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11 months ago

Physics Dual Nature of Radiation and Matter

An ideal fluid of density 800kgm^-3, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to $\frac{a}{2} .$ The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is $\frac{\sqrt[]{x}}{6} m s^{- 1}$ for x =________. (Given g = 10 ms^-²)

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Payal Gupta

Contributor-Level 10

$ρ = 800 k g / m^{3}$

P₁ – P₂ = 4100 Pa

Bernoulli's question b/w (1) & (2)

$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + z_{2}$

$\frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{v_{2}^{2}}{2 g} + 0$ -(1)

Equation of continuity

A₁v₁ = A₂v₂

v₂ = 2v₁-(2)

from (1) & (2)

$\Rightarrow \frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{{(2 v_{1})}^{2}}{2 g}$

$\frac{4100}{800 * 10} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{4 v_{1}^{2}}{2 g}$

$\frac{12100}{8000} = \frac{3 v_{1}^{2}}{2 g}$

$\Rightarrow v_{1}^{2} = \frac{2 * 10}{3} * \frac{12100}{8000}$

$= \frac{2}{3} * \frac{121}{8}$

x = 363

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11 months ago

Physics Thermodynamics

A 110 V, 50Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance $55 Ω,$ at resonance in the circuit, will be__________A.

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Payal Gupta

Contributor-Level 10

At Resonance

X_L = X_C

then l_L = l_C

Now phasor diagram

for L & C

So, Net current = zero

$= (l_{L} + l_{C})$

Therefore current through R circuit at resonance will be zero

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11 months ago

Physics Electromagnetic Induction

As per the given circuit, the value of current through the battery will be__________A.

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Payal Gupta

Contributor-Level 10

$\begin{array}{l} D_{1} \\ D_{3} \end{array}}$ Forward biased offer zero Resistance

D₂} Reversed biased offers Infinite Resistance

$I = \frac{v}{R} = \frac{10}{10} = 1 A m p$

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11 months ago

Physics Oscillations

In a vernier calipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be__________* 10^-2mm.

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Payal Gupta

Contributor-Level 10

Let 'x' he the value of one division of main scale

$x = \frac{1}{20} c m = 0.05 c m$

Let y be value of one division on venire scale given

10 y = 9 x

$y = \frac{9 x}{10}$

Least count = $x - \frac{9 x}{10} = \frac{x}{10} = \frac{0.05}{10}$

= 0.005 cm

= 5 * 10^-2 mm

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11 months ago

Physics Ray Optics and Optical Instruments

A light ray is incident, at an incident angle q₁, on the system of two plane mirrors M₁ and M₂ having an inclination angle 75° between them (as shown in figure). After reflecting from mirror M₁ it gets reflected back by the mirror M₂ with an angle of reflection 30°. The total deviation of the ray will be ___________ degree.

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Payal Gupta

Contributor-Level 10

$δ_{t o t a l} = 360 ° - 2 θ$

= 360 – 2 * 75° = 210°

Total deviation = $δ = 150 °$

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 $\overset{o}{A}$ is 0.42 V. If the threshold frequency is x * 10¹³ /s, where x is ________(nearest integer).

(Given, speed light = 3 * 10⁸ m/s, Planck's constant = 6.63 * 10^-34 Js)

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Vishal Baghel

Contributor-Level 10

$e V_{0} + h υ_{0} = \frac{h C}{λ}$

$\Rightarrow 1.6 * 1 0^{- 19} * 0.42 + 6.63 * 1 0^{- 34} υ_{0} = \frac{6.63 * 1 0^{- 34} * 3 * 1 0^{8}}{66 3^{0} * 1 0^{- 10}}$

$\Rightarrow 6.63 * 1 0^{- 34} υ_{0} = 1 0^{- 19} (3 - 1.6 * 0.42)$

$6.63 * 1 0^{- 34} υ_{0} = 2.328 * 1 0^{- 19}$

$υ_{0} = 35.11 * 1 0^{13} = 35$

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1cm on main scale and 50 Vernier scale divisions are equal to 49 main scale division, then least count of the travelling microscope is __________* 10^-6 m.

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Vishal Baghel

Contributor-Level 10

Let 'x' be the value of one division on main scale

$x = \frac{1}{40} c m$

Let 'y' be the value of one division on vernier

Now given

50y = 49 x

$y = \frac{49 x}{50}$

$\begin{array}{l} = 5 * 1 0^{- 4} c m \\ = 5 * 1 0^{- 6} m \\ 5 \end{array}$

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11 months ago

Physics Electric Charges and Fields

A $10 Ω$ 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 $μ s .$ The average e.m.f. induced in the coil is__________V.

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Payal Gupta

Contributor-Level 10

Decay of current in Inductor is given by,

$i = i_{0} e^{- t / τ} [W h e r e τ = \frac{L}{R}; i_{0} = \frac{v}{R} = \frac{20}{10} = 2]$

At t = 100 $μ s$

i = 0

i.e. i = i0 $e^{- 100 / τ} = 0$ -(1)

e.m.f induced

$e = \frac{- L d i}{d t} = - L \frac{d}{d t} [i_{0} e^{- t / τ}]$

= $= - L i_{0} \frac{- 1}{2} e^{- t / τ}$

$e = \frac{L i_{0}}{τ} e^{- t / τ}$

$e_{a v g} = \frac{\int_{0}^{200 μ s} e d t}{\int_{0}^{200 μ s} d t} = \frac{\int_{0}^{200 μ s} \frac{L i_{0}}{τ} e^{- t / τ} d t}{\int_{0}^{200 μ s} d t}$

$= \frac{L i_{0}}{200 * 1 0^{- 6} s e c} = \frac{20 * 1 0^{- 3} * 2}{200 * 1 0^{- 6}} = 400$

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt[]{7} m .$ The refractive index of water is $\frac{4}{3} .$ The area of the surface of water through which light from the bulb can emerge out is xπm². The value of x is_________.

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Vishal Baghel

Contributor-Level 10

by snell's law

$\frac{4}{3} s i n θ_{C} = μ_{1} s i n 90 °$

$s i n θ_{C} = \frac{3}{4} = \frac{r}{\sqrt[]{r^{2} + {(\sqrt[]{7})}^{2}}}$

$\frac{3}{4} = \frac{r}{\sqrt[]{r^{2} + 7}}$

Squaring both side

$\frac{9}{16} = \frac{r^{2}}{r^{2} + 7}$

$A = r^{2} = π * 9 = 9 π$

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

Two 10cm log, straight wires, each carrying a current of 5A are kept parallel to each other. If each wire experienced a force of 10^-5N, then separation between the wires is__________cm.

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Vishal Baghel

Contributor-Level 10

$F = \frac{μ_{0} * 2}{4 π} \frac{i_{1} i_{2}}{d} l$

$1 0^{- 5} = \frac{1 0^{- 7} * 2 * 5 * 5}{d} = \frac{10}{100}$

$d = 5 * 1 0^{- 2} m$

d = 5 cm

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