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The net force (\(F_{\text{net}}\)) is the weight (\(W\)) minus the total upward forces (upthrust (\(U\)) and viscous drag (\(F_{\text{drag}}\))), and this net force causes the acceleration (\(a\)) according to Newton's second law (\(F_{\text{net}}=ma\)): \(F_{\text{net}} = W - U - F_{\text{drag}}\) \(ma = (m \times g) - U - F_{\text{drag}}\) \((10\text{ kg}) \times a = (10\text{ kg} \times 10\text{ m/s}^{2}) - 10\text{ N} - 15\text{ N}\) \(10a = 100\text{ N} - 25\text{ N}\) \(10a = 75\text{ N}\) \(a = \frac{75\text{ N}}{10\text{ kg}}\) \(a = 7.5\text{ m/s}^{2}\).

a. Surface tension: Surface tension is a property of a liquid whereby its surface behaves as if it were covered by an elastic skin. Or it is the tangential force of a liquid. b. The surface “skin” of a liquid exists because of the attractive forces between its molecules. A molecule inside the liquid is pulled equally in all directions by surrounding molecules, so the net force on it is zero. However, a molecule at the surface has more neighboring molecules below than above, creating a net inward (downward) pull. This inward force tends to minimize the surface area, causing the surface to behave like a stretched elastic skin. This effect is known as surface tension. See diagram above. Aim: To demonstrate surface tension in a liquid Apparatus: A needle, water, a beaker, and a blotting paper Procedure: The beaker of water is filled as shown above. A needle is now placed on a blotting paper and then pushed gently to the middle of the water surface. The paper wets and sinks, but the needle remains floating on the water's surface. Conclusion: The surface acts like a skin covering the liquid, which can support the weight of the needle. c. (i). When mercury spills on a glass sheet, it forms a spherical droplet (ii). A flowing needle on water (iii). A toy duck on liquid. d. The cohesion force between molecules of water is less than its adhesion to glass, therefore when water is allowed to spill on glass, it wets. Conversely, the cohesion between the molecules of mercury is greater than its adhesion to glass. Hence, it does not wet glass. e. (i). By adding detergents, (ii). By heating the liquid (ii). By addition of camphor

Answer: C Fraction immersed = \(\frac{\text{density of block}}{\text{density of liquid}}\) = \(\frac{400}{1600}\) = 0.25.

Upthrust = weight of fluid displaced = mg = 0.3 x 10 = 3.0N

The elastic potential energy \( E \) is proportional to the square of the force: \(E \propto F^2\) (for the same spring). The relationship between the energies for different forces is given by: \(\frac{E_2}{E_1} = \left( \frac{F_2}{F_1} \right)^2\) Now, substituting the known values: \(E_2 = 0.16 \times \left( \frac{700}{200} \right)^2 = 0.16 \times (3.5)^2 = 0.16 \times 12.25 = 1.96 \, \text{J}\)

Net force in the horizontal (x) direction: \(F_x = 8 \, \text{N} - 4 \, \text{N} = 4 \, \text{N} \quad \text{(to the right)}\) Net force in the vertical (y) direction: \(F_y = 15 \, \text{N} - 12 \, \text{N} = 3 \, \text{N} \quad \text{(3 N upward)}\) Magnitude of the resultant force: \(R = \sqrt{F_x^2 + F_y^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{N}\)

Electric current is a fundamental quantity in physics while the rest are not.

At constant pressure, volume ∝ absolute temperature (Charles’ law). \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) \(\frac{200}{273} = \frac{V_2}{546}\) v\(_2\) = \(\frac{200 \times 546}{273}\) = 400cm\(^3\)

The pressure exerted by the water at the bottom of the reservoir is given by: \(P = \frac{F}{A}\) Given: \( F = 5000 \, \text{N} \) (weight of water), \( L = 10 \, \text{m} \) (length), \( B = 5 \, \text{m} \) (breadth) Calculate the area \( A \): \(A = L \times B = 10 \, \text{m} \times 5 \, \text{m} = 50 \, \text{m}^2\) Calculate the pressure \( P \): \(P = \frac{F}{A} = \frac{5000 \, \text{N}}{50 \, \text{m}^2} = 100 \, \text{Pa}\)

Resistance thermometers respond faster because they have small sensor mass and use direct electrical detection. Liquid-in-glass and gas thermometers are slower due to thermal expansion and larger thermal inertia, often taking minutes to equilibrate.

Using Charles' law: \(\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad \text{(T in Kelvin, constant P)}\) Calculate \( T_1 \): \(T_1 = 57 + 273 = 330 \, \text{K}\) Since \( V_2 = \frac{1}{5} V_1 \) (volume shrinks to one-fifth), we find \( T_2 \): \(T_2 = T_1 \times \frac{V_2}{V_1} = 330 \times \frac{1}{5} = 66 \, \text{K}\) In ^\circC, this is: \(T_2 = 66 - 273 = -207 \, ^\circC\)

100º → 100 70º → x -20º → 0 Using the interpolation method \(\frac{70 -(-20)}{100 -(-20)}\) = \(\frac{x - 0}{100 - 0}\) \(\frac{90}{120} = \frac{x}{100}\) 120x = 90 x 100 = 9000 x = \(\frac{9000}{120}\) = 75.0ºC

L = 2 m; \alpha = 1.8 \times 10\(^{-5}\) K\(^{-1}\)); \Delta\theta = 100 K \alpha = \(\frac{\DeltaL }{ (L\Delta\theta)}\) \DeltaL = \alpha L \Delta \theta \DeltaL = 1.8 \times 10\(^{-5}\) x 2 x 100 = 3.6 x 10\(^{-3}\) \(\therefore\) \(\Delta\) L = 0.0036m.

P\(_1\) = 20 m + 10 m = 30 m = 3000 cm V\(_1\) = 2 cm\(^3\) P\(^2\) = 10 m = 1000 cm V\(_2\) = ? P\(_1\)V\(_1\) = P\(_2\)V\(_2\) 3000 \times 2 = 1000 \times V\(_2\) V\(_2\) = \(\frac{3000 \times 2}{1000}\) ∴ V\(_2\) = 6 cm\(^3\)

P\(_1\) = 200 Nm\(^{-2}\); P₂ = 100 Nm\(^{-2}\); T\(_1\) = 127 + 273 = 400 K; T\(_2\) = -73 + 273 = 200 K; \(\frac{V_2}{V_1}\) = ? \(\frac{P_1V_1 }{ T_1} = \frac{P_2V_2 }{T_2}\) \(\frac{V_2}{V_1}\) = \(\frac{P_1 T_2}{P_2 T_1}\) \(\frac{V_2}{V_1}\) = \(\frac{200 \times 200}{100 \times 400}\) = \(\frac{1}{1}\) = 1: 1