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
Menu
Grade: 11
        
A faulty thermometer has LFP=10degree and UFP=110degree. What is actual temperature of an object if this thermometer reads 40degree celcius
 
3 months ago

Answers : (2)

Arun
23032 Points
							
 
Here faulty reading (x)=40
Lower fixed point of faulty readind (L.F.P)=10
Upper fixed point of faulty reading (U.F.P.)=110
Fundamental interval (F.I.)=110- 10 = 100
Correct reading(C)=?
c/100*(x-L.F.P)/F.I.
C/100=40/100 =0.4
C=0.4*100=40 celsius scale
3 months ago
Vikas TU
9499 Points
							
Dear student:
usually the thermometer fixed points basically are:
Lower point =0°c
Upper point=100°c
so in between there are 100 divisions.
each one division corresponds to 1°c

But in the above mentioned case :
Lower point=10°c
upper point=110°c
so total division=110-10=100
thus, 1 degree= 1 division=100/100
Now let us find the given temperature:
measured temperature 40 °c
=100x40/100
=40°c
Hope it is clear.
3 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

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


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details