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An absolute thermometer, such as a gas thermometer, often uses helium as its thermometric substance because it has favorable properties for measuring temperature accurately over a wide range. Helium behaves nearly ideally at low temperatures and pressures, making it suitable for measuring temperatures close to absolute zero. While mercury is useful for many traditional thermometers due to its liquid state and thermal properties, it is not suitable for absolute thermometers, which require a substance that can accurately measure extremely low temperatures and behave ideally across a wide range of conditions. Thus, helium is the correct choice for an absolute thermometer.

If F = 85.5Hz, V = 342ms-1, Then wavelength = V/F = 342/85.5 = 4m Then if wavelength = 4m => wavelength/2 = 4/2 = 2m = NN, Thus NA = wavelength/4 = 4/4 = 1m.

For a string fixed at both ends and plucked at the middle, the fundamental frequency is given by f0 = v/λ, where v = velocity of sound, λ = wave length. At its fundamental, note, the the length of the string is given l = λ/2, => λ0 = 2l, where λ0 = wave length fundamental note. for the first over tone, λ1 = l. For the second over tone, λ2 = 21/3 = (2 x 24)/3 = 16cm. i.e. λ2 = 16cm.

Efficiency = \(\frac{power output}{power input}\) = \(\frac{R}{R + r}\) Given efficiency = 75% = 0.75, R = 6.0\Omega, r = ? Efficiency = \(\frac{R}{R + r}\) 0.75 = \(\frac{6.0}{6.0 + r}\) 0.75(6.0 + r) = 6.0 4.5 + 0.75r = 6.0 0.75r = 6.0 - 4.5 = 1.5 r = \(\frac{1.5}{0.7}\) = 2\Omega The internal resistance r of the cell is 2.0 ohms.

When you insert a dielectric material (like glass) between the plates, the permittivity ε increases because the dielectric constant k of glass is greater than that of air or vacuum. The new capacitance when a dielectric is inserted is given C′ = \(\frac{\text{kεoA}}{\text{d}}\) where k is the dielectric constant of the glass, and ε\(_0\) is the permittivity of free space. Since k >1 for glass, inserting it increases the capacitance: C ′ = kC Thus, with the glass block inserted, the capacitance C′ will be greater than the original capacitance C.

Electrostatic force = F\(_E\) = \(\frac{e^2}{4\piε_od^2}\) Gravitational force = F\(_G\) = \(\frac{Gm^2}{d^2}\) \(\frac{F_E}{F_G}\) = \(\frac{\frac{e^2}{4\piε_od^2}}{\frac{Gm^2}{d^2}}\) = \(\frac{e^2}{4\piε_oGm^2}\) NOTE: (d\(^2\)s canceled out)

let the internal resistance of the cell = r, therefore E = I (R+r) (i) E = 0.4(4+r), (ii) E = 0.2 (10+r) therefore 0.4 (4+r)= 0.2 (10+r) 1.6 + 0.4r = \(2.0 \times 0.2\text{r}\) 0.4r - 0.2r = 2.0 - 1.6 therefore 0.2r = 0.4 => r = \(\frac{0.4}{0.2}\) = 2Ω therefore r = 2Ω

3919K 3919K. Since neither the mass number (39), nor the atomic number (19) is affected by the emission,the particles emitted can only be gamma ray, which is not a charge particle, but simply a radiation.

Electricity conduction through gases is usually done under low pressure (I) and high voltage (III). Thus OPTION B is correct.

I = \(\frac{V}{R}\) = \(\frac{E}{R+r}\) , if r = 2Ω V = \(\frac{3}{5}\)E => \(\frac{3E}{5R} = \frac{E}{R+2}\) i.e 3E(R+2) = 5RE 3RE + 6E = 5RE ( the Es will cancel out) i.e 3R + 6 = 5R 6 = 5R - 3R => 2R = 6 i.e R = \(\frac{6}{2}\) = 3Ω

E = \(\frac{V}{d}\) E is the electric field (in volts per meter, V/m), V is the potential difference (in volts, V), d is the distance between the two points (in meters, m) Given: Potential difference V = 10\(^7\) Height d = 4m That is, E = \(\frac{10^7}{4} = 2.5\times 10^6 NC^{-1}\)

Efficiency = \(\frac{\text{power output}}{\text{power Input}}\) => \(\frac{80}{100} = \frac{\text{power output}}{1.5\times 250}\) i.e power output = \(\frac{1.5\times250\times80}{100}\) = 300W

Total power in watt = \(60\times5 + 4\times100\) = 300 + 400 = 700W Total power in kilo watt = 700/1000 = 0.7kW Unit of electricity in kilowatt-hour = 0.7 * 20hrs = 14 okwh i.e If I kwh costs # 10.00 14kwh = \(10.00 \times14\) = #140

If I = 2 Sin wt; but in a.c. circuit current I = I\(_0\) Sin wt; thus comparing the two equations, it implies that I\(_0\) = 2A; Again, the root means square (I(rms)) of an a.c. current is the value of the d.c. current that will dissipate the same amount of heat in a given resistance as the a.c. and I(rms) = \(\frac{I_0}{\sqrt{2}}\) = \(\frac{2.0A}{\sqrt{2}}\) = \(\sqrt{2}\)A (by rationalizing the denominator \(\frac{2.0A}{\sqrt{2}}\))

For the transformer, V ∝ ∩ => V s ∝ ∩s; Vp ∝ ∩p => \(\frac{V_p}{V_s} = \frac{∩_p}{∩_s}\), therefore \(\frac{120}{1200} = \frac{∩_p}{1000}\) therefore ∩\(_p \times 1200 = 120\times 1000\) ∩\(_p = 120\times\frac{1000}{1200}\) = 100