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C. 6 kg Initial velocity \( u = 4 \, \text{m/s} \), Acceleration \( a = 5 \, \text{m/s}^2 \), Time \( t = 2 \, \text{s} \) Final velocity: \(v = u + at = 4 + 5 \times 2 = 14 \, \text{m/s}\) Change in momentum (impulse): \(\Delta p = m(v - u) = 60\) Substituting for \( v \): \(m(14 - 4) = 60\) Simplifying: \(m \times 10 = 60\) Thus, \(m = 6 \, \text{kg}\)

A. Q In a simple pendulum, kinetic energy is maximum at the lowest point (equilibrium position) where potential energy is minimum and speed is maximum.Here, Q is the control point (mean/equilibrium position), so maximum kinetic energy occurs at Q. At extremes P and S, kinetic energy is zero (velocity = 0). At R, it is between the extreme and the equilibrium.

The frequency \( f \) of a simple pendulum is: \(f = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \quad \text{where } L = 1.2 \, \text{m}, \, g = 10 \, \text{m/s}^2, \, \pi \approx \frac{22}{7}\) Calculate: \(\frac{g}{L} = \frac{10}{1.2} = \frac{25}{3}, \quad \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}}\) \(f = \frac{1}{2 \times \frac{22}{7}} \times \frac{5}{\sqrt{3}} = \frac{35}{44\sqrt{3}}\) Approximating \( \sqrt{3} \approx 1.732 \): \(44 \times 1.732 \approx 76.208 \quad \Rightarrow \quad \frac{35}{76.208} \approx 0.459 \, \text{Hz}\) Thus, the answer is \( D. \, 0.5 \) Hz.

Period of oscillation = \(\frac{\text{ time}}{\text{no. of oscillations}}\) = \(\frac{38}{20}\) = 1.9s

Linear speed = \(\omega\) r \(\omega\) = \(\frac{\text{v}}{\text{r}}\) \(\omega\) = \(\frac{10}{4}\) = 2.5 rads\(^{-1}\)

The frequency (f) of a simple pendulum is the number of oscillations it completes per second and is given by the formula: f = \(\frac{1}{2\pi}\)\(\sqrt{\frac{\text{g}}{\text{l}}}\) where l is the length of the string and g is the acceleration due to gravity. Because the length l is in the denominator, the frequency is inversely proportional to the square root of the length. Therefore, increasing the length results in a slower swing rate, effectively reducing the frequency.

Answer: B. 0.015 (Net F = ma = 40\times0.3 = 12 N; friction = 18 - 12 = 6 N; \mu = \(\frac{\text{friction}}{\text{mg}}\) = \(\frac{6}{(40\times10)}\) = 0.015.)

Frictional force is the tangential force opposing an object's free sliding on an adjacent surface. It originates from surface irregularities and intermolecular attractions between materials, resisting relative motion or its initiation.

Given: g = 9.8 m/s\(^2\), a = 10 m/s, v = 0( come to rest), s = 25m v\(^2\) = u\(^2\) + 2as - 100 = 2 x 25 x a a = \(\frac{-100}{50}\) = - 2 deceleration is caused by kinetic friction But, \(\mu\) = \(\frac{\text{f}}{\text{R}}\) = \(\frac{\text{ma}}{\text{mg}}\) = \(\frac{-2}{9.8}\) ≈ 0.2.

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: \(R = \sqrt{F_x^2 + F_y^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{N}\)