Physics

Get insights from 5.6k questions on Physics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics

Follow Ask Question
5.6k

Questions

0

Discussions

31

Active Users

0

Followers

New question posted

10 months ago

0 Follower 1 View

New answer posted

10 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- R= δlA for greater value of R, A should be less and its possible value when connected across 1cm * 1/2cm faces.

New answer posted

10 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

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

molar mass = mass of avogadro's number of atoms= 6.023 * 10 23 atoms.

197 g of gold contains =6.023 * 10 23 a t o m s

1g of gold contain= 6.023 * 10 23 197 atoms

39.4 g of gold atoms = 6.023 * 10 23 * 39.4 197 = 1.2 * 10 23 atoms

New answer posted

10 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation-The potential drop along the wires of potentiometer should be greater than emfs of cells.
In a potentiometer experiment, the emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. Here, values of emfs of two cells are given as 5 V and 10 V, therefore, the potential drop along the potentiometer wire must be more than 10 V.

New answer posted

10 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (c)

Explanation – R/S= (l1/100-l1)= 100 (2.9/100-2.9)= 100/97.1=2.98ohm

So he should change S to almost 3 ohm and repeat the experiment.

New answer posted

10 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

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

Consider the diagram

let n =number of molecules per unit volume

Vrms= rms speed of gas molecule

When block is moving with speed vo, relative speed of molecules w.r.t front face =v+vo

Coming head on, momentum transferred to block per collision =2m (v+vo)

Number of collisions in time ? t = 1 2 (v+vo)n ? t A  where A is the area of cross section.

So momentum transferred in time ? t =m (v+vo)2nA ? t   this is from front surface

Similarly momentum transferred in time ? t = m (v-vo)2nA ? t ) this is from back surface

Drag force = mnA (v+vo)2- (v-vo)2)

= mnA (4wo)=4mnAvvo

= 4 ρ A vvo

So ρ =mn/V=M/

...more

New answer posted

10 months ago

0 Follower 15 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- eeq= e2r1+e2r1r1+r2

New answer posted

10 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation- as we know that J=E, and current density is directly proportional to electric field, so electric field produced by charges accumulated on the surface of wire.

New answer posted

10 months ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

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

Given volume V = 1m3

area = 0.01mm2

= 8.01 * 10 - 6 m2= 10 - 8 m2

Temperature both inside and outside are equal

So initial pressure inside the box = 1.50atm

Final pressure inside the box= 0.1atm

Assuming Vx= speed of nitrogen molecule in x direction

ni = number of molecules per unit volume in a time interval of ? T

Let area of the wall, number of particles colliding in time

? t = 1 2 n i (vx ? t )A , here we use ½ because particle moves both in positive and negative direction.

Vx2+ Vy2+ Vz2= Vrms2

Vx2= Vrms2/3 if all three velocities are equal.

½ mvrms2= 3/2KBT/m

New answer posted

10 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

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

Explanation- according to ohms law

I= VeffReff+R

If voltage and resistance increase

V'= nVeff, R'= nReff

I'= nVeffnReff+nR= VeffReff+R =

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1850k Answers

Share Your College Life Experience

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.