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Physics 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

Physics 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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Physics 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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Physics 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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Physics 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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Physics Ray Optics and Optical Instruments

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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Physics physics ncert exemplar solutions class 12th chapter two

The SI unit of a physical quantity is pascal-second. The dimensional formula of this quantity will be:

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Payal Gupta

Contributor-Level 10

S. I. unit Pascal – second $= \frac{N}{m^{2}} - s e c$

= $\frac{M L T^{- 2} T}{L^{2}}$

= $M L^{- 1} T^{- 1}$

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Physics Units and Measurement

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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alok kumar singh

Contributor-Level 10

1° ® 60'

Þ 60' ® 10°

L = 1.5 * 10¹¹ m

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Physics Class 12th

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Physics Electrostatic Potential and Capacitance

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$(E_{2}$ - $E_{1}) . \hat{n} = \frac{σ}{ε_{0}}$

Where $\hat{n}$ is a unit vector normal to the surface at a point and s is the surface charge density at that point. (The direction of $\hat{n}$ is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is $σ \hat{n}$ / $ε_{0}$ .

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

[Hint: For (a), use Gauss's law. For, (b) use the fact that work done by electrostatic field on a closed lo

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(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$(E_{2}$ - $E_{1}) . \hat{n} = \frac{σ}{ε_{0}}$

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

[Hint: For (a), use Gauss's law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

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alok kumar singh

Contributor-Level 10

2.16 Electric field on one side of a charged body is $E_{1}$ and electric field on the other side of the same body is $E_{2} .$ If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

$\overset{?}{E_{1}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(i)

Where,

$\overset{?}{n}$ = unit vector normal to the surface at a point

$σ$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\overset{?}{E_{2}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(ii)

$E l e c t r i c f i e l d a t a n y p o i n t$ due to the two surfaces,

$\overset{?}{E_{2}} - \overset{?}{E_{1}} = \frac{σ}{2 ε_{0}} \overset{?}{n} + \frac{σ}{2 ε_{0}} \overset{?}{n} = \frac{σ}{ε_{0}} \overset{?}{n}$ ……(iii)

$S i n c e i n s i d e a c l o s e d c o n d u c t o r, \overset{?}{E_{1}} = 0,$

$\overset{?}{E}$ = $\overset{?}{E_{2}}$ = $\frac{σ}{ε_{0}} \overset{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{σ}{ε_{0}} \overset{?}{n}$

...more

$\overset{?}{E_{1}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(i)

Where,

$\overset{?}{n}$ = unit vector normal to the surface at a point

$σ$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\overset{?}{E_{2}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(ii)

$E l e c t r i c f i e l d a t a n y p o i n t$ due to the two surfaces,

$\overset{?}{E_{2}} - \overset{?}{E_{1}} = \frac{σ}{2 ε_{0}} \overset{?}{n} + \frac{σ}{2 ε_{0}} \overset{?}{n} = \frac{σ}{ε_{0}} \overset{?}{n}$ ……(iii)

$S i n c e i n s i d e a c l o s e d c o n d u c t o r, \overset{?}{E_{1}} = 0,$

$\overset{?}{E}$ = $\overset{?}{E_{2}}$ = $\frac{σ}{ε_{0}} \overset{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{σ}{ε_{0}} \overset{?}{n}$

When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

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