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6 months ago

Class 12th 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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6 months ago

Class 12th 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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6 months ago

Class 12th 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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6 months ago

Class 12th 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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6 months ago

Class 12th 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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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

A ball of mass 0.15 kg hits the wall with its initial speed of 12 ms^-1and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100 N, calculate the time during of the contact of ball with the wall.

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alok kumar singh

Contributor-Level 10

$F = \frac{Δ ρ}{Δ t} = \frac{1.8 - (- 1.8)}{Δ t} = 100$

$\Rightarrow Δ t = \frac{3.6}{100} = 0.036 s e c$

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6 months ago

Class 12th Motion in a Straight Line

A scooter accelerates from rest for time t₁ at constant rate a₁ and then retards at constant rate a₂ for time t₂ and comes to rest. The correct value of $\frac{t_{1}}{t_{2}}$ will be

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Payal Gupta

Contributor-Level 10

$t a n α = \frac{v_{m a x}}{t_{1}} = a_{1}$

$t a n β = \frac{v_{m a x}}{t_{2}} = a_{2}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{t_{2}}{t_{1}} \Rightarrow \frac{t_{1}}{t_{2}} = \frac{a_{2}}{a_{1}}$

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6 months ago

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

In a Vernier Calipers, 10 divisions of Vernier scale is equal to the 9 divisions of main scale. Then both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4^th Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1mm. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and 6^th Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical bo

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alok kumar singh

Contributor-Level 10

1 MSD = 1mm

9 MSD = 10 VSD

1VSD = 0.9 MSD = 0.9 mm

Least count = 1 MSD – 1 VSD

= 0.1 mm = 0.01 cm

Zero Error = (10 – 4) * 0.1 = 0.6 mm

Reading = MSR + VSR – Zero error

= 3 cm + 6 * 0.01 – (0.06)

= 3.12 cm $?$ 3.10 cm

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6 months ago

Class 12th Oscillations

A particle executes S.H.M., the graph of velocity as a function of displacement is

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Payal Gupta

Contributor-Level 10

$v = ω \sqrt[]{A^{2} - x^{2}} \Rightarrow \frac{v^{2}}{A^{2} ω^{2}} + \frac{x^{2}}{A^{2}} = 1$ Path is ellipse.

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Four

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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