Question Library

Using \(\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\) \(P_1\) = 80Nm\(^{2}\) \(T_1\) = 47 + 273 = 320k \(T_2\) = 27 + 273 = 300k \(P_2\)? \(\therefore\) = \(P_2\) = \(\frac{P_{1}T_{2}}{T_{1}}\) = \(\frac{80 \times 300}{320}\) = 75.Nm\(^{2}\)

\(p_1\) = \(P_0\), \(V_1\) = \(V_0\), \(V_2\) = \(\frac{1}{5}\)\(V_0\) \(P_2\) = ? \(P_1\)\(V_1\) = \(P_2\)\(V_2\) \(P_2\) = \(\frac{P_{1}V_{1}}{V_{1}}\) = \(\frac{\not P_{o}V_{o}}{\frac{1}{5}P_{o}}\) = 5\(\not P_{o}\)

Heat involved (H) = 88,000J, Mass(M)= 0.5kg, specific heat capacity of water(C) = 4200JkgºC, Specific latent heat of fusion(L) = ?, Temperature change(\Delta\theta) = \theta2 - \theta1 = (10 - 0)^\circ = 10^\circ H = MC\Delta\theta + ML or H = M(C\Delta\theta + L) -->\(\frac{H}{M}\) = C\Delta\theta + L : L = \(\frac{H}{M}\) - C\Delta\theta = \(\frac{88,000}{0.5}\) - 4200 \(\times\) 10 L = 176,000 - 42000 = 134,000 L = 134,000 or 134kj/kg

H = mc\(\bigtriangleup\theta\) + mL H = ? m = 50g, c = 4.23J/gk,L = 2260, \(\bigtriangleup\theta\) = 80ºC H = 50 [4.23 x 80 + 2260] = 129920 \(\approx\) 130,000J

Using the law of conservation of energy mc\(\bigtriangleup\theta\) = mgh \(\bigtriangleup\theta\) = \(\frac{gh}{c} = \frac{10 \times 20}{450} = \frac{4}{9}\)ºC

A hygrometer is an instrument that is used to measure water vapor in the atmosphere. This is also commonly referred to as relative humidity in the atmosphere.

The boiling point of a liquid depends on temperature, atmospheric pressure, and the vapor pressure of the liquid. When the atmospheric pressure is equal to the vapor pressure of the liquid, boiling will begin

Given; L = 2H F = \(\frac{50}{\pi}\) \(H_2\) \(X_L\) = ? \(X_L\) = 2\(\pi\)fL = 2\(\pi\) x \(\frac{50}{\lambda}\) x 2 = 200\(\Omega\)

Given f = 1.6 x10\(^{16}\)\(H_3\), h = 6.63 x 10\(^{-39}\)Js e = 1.6 x 10\(^{-19}\)C, V = ? V = \(\frac{w}{e} = \frac{hf}{e}\) = \(\frac{1.6 \times 10^{16} \times 6.63 \times 10^{34}}{1.6 \times 10^{-19}}\) = 66.3V

Both the electric field and magnetic field affect the \beta-particles.

Objects distance (u) = image distance (v) object distance = 3m |uv| = 3 + 3 = 6m

Concave Mirror: It is a spherical mirror whose outer surface is polished and inner or concave side is reflecting surface. Convex Mirror: It is a spherical mirror whose inner surface is polished and whose outer side or convex side is the reflecting surface. f = \(\frac{m}{4}\) = f = \(\frac{d}{4}\) f = \(\frac{20}{4}\) = 5cm (For metal, it is the polished side that will be the reflecting surface)

\(\frac{\text {sin i}}{\text {sin r}}\) = \(\frac{\text {velocity in air}}{\text {velocity in glass}}\) \(\frac{\text {sin i}}{\text {sin 30}}\) = \(\frac{3 \times 10^{8}}{2 \times 10^{8}}\) Sin i = \(\frac{3}{4}\) = 0.75

Using y = A sin (wt + 0) Given y = 0.25 x 10\(^{-3}\)sin (500t - 0.025x) By comparing, w = 5.00 x 10\(^{2}\)rads\(^{-1}\)

V = \(\frac{\lambda}{T}\) : v = 330ms\(^{-1}\) \(\lambda\) = 1.65m T = ? 330 = 1.65 T = \(\frac{1.65}{330}\) = 0.005s