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New answer posted

7 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation-

 total resistance = 2+8= 10 ohm

I= 6-42+8 = 0.2A

The direction of flow of current is always from high potential to low potential

If VB > VA

VB-4V-0.2 * 8 = VA

VB-VA= 3.6V

New answer posted

7 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation – in series combination of resistors current I =e/R+r

10I= eR+rn

1+n1+1n  10 = ( 1+n1+n )n  n=10

New answer posted

7 months ago

0 Follower 16 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (c) Q1= W+Q2

W=Q1-Q2>0

Q1>Q2>0

We can also write Q21

New answer posted

7 months ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) the given process is a cyclic process i.e returns to the original state 1

Hence change in internal energy dU =0

dQ= dU+dW=0+dW

hence total heat supplied is converted to work done by the gas which is not possible by second law of thermodynamics.

(c) When the gas expands adiabatically from 2 to 3 . it is not possible to return to the same state without being heat supplied hence 3 to 1 cannot be adiabatic.

New answer posted

7 months ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c) Change in internal energy for process A to B

dU=nCvdT=nCv (dT)=nCv (TB-TA)

work done from A to B  = area under the PV curve which is maximum for path I

New answer posted

7 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) For isothermal dT= 0 so T=constant

For an ideal gas dU = change in internal energy = nCvdT=0

From first law of thermodynamics dQ= dU+dW

dQ= dW

New answer posted

7 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (b), (d) When the rod is hammered the external work is done on the rod which increases its temperature.

Heat is transferred to the gas in the small container by big reservoir at temperature T2

As the weight is added to the cylinder arrangement in the form of external pressure so it cannot reversed.

New answer posted

7 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- according to the relation V=e-Ir and I=e/r+R

The graphical relation between voltage and resistance.

New answer posted

7 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) Let us assume that T1, T2T3

According to questions there is no loss of heat in the surroundings

Heat lost by M3 = heat gained by M1+ heat gained by M2

M3s (T3-T)= M1s (T-T1)+M2s (T-T2)

T [M1+M2+M3]= M3T3+M1T1+M2T2

T = M 1 T 1 + M 2 T 2 + M 3 T 3 M 1 + M 2 + M 3

New answer posted

7 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- If the current in auxiliary circuit (lower circuit containing primary cell) decreases, and potential difference across A and jockey/increases. Then deflection in galvanometer is one sided and the deflection decreased, while moving from one end 'A ' of the wire to the end 'S'.
And clearly this is possible only when positive terminal of the cell E1 is connected at X and E1>E.
(ii) If the current in auxiliary circuit increases, and potential difference across A and jockey J increases. Then also deflection in galvanometer shows one sided deflection.
And th

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