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New answer posted

a year 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 – In the circuit when an electron approaches a junction, in addition to the uniform E that faces it normally (which keep the drift velocity fixed), as drift velocity (vd) is directly proportional to Electric field (E). That's why there are accumulation of charges on the surface of wires at the junction.
These produce additional electric fields. These fields alter the direction of momentum. Thus, the motion of a charge across junction is not momentum conserving

New answer posted

a year ago

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P
Payal Gupta

Contributor-Level 10

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

Height of stairs h= 10m

Energy produced by burning 1 kg of fat = 7000Kcal

Energy produced by burning 5kg of fat = 5 * 7000 = 35000 K c a l

Energy utilised in going up and down one time

= mgh + 1 2 m g h = 3 2 m g h

= 3 2 * 60 * 10 * 10

= 9000J= 9000/4.2=3000/1.4cal

Number of times, the person has to go up and down the stairs

= 35 * 10 6 3000 1.4 = 3.5 * 1.4 * 10 6 3000  = 16.3 * 10 3 times

New answer posted

a year ago

0 Follower 43 Views

A
alok kumar singh

Contributor-Level 10

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

Explanation- applying kirchhoff's junction rule I1 = I+I2

Applying kirchhoff's rule in outer loop containing 10V cell

10= IR+10I1……………. (1)

Applying kirchhoff's rule in outer loop containing 2V cell

2= 5I2-RI= 5 (I1-I)-RI

4= 10I1-10I-RI………… (2)

From 1 and 2

6=3RI+10I

2=I (R+10/3)

V= I (R+Reff)

After comparing V=2V, Reff= 10/3 ohm

Since effective internal resistance Reff of two cells 10/3 ohm, being the parallel combination 5 ohm and 10 ohm . the equivalent circuit is

 

New answer posted

a year ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

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

Temperature of the source T1= 500K and sink T2= 300K

Work done W= 1000J

Efficiency of Carnot engine = 1-T2/T1= 1-300/500= 200/500= 2/5

Efficiency = W/Q1

So Q1= W/efficiency = 1000 5 2 = 2500 J

New answer posted

a year ago

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P
Payal Gupta

Contributor-Level 10

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

During driving temperature of the gas increases while volume remains constant. So according charle's law, at constant volume V.

Pressure is directly proportional to temperature. Therefore pressure of gas increases.

New answer posted

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

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

Explanation- according to ohm's law V= IR

I= 6/6 = 1A

I= AneVd  or Vd= i/neA

New answer posted

a year ago

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P
Payal Gupta

Contributor-Level 10

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

Yes during adiabatic compression the temperature of a gas increases while no heat is

In adiabatic compression dQ=0

From the first law of thermodynamics dU= dQ-dW

dU=-dW

in compression work is done on the gas i.e work done is negative

dU=positive

hence internal energy of the gas increases due to which its temperature increases.

New answer posted

a year ago

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Payal Gupta

Contributor-Level 10

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

If a refrigerator's doors is kept open, then room will become hot, because amount of heat removed would be less than the amount of heat released in the room.

New answer posted

a year ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

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

Explanation – let R' be the resistance of potentiometer wire.

Effective resistance of potentiomter and variable resistor r=50ohm is 500+R'

Effective voltage across potentiometer = 10V

The current through main circuit I= V 50 + R = 10 50 + R

Potential difference across wire of potentiometer

 IR'= 10 R ' 50 + R

Since with 50 ohm resistor, null point is not obtained it is possible when

10 * R ' 50 + R < 8

10R' < 400 + 8 R '

2R'<400 or R'<200 ohm

Similarly with 10 ohm resistor, null point is obtained its is only possible when

10 * R ' 50 + R < 8

2R'>40

R'>40

10 * 3 4 R ' 10 + R ' < 8

7.5R'<80+8R'

R'>160

160

Any R' between 160 ohm and 200 ohm will achieve.

Since the null point on the last 4th segment of

...more

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

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

For path1

Heat Q1= 1000J

Work done =W1

For path 2

Work done W2= W1-100

As change in internal energy is same

dU=Q1-W1=Q2-W2

1000-W1=Q2-W1+100

Q2= 1000-100= 900J

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