 ×     #### 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
```
If error in measurement of speed is 10% then find the error in measureent of kinetic energy.

```
11 years ago 17 Points
```							Kinetic Energy K = (1/2) m v2
Take log of both sides.
log(K) = log(1/2) + log(m) + 2 log(v)
Now, differentiate both sides.
Differential of log(x) is 1/x dx.
x can be anything.
So, here, we have
(1/K) dK = 0 + (1/m) dm + 2 (1/v) dv    [Because log(1/2) is a constant, and derivative of any constant is zero]
Now, dK is nothing but delta(K).
So, dK is the error in measurement of K.
dK/K is the relative error in measurement of K.
Percentage error is relative error x 100
Hence,
dK/K = dm/m + 2 dv/v
We are given that dv/v = 10% = 0.1
dm/m is not given. So we will assume there is no error in measurement of m.
Therefore, Relative error in measurement of K = dK/K = 0 + 2(0.1) = 0.2
Percentage error = 0.2 x 100 = 20 %

```
11 years ago
```							error in KE turns out to be dm/m+2dv/v when we differentiate the expression for KE.
Assuming no error in mass measurement the error in KE should be 2*10 percent=>20%.
please specify doubts specifically and which part of question did you stuck while solving the problem and attempt questions well before you give up.!!!
```
11 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on General Physics

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

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