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9 months ago

Physics Work, Energy and Power

A car cover AB distance with first one-third at velocity $υ_{1} m s^{- 1},$ second one-third at $υ_{2} m s^{- 1}$ and last one-third at $υ_{3} m s^{- 1} .$ If then the average velocity of the car is__________ ms^-1.

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alok kumar singh

Contributor-Level 10

$t_{1} = \frac{L}{3 V_{1}}$

$t_{2} = \frac{L}{3 V_{2}}$

$t_{3} = \frac{L}{3 V_{3}}$

$V_{a m g} = \frac{L}{t_{1} + t_{2} + t_{3}} = \frac{L}{\frac{L}{3 V_{1}} + \frac{L}{3 V_{2}} + \frac{L}{3 V_{3}}}$

$V_{a m g} = \frac{3}{\frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{1}{v_{3}}}$

$= \frac{3}{\frac{1}{11} + \frac{1}{2 * 11} + \frac{1}{3 * 11}}$

= $\frac{3 * 11 * 6}{6 + 3 + 2}$

= 3 * b

= 18 m/sec

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9 months ago

Physics Units and Measurement

A uniform disc with mass M = 4 kg and radius R = 10 cm is mounted on a fixed horizontal axle as shown in figure. A block with mass m = 2kg hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is_________ N.

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alok kumar singh

Contributor-Level 10

NLM² for (1)

mg – T = ma

Þ 2 * 10 – T = 2a

Þ T + 2a = 20 - (1)

Rotation equestion for (2)

T * R = I_cma

Þ T R = $I_{C M} \frac{a}{R}$

$[α = \frac{a}{R}, N o s l i p p i n g c o n d i t i o n]$

T = $\frac{M a}{2}$

$T = \frac{4}{2} a$

T = 2a - (2)

from (1) & (2)

2T = 20

T = 10N

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9 months ago

Physics Nuclei

A liquid of density 750 kgm^-3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from A₁ = 1.2 * 10^-2m² to A₂ = $\frac{A_{1}}{2} .$ The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is_________ * 10^-3m³s^-1.

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alok kumar singh

Contributor-Level 10

Given A₂ = $\frac{_{}}{}$ $\frac{A_{1}}{2}$

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

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

[z₁= z₂]

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

[Q = A₁V₁ = A₂V₂]

$\frac{4500}{750} = \frac{{(\frac{Q}{A_{2}})}^{2}}{2} - {(\frac{Q}{A_{1}})}^{2}$

$12 = Q^{2} [\frac{1}{A_{2}^{2}} - \frac{1}{A_{1}^{2}}]$

$Q = 2 A_{1} = 2 * 1.2 * 1 0^{- 2}$

= 2.4 * 10^-2 m³ /sec

= 24 * 10^-3 m³ /sec

= 24

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9 months ago

Physics Class 12th

A running fork of frequency 340 Hz resonates in the fundamental mode with a air column of length 125cm in a cylindrical tube closed at one end. When water is slowly poured in it, the minimum height of water required for observing resonance once again is __________cm.

(Velocity of sound in air is 340 ms^-1)

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alok kumar singh

Contributor-Level 10

First resonating length = $λ = \frac{V}{f} = \frac{340}{340} = 1 m$ $\frac{}{}$

Second resonating length = $\frac{3 λ}{4} = 75 c m$ $\frac{}{}$

Third resonating length = $\frac{5 λ}{4} = 125 c m$ $\frac{}{}$

Height of water required = 125 – 75 = 50cm

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9 months ago

Physics Class 12th

A capacitor C₁ of capacitance $5 μ F$ is charged to a potential of 30V using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor C₂ of capacitance $10 μ F$ as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor C₂ is__________ $μ C .$

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alok kumar singh

Contributor-Level 10

Q = C₁v

...more

Q = C₁v

= 5 * 30

Q = 150 $μ c$

Now it is connected to an unchanged capacitor C₂

Let change on C₁ ® q₁

& 4 C₂ ® q₂

Apply kvL

$\frac{q_{1}}{q} - \frac{q_{2}}{C_{2}} = 0$

Þ q₂ = 2q₁ – (1)

& conservation of change

q₁ + q₂ = 150 – (2)

Þ from (1) & (2)

3q₁ = 150

q₁ = 50 $μ c$

$q_{2} = 100 μ c$

$q_{2} = 100 μ c o n c_{2}$

less

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9 months ago

Physics Class 12th

All resistances in figure are $1 Ω$ each. The value of current 'I' is $\frac{a}{5} A .$ The value of a is___________.

