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Physics NCERT Exemplar Solutions Class 12th Chapter Four

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4 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Moving Charges and Magnetism

The I-V characteristics of p-n junction diode in forward bias is shown in the figure. The ration of dynamic resistance, corresponding to forward bias voltage of 2 V and 4 V respectively, is

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Payal Gupta

Contributor-Level 10

Dynamic Resistance at 2v

= $\frac{Δ v}{Δ I} = \frac{2.1 - 2}{(10 - 5) * 1 0^{- 3}} = \frac{0.1}{5} * 1 0^{3}$

Dynamic Resistance at 4 v

= $\frac{Δ v}{Δ I} = \frac{4.2 - 4}{(250 - 200)} = \frac{0.2}{50 * 1 0^{- 3}} = \frac{0.2}{50} * 1 0^{3}$

$\frac{R_{2 v}}{R_{4 v}} = \frac{0.1 * 1 0^{3} * 50}{5 * 0.2 * 1 0^{3}} = 5 : 1$

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4 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Motion in a Plane

Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerated with $\frac{g}{6}$ vertically upwards then the time period will be:

(Where g = acceleration due to gravity)

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Payal Gupta

Contributor-Level 10

$g_{e f f} = g + a$

$= g + \frac{g}{6}$

$= \frac{7 g}{6}$

$T' = 2 π \sqrt[]{\frac{l}{g_{e f f}}}$

$T' = \sqrt[]{\frac{6}{7}} T$

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Physics NCERT Exemplar Solutions Class 12th Chapter Four Motion in a Plane

Two vectors $\vec{A} a n d \vec{B}$ have equal magnitudes. If magnitude of $\vec{A} + \vec{B}$ is equal to two times the magnitude of $\vec{A} - \vec{B},$ then the angle between $\vec{A} a n d \vec{B}$ will be:

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alok kumar singh

Contributor-Level 10

$| \vec{A} + \vec{B} | = 2 | \vec{A} - \vec{B} |$

Given A = B

Squaring equation (1) both side.

${| \vec{A} + \vec{B} |}^{2} = {(2 | \vec{A} - \vec{B} |)}^{2}$

$A^{2} + B^{2} + 2 \vec{A} . \vec{B} = 4 (A^{2} + B^{2} - 2 \vec{A} . \vec{B})$

2A² + 2A² cosq = 4 (2A² – 2A² cosq)

2A² + 2A² cosq = 8A² – 8A² cosq

10A² cosq = 6A²

cosq = $\frac{3}{5}$

q = cos^-1 (3/5)

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4 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Physics Moving Charges and Magnetism

A charge particle moves along circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charge particle increases to 4 times its initial value. What will be the ratio of new radius to the original radius of circular path of the charge particle:

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alok kumar singh

Contributor-Level 10

$\sqrt[]{k \cdot E}$

$\frac{r_{f}}{r_{0}} = \frac{\sqrt[]{k . E_{f}}}{k . E_{i}}$

$\frac{r_{f}}{r_{0}} = \frac{\sqrt[]{4 k . E_{i}}}{\sqrt[]{k . E_{i}}}$

$\frac{r_{f}}{r_{0}} = \frac{2}{1} = 2 : 1$

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Physics NCERT Exemplar Solutions Class 12th Chapter Four Motion in a Plane

If $\bar{A} = (2 \hat{i} + 3 \hat{j} - \hat{k}) m a n d \vec{B} = (\hat{i} + 2 \hat{j} + 2 \hat{k}) m .$ The magnitude of component of vector $\vec{A}$ along vector $\vec{B}$ will be________m.

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alok kumar singh

Contributor-Level 10

Component of $\vec{A}$ along $\vec{B}$ = $\frac{\vec{A} . \vec{B}}{| \vec{B} |}$

= 2

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Physics NCERT Exemplar Solutions Class 12th Chapter Four Motion in a Plane

If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory is y = 5x(1 – x). The y component vector of the initial velocity is ________ $\hat{j}$ .

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Vishal Baghel

Contributor-Level 10

y = x5 (1 - x) = x tanθ $(1 - \frac{x}{R})$

tan = 5, R = 1

$s i n θ = \frac{5}{\sqrt[]{26}}, c o s θ = \frac{1}{\sqrt[]{26}}$

$R = \frac{u^{2} s i n 2 θ}{g} = 1$

$\Rightarrow u^{2} = 26 \Rightarrow u = \sqrt[]{26} m / s$

y - component of initial velocity

= u sinθ

= $= \sqrt[]{26} * \frac{5}{\sqrt[]{26}}$

= 5 m/s

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Physics NCERT Exemplar Solutions Class 12th Chapter Four Motion in a Plane

Two projectiles thrown at $30 ° a n d 45 °$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is:

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Vishal Baghel

Contributor-Level 10

H₁ = H₂

$\frac{u_{1}^{2} s i n^{2} θ_{1}}{2 g} = \frac{u_{2}^{2} s i n^{2} θ_{2}}{2 g}$

$\Rightarrow u_{1}^{2} {(s i n 30 °)}^{2} = {u_{2}}^{2} {(s i n 45 °)}^{2}$

$\Rightarrow \frac{u_{1}}{u_{2}} = \sqrt[]{2}$

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4 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Motion in a Plane

Two projectiles are thrown with same initial velocity making an angle of $45 °$ and $30 °$ with the horizontal respectively. The ration of their respectively ranges will be:

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alok kumar singh

Contributor-Level 10

$R_{1} = \frac{u^{2} s i n 90}{g}$

$R_{2} = \frac{u^{2} s i n 60}{g}$

$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{2}{\sqrt[]{3}}$

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4 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Motion in a Plane

The trajectory of a projectile in an vertical plane is y = ax - bx², where a and b are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection q and the maximum height attained H are respectively given by

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Payal Gupta

Contributor-Level 10

$y = α x - β x^{2}$

$\Rightarrow \frac{d y}{d x} = α - 2 β x = 0$

$\Rightarrow x = \frac{α}{2 β}$

$y_{m a x} = α * \frac{α}{2 β} - β * {(\frac{α}{2 β})}^{2} = \frac{α^{2}}{4 β}$

Range = $2 x = \frac{α}{β} = \frac{2 u^{2} s i n θ . c o s θ}{g}$

On comparing with

y = x tan θ- $\frac{g x^{2}}{2 u^{2} c o s^{2} θ}$

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4 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Moving Charges and Magnetism

An infinitely long hollow conducting cylinder with radius R carries a uniform current along its surface. Choose the correct representation of magnetic field (B) as a function of radial distance (r) from the axis of cylinder.

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Payal Gupta

Contributor-Level 10

Using Ampere's law for long hollow cylinder carrying current on its surface

B_in = 0

$B_{o u t} = \frac{μ_{0} i}{2 π r}$

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