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\(\frac{P_1V_1}{T_1}\) = \(\frac{P_2V_2}{T_2}\) P\(_2\) = 2P\(_1\) ; T\(_1\) = 2T\(_2\) V\(_2\) = \(\frac{P_1 \times V_1 \times T_2}{T_1 \times P_2}\) = \(\frac{P_1 \times 100 \times T_2}{2T_2 \times 2P_1}\) = \(\frac{100}{4}\) = 25cm\(^3\)

Using V = at + u V = (-4 x 10) + 44 = 4ms-1

Power = \(\frac{\text{force x distance}}{\text{time}}\) = \(\frac{50 X 10 X 5}{5}\)= 500W

No expanded working has been attached for this question yet. The verified answer is Option C.

Power = \(\frac{Work done}{time}\) = \(Js^{-1}\) Power = \(current \times voltage\) = \(Ampere volt\) Power = \(\frac{voltage^{2}}{resistance}\) = \(\frac{volts^{2}}{ohm}\)

F = ma = \(\frac{m(u - v)}{t}\) 16 = \(\frac{4(v - 0)}{5}\) v = 20ms-1

Mass of one oil = \(\frac{5 \times 10^{-4}}{1000}\) = 5 x 10\(^{-7}\)kg Density = \(\frac{\text {mass}}{\text{area x thickness}}\) 500 = \(\frac{5 \times 10^{-7}}{\text{0.5 x thickness}}\) thickness = 2 x 10\(^{-9}\)m

Efficiency = M.A x 100% = \(\frac{200}{50}\) x \(\frac{1}{5}\) x 100 = 80%

Density of H2O = 1gcm3 Density x pressure = constant D1 x P1 = D2 x P2 13.6 x 0.76 = 1 x P2 P2 = 10m

Heat energy = Electrical energy mc\(\Delta\)\(\theta\) = I\(^2\)Rt 1000 x c x 10 = 3 x 3 x 20 x 10 0.18Jg\(^{-1}\)ºC

To find the internal resistance of the cell, we can use the formula: \(V = E - I r\) Where: - \( V \) = terminal voltage (2 V) - \( E \) = electromotive force (emf) of the cell (2.2 V) - \( I \) = current through the circuit - \( r \) = internal resistance of the cell {Step 1: Calculate the Current} Using Ohm's law, the current (\( I \)) flowing through the resistor is given by: \(I = \frac{V}{R}\) Where: - \( V = 2 \, \text{V} \) (terminal voltage) - \( R = 4 \, \Omega \) Thus, \(I = \frac{2 \, \text{V}}{4 \, \Omega} = 0.5 \, \text{A}\) {Step 2: Substitute Values into the Formula} Now substituting \( I \), \( V \), and \( E \) into the formula: \(2 = 2.2 - 0.5 r\) {Step 3: Solve for Internal Resistance \( r \)} Rearranging the equation gives: \(0.5 r = 2.2 - 2\) \(0.5 r = 0.2\) \(r = \frac{0.2}{0.5} = 0.4 \, \Omega\) Thus, the internal resistance of the cell is: \(\boxed{0.4 \, \Omega}\)

No expanded working has been attached for this question yet. The verified answer is Option C.

Field intensity = \(\frac{force}{charge}\) \(\frac{4}{0.2}\) = 20N/C

For water to begin flowing from the tank through an inverted U-tube with an open end at point P, the U-tube must be completely filled with water, and the pressure at point O inside the tube should be lower than the water level W in the tank.

Two current-carrying conductors attract each other when the current is in the same direction and repel each other when the current is in the opposite direction. This can be verified using Fleming's left-hand rule.