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Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Twelve Thermodynamics

A 110 V, 50Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance $55 Ω,$ at resonance in the circuit, will be__________A.

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Payal Gupta

Contributor-Level 10

At Resonance

X_L = X_C

then l_L = l_C

Now phasor diagram

for L & C

So, Net current = zero

$= (l_{L} + l_{C})$

Therefore current through R circuit at resonance will be zero

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Physics NCERT Exemplar Solutions Class 12th Chapter Twelve Thermodynamics

A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is

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Payal Gupta

Contributor-Level 10

$r = \frac{m v}{q B}$

$r_{α} = \frac{m_{α} v}{q_{α} B}$

$r_{P} = \frac{m_{P} v}{q_{P} B}$

$\frac{r_{α}}{r_{P}} = \frac{m_{α} q_{P}}{m_{P} q_{α}} = \frac{m_{α}}{m_{P}} (\frac{q_{P}}{q_{α}})$

$= \frac{4 m}{m} (\frac{q}{2 q}) = 2 : 1$

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Physics NCERT Exemplar Solutions Class 12th Chapter Twelve Thermodynamics

300 cal. Of heat is given to a heat engine and it rejects 225 cal. of heat. If source temperature is 227°C, then the temperature of sink will be__________°C.

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alok kumar singh

Contributor-Level 10

T₁ = 227°C

= 500k

T₂ =?

$\frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$

$\frac{300}{500} = \frac{225}{T_{2}}$

$T_{2} = \frac{500 * 225}{300}$

= 5 * 75

...more

T₁ = 227°C

= 500k

T₂ =?

$\frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$

$\frac{300}{500} = \frac{225}{T_{2}}$

$T_{2} = \frac{500 * 225}{300}$

= 5 * 75

T₂ = 375k

T₂ = 102°C

less

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Twelve Thermodynamics

A cylinder of fixed capacity of 44.8 litres contains helium gas at standard gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0°C will be:

(Given gas constant R = 8.3 JK^-¹ -mol^-¹)

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alok kumar singh

Contributor-Level 10

44.4 litres of He $\frac{44.8}{22.4} m o l e s$

$η = 2$ moles of He

Q = $η C_{v} Δ T$ (for fixed capacity (v = constant)

= $2 * \frac{R}{γ - 1} Δ T$

= $\frac{2 * R}{\frac{5}{3} - 1} * 20$

= $= 2 * \frac{3 R}{2} * 20$

Q = 60R = 60 * 8.3

Q = 498 J

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Twelve Thermodynamics

A gas has n degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be:

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alok kumar singh

Contributor-Level 10

$C v = \frac{n * R}{2}$

$C p = C v + R = \frac{(n + 2) R}{2}$

$\Rightarrow \frac{C v}{C P} = \frac{n}{(n + 2)}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Twelve Thermodynamics

1 mole of rigid diatomic gas performs a work of $\frac{Q}{5}$ when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is $\frac{x R}{8} .$ The value of x is

[R = universal gas constant]

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Payal Gupta

Contributor-Level 10

$Δ Q = Δ U + Δ W$

$\Rightarrow Q = Δ U + \frac{Q}{5} \Rightarrow Δ U = \frac{4 Q}{5} = n C_{v} Δ T \Rightarrow \frac{4 Q}{5} = \frac{5 R}{2} Δ T \Rightarrow Δ T = \frac{8 Q}{25 R}$

$Q = n c Δ T = 1 * C * \frac{80}{25 R} \Rightarrow C = \frac{25 R}{8} \Rightarrow x = 25$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Twelve Thermodynamics

A monoatomic gas at pressure P and volume V is suddenly compressed to one eight of its original volume. The final pressure at constant entropy will be:

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Vishal Baghel

Contributor-Level 10

Constant Entropy means Adiabatic process

$P v^{y} = C$

$P_{1} {v_{1}}^{y} = P_{2} {v_{2}}^{y}$

$P_{2} = P_{1} {(\frac{v_{1}}{v_{2}})}^{y} = P {(\frac{v}{\frac{1}{8} v})}^{y}$

$P_{2} = P (8) \frac{5}{3}$

P₂ = 32 P

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Twelve Thermodynamics

The volume V of a given mass of monoatomic gas changes with temperature T according to the relation $V = k T^{\frac{2}{3}} .$ The work done when temperature changes by 90 K will be xR. The value of x is…….

[R = universal gas constant]

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Payal Gupta

Contributor-Level 10

$V = k T^{\frac{2}{3}}$

$T V^{- \frac{3}{2}} = K$

$\Rightarrow γ - 1 = - 3 / 2$

$\Rightarrow γ = - 1 / 2$

Work done = $\frac{n R Δ T}{- y + 1} = \frac{1 * R * 90}{3 / 2} = 6 R$

n = 60

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Twelve Thermodynamics

7 mol of a certain monoatomic ideal gas undergoes a temperature increase of 40K at constant pressure. The increase in the internal energy of the gas in this process is:

(Given R = 8.3 JK^-1 mol^-1)

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Vishal Baghel

Contributor-Level 10

$Δ u = n C_{V} Δ T$

$= n \frac{3 R}{2} Δ T$

$= 7 * \frac{3}{2} * 8.3 * 40$

$= 3486 J$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Twelve Atoms

The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be

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Payal Gupta

Contributor-Level 10

$Δ E = 13.6 (\frac{1}{1^{2}} - \frac{1}{5^{2}}) = 13.6 * \frac{24}{25} e V$

$\Rightarrow \frac{h c}{λ} = 13.6 * \frac{24}{25} e V . . . . . . . . . . (1)$

With the help of conservation of linear momentum, we can write

$\frac{h}{λ} = m_{H} v_{H} \Rightarrow \frac{h c}{λ} = c m_{H} v_{H} \Rightarrow v_{H} = \frac{\frac{h c}{λ}}{c m_{H}} = \frac{13.6 * \frac{24}{25} * 1.6 * 1 0^{- 19}}{3 * 1 0^{8} * 1.67 * 1 0^{- 27}} = 4.17 m / s$

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