^{-2}\)

The property described in this question is called Elasticity which is the ability of a substance to regain its original shape and size after the removal of the distorting force.

F = ke k = \(\frac{F_1}{e_1}\) F\(_1\) = 10N, e\(_1\) = 0.02m F\(_2\) = 40N, e\(_2\) = ? \(\frac{F_1}{e_1}\) = \(\frac{F_2}{e_2}\) \(\frac{10}{0.02}\) = \(\frac{40}{e_2}\) e\(_2\) = \(\frac{40 \times 0.02}{10}\) = 0.08m The new length = 1 + 0.08 = 1.08m

The term with a different dimension from the others is D. Torque, as its dimensions ML\(^2\)T\(^{−2}\) which differ from the dimensions of stress and pressure, which are ML\(^{−1}\)T\(^{−2}\)

(a) SURFACE TENSION = \(\frac{\text{force}}{\text{length}}\) FORCE = mass x accel. due to gravity Dimension of force = M.LT\(^{-2}\) Dimension of length = L Dimension of surface area = \(\frac{MLT^{-2}}{L}\) = MT\(^{-2}\) (b) The instruments used for measuring the force of gravity at a place are: (1) Gravimeter, (2) Spring balance, (3) Gravitational accelerometer.