Class 11th

Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th

Follow Ask Question
8k

Questions

0

Discussions

28

Active Users

0

Followers

New question posted

4 months ago

0 Follower 2 Views

New answer posted

4 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

While the particle moves from mean position to displacement, half of its amplitude, its phase changes by π/6 rad. So,
Time taken, t = (π/6)/ω = T/12 = (2/12)s = (1/6)s
a = 6

New answer posted

4 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Using Conservation of Mechanical Energy at point-A and at point-B, we can write
K_B = U_A - U_B [Since K_A = 0]
⇒ (1/2)mv_B² = mg (h_A - h_B)
⇒ v_B = √ (2 * 10 * (10 - 5) = 10m/s

New answer posted

4 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

For Wire-A
F / (πr_A²) = Y (l_A / 2) . (1)
For Wire-B
F / (πr_B²) = Y (l_B / 4) . (2)
From equations (1) and (2), we can write
(l_A / (2r_A²) = (l_B / (4r_B²) ⇒ l_B/l_A = 2 (r_B/r_A)² ⇒ x = l_B = 32

New answer posted

4 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Angle from direction of flow = 90° + θ = 120°
⇒ θ = 30°
sin 30° = u/v ⇒ u/10 = 1/2 ⇒ u = 5 m/s

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

The reaction is: FeCl? + 3H? C? O? + 3KOH → K? [Fe (C? O? )? ] (A) + 3HCl + 3H? O

In the complex [Fe (C? O? )? ]³? , the oxalate ion (C? O? )²? is a bidentate ligand.

There are three bidentate ligands, so the coordination number, or secondary valency, of Fe is 3 * 2 = 6.

New answer posted

4 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

The cell constant (G) is given by G = κ * R, where κ is conductivity and R is resistance.

Since the cell constant is constant: (κ * R)KCl = (κ * R)HCl

0.14 Sm? ¹ * 4.19 Ω = κ_HCl * 1.03 Ω

κ_HCl = (0.14 * 4.19) / 1.03 = 0.569 Sm? ¹

This is equivalent to 56.9 * 10? ² Sm? ¹.

The answer, rounded off, is 57.

 

New answer posted

4 months ago

0 Follower 12 Views

R
Raj Pandey

Contributor-Level 9

Apply conservation of momentum along y-axis, we can write
10v? - 10v? sin 30° = 0
⇒ v? = 20m/s

New answer posted

4 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Deceleration, a = u² / 2S = 10² / (2 * 0.5) = 100 m/s².
Retarding force, F = MA = 0.1 * 100 = 10N

New answer posted

4 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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
  • 681k Reviews
  • 1800k 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.