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New answer posted

10 months ago

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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

10 months ago

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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

10 months ago

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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

10 months ago

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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

10 months ago

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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

10 months ago

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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

...more

New answer posted

10 months ago

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P
Payal Gupta

Contributor-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= ( 2 V o V o )yPo= 2yPo

Hence ratio of final pressure = 2 γ P o 2 P o = 2 γ - 1

New answer posted

10 months ago

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A
alok kumar singh

Contributor-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 ago

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A
alok kumar singh

Contributor-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 ago

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P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) Work done ABCD = area of rectangle ABCDA

= AB * B C = (3Vo-Vo) * (2po-po)

= 2V0 * Po= 2poVo

And work done by the gas =- 2poVo

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