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Physics Class 12th

Varying velocity and varying acceleration

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alok kumar singh

Contributor-Level 10

A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration. It is because direction of both velocity as well as acceleration will change continuously.

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8 months ago

Physics Class 12th

A carrier wave with amplitude of 250 V is amplitude modulated by a sinusoidal base band signal of amplitude 150V. The ratio of minimum amplitude to maximum amplitude for the amplitude modulated wave is 50 : x, then value of x is__________.

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alok kumar singh

Contributor-Level 10

$A_{m a x} = A_{c} + A_{m}$

$A_{m i n} = A_{c} - A_{m}$

$\frac{A_{m i n}}{A_{m a x}} = \frac{A_{c} - A_{m}}{A_{c} + A_{m}} = \frac{250 - 150}{250 + 150}$

$\Rightarrow \frac{100}{400} = \frac{50}{200}$

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8 months ago

Physics Work, Energy and Power

An engine is attached to a wagon through a shock absorber of length 1.5m. The system with a total mass of 40,000kg is moving with a speed of 75 kmh^-1 when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0m. If 90% of energy of the wagon is lost due to friction, the spring constant is________* 10⁵ N/m.

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alok kumar singh

Contributor-Level 10

By work energy theorem

Work done = change in K.E.

Work done by friction work done by spring

$= 0 - \frac{1}{2} m V^{2}$

As 90% of K.E. is losed by friction so that

$- \frac{90}{100} (\frac{1}{2} m V^{2}) - \frac{1}{2} K x^{2} = - \frac{1}{2} m V^{2}$

-K -> -16 * 10⁵

K = 16 * 10⁵

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8 months ago

Physics Physics System of Particles and Rotational Motion

A 2kg steel rod of length 0.6m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the firction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is________ms^-1.

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alok kumar singh

Contributor-Level 10

$l = \frac{m l^{2}}{3}$

Energy conservation Low

$m g l = \frac{1}{2} \frac{m l^{2}}{3} ω^{2} . . . . . (i)$

And speed V = $ω r = ω l$

then $V = \sqrt[]{6 g l} = \sqrt[]{6 * 10 * . 6}$

->6 m/s

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8 months ago

Physics Laws of Motion

When a body slides down from rest along a smooth inclined plane making an angle of 30° with the horizontal, it takes time T. When the same body slides down from the rest along a rough inclined plane making the same angle and though the same distance, it takes time aT, where a is a constant greater than 1. The co-efficient of friction between the body and the rough plane is $\frac{1}{\sqrt[]{x}} (\frac{α^{2} - 1}{α})$ where x =___________.

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alok kumar singh

Contributor-Level 10

for smooth surface

$a = g s i n 30 ° = \frac{g}{2}$

$S_{1} = u t + \frac{1}{2} a t^{2}$

$S_{1} = \frac{1}{2} \frac{g}{2} t^{2} = \frac{g}{4} t^{2} . . . . . . . (i)$

for rough Surface

$S = \frac{1}{2} \frac{g}{2} (1 - μ \sqrt[]{3}) α^{2} t^{2} . . . . . . . . (i i)$

By (i) and (ii)

$μ = \frac{1}{\sqrt[]{3}} (\frac{α^{2} - 1}{α^{2}}) x = \sqrt[]{3}$

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8 months ago

Physics Gravitation

The satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hour respectively. The radius of the orbit of nearer satellite is 2 * 10³ km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $\frac{π}{x} r a d h^{- 1}$ where x is__________.

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alok kumar singh

Contributor-Level 10

For A satellite T₁ = 1hour

$S o ω_{1} = 2 π r e d / h o u r$

for B satellite T₂ = 8 hour

given R₁ = 2 * 10³ Km

Relative $ω = \frac{V_{1} - V_{2}}{R_{2} - R_{1}} = \frac{2 π * 1 0^{3}}{6 * 1 0^{3}}$

$\Rightarrow \frac{π}{3}$ rad/hour

x = 3

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8 months ago

Physics Class 11th

The average translational kinetic energy of N₂ gas molecules at_________°C becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given k_B = 1.38 * 10^-23 J/K) (Fill the nearest integer).

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alok kumar singh

Contributor-Level 10

K.E. energy of electron = eV

Translational K.E. of N₂ = $\frac{3}{2} K T$

So eV = $\frac{3}{2} K T$

$1.6 * 1 0^{- 19} * 0.1 = \frac{3}{2} * 1.38 * 1 0^{- 23} * T$

T = 773 – 273 = 500°C

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8 months ago

Physics Thermodynamics

The temperature of 3.00 mol of an ideal diatomic gas in increased by 40.0°C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of work done by the gas is $\frac{x}{10} .$ Then the value of x (round off to the nearest integer) is________.

(Given R = 8.31 J mol^-1K^-1)

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alok kumar singh

Contributor-Level 10

Since process is isochoric

So $Δ U = n C_{v} Δ T$

$Δ U = n (\frac{5}{2} R) Δ T - - (i) [C_{V} = \frac{5}{2} R]$

And external work

$Δ W = n R Δ T - - (i i)$

$\frac{5}{2} = \frac{x}{10} \Rightarrow x = 25.00$

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8 months ago

Physics Class 12th

A uniform heating wire of resistance 36 $Ω$ is connected across a potential difference of 240V. The wire is then cut into half and a potential difference of 240 V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : x, where x is________.

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alok kumar singh

Contributor-Level 10

In first condition R₁ = 36 $Ω$

In second condition R₂ = 18 $Ω$

$P_{1} = \frac{V^{2}}{R_{1}} = \frac{{(240)}^{2}}{36}$

$P_{2} = \frac{V^{2}}{R_{2}} + \frac{V^{2}}{R_{2}} = \frac{{(240)}^{2}}{18} + \frac{{(240)}^{2}}{18}$

$P_{2} = \frac{{(240)}^{2}}{9}$

So $\frac{P_{1}}{P_{2}} = \frac{{(240)}^{2} / 36}{{(240)}^{2} / 9} = \frac{1}{4}$

x = 4

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8 months ago

Physics Physics Thermal Properties of Matter

A steel rod with y = 2.0 * 10¹¹ Nm^-2 and a = 10^-5 °C^-1 of length 4m and area of cross-section 10cm² is heated from 0°C to 400°C without being allowed to extend. The tension produced in the rod is x * 10⁵ N where the value of x is_________.

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alok kumar singh

Contributor-Level 10

Force = Aya $Δ T$

Force $= (10 * 1 0^{- 4}) * (2 * 1 0^{11}) * 1 0^{- 5} * 400$

$F = 8 * 1 0^{5} N$

$x * 1 0^{5} = 8 * 1 0^{5}$

x = 8

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