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i.e V\(_L\) - V\(_C\) = -ve (V\(_C\) - V\(_L\) = +ve) From Tan \(\phi\) = \(\frac{V_L - V_C}{\text{R}}\) It shows that V\(_L\) - V\(_C\) will be inverted on the y-axis V\(_C\) > V\(_L\)

I = 0.05 A, p.d across R\(_L\) = V\(_L\) = I R\(_L\) = 0.05 \times 50 V\(_L\) = 2.5 V

X\(_C\) = \(\frac{1}{(2\pifC)}\) = \(\frac{1}{(2\pi \times (\frac{200}{\pi}) \times 25 \times 10^{-6})}\) = \(\frac{1}{(2 \times 200 \times 25 \times 10^{-6})}\) = \(\frac{1}{10^{-2}}\) = 10\(^2\) = 100 \Omega \(\therefore\) X\(_C\) = 100 \Omega

X\(_o\) equals PQ because it is the amplitude or maximum current.

R\(_L\) = 2\(\pi\)fL = 2 \(\pi\) x \(\frac{100}{\pi}\) x 0.25 = 50.0\(\Omega\)

Electrical conduction in semiconductors occurs via electrons and holes.

Answer: A \lambda = \(\frac{ln2}{T½ }\) ⇒ \(\frac{0.693}{138.5}\) ≈ 5.0 \times 10\(^{-3}\) s\(^{-1}\).

A. Electrophorus. It generates static electricity via induction and is not a radiation detector. The others are: Scintillation chamber: Detects radiation via light flashes. Geiger-Muller counter: A gas-filled detector for ionizing radiation. Film badge: Records cumulative radiation exposure on photographic film.

A. Less than 20 cm For a concave mirror, an erect image is virtual (behind the mirror) and forms when the object is placed between the pole and focus (\( u < f \)). Magnification: \(m = +4 \quad (\text{erect, so positive})\) Using the mirror formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) Given \( f = -20 \, \text{cm} \) (concave), \( u \) is negative. Since \( m = -\frac{v}{u} \): \(v = -m u = -(+4)u = -4u\) Substituting into the mirror formula: \(\frac{1}{-4u} + \frac{1}{u} = \frac{1}{-20}\) Combining the fractions: \(\frac{3}{4u} = \frac{1}{-20}\) Thus, solving for \( u \): \(3 \cdot -20 = 4u \quad \Rightarrow \quad u = -15 \, \text{cm}\) Object distance \( |u| = 15 \, \text{cm} < 20 \, \text{cm} \). Thus, it is less than 20 cm. OR A concave mirror only gives an erect, magnified image when the object is placed between the mirror and its focal point. Since the focal length is 20 cm, the object must be closer than 20 cm to the mirror.

D. 1.25 m In a pinhole camera, the relationship between image size and object size is given by: \(\frac{\text{Image Size}}{\text{Object Size}} = \frac{v}{u}\) where \( v \) is the camera length. Given \( v = 25 \, \text{cm} = 0.25 \, \text{m} \). The image is one-fifth the object size, so the magnification is: \(\text{Magnification} = \frac{1}{5} = \frac{v}{u}\) Thus, we can rearrange this to find \( u \): \(u = 5v = 5 \times 0.25 = 1.25 \, \text{m}\).

In a kaleidoscope (two plane mirrors inclined at an angle ( \theta )): \(\text{Number of images of one object} = \frac{360^\circ}{\theta} - 1\) Given: \(\theta = 30^\circ\) \(\frac{360}{30}\) = 12 So, images formed by one object: 12 - 1 = 11 For 10 objects (tinsels): 10 x 11 = 110 images formed.

Annular Eclipse: Occurs when the Sun, Moon, and Earth are aligned in a straight line, but the Moon is at a distance where it does not completely cover the Sun.

Pinhole camera, shadow formation, and the occurrence of an eclipse are a result of the rectilinear propagation of light.

The moon is not self-luminous. Self-luminous objects are those that can emit light on their own. The moon does not emit light; it only reflects light from the sun.

When the object is placed exactly at the centre of curvature, the image forms at the same location (centre of curvature), real, inverted, and the same size.