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6 months ago

Class 12th Oscillations

The equation of a particle executing simple harmonic motion is given by x = sin $(t + \frac{1}{3}) m$ . At t = 1s, the speed of particle will be

(Given : p = 3.14)

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Payal Gupta

Contributor-Level 10

$x = s i n π (t + \frac{1}{3})$

$v = \frac{d x}{d t} = c o s [π (t + \frac{1}{3})] * π$

$= c o s (π + \frac{π}{3}) * π$

$= - c o s \frac{π}{3} * (π)$

$= - \frac{π}{2} = - 1.57 m / s e c$

Speed = 157 cm/sec

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Class 12th Thermodynamics

A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of bullet is:

(Given, initial temperature of the bullet = 127°C,

Melting point of the bullet = 327°C,

Latent heat of fusion of lead = 2.5 * 10⁴ J kg^-¹,

Specific heat capacity of lead = 125 J/kg K)

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Payal Gupta

Contributor-Level 10

40% of K.E. = $m c Δ T + m L$

$0.4 * \frac{1}{2} * m v^{2} = m [125 (327 - 127) + 2.5 * 1 0^{4}]$

$\Rightarrow 0.2 v^{2} = [250 * 1 0^{2} + 25 * 1 0^{3}]$

$v^{2} = \frac{2 * 25 * 1 0^{3}}{0.2}$

v = 5 * 100

v = 500 m/sec

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6 months ago

Class 12th Kinetic Theory

According to kinetic theory of gases,

A. The motion of the gas molecules freezes at 0°C.

B. The mean free path of gas molecules decreases if the density of molecules is increased.

C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.

D. Average kinetic energy per molecules per degree of freedom is $\frac{3}{2} k_{B} T$ (for monoatomic gases).

Choose the most appropriate answer form the options given below

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Payal Gupta

Contributor-Level 10

$\bar{x}$ (mean free path) = $\frac{1}{\sqrt[]{2} π d^{2} n_{v}}$

$n_{v} = \frac{n N_{A}}{v},$ n → No. of moles in volume v N_A ® Avogadro's Number

$\frac{n}{v} = \frac{P}{R_{T}}$

$\bar{x} \propto v \propto \frac{1}{ρ (d e n s i t y} & \bar{x} α T a t c o n s t a n t P)$

^{$ρ ↑ \Leftrightarrow \bar{x} ↓ | \bar{x} ↑ \Leftrightarrow T ↑$}

The motion of the gas molecules freezes at 0K not 0° C

Average kinetic Energy per molecule per degree of freedom is $= \frac{1}{2} k_{B} T$ (for Mono atomic gases)

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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

For a perfect gas, two pressures P1 and P2 are shown in figure. The graph shows:

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Payal Gupta

Contributor-Level 10

At constant Temp

$P α \frac{1}{v}$

v₂ > v₁

⇒ P₁ > P₂

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Class 12th Gravitation

Four spheres each of mass m form a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is

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Payal Gupta

Contributor-Level 10

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Class 12th System of Particles and Rotational Motion

A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is $\sqrt[]{x (h^{2} - g L)} .$ The value of x is

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Payal Gupta

Contributor-Level 10

Conservation of Energy b/w (1) & (2)

$\frac{1}{2} m v^{2} + m g l = \frac{1}{2} m u^{2}$

$v = \sqrt[]{u^{2} - 2 g l}$

$\vec{Δ v} = v_{\hat{j}} - u_{\hat{i}}$

$| Δ \vec{v} | = \sqrt[]{{(u)}^{2} + {(\sqrt[]{u^{2} - 2 g l})}^{2}}$

$= \sqrt[]{u^{2} + u^{2} - 2 g l}$

$= \sqrt[]{2} (u^{2} - g l)$

x = 2

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

One end of a massless spring of spring constant k and natural length $l_{0}$ is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $ω$ about an axis passing through fixed end, then the elongation of the spring will be

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Payal Gupta

Contributor-Level 10

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6 months ago

Class 12th Motion in a Straight Line

When a ball is dropped into a lake form a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is

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Payal Gupta

Contributor-Level 10

Total time (T) = 4 sec

Given u = 0 m/sec

a = g = 9.8 m/sec²

h = 4.9 m, t =?

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

$4.9 = 0 . t + \frac{1}{2} * 9.8 t^{2}$

t = 1 sec

'v' be the velocity with which ball hits the water v = u + at

= 0 + 9.8 * 1 = 9.8 m/sec

Time taken to reach the bottom of the lake from surface of the lake

= 4 – 1 = 3 sec

v = 9.8 m/sec

$H = u t + \frac{1}{2} g t^{2} = 9.8 * 3 + \frac{1}{2} * 9.8 * 9$

29.4 + 4.9 * 9 = 29.4 + 44.1

H = 73.5 m

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The distance of the Sun from earth is 1.5 * 10¹¹ m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be

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Payal Gupta

Contributor-Level 10

1° → 60'

⇒ 60' → 10°

$60' \to {\frac{π}{180}}^{C}$

$θ = \frac{π * 2 * 1 0^{3}}{180 * 3600}$

L = 1.5 * 10¹¹ m

$D = L θ = 1.5 * 1 0^{11} * \frac{π}{180 * 3600} * 2 * 1 0^{3} = 1.45 * 1 0^{9} m$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The distance of the Sun from earth is 1.5 × 10¹¹ m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be

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