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New answer posted
10 months agoContributor-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
10 months agoContributor-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
10 months agoContributor-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
10 months agoContributor-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
10 months agoContributor-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 =
New answer posted
10 months agoContributor-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
New answer posted
10 months agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
(a) Container A is isothermal and container B is adiabatic

For isothermal process P1V1=P2V2
Po (2Vo)= P2 (Vo)
P2= 2Po
for adiabatic process
P1V1y= P2V2y
Po (2Vo)y=P2 (Vo)y
P2= ( )yPo= 2yPo
Hence ratio of final pressure =
New answer posted
10 months agoContributor-Level 10
This is a Short Answer Type Questions as classified in NCERT Exemplar
Explanation- If the value of R is increased, the current through the wire will decrease which in turn decreases the potential difference across AB, and hence potential gradient (k) across AB decreases.
Since, at neutral point, for given emf of cell, l increases as potential gradient (k) across AB has decreased because E' = kl
Thus, with the increase of l, which will result in increase in balance length. So, jockey J will shift towards B
New answer posted
10 months agoContributor-Level 10
This is a Short Answer Type Questions as classified in NCERT Exemplar
Explanation – power consumed through wire is P=V2/R =I2R . for transmitting the power there are two types either current is large or voltage is less or vice versa. But when current in the circuit is higher then power loss is also higher . so we transmit the power through high voltage circuits.
New answer posted
10 months agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
(b) Work done ABCD = area of rectangle ABCDA
= AB = (3Vo-Vo) (2po-po)
= 2V0 Po= 2poVo
And work done by the gas =- 2poVo
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