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Previous Year Question Papers NEET 2016 Phase II - Physics Page: 3
13903.The potential differences across the resistance, capacitance and inductance are 80 V, 40 V and 100 V respectively in an L–C–R circuit. The power factor of this circuit is
1.0
0.4
0.5
0.8
Explanation:
Power factor = $\dfrac{R}{z} = \dfrac{iR}{iz} = \dfrac{80}{\sqrt{80^2 + 60^2}} = \dfrac{80}{100}$ = 0.8
13904.A 100 Ω resistance and a capacitor of 100 Ω reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is
11$\sqrt{2}$A
2.2 A
11 A
4.4 A
Explanation:

$z = \sqrt{R^2 + X_c^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2}$

$i_{max} = \dfrac{V_{max}}{z} = \dfrac{220\sqrt{2}}{100\sqrt{2}} = 2.2$

13905.Two identical glass (μg = 3/2) equiconvex lenses of focal length f each are kept in contact. The space between that two lenses in filled with water (μw = 4/3). The focal length of the combination is
3f / 4
f / 3
f
4f / 3
Explanation:
equiconvex lens

$\dfrac{1}{f} = \left(\dfrac{3}{2}–1\right)\dfrac{2}{R} = \dfrac{1}{R}$

$\dfrac{1}{f} = \left(\dfrac{4}{3}–1\right)\left\{–\dfrac{2}{R}\right\} = –\dfrac{2}{3R}$

$\dfrac{1}{f_{eq}} = \dfrac{1}{f}–\dfrac{2}{3f} + \dfrac{1}{f} = \dfrac{3 – 2 + 3}{3f} = \dfrac{4}{3f}$

$f_{eq} = \dfrac{3f}{4}$

13906.An air bubble in a glass slab with refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is
16
8
10
12
Explanation:
air bubble in glass slab

$\dfrac{X}{\mu} + \frac{(l–X)}{\mu} = 3 + 5$

$\dfrac{l}{\mu} = 8$

$l = 8 \times \dfrac{3}{2} = 12 cm$

13907.The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio $\dfrac{I_{max} – I_{min}}{I_{max} + I_{min}}$ will be
$\dfrac{2\sqrt{n}}{(n + 1)^2}$
$\dfrac{\sqrt{n}}{(n + 1)}$
$\dfrac{2\sqrt{n}}{(n + 1)}$
$\dfrac{\sqrt{n}}{(n + 1)^2}$
Explanation:

$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{nI_1} + \sqrt{I})^2 = (\sqrt{n} + 1)^2I$

$I_{min} = (\sqrt{n} – 1)^2I$

$\dfrac{I_{max} – I_{min}}{I_{max} + I_{min}}$ = $\dfrac{(\sqrt{n} + 1)^2I – (\sqrt{n} – 1)^2I}{(\sqrt{n} + 1)^2I + (\sqrt{n} – 1)^2I}$

$= \dfrac{n + 1 + 2\sqrt{n} – n – 1 + 2\sqrt{n}}{(n + 1 + 2\sqrt{n}) + (n + 1 – 2\sqrt{n})} = \dfrac{4\sqrt{n}}{2(n + 1)} = \dfrac{2\sqrt{n}}{(n + 1)}$

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