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To find the ratio of the final volume to the initial volume of a gas when both pressure and temperature change, we can use the Ideal Gas Law, which states that \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) P\(_1\) = 300N/m\(^2\) V\(_1\) = ?, V\(_2\) = ?, T\(_1\) = 127^\circC =127+273 = 400K, T\(_2\) = −73^\circC = −73 + 273 = 200K P\(_2\) = 120N/m\(^2\) Using the Ideal Gas Law, we can rearrange it to find the ratio of volumes: = \(\frac{V_2}{V_1} = \frac{P_1T_2}{P_2T_1}\) = \(\frac{V_2}{V_1} = \frac{300 \times 200}{120 \times 400}\) = \(\frac{5}{4}\) Therefore, V\(_2\) : V\(_1\) = 5 : 4.

If the bob is fully immersed in the water, then the volume of water displaced will be equal to the volume of the bob = 4/3πr3, where r is the radius of the bob. This volume of water in the cylinder = πr2l, where r = radius of the cylinder, and l = level of water rose in the cylinder. Therefore, π x r2 x l = 4/3 x π x 33 Then r2 = 36 r = √36 r = 6cm

The resultant of two forces acting on an object is maximum when the angle between them is zero, during which the two forces are parallel and pointing in the same direction

In real-world machines, due to factors such as friction, deformation, and other losses, the mechanical advantage is always less than the velocity ratio: MA < VR This means that the work output (useful work done) is always less than the work input (energy supplied), leading to efficiency being less than 100%.

Velocity: Vector quantity (has both magnitude and direction). Displacement: Vector quantity (change in position). Reaction: Typically refers to a reaction force, which is a vector quantity.(OPTION A)

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