Thank you for registering.

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

Please check your email for login details.
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 tightly coiled spring having 75 coils,each 3.5 cm in diameter is made of insulated metal wire 3.25 mm in diameter. An ohm meter connected across its opposite ends reads 1.74ohm . what is the reseptivity of metal ?

A tightly coiled spring having 75 coils,each 3.5 cm in diameter is made of insulated metal wire 3.25 mm in diameter. An ohm meter connected across its opposite ends reads 1.74ohm . what is the reseptivity of metal ? 

Grade:12

1 Answers

Md Mushahid
21 Points
2 years ago
let us consider given 75 coils 
diameter of coiled spring  dc = 3.50cm = 3.5\times 10^{-2}m
diameter of insulated metal wire dm = 3.25mm =3.25\times 10^{-3}m
R =1.74\Omega
\rho = \frac{R\times A}{L}
First we need to find the value of  L,A 
L=75\times \pi \times 3.50\times 10^{-2}
L =8247m 
A = \pi \times r ^{2} = \frac{\pi }{4}\times d ^{2}
A = \frac{3.142 }{4}\times\left ( 3.25\times 10^{-3} \right )^{2}
A = 8.296\times10 ^{-6} m
 
\rho = \frac{R\times A}{L}
\rho = \frac{1.74\times 8.296\times 10^{-6}}{8.247}
 
\rho = 1.7540\times 10^{-6} \Omega m
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free