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12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The refracting angle of a prims is A and refractive index of the material of the prism is cot (A/2). Then the angle of minimum deviation will be

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Payal Gupta

Contributor-Level 10

At minimum deviation –

$δ = 2 i - A \Rightarrow i = \frac{δ + A}{2}, r = \frac{A}{2}$

Using snell's low : - $s i n i = μ = s i n r$

$\Rightarrow \frac{δ + A}{2} = \frac{π - A}{2} \Rightarrow δ = π - 2 A$

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12 months ago

Physics Electromagnetic Waves

A radar sends an electromagnetic single of electric (E₀) = 2.25 V/m and magnetic field (B₀) = 1.5 * 10^-⁸ T which strikes a target on line of sight at a distance of 3km in a medium. After that, a part of signal (echo) reflects back towards the radar with same velocity and by same path. If the signal was transmitted at time t = 0 from radar, then after how much time echo will reach to the radar?

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Payal Gupta

Contributor-Level 10

$E_{0} = 2.25 \frac{v}{m}$ Speed of declromagnetic signal, $v = \frac{E_{0}}{B_{0}} = \frac{2.25 v / m}{1.5 * 1 0^{- 8} T}$

$B_{0} = 1.5 * 1 0^{- 8} T \Rightarrow v = \frac{225}{150} * 1 0^{8} = 1.5 * 1 0^{8} m / s$

Time to reach each to the same

radar, $t = \frac{2 * 3 * 1 0^{3} m}{1.5 * 1 0^{8} m / s} = 4 * 1 0^{- 5} S$

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12 months ago

Physics physics ncert exemplar solution class 12th chapter one

The three charges q/2, q and q/2 are placed at the corners A, B and C of a square of side 'a' as shown in figure. The magnitude of electric field (E) at the corner D of the square, is

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Payal Gupta

Contributor-Level 10

$E = \frac{k q}{2 a^{2}}$

$E' = \frac{k q}{{(\sqrt[]{2} a)}^{2}} = \frac{k q}{2 a^{2}}$

$E_{N e t} = 2 E c o s 45 ° + E'$

$= \frac{k q}{a^{2}} (\frac{1}{\sqrt[]{2}} + \frac{1}{2})$

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12 months ago

Physics physics ncert exemplar solution class 12th chapter one

Given below are two statements:

Statement – I : A point change is brought in an electric field. The value of electric field at a point near to the change may increase if the charge is positive.

Statement – I : An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero.

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Payal Gupta

Contributor-Level 10

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12 months ago

Physics Thermodynamics

Given below are two statements:

Statement – I : When $μ$ moment of an ideal gas undergoes adiabatic change from state $(P_{1}, V_{1}, T_{1})$ to state $(P_{2}, V_{2}, T_{2})$ then work done is w = $\frac{μ R (T_{2} - T_{1})}{1 - γ},$ where $γ = \frac{C_{p}}{C_{v}} a n d$ R = universal gas constant.

Statement – II : In the above case, when work done on the gas, the temperature of the gas would rise.

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Payal Gupta

Contributor-Level 10

In case of adiabatic process

$W o r k d o n e, W = \frac{η R (T_{2} - T_{1})}{1 - Y}$

$A s, Δ Q = Δ U + W$

for adiabatic process :- $Δ Q = 0$

$\Rightarrow Δ U = - W$

So, when work is done by the gas, temperature decreases and when work is done on the gas, temperature rises.

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12 months ago

Physics Gravitation

Two planets A and B of equal mass are having their period of revolutions T_A and T_B such that T_A = 2T_B. These planets are revolving in the circular orbits or radii r_A and r_B respectively. Which out of the following would be the correct relationship of their orbits?

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Payal Gupta

Contributor-Level 10

According to kepler's third law of time period –

${(\frac{T_{A}}{T_{B}})}^{2} = {(\frac{r_{A}}{r_{B}})}^{3} \Rightarrow \frac{r_{A}}{r_{B}} = 4^{1 / 3}$

$\Rightarrow r_{4}^{3} = 4$ $r$ $B$ $3$

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12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Match List – I with List – II

	List – I		List – II
A.	Moment of inertia of solid sphere of radius R about and tangent.	I.	5/3MR²
B.	Moment of inertia of hollow sphere of radius (R) about any tanget	II.	7/5MR²
C.	Moment of inertia of circular ring of radius (R) about its diameter.	III.	1/4MR²
D.	Moment of inertial of circular disc of radius (R) about any diameter.	IV.	1/2MR²

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Payal Gupta

Contributor-Level 10

Use formula for M.I.

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12 months ago

Physics Motion in a Plane

Motion of a particle in x – y plane is described by a set of following equations x = 4sin $(\frac{π}{2} - ω t)$ m and y = 4sin $(ω t) m .$ The path of a particle will be

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Payal Gupta

Contributor-Level 10

$x = 4 s i n (\frac{π}{2} - ω t)$ - (i)

$y = 4 s i n (ω t)$ - (ii)

From (i) and (ii) cos2 $ω t + s i n^{2} ω t = {(\frac{x}{4})}^{2} + {(\frac{y}{4})}^{2} = 1$

$\Rightarrow x^{2} + y^{2} = {(4)}^{2}$

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12 months ago

Physics Laws of Motion

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k² rt², where k is a constant. The power delivered to the particle by the force acting on it is given as

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Payal Gupta

Contributor-Level 10

a = k²rt²

$\Rightarrow \frac{v^{2}}{r} = k^{2} r t^{2}$

$\Rightarrow v = k r t$

$a_{t} = \frac{d v}{d t} = k r$

⇒ tangential force, F_t = ma_t = mkr

$∴ P o w e r d e l i v e r e d, P = F_{t} v = (m k r) (k r t) = m k^{2} r^{2} t$

Note → Power delivered by centripetal force will be zero.

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12 months ago

Physics Mechanical Properties of Fluids

Given below are two statements : One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : Product of Pressure (P) and time (t) has the same dimension as that of coefficient of viscosity.

Reason R : Coefficient of viscosity = $\frac{F o r c e}{V e l o c i t y g r a d i e n t}$

Choose the correct answer from the options given below:

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Payal Gupta

Contributor-Level 10

Pressure * time $= \frac{F t}{A} = [M L T^{- 2}] [T] [L^{- 2}] = [M L^{- 1} T^{- 1}]$

Coefficient of viscosity, $η = [M L^{- 1} T^{- 1}]$

$F_{v i s c o u s} = - η A \frac{Δ v}{Δ y}$

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