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physics ncert exemplar solutions class 12th chapter two

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Ray Optics and Optical Instruments

Light enters form air into a given medium at angle of 45° with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of 15° from its original direction. The refractive index of the medium is

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Payal Gupta

Contributor-Level 10

Using snell's law –

sin I = $μ s i n r$

sin45° = $μ s i n 30 °$

$μ$ $=$ $\frac{s i n 45}{s i n 30}$ $=$ $\frac{1}{2}$ $*$ $2$ $=$ √2

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physics ncert exemplar solutions class 12th chapter two Class 11th

The root mean square speed of smoke particles of mass 5 * 10^-17kg in their Brownian motion in air at NTP is approximately. [Given k = 1.38 * 10^-23 JK^-1]

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Payal Gupta

Contributor-Level 10

m = 5 * 10^-17kg ; k = 1.38 * 10^-23 Jk^-1

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}} = \sqrt[]{\frac{3 N k T}{N m}} = \sqrt[]{\frac{3 k T}{m}} = \sqrt[]{\frac{3 * 1.38 * 1 0^{- 23} * 300}{5 * 1 0^{- 17}}}$

$= \sqrt[]{18 * 13.8 * 1 0^{- 6}}$

$= 15.7 * 1 0^{- 3} m / s$

$= 15 m m / s$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Magnetism and Matter

The vertical component of the earth's magnetic field is 6 * 10^-5 T at any place where the angle of dip is 37°. The earth's resultant magnetic field at that place will be (Given tan37° = $\frac{3}{4}$ )

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Payal Gupta

Contributor-Level 10

B_v = 6 * 10^-5 T

B_v = B_N sin37°

$∴ B_{N} = \frac{B_{v}}{s i n 37 °} = \frac{6 * 1 0^{- 5}}{3 / 5} = 1 0^{- 4} T$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Thermodynamics

A thermodynamic system is taken from an original state D to an intermediate sate E by the linear pressure shown in the figure. Its volume is then reduced to the original volume from E to F by an isobaric process. The total work done by the gas from D to E to F will be

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Payal Gupta

Contributor-Level 10

W_total = W_D→E + W_E→F = Area of triangle DEF

= $\frac{1}{2} * 3 * 300 = 450 J .$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Class 11th

The torque of a force $5 \hat{i} + 3 \hat{j} - 7 \hat{k}$ about the origin is $τ$ . If the force acts on a particle whose position vector is $2 \hat{i} + 2 \hat{j} + \hat{k},$ then the value of $τ$ will be

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Payal Gupta

Contributor-Level 10

$\vec{F} = 5 \hat{i} + 3 \hat{j} - 7 \hat{k}$

$\vec{r} = 2 \hat{i} + 2 \hat{j} + \hat{k}$

$\vec{τ} = | \vec{r} * \vec{F} | = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 \\ 5 & 3 & - 7 \end{matrix} | \begin{matrix} = \hat{i} (- k_{1} - 3) - \hat{j} (- 14 - 5) + \hat{k} (6 - 10) & = - 17 \hat{i} + 19 \hat{j} - 4 \hat{k} \end{matrix}$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Work, Energy and Power

If momentum of a body is increased by 20%, then its kinetic energy increases by

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Payal Gupta

Contributor-Level 10

$\frac{Δ K}{K_{i}} * 100 = \frac{K_{f} - K_{i}}{K_{i}} * 100$

$= \frac{\frac{P_{f}^{2}}{2 m} - \frac{P_{i}^{2}}{2 m}}{\frac{P_{i}^{2}}{2 m}} * 100 = \frac{P_{f}^{2} - P_{i}^{2}}{P_{i}^{2}} * 100$

$= \frac{{(1.2 P_{i})}^{2} - {(P_{i})}^{2}}{P_{i}^{2}} * 100 = \frac{1.44 P_{i}^{2} - P_{i}^{2}}{P_{i}^{2}} * 100$

= 44%

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Laws of Motion

Two bodies of masses m₁ = 5kg and m₂ = 3kg are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass m₁ will be: [Take g = 10 ms^-2]

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Payal Gupta

Contributor-Level 10

m₁ = 5kg

m₂ = 3kg

As m₁ is in rest

T = m_gg = 30

N = m₁ g cos

T = $m_{1} g s i n θ \Rightarrow 50 s i n θ = 30$

sin $= \frac{3}{5} \Rightarrow θ = 37 °$

$∴ N = 50 c o s θ = 50 c o s 37 ° = 50 * \frac{4}{5} = 40 N$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Motion in Straight Line

A ball is released from a height h. If t₁ and t₂ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between t₁ and t₂.

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Payal Gupta

Contributor-Level 10

$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$

h = $\frac{1}{2} g {(t_{1} + t_{2})}^{2}$

from (i) and (ii) :-

$\frac{1}{2} g t_{1}^{2} = \frac{g}{4} {(t_{1} + t_{2})}^{2}$

$2 t_{1}^{2} = {(t_{1} + t_{2})}^{2}$

$\sqrt[]{2} t_{1} = t_{1} + t_{2}$

$t_{1} = \frac{t_{2}}{\sqrt[]{2} - 1} = (\sqrt[]{2} + 1) t_{2}$

$t_{2} = (\sqrt[]{2} - 1) t_{1}$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Class 11th

An object of mass 1 kg is taken to a height form the surface of earth which is equal to three times that radius of earth. The gain in potential energy of the object will be

[If, g = 10ms^-2 and radius of earth = 6400km]

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Payal Gupta

Contributor-Level 10

m = 1kg ; $Δ U = \frac{- G m_{e} m}{(R + h)} - (\frac{G m_{e} m}{R})$

$Δ U = \frac{G M m}{R} - \frac{G M m}{R + h}$

$= m g_{0} R - m g_{0} \frac{R}{(1 + \frac{h}{R})} = m g_{0} {1 - \frac{1}{(1 + \frac{3 R}{R})}}$

$= m g_{0} R (\frac{3}{4})$

$= \frac{3}{4} * 1 * 10 * 6400 k m$

$= 48000 * 1 0^{3} J .$

$= 48 M J .$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Class 12th

Read the following statements:

(A) Volume of the nucleus is directly proportional to the mass number.

(B) Volume of the nucleus is independent of mass number.

(C) Density of the nucleus is directly proportional to the mass number.

(D) Density of nucleus is directly proportional to the cube root of the mass number.

(E) Density of the nucleus is independent of the mass number.

Choose the correct option from the following options.

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Payal Gupta

Contributor-Level 10

Radius, R = R₀A^1/3

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