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Physics Physics NCERT Exemplar Solutions Class 12th Chapter Six

A body mass 8kg and another of mass 2kg are moving with equal kinetic energy. Thee ratio of their respective momenta will be:

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alok kumar singh

Contributor-Level 10

$K E = \frac{ρ^{2}}{2 m}$

$\Rightarrow ρ^{2} α m$

$\Rightarrow {(\frac{P_{1}}{P_{2}})}^{2} = \frac{8}{2} = \frac{4}{1}$

$\Rightarrow \frac{P_{1}}{P_{2}} = \frac{2}{1}$

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a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Ten

In Young's double slit experiment, the fringe width is 12mm. If the entire arrangement is placed in water refractive index $\frac{4}{3}$ , then the fringe width becomes (in mm):

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Vishal Baghel

Contributor-Level 10

$β = 12 m m = 12 * 1 0^{- 3} m$

$\begin{array}{l} β' = \frac{β}{μ} = \frac{12 * 1 0^{- 3}}{4 / 3} = 9 * 1 0^{- 3} m \\ β' = 9 m m \end{array}$

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a year ago

Physics Ray Optics and Optical Instruments

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

Assertion A : For a simple microscope, the angular size of the object equals the angular size of the image.

Reason R : Magnification is achieved as the small object can be kept much closer to the eye that 25cm and hence it subtends a large angle.

In the light of the above statements, choose the most appropriate answer from the options given below

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New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

A parallel beam of light of wavelength 900nm and intensity 100Wm^-2 is incident on a surface perpendicular to the beam. The number of photons crossing 1cm² area perpendicular to the beam in one second is:

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Vishal Baghel

Contributor-Level 10

$E = I A$

$= 100 * 1 * 1 0^{- 4}$

$= 1 0^{- 2} W$

$E = n h \frac{c}{λ}$

$1 0^{- 2} = \overset{?}{n} * \frac{6.64 * 1 0^{- 34} * 3 * 1 0^{8}}{900 * 1 0^{- 9}}$

$\overset{?}{n} = \frac{1 0^{- 2} * 9 * 1 0^{- 7}}{6.64 * 1 0^{- 34} * 3 * 1 0^{8}}$

$\overset{?}{n} = 4.5 * 1 0^{16}$

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a year ago

Physics Electromagnetic Induction

An aeroplane, with its wings spread 10m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5 * 10^-4Wb/m² and the angle of dip is 60°. The emf induced between the tips of the plane wings will be

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Payal Gupta

Contributor-Level 10

B_v = B sin 60°

$B_{v} = 2.5 * 1 0^{- 4} * \frac{\sqrt[]{3}}{2}$

$E m f = B_{v} * v * l = 2.5 * 1 0^{- 4} * \frac{\sqrt[]{3}}{2} * 180 * \frac{5}{18} * 1 = 108.25 * 1 0^{- 3} v o l t s$

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a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is:

(Take log₁₀ 1.88 = 0.274)

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Vishal Baghel

Contributor-Level 10

$A = A o e^{- λ t}$

$\Rightarrow 2250 = 4250 e^{- λ t}$

$\Rightarrow λ * 10 * 0.434 = 0.274$

$\Rightarrow λ = \frac{0.274}{10 * 0.434} = 0.63 m i n^{- 1}$

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a year ago

Physics Thermodynamics

The internal energy (U), pressure (P) and volume (V) of an ideal gas are related as U = 3PV + 4. The gas is

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Payal Gupta

Contributor-Level 10

U = 3PV + 4

$\Rightarrow n C_{v} T = 3 P V + 4$

$\Rightarrow n * \frac{f R T}{2} = 3 P V + 4$

$\Rightarrow f = 6 + \frac{8}{P V}$

$\Rightarrow f > 6 \Rightarrow p o l y a t o m i c$

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a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

A radio can tune to any station in 6 MHz to 10 MHz band. The value of corresponding of wavelength bandwidth will be:

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Vishal Baghel

Contributor-Level 10

$λ_{1} = 6 M H z = 6 * 1 0^{6} H z$

$λ_{1} = \frac{C}{υ_{1}} = \frac{3 * 1 0^{8}}{6 * 1 0^{6}} = 50 m$

$λ_{2} = \frac{C}{υ_{2}} = 10 * 1 0^{6} = \frac{3 * 1 0^{8}}{10 * 1 0^{6}} = 30 m$

Wavelength bandwidth

$= | λ_{1} - λ_{2} | = 20 m$

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a year ago

Physics physics ncert exemplar solutions class 12th chapter two

If 'C' and 'V' represent capacity and voltage respectively then what are the dimensions of $λ$ where C/V = $λ ?$

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Payal Gupta

Contributor-Level 10

$λ = \frac{c}{v} = \frac{Q^{3}}{W^{2}} = \frac{I^{3} T^{3}}{M^{2} L^{4} T^{- 4}} \Rightarrow [λ] = [M^{- 2} L^{- 4} T^{7} l^{3}]$

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a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Five

Three masses M = 100kg, m₁ = 10kg and M₂ = 20kg are arranged in a system as shown in figure. All the surface are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force F is applied on the system so that the mass m₂ moves upward with an acceleration of 2ms^-2. The value of F is:

(Take g = 10ms^-2)

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Vishal Baghel

Contributor-Level 10

Given Acceleration of m₂ is 2m/s²

(upward)

N = 20a T – m₂g = m₂a

T – 20g = 20a

T = 200 + 20 * 2

T = 240 N

T = m₁ a₁

240 = 10a₁

a₁ = 24 m/s²

$\Rightarrow T a + T a_{1} - T a_{2} = 0$

$a_{2} = 24 + 2 = 26 m / s^{2}$

F – T – N = Ma₂

F – 240 = 100 * 26 + 20 * 26

F = 3360 N

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