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6 months ago

Physics Physics Ncert Solutions Class 12th

A telegraph line of length 100km has a capacity of $0.01 μ F / k m$ and it carries an alternating current at 0.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that need to be introduced in series is _________mH.

(if p = $\sqrt[]{10}$ )

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Raj Pandey

Contributor-Level 9

C = 10^-6 f

f = 0.5 * 10³ = 500

$z = \sqrt[]{x_{R}^{2} + {(x_{L} - x_{C})}^{2}}$

for minimum impedance :

x_L = x_C

$\Rightarrow ω^{2} = \frac{1}{L C}$

$\Rightarrow L = \frac{1}{π^{2}} = \frac{1}{10} * 1000 = 100 m H$

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6 months ago

Physics Class 12th

(if p = $\sqrt[]{10}$ )

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6 months ago

Physics Physics Moving Charges and Magnetism

A singly ionized magnesium atom (A = 24) ion is accelerated to kinetic energy 5 ke V, and is projected perpendicularly into a magnetic field B of the magnitude 0.5 T. The radius of path formed will be ________cm.

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Raj Pandey

Contributor-Level 9

$r = \frac{m v}{q B} = \frac{\sqrt[]{2 k m}}{q B}$

$\frac{\sqrt[]{2 * 5 * 1 0^{3} * 1.6 * 1 0^{- 19} * 24 * 1.66 * 1 0^{- 27}}}{1.6 * 1 0^{- 19} * \frac{1}{2}}$

= 9.975 * 10^-2 cm

= 9.975 cm

$r \approx 10$ cm

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6 months ago

Physics Physics Thermodynamics

The total internal energy of two mole mono atomic ideal gas at temperature T = 300K will be __________J. (Given R = 8.31 J/mol.K)

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Raj Pandey

Contributor-Level 9

$U = n c_{v} T$

$\begin{array}{l} = 2 * \frac{3}{2} * R * 300 \\ = 7479 J \end{array}$

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6 months ago

Physics Class 11th

A hanging mass M is connected to a four times bigger mass by using a string-pulley arrangement, as shown in the figure. The bigger mass is placed on a horizontal ice-slab and being pulled by 2 Mg force. In this situation, tension in the string is $\frac{x}{5} M g$ for x_________. Neglect mass of the string and fiction of the block (bigger mass) with ice slab.

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Raj Pandey

Contributor-Level 9

Using newton's law:

2mg – T = 4ma - (1)

T – mg = ma - (2)

From (1) & (2) :

$m g = 5 m a \Rightarrow a = \frac{g}{5} = 2 m / s^{2}$

$T = m g + \frac{m g}{5} = \frac{6 m g}{5}$

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6 months ago

Physics Class 11th

A man of 60kg is running on the road and suddenly jumps into stationary trolly car of mass 120kg. Then, the trolly car starts moving with velocity 2ms^-¹. The velocity of the running man was _________ ms^-¹, when he jumps into the car:

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Raj Pandey

Contributor-Level 9

Using conservation of momentum:

60 * v = (60 + 120) * 2

⇒v = 6m/s

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6 months ago

Physics Class 11th

The position vector of 1kg object is $\vec{r} = (3 \hat{i} - \hat{j}) m$ and its velocity $\vec{v} = (3 \hat{j} + \hat{k}) m s^{- 1} .$ The magnitude of is angular momentum is $\sqrt[]{x} N m$ where x is________.

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Raj Pandey

Contributor-Level 9

$\overset{?}{L} = \overset{?}{r} * m \overset{?}{v}$

= $(3 \hat{i} ? \hat{j}) * (3 \hat{j} + \hat{k})$

$= 9 \hat{k} + 3 (? \hat{j}) ? \hat{i}$

$= ? \hat{i} ? 3 \hat{j} + 9 \hat{k}$

$| \overset{?}{L} | = \sqrt[]{1 + 9 + 81} = \sqrt[]{91}$

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6 months ago

Physics Physics Alternating Current

An AC source is connected to an inductance of 100 mH, a capacitance of 100 $μ F$ and a resistance of 120 $Ω$ as shown in figure. The time in which the resistance having a thermal capacity 2 J/°C will get heated by 16°C is________ s.

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Raj Pandey

Contributor-Level 9

$x_{R} = 120 Ω$

$x_{L} = ω L = 100 * 100 * 1 0^{- 3} 10 Ω$

$x_{C} = \frac{1}{ω c} = \frac{1}{100 * 100 * 1 0^{- 6}} = 100 Ω$

$z = \sqrt[]{12 0^{2} + {(100 - 10)}^{2}} = \sqrt[]{12 0^{2} + 9 0^{2}}$

= 150 $Ω$

$\Rightarrow 32 = \frac{160}{75} * t \Rightarrow t = \frac{32 * 75}{160} = 15$

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6 months ago

Physics Class 12th

Current measured by the ammeter (A) in the reported circuit when no current flows through $10 Ω$ resistance, will be A.

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Raj Pandey

Contributor-Level 9

Using whetstone

$\frac{R}{3} = \frac{4}{6} \Rightarrow R = 2 Ω$

$R_{e q u} = \frac{54}{15} Ω$

$i = \frac{36}{\frac{54}{15}} = 10 A$

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6 months ago

Physics Physics Current Electricity

A meter bridge setup is shown in the figure. It is used to determine an unknown resistance R using a given resistor of 15 $Ω$ . The galvanometer (G) shows null deflection when tapping key is at 43cm mark from end A. If the end correction for end A is 2cm. Then the determined value of R will be ________ $Ω$ .

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Raj Pandey

Contributor-Level 9

At null point:

$\frac{15}{43 + 2} = \frac{R}{100 - 43}$

$\Rightarrow R = 19 Ω$

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