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W = mg = 0.5 x 10 = 5 N Since it's light, neglect the weight of the metre rule. The effective tension T acting in the vertical direction = T sin 30^\circ From the second condition of equilibrium, sum of clockwise moments equal sum of anticlockwise moments Taking moment at E ⇒ T sin 30^\circ x 30 = 5 x 50 ⇒ ∴T=250 T = \(\frac{250}{15}\) = 16.67N

The working of the beam balance is based on the principle of moments. The principle of moments states that the sum of the clockwise moments on a beam must be equal to the sum of the counterclockwise moments on the beam. This means that the weight of the object on the left side of the beam must be multiplied by the distance from the fulcrum to the left side of the beam, and the weight of the object on the right side of the beam must be multiplied by the distance from the fulcrum to the right side of the beam. If these two products are equal, then the beam will be in equilibrium.

A couple is a pair of forces that are equal in magnitude but opposite in direction, and that are applied to a body at different points. The forces of a couple do not produce any translation, but they do produce a rotation.

U= 75 \(ms^-1\) =22º, g= 10 \(ms^-2\) , R=? R= \(\frac{U^2sin2\theta}{g}\) ⇒R= \(\frac{75^2\timessin2(22)}{10}\) ⇒R= \(\frac{5625\timesSin44}{10}\) ⇒R= \(\frac{5625\times0.6947}{10}\) ⇒R= \(\frac{3907.69}{10}\) ⇒ 391m

Object A will undergo unstable equilibrium as it can move farther from its original position Object B will undergo stable equilibrium as it can return to its original position Object C will undergo neutral equilibrium as it can remain in a new position

m=2kg, r=1.2m, f=120rev/min ; T=? f= \(\frac{120rev}{1min}\) × \(\frac{1min}{60s}\) = 2rev\s w=2πf= 2×π×2= 4πrad\s T= \(\frac{mv^2}{r}\), but v = wr ⇒ T = \(\frac{(m)(wr)^2}{r}\) ⇒ T = \(mw^2r\) = T = 2× (4π)^2 × 1.2 T = 379N ( to 3 s.f)

Let the original length L=xm ;New length =( x -3 ) m \(T_1\) = 5.77s; \(T_2\) = 4.60s, \(T^2\) \alpha L ⇒\(T^2\) = kL where K is constant ⇒ K = \(\frac{T^2_1}{L_1} = \frac{T^2_2}{L_2}\) ⇒\(\frac{5.77^2}{x}\) = \(\frac{4.60^2}{x-3}\) ⇒ \(\frac{33.29}{x}\) = \(\frac{4.60^2}{x-3}\) ⇒ 33.29(x-3) = 21.16x ⇒ 33.29x - 99.87 =21.16x ⇒12.13x = 99.87 ;x =\(\frac{99.87}{12.13}\) = 8.23m The new length of the pendulum =x-3 = 8.23-3 =5.23m

W = 400 N; P = 100 N; \theta = 30o; \mu = ? Frictional force (Fr) = \muR (where R is the normal reaction) The forces acting along the horizontal direction are Fr and Px ∴ Pcos 30^\circ - Fr = ma (Pcos 30^\circ is acting in the +ve x-axis while Fr in the -ve x-axis) ⇒ 100cos 30^\circ - \muR = ma Since the box is moving at constant speed, its acceleration is zero ⇒ 100cos 30^\circ - \muR = 0 ⇒ 100cos 30o = \muR ----- (i) The forces acting in the vertical direction are W, Py and R ∴ R - Psin 30^\circ - W = 0 (R is acting upward (+ve) while Py and W are acting downward (-ve) and they are at equilibrium) ⇒ R - 100sin 30^\circ - 400 = 0 ⇒ R = 100sin 30^\circ + 400 ⇒ R = 50 + 400 = 450 N From equation (i) ⇒ 100cos 30^\circ = 450\mu ⇒\mu=100cos30^\circ N = \(\frac{100cos30^\circ}{450}\) = \mu = 0.19

Surface tension is the force that acts on the surface of a liquid, tending to minimize the surface area. It is caused by the cohesive forces between the molecules of the liquid. The stronger the cohesive forces, the higher the surface tension. Mercury has the highest surface tension of all the liquids listed. This is because the atoms of mercury are very close together, and they have a strong attraction for each other. This strong cohesive force results in a high surface tension.

Adhesive surfaces are surfaces that are wetted by water. This means that the water molecules are attracted to the wood surface, and they spread out to form a thin layer in order to maximize their contact with the surface. The adhesive forces between the water molecules and the wood surface are stronger than the cohesive forces between the water molecules, resulting in a thin layer.

Let mb=mass of empty bottle, \(m_w\)=mass of water only and \(m_a\)= mass of alcohol only given; \(m_b\)=19g \(m_b\) + \(m_w\) = 66g \(m_b\) + \(m_a\) = ? R.d=0.8 R.d=mass of alcohol ⇒\(\frac{mass of alcohol}{mass of equal volume of water}\) ⇒ mass of equal volume of water = \(m_w\)=66-19=47g ⇒0.8 = \(\frac{m_a}{47}\) ⇒\(m_a\)=0.8\times47 =37.6g ; \(m_b\) + \(m_a\) = 19+37.6=56.6g

r=0.2mm = \(2 \times {10^{-4}}\) e= 0.05% of L,=0.005L, m=1.5kgg=\(10 ms^{-2}\) , y=?, (1000 mm = 1m) Y = \(\frac{FL}{Ae}\) w = mg = 1.5 \times 10 = 15 A = \(\pi r^2\) = \(\frac{22}{7}\times(2\times10^{-4})^2\) = \(1.26\times10^{-7}(m^2)\) ⇒ Y = \(\frac{15L}{1.26\times10^{-7}\times0.005L}\) ⇒ Y = \(\frac{15}{1.26\times10^{-7}\times0.005}\) ⇒ Y = \(\frac{15}{6.29\times10^{-10}}\) Y = \(2.4\times10^{10}(Nm^{-2})\)

Relative Humidity= \(\frac {partial pressure of H2O}{saturated vapor pressure of H2O}\)\times100% S.v.p. of water at 40^\circc = 55.3 ⇒ Relative Humidity= \(\frac{38.8}{55.3}\)\times100% Relative Humidity= 70%

h=32·8m, p= \(1\times10^3kgm^-3\) g=9·8 \(ms^-2\) , \(P_a\)=101·3KPa \(P_T\)= ? P=hpg ⇒\(32·8\times1\times10^3 \times 9·8\)=321440 \(P_a\) ⇒321·44KPa ⇒ \(P_T\) =321.44+101.3 ;\(P_T\)=422·7KPa

\(r_1\) = 4.5cm , \( P_1\) =is the total pressure on the bubble at a depth of 12m from the surface. \(P_1\) = 12 + 10.34 =22.34m \(V_1\) = \(\frac{4}{3}\)\pi\times \(r^3_1\) = \(\frac{4}{3}\times \pi\times{4.5^3cm^3}\) \(P_2\) = 10.34m \(V_2\) = \(\frac{4}{3} {\pi}{r^3_2}\) from boyles law: \(P_1V_1\) = \(P_2V_2\) ⇒ 22.34\times \(\frac{4}{3}\times \pi\times{4.5^3}\) = 10.34 \times \(\frac{4}{3}\times\pi\times{r^3_2}\) ⇒ 22.34 \times \(4.5^3\) = 10.34 \times \(r^3_2\) ⇒ \(r^3_2 = \sqrt[3]{196.88}\) ⇒ \(r_2\) = 5.82cm