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8 months ago

Physics Class 12th

In a given circuit diagram, a 5 V zener diode along with a series resistance is connected across a 50 V power supply. The minimum value of the resistance required, if the maximum zener current is 90 mA be________ $Ω$ .

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Vishal Baghel

Contributor-Level 10

For 90 mA in zener diode, current in R should be greater than 90 mA.

$\frac{45}{R} = 90 * 1 0^{- 3} \Rightarrow R = \frac{45}{90 * 1 0^{- 3}} = 500 Ω$

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New answer posted

8 months ago

Physics Class 12th

In an electric circuit, a cell of certain emf provides a potential difference of 1.25 V across a load resistance of $5 Ω$ . However, it provides a potential difference of 1V across a load resistance of $2 Ω .$ The emf of the cell is given by $\frac{x}{10} V .$ Then the value of x is_________.

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Vishal Baghel

Contributor-Level 10

$i = \frac{ε}{R + r}$

$v_{R} = \frac{ε R}{R + r}$

$\frac{1.25 = \frac{ε (5)}{5 + r}}{1 = \frac{ε (2)}{2 + r}}$

r = 1

Using above equ : $ε = \frac{3}{2} = \frac{15}{10} V$

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New answer posted

8 months ago

Physics Mechanical Properties of Solids

The area of cross – section of a railway track is 0.01 m². The temperature variation is 10°C, Coefficient of linear expansion of material of track is 10^-⁵ / °C. The energy stored per meter in the track is_________J/m.

(Young's modulus of material of track is 10¹¹ Nm^-²)

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Vishal Baghel

Contributor-Level 10

Energy density (u) = $\frac{1}{2} * σ * ε = \frac{1}{2} * Y ε * ε = \frac{1}{2} ε^{2} Y$ $\frac{}{} \frac{}{} \frac{}{}^{}$

$\Rightarrow e n e r g y s t o r e d / m^{2} = \frac{1}{2} ε^{2} Y A$

Strain $(ε) = \frac{Δ l}{l_{0}} = α Δ T = 1 0^{- 5} * 10 = 1 0^{- 4}$ $\frac{}{_{}}^{}^{}$

$\Rightarrow e n e r g y s t o r e d / m^{2} = \frac{1}{2} * 1 0^{- 8} * 1 0^{11} * 1 0^{- 2} = 5$

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New answer posted

8 months ago

Physics Thermodynamics

In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is__________°C.

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Vishal Baghel

Contributor-Level 10

from newtan's law of cooling :

$\frac{d T}{d t} = - k (T - T_{0})$

Using average value :

$\frac{75 - 65}{5} = - k [\frac{(75 + 65) 2}{2} - 25]$

$\frac{(65 - T)}{5} = - k [\frac{(65 + T)}{2} - 25]$

$\Rightarrow \frac{10}{65 - T} = \frac{45}{\frac{65 + T}{2} - 25}$

$\Rightarrow T = 57 ° C$

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New answer posted

8 months ago

Physics Units and Measurement

Three students S₁, S₂ and S₂ perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.

Student No.	Length of Pendulum (cm)	No. of oscillations (n)	Total for n oscillations	Time period (s)
1	64.0	8	128.0	16.0
2	64.0	4	64.0	16.0
3	20.0	4	36.0	9.0

Least count of length = 0.1 cm

least count for time = 0.1 s)

If E₁, E₂ and E₃ are the percentage errors 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no.___________.

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Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{l}{g}}$

$g = 4 π^{2} \frac{l}{T^{2}}$

$\Rightarrow \frac{Δ g}{g} = \frac{Δ l}{l} + \frac{2 Δ T}{T}$

Student 1 will get least error because it has more number of oscillation

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New answer posted

8 months ago

Physics System of Particles and Rotational Motion

The centre of a wheel rolling on a plane surface moves with a speed $υ_{0} .$ A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt[]{x} υ_{0} .$ Then the value of x is__________.

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Vishal Baghel

Contributor-Level 10

$V_{P} = \sqrt[]{V_{0}^{2} + V_{0}^{2}} = \sqrt[]{2} V_{0}$

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New answer posted

8 months ago

Physics Class 12th

A ray of light passing through a prism $(μ = \sqrt[]{3})$ surface minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is________(in degrees).

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Vishal Baghel

Contributor-Level 10

At minimum deviation :

i = i'

r = r'

$r + r' = A \Rightarrow r = \frac{A}{2}$

Snell's law :

1 sin I = $\sqrt[]{3} s i n r = \sqrt[]{3} s i n \frac{i}{2}$ $\frac{}{}$

$\Rightarrow 2 c o s \frac{i}{2} = \sqrt[]{3}$

$\Rightarrow i = 60 °$

$\begin{array}{l} i = 2 r \Rightarrow r = 30 ° \\ \Rightarrow A = 60 ° \end{array}$

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New answer posted

8 months ago

Physics Class 12th

The total charge enclosed in an incremental volume of 2 * 10^-⁹ m³ located at the origin is _________nC, if electric flux density of its field is found as

D = e^-^x sin $y \hat{i} - e^{- x} c o s y \hat{j} + 2 z \hat{k} C / m^{2} .$

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Vishal Baghel

Contributor-Level 10

Using gauss law in differential form

$\vec{\nabla} . \vec{E} = \frac{ρ}{ε_{0}}$

$\vec{\nabla} . (ε_{0} \vec{E}) = ρ$

$\Rightarrow \vec{\nabla} . \vec{D} = ρ$

$\Rightarrow (\frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j} + \frac{\partial}{\partial z} \hat{k}) . \vec{D} = ρ$

$\begin{array}{l} \Rightarrow ρ = - e^{- x} s i n y + e^{- x} s i n y + 2 = 2 \\ \Rightarrow Q = 2 (V o l u m e) = 4 * 1 0^{- 9} c = 4 n c \end{array}$

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New answer posted

8 months ago

Physics Electric Charges and Fields

An electric dipole is placed on x – axis in proximity to a line charged of linear density 3.0 * 10^-⁶ C/m. Line charge is placed on z – axis and positive and negative charge of dipole is at a distance of 10 mm and 12 mm from the origin respectively. If total force of 4N is exerted on the dipole, find out the amount of positive charge of the dipole.

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Vishal Baghel

Contributor-Level 10

$F = [\frac{2 k λ}{r_{1}} - \frac{2 k λ}{r_{2}}] q = 2 k λ q [\frac{1}{r_{1}} - \frac{1}{r_{2}}]$

$2 * 9 * 1 0^{9} * 3 * 1 0^{- 6} [\frac{1000}{10} - \frac{1000}{12}] q$

$\begin{array}{l} \Rightarrow 4 = 9 * 1 0^{5} q \\ \Rightarrow q = 4.44 μ C \end{array}$

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New answer posted

8 months ago

Physics Class 12th

A Copper (Cu) rod of length 25cm and cross – sectional area 3 mm² is joined with a similar aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends A and B.

(Take Resistivity of Copper = 1.7 * 10^-⁸ $Ω m$

Resistivity of Aluminium = 2.6 * 10^-⁸ $Ω m$ )

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Vishal Baghel

Contributor-Level 10

$R_{1} = ?_{1} \frac{l}{A} R_{2} = ?_{2} \frac{l}{A}$

$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{A}{l} [\frac{1}{?_{1}} + \frac{1}{?_{2}}]$

$= \frac{3 * 1 0^{? 6}}{\frac{1}{4}} [\frac{1}{1.7 * 1 0^{? 8}} + \frac{1}{2.6 * 1 0^{? 8}}]$

$R = 0.858 * 1 0^{? 3} ?$

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