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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

A gas mixture consists of 2.0 moles of oxygen and 4.0 moles of neon at temperature T. Neglecting all vibrational modes, calculate the total internal energy of the system. (Oxygen has two rotational modes.)

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

Oxygen gas having 5 degrees of freedom

Energy per mole of the gas =5/2RT

For 2 moles of the gas total internal energy =2 $*$ 5/2RT=5RT

Neon is a monoatomic gas having 3 degrees of freedom

Energy per mole =3/2RT

We have 4 moles of Ne

Energy = 4 $*$ 3/2RT=6RT

Total energy =5RT+6RT=11RT

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

In a meter bridge, the point D is a neutral point (figure).
(a) The meter bridge can have no other neutral. A point for this set of resistances.
(b) When the jockey contacts a point on meter wire left of D, current flows to B from the wire
(c) When the jockey contacts a point on the meter wire to the right of D, current flows from B to the wire through galvanometer
(d) When R is increased, the neutral point shifts to left

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, c)

Explanation- In case of meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, bridge is balanced. If in balanced position of bridge AB = l, BC = (100 – l) so that Q/P= (100-l)/l. Also P/Q=R/S=>S= (100-l)/l R
When there is no deflection in galvanometer there is no current across the galvanometer, then points B and D are at same potential. That point at which galvanometer shows no deflection is called null point, then potential at B and neutral point D are same. When the jockey contacts a point on the meter wire to

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

Two molecules of a gas have speeds of 9 x 10⁶ m s^-1 and 1 x 10⁶ ms⁻¹ , respectively. What is the root mean square speed of these molecules.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

The measurement of an unknown resistance R is to be carried out using Wheatstone bridge as given in the figure. Two students perform an experiment in two ways. The first student takes R₂= 10 Ω and R₁ = 5 Ω. The other student takes R₂ = 1000 Ω, and R₁ = 500 Ω. In the standard arm, both take R₃ = 5 Ω.
Both find R = R₂/R₁, R₃ = 100 Ω within errors.
(a) The errors of measurement of the two students are the same
(b) Errors of measurement do depend on the accuracy with which R₂ and R₁ can be measured
(c) If the student uses large values of R₂ and R₁, the currents through the arms will be feeble

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alok kumar singh

Contributor-Level 10

Answer – (b, c)

Explanation – According to the relation that p/q=r/s if galvanometer shows no deflection. In the above equation $\frac{R 1}{R 2} = \frac{R 3}{R 4}$ the value of R4 depends upon R1 and R2 . if the value of R1 and R2 will be feeble the value of R4 would be affected.

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

The molecules of a given mass of a gas have root mean square speeds of 100ms^-1 at 27 C° and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

V_rms= $\sqrt[]{\frac{3 R T}{M}}$

V_rms= $\sqrt[]{T}$

$\frac{V_{r m s 1}}{V_{r m s 2}} = \sqrt[]{\frac{T_{1}}{T_{2}}}$

T₁=27^oC= 27+273=300K

T₂=127^oC= 127+273= 400K

$\frac{100}{V_{r m s 2}} = \sqrt[]{\frac{300}{400}} = \frac{\sqrt[]{3}}{2}$

V_rms²= $\frac{2 * 100}{\sqrt[]{3}} = \frac{200}{\sqrt[]{3}} m / s$

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

Temperature dependence of resistivity ρ(T) of semiconductors, insulators and metals is significantly based on the following factors:
(a) Number of charge carriers can change with temperature T.
(b) Time interval between two successive collisions can depend on T.
(c) Length of material can be a function of T.
(d) Mass of carriers is a function of T.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- according to the relation $ρ$ = $\frac{m}{n e 2 t}$ also T $\propto \frac{1}{t}$

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

Consider a simple circuit shown in figure stands for a variable resistance R'. R' can vary from R₀to infinity, r is internal resistance of the battery (r <(a) Potential drop across AB is nearly constant as R' is varied.
(b) Current through R' is nearly a constant as R' is varied.
(c) Current I depends sensitively on R'
(d) I ≥ V/r + R always;

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, d)

Explanation- potential drop across AB is independent to R' also when we decrease the value of R the current will also very large. According to the relation I=e/R+r

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

Kirchhoff's junction rule is a reflection of
(a) Conservation of current density vector.
(b) Conservation of charge.
(c) The fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction.
(d) The fact that there is no accumulation of charges at a junction.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (b, d)

Explanation-Algebraic sum of the currents flowing towards any point in an electric network is zero. It also tell us about the conservation of charge.

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

The volume of a given mass of a gas at 27°C, 1 atm is 100 cc. What will be its volume at 327°C?

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

V $\propto T$

V/T = constant

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

T₁=273+27=300K

T₂= 273+327= 600K

V₁= 100cc

V₂=V₁ (600/300)

V₂=2V₁

V₂= 2 (100)=200cc

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10 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

Which of the following characteristics of electrons determines the current in a conductor?
(a) Drift velocity alone
(b) Thermal velocity alone
(c) Both drift velocity and thermal velocity
(d) Neither drift nor thermal velocity

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- I=AneV_d , current is directly proportional to drift velocity.

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