×

#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
A rod with rectangular cross section oscillates about a horizontal axis passig through one of its ends and it behave like a second pendulum. Determine it`s length

```
3 years ago

Vikas TU
13784 Points
```							Dear student T = 2*pie sqrt(I/mgd) - It is Rigid body By parallel axis theorem I = ML^2/12 + M(L/2)^2 = Ml^2/3 d = l/2 T = 2*pie sqrt(2ML^2/3MgL) = 3/2 *1 = 1.5m
```
one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Engineering Entrance Exams

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

Post Question