Class 12th

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New answer posted

a year ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

In non-polar molecules, centre of +ve charge coincides with centre of –ve charge, Hence, net dipole moment becomes zero.

When non-polar material is placed in external field, centre of charge does not coincide, hence give non-zero moment.

New answer posted

a year ago

0 Follower 13 Views

R
Raj Pandey

Contributor-Level 9

After switch 'S' is closed

Q 1 + Q 2 = C 1 V -    (1)

              Using KVL

              Q 1 C 1 − Q 2 C 2 = 0  

              Q 1 = Q 2 C 1 C 2                       - (2)

              from (1) & (2)

              ⇒ Q 2 [ C 1 + C 2 C 2 ] = C 1 V ⇒ Q 2 = ( C 1 C 2 C 1 + C 2 ) V

 

 

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

According to Lorentz's Force, we can write

  F → = F → E l e c t r i c + F → M a g n e t i c = q E → + q ( v → * B → ) ,     s o

Statement I is correct but Statement II is incorrect

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Mid point of BC is 1 2 ( 5 i ^ + ( α − 2 ) j ^ + 9 k ^ )  

A B ¯ = i ^ + ( α − 4 ) j ^ + k ^  

A C ¯ = i ^ + ( − 2 − α ) j ^ + k ^

For a = 1,   A B ¯ and A C ¯  will be collinear. So for non collinearity

a = 2

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

x − 1 1 = y − 2 2 = z − 1 2 = − 2 ( 1 + 4 + 2 − 1 6 ) 1 + 2 2 + 2 2  

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

| x y z + 1 3 6 6 1 1 2 | = 0

->2x – z = 1

option (B) satisfies.

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

According to question, we can write

  d = 2 h R ⇒ d 2 d 1 = h 2 h 1 ⇒ h 2 = ( d 2 d 1 ) 2 h 1 = ( 2 1 ) 2 * 1 2 5 = 5 0 0 m

Increment in height of tower = h2 – h1 = 500 – 125 = 375 m

New answer posted

a year ago

0 Follower 21 Views

A
alok kumar singh

Contributor-Level 10

Let AB ≡ x − 2 y + 1 = 0  

AC  ≡ 2 x − y + 1 = 0  

So vertex A = (1, 1)

altitude from B is perpendicular to AC and passing through

orthocentre.

So, BH = x + 2y – 7 = 0

CH = 2x + y – 7 = 0

now solve AB & BH to get B (3, 2) similarly CH and AC to get C (2, 3) so centroid is at (2, 2)

 

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

B P = μ 0 I 2 π r * 2

⇒ 3 0 0 * 1 0 − 6 = 4 π * 1 0 − 7 2 π * 4 * 1 0 − 2 * 2                                                                         

=> I = 30 A

New answer posted

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Line ⊥  to the normal

->3p + 2q – 1 = 0

( 2 , − 1 , − 3 ) lies in the plane 2p + q = 8

From here p = 15, q = -22

Equation of plane 15x – 22y + z – 5 = 0

Distance from origin =  | 5 1 5 2 + ( − 2 2 ) 2 + 1 2 | = 5 1 4 2  

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Process-AB ⇒ Isobaric,                                                           

Process-AC ⇒ Isothermal, and

Process-ADÞAdiabatic

W2 < W1 < W3

 

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