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\(r_1\) = 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

The electric force between two charged particles is always directed from the positive particle to the negative particle. This is because opposite charges attract each other, while like charges repel each other.

When a negatively charged rod is brought close to a metal sphere, the free electrons in the sphere are repelled from the rod and move to the other end of the sphere. This creates a region of positive charge on the side of the sphere closest to the rod, and a region of negative charge on the opposite side. The process of charge distribution stops when the net force on the free electrons inside the metal is equal to zero. If the sphere is then earthed, the free electrons will flow from the sphere to the ground, leaving the sphere with a net positive charge.

The angle of dip is the angle between the direction of the intensity of the total magnetic field of Earth and a horizontal line in the magnetic meridian. It is also known as the magnetic inclination. The angle of dip ranges from 0 degrees at the magnetic equator to 90 degrees at the magnetic poles.

The polarities at X and Y would be north and north.

Ep = 2.2kV = 2200V; Es=110V; Ns=25; Np=? =\(\frac{E_p}{E_s}\) = \(\frac{N_p}{N_s}\) = \(\frac{2200}{110}\) = \(\frac{N_p}{25}\) =110 \times Np = 2200 \times 25 =\(N_p\) = \(\frac{55,000}{110}\) ∴\(N_p\) = 500 turns

An AC generator uses slip rings to transfer the induced current smoothly to the circuit. A DC generator uses split rings to transfer the induced current to the circuit and also convert the induced AC into pulsating DC. So, to convert an AC generator into a DC generator, the slip rings needs to be replaced with split rings.

ε = 24.0 V, r=1\Omega, R=5.0\Omega terminal p·d = ? ε = IR+Ir = I(R+r) ⇒ IR+Ir = I(R+r) ⇒ I = \(\frac{24}{5+1}\) ⇒ I =\(\frac{24}{6}\) ⇒ I = 4 A terminal p·d = 4 \times 5 terminal p·d = 20.0 V

A= \(0.8m^2\) d= 20mm =\(\frac{20}{1000}\) = 0.02m v =120v; \(ε_oA\)= \(8.85 \times 10^-12 fm^-1\) C = \(\frac{ε_oA}{d}\) C =\(\frac{8.85 \times 10^-12\times0.8}{0.02}\) C= \(3.54 \times 10^-10 F\) Q= CV ⇒ \(3.54 \times 10^-10\) \times 120 = \(4.25\times10^-8c\) Q=\(42.5 \times 10^-9c\) = 42.5nC