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The diagram is that of a plano-convex lens; a type of converging lens.

Upward force = downward force = weight of the Bar 20N + 20N = Weight of the BAR Therefore, the weight of the bar = 40N

When the object is placed between the optical centre of the lens and the focal length, the image formed is virtual, erect and magnified.

This means that when one Point is at its highest point, the other is at its lowest point, resulting in a 180-degree phase difference.

\(F= \frac{Gm_{1}m_{2}}{r^{2}}\) \(m_{1} = 100kg\) \(m_{2} = 20kg; r =1.2m\) \(G= 6.67\times 10^{-11}\) \(F = \frac{6.67\times 10^{-11}\times 100\times 20}{(1.2)^{2}} = 9.26\times 10^{-8}\) Gravitational force is always attractive. Therefore, the net force acting on object Q will pull it directly towards the larger mass, P.

The ideal transformer (either step-up or step-down) works with the equation: \(\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} = \frac{I_{p}}{I_{s}}\) where \(V_{s}\) and \(V_{p}\) are the voltages applied to the secondary and the primary respectively \(N_{s}\) and \(N_{p}\) are the number of turns on the secondary and primary coils respectively \(I_{p}\) and \(I_{s}\) are the currents flowing in the primary and secondary coils respectively. \(\frac{V_{s}}{100} = \frac{200}{100}\) \(V_{s} = \frac{200\times 100}{100} = 200V\)

The velocity ratio and therefore, the mechanical advantage of an inclined plane system increases as \(\theta\) decreases and vice-versa.

Given \(K_{1}\) = \(\frac{mv^{2}}{2}\). When v is halved, we have \(K_{2}\) = \(\frac{m\times (\frac{v}{2})^{2}}{2}\) = \(\frac{K_{1}}{4}\) Hence, the initial kinetic energy is quartered.

The fundamental frequency \((f_{0})\) = \(\frac{v}{2l}\) \(f_{0} = 150Hz; l =400cm = 4m\) \(150 = \frac{v}{2\times 4}\) \(v = 150 \times 8 = 1200ms^{-1}\)

Number of wavelengths = Number of oscillations Number of wavelengths = \(\frac{1}{4} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} +\frac{1}{4}\) = 2\(\frac{1}{2}\)

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

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

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

Capacitors pass AC but block DC signals (when charged up to the applied DC voltage), they are often used to separate the AC and DC components of a signal. Capacitors are also primarily used to store energy.

Ammeters are connected in series so, A3= A2 + A1 A1 = A3 - A1 = 5A - 2A = 3A