Vardaan Learning Institute

Given: Height of object $h_o = +5$ cm.
Focal length $f = -10$ cm (concave mirror).
Object distance $u = -20$ cm.
Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} + \frac{1}{-20} = \frac{1}{-10} \Rightarrow \frac{1}{v} = \frac{1}{-10} + \frac{1}{20}$
$\frac{1}{v} = \frac{-2 + 1}{20} = \frac{-1}{20} \Rightarrow v = -20$ cm.
Magnification $m = -\frac{v}{u} = -\frac{-20}{-20} = -1$.
$m = \frac{h_i}{h_o} \Rightarrow -1 = \frac{h_i}{5} \Rightarrow h_i = -5$ cm.
Position: 20 cm from mirror on same side.
Nature: Real and Inverted. Size: Same size as object (5 cm).

Given: Focal length $f = +15$ cm (convex lens).
Image distance $v = -10$ cm (As $f=15$ cm and if image is real it would be $>15$ cm usually, but here $v=10$ cm indicates virtual image on same side).
Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{-10} - \frac{1}{u} = \frac{1}{15} \Rightarrow -\frac{1}{u} = \frac{1}{15} + \frac{1}{10}$
$-\frac{1}{u} = \frac{2+3}{30} = \frac{5}{30} = \frac{1}{6} \Rightarrow u = -6$ cm.
Magnification $m = \frac{v}{u} = \frac{-10}{-6} = \frac{5}{3} \approx 1.67$.

Refractive index $n_g = 1.50$.
Speed of light in vacuum $c = 3 \times 10^8 \text{ ms}^{-1}$.
Formula: $n_g = \frac{c}{v}$ where v is speed in glass.
$v = \frac{c}{n_g} = \frac{3 \times 10^8}{1.50} = 2 \times 10^8 \text{ ms}^{-1}$.

Diameter $d = 0.5$ mm = $5 \times 10^{-4}$ m. Radius $r = 2.5 \times 10^{-4}$ m.
Resistivity $\rho = 1.6 \times 10^{-8} \Omega \text{ m}$. Resistance $R = 10 \Omega$.
Area $A = \pi r^2 = 3.14 \times (2.5 \times 10^{-4})^2 = 3.14 \times 6.25 \times 10^{-8} \approx 19.625 \times 10^{-8} \text{ m}^2$.
$R = \rho \frac{l}{A} \Rightarrow l = \frac{R A}{\rho}$.
$l = \frac{10 \times 19.625 \times 10^{-8}}{1.6 \times 10^{-8}} = \frac{196.25}{1.6} \approx 122.7$ m.
If diameter is doubled, Area becomes 4 times ($A \propto d^2$).
Since $R \propto \frac{1}{A}$, Resistance becomes $\frac{1}{4}$th.

Resistance $R = 8 \Omega$. Current $I = 15$ A.
Rate at which heat is developed is Power $P$.
$P = I^2 R = (15)^2 \times 8 = 225 \times 8 = 1800$ W = 1800 J/s.

Power $P_1 = 100$ W, $P_2 = 60$ W. Voltage $V = 220$ V.
Current $I_1 = \frac{P_1}{V} = \frac{100}{220} = \frac{10}{22}$ A.
Current $I_2 = \frac{P_2}{V} = \frac{60}{220} = \frac{6}{22}$ A.
Total Current $I = I_1 + I_2 = \frac{10+6}{22} = \frac{16}{22} = \frac{8}{11} \approx 0.727$ A.