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alok kumar singh

Contributor-Level 10

I = $\frac{3 * 8}{15 R} = \frac{8}{5 R} = \frac{a}{5}$

$\Rightarrow \frac{8}{5 * 1} = \frac{a}{5}$

a = 8

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9 months ago

Physics Class 12th

In the given circuit, the magnitude of V_L and V_C are twice that of V_R. Given that f = 50 Hz. the inductance of the coil is $\frac{1}{K π} m H .$ The value of K is_________.

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alok kumar singh

Contributor-Level 10

Given

V_L = V_C= 2V_R, f = 50Hz

$L = \frac{1}{k π} m H$

since, V_L = V_C.

then

$V_{n e t} = \sqrt[]{{(V_{L} - V_{C})}^{2} + {(V_{R})}^{2}}$

V_R = V_Net= 220V

I = $\frac{V_{N e t}}{Z} = \frac{220}{\sqrt[]{{(X_{L} - X_{C})}^{2} + 1 2^{2}}}$

$I = 44 A$

$V_{L} = I x_{L} = 2 v_{R}$

= I _xL = 440

x_L = 10

WL = 10

2πfL = 10

$L = \frac{1}{\frac{1}{100} π} k = \frac{1}{100} = 0.01 \approx 0$

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New answer posted

9 months ago

Physics Physics Moving Charges and Magnetism

In a Young's double slit experiment, an angular width of the fringe is 0.53° on a screen place at 2m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is $\frac{1}{α} .$ The value of a is__________.

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alok kumar singh

Contributor-Level 10

we know, = $\frac{λ}{d}$ $μ = \frac{C}{V} = \frac{C}{υ λ}$

$λ_{1} = \frac{C}{μ_{1} υ_{1}}$ [with change in medium frequency refnain constaint]

$λ_{2} = \frac{C}{μ_{2} υ_{2}}$

$\frac{λ_{2}}{λ_{1}} = \frac{μ_{1}}{μ_{2}} = \frac{5}{7}$ $λ_{2} = \frac{5}{7} λ_{1}$

$θ_{1} = \frac{λ_{1}}{d_{1}}$

2 = $\frac{λ_{2}}{d_{2}}$

$\frac{θ_{2}}{θ_{1}} = \frac{λ_{2}}{λ_{1}} [? d_{1} = d_{2}]$

2 $= \frac{5}{7} * θ_{1}$

$θ_{2} = \frac{5}{7} * 0.35$

$θ_{2} = \frac{1}{4} = \frac{1}{α}$

= 4

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New answer posted

9 months ago

Physics Class 12th

In the following unclear reaction,

D $\overset{α}{\to} D_{1} \overset{β^{-}}{\to} D_{2} \overset{α}{\to} D_{3} \overset{γ}{\to} D_{4}$

Mass number of D is 182 and atomic number is 74. Mass number and atomic number of D₄ respectively will be________

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alok kumar singh

Contributor-Level 10

Please find the solution below:

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New answer posted

9 months ago

Physics Current Electricity Analysis

A zener of breakdown voltage V_Z = 8V and maximum zener current, I_ZM = 10mA is subjected to an input voltage V_i = 10V with series resistance R = 100 $Ω$ In the given circuit R_L represents the variable load resistance. The ratio maximum and minimum value of R_L is__________.

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alok kumar singh

Contributor-Level 10

Across zener diode & RL

8 = I_LR_L

8 = $(20 - 10) R_{L (m a x)}$

$R_{L m a x} = \frac{8}{10} k Ω$

At max current in loop (1)

10 = i_max R + 8

i_max= $\frac{2}{R} = \frac{2}{100} = 0.02 A$

= 20 mA

R_L (max) = $\frac{8}{10} k Ω$

At minimum curre3nt through zener

i_max R_L (minimum) = 8

R_L (minimum) = $\frac{8}{20} k Ω$

$\frac{R_{L m a x}}{R_{L m i n}} = 2$

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