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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

A pyrometer is a type of remote-sensing thermometer used to measure the temperature of a surface. It does this by measuring the thermal radiation or infrared energy being emitted from the object. Therefore, a pyrometer thermometer measures temperature from the thermal radiation emitted by objects.

density (ρ) = 13600 \(kgm^{-3}\); g =\(9.8 ms^{-2}\) \(P_{atm}\) =101,325 Pa; ρ=13600 \(kgm^{-3}\) Pabs=\(P_{atm}\)+ρgh ⇒Pabs=101,325+13600 x 9.8 x 0.25 ⇒Pabs=101,325+33320 ⇒ Pabs=134,645Pa

The sensitivity of a thermometer is defined as the smallest temperature change that can be detected or measured, NOT how quickly that temperature change can be detected. In contrast, the difference between the maximum and the minimum temperature is the range of a thermometer

L=1.5m, A=\(620m^2\), ø=25-15=10^\circc K=0.6071 , \(Wm^{-1}k^{-1}\) , \(\frac{q}{t}\) =? \(\frac{q}{t}\) =KA, \(\frac{►t}{L}\) where k is the thermal conductivity constant and \(\frac{q}{t}\) is the rate of heat transfer =\(\frac{q}{t}\)=0.6071\times620\times\(\frac{10}{1.5}\) \(\frac{q}{t}\) = 2509.35w ≈ 2.5kw

Quantity of heat required to raise the temperature of the ice cube from \(-10^oC\) to \(0^oC\) ⇒H=mc►ø = 0.02 \times 2100 \times 0 -(-10) ⇒ H = 0.02 \times 2100 \times (0 + 10) = 0.02 \times 2100 \times 10 ⇒ H = 420 J quantity of heat required to melt ice at \(0^oC\) ⇒ H = mL = 0.02 \times 3.34 \times \(10^5\) ⇒H=6680 j Quantity of heat required to raise the temperature of the melted ice cube (water) from \(0^oC\) to \(-10^oC\) ⇒H=mc►ø = 0.02 \times 4200 \times (10-0) ⇒H = 0.02 \times 4200 \times 10 ⇒H = 840j ;420 + 6680 + 840 = 7940 j

The S.I unit of heat capacity is JK-1 and it is an extensive property as it depends on the amount of substance present. Therefore, (i) and (iv) only are not true