Temperature coefficient of resistance of platinum is 4- 10-3K-1 at 0-C- The temperature at which the increase in the resistance of platinum wire is 10

R_t=R_0[1+α t] R_t=R_0+10/100R_0=1.1R_0 1.1R_0=R_0[1+α t] 1.1 1=α t 0.1=4× 10^ 3t t=100/4=25^∘C ...

166

You visited us 166 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Dependence of Resistance on Temperature

Temperature c...

Question

Temperature coefficient of resistance of platinum is $4 \times 10^{- 3} K^{- 1}$ at $0^{\circ}$ C. The temperature at which the increase in the resistance of platinum wire is 10% of its value at $0^{\circ}$ C.

250^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

500^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

50^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25^{\circ} C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $25^{\circ} C$
$R_{t} = R_{0} [1 + α t] R_{t} = R_{0} + \frac{10}{100} R_{0} = 1.1 R_{0} 1.1 R_{0} = R_{0} [1 + α t] 1.1 - 1 = α t 0.1 = 4 \times 10^{- 3} t t = \frac{100}{4} = 25^{\circ} C$

Suggest Corrections

Similar questions

Temperature coefficient of resistance of platinum is $4 \times 10^{- 3} K^{- 1}$ at $0^{\circ}$ C. The temperature at which the increase in the resistance of platinum wire is 10% of its value at $0^{\circ}$ C.

Q. Temperature coefficient of resistance of platinum is

4 \times 10^{- 3} / K

20^{\circ} C

. Temperature at which increase in resistance of platinum is

10

% is value at

20^{\circ} C

Q. The temperature coefficient of resistance of platinum is

α = 3.92 \times 10^{- 1} K^{- 1}

20^{o} C

. Find the temperature at which the increase in the resistance of platinum wire is

10

of its value at

20^{o} C

Q. A platinum wire has a resistance of

10 Ω

0^{\circ} C

and

20 Ω

273^{\circ} C

. Find the value of temperature coefficient of platinum.

Join BYJU'S Learning Program

Temperature coefficient of resistance of platinum is 4×10−3K−1 at 0∘C. The temperature at which the increase in the resistance of platinum wire is 10% of its value at 0∘C.

The correct option is D 25∘C Rt=R0[1+αt]Rt=R0+10100R0=1.1R01.1R0=R0[1+αt]1.1−1=αt0.1=4×10−3tt=1004=25∘C

Temperature coefficient of resistance of platinum is $4 \times 10^{- 3} K^{- 1}$ at $0^{\circ}$ C. The temperature at which the increase in the resistance of platinum wire is 10% of its value at $0^{\circ}$ C.

The correct option is D $25^{\circ} C$
$R_{t} = R_{0} [1 + α t] R_{t} = R_{0} + \frac{10}{100} R_{0} = 1.1 R_{0} 1.1 R_{0} = R_{0} [1 + α t] 1.1 - 1 = α t 0.1 = 4 \times 10^{- 3} t t = \frac{100}{4} = 25^{\circ} C$