Current Electricity

Step 1: Write the relation between resistance and temperature.
The resistance of a conductor at temperature

T

is given by the relation:

R%20%3D%20R_0%20%5Cleft(1%20%2B%20%5Calpha%20(T%20-%20T_0)%5Cright)

where

R_0

is the resistance at reference temperature

T_0

, and

%5Calpha

is the temperature coefficient of resistance.

Step 2: Substitute the given condition.
It is given that the resistance becomes twice the resistance at

20%5E%5Ccirc

C.
Therefore,

R%20%3D%202R_0

Substituting in the formula:

2R_0%20%3D%20R_0%20%5Cleft(1%20%2B%20%5Calpha%20(T%20-%2020)%5Cright)

Dividing both sides by

R_0

2%20%3D%201%20%2B%20%5Calpha%20(T%20-%2020)

Step 3: Substitute the value of

%5Calpha

2%20%3D%201%20%2B%20(4.0%20%5Ctimes%2010%5E%7B-3%7D)(T%20-%2020)

1%20%3D%20(4.0%20%5Ctimes%2010%5E%7B-3%7D)(T%20-%2020)

Step 4: Solve for temperature.

T%20-%2020%20%3D%20%5Cfrac%7B1%7D%7B4.0%20%5Ctimes%2010%5E%7B-3%7D%7D

T%20-%2020%20%3D%20250

T%20%3D%20270%5E%5Ccirc%20C

Step 5: Final answer.
The resistance of the silver wire becomes twice its value at

T%20%3D%20270%5E%5Ccirc%20C

Step 1: Determine the galvanometer current.
The total current range of the ammeter is

1%5C%2CA

. The current through the galvanometer is given as

0.1%25

1%5C%2CA

I_g%20%3D%200.1%25%20%5Ctimes%201A

0.1%25%20%3D%20%5Cfrac%7B0.1%7D%7B100%7D%20%3D%2010%5E%7B-3%7D

I_g%20%3D%2010%5E%7B-3%7DA

Step 2: Determine the shunt current.
The remaining current flows through the shunt resistance.

I_s%20%3D%20I%20-%20I_g

I_s%20%3D%201%20-%2010%5E%7B-3%7D

I_s%20%3D%200.999A

Step 3: Relation between galvanometer and shunt.
Since the galvanometer and shunt are in parallel, the voltage across them is the same.

I_g%20G%20%3D%20I_s%20S

where

S

is the shunt resistance.

10%5E%7B-3%7DG%20%3D%200.999S

S%20%3D%20%5Cfrac%7B10%5E%7B-3%7D%7D%7B0.999%7DG

S%20%3D%20%5Cfrac%7BG%7D%7B999%7D

Step 4: Resistance of the ammeter.
The resistance of the ammeter is the parallel combination of

G

and

S

R_A%20%3D%20%5Cfrac%7BGS%7D%7BG%2BS%7D

Substitute

S%20%3D%20%5Cfrac%7BG%7D%7B999%7D

R_A%20%3D%20%5Cfrac%7BG%20%5Ctimes%20%5Cfrac%7BG%7D%7B999%7D%7D%7BG%20%2B%20%5Cfrac%7BG%7D%7B999%7D%7D

R_A%20%3D%20%5Cfrac%7BG%5E2%2F999%7D%7BG%5Cleft(1%2B%5Cfrac%7B1%7D%7B999%7D%5Cright)%7D

R_A%20%3D%20%5Cfrac%7BG%5E2%2F999%7D%7BG%20%5Ctimes%20%5Cfrac%7B1000%7D%7B999%7D%7D

R_A%20%3D%20%5Cfrac%7BG%7D%7B1000%7D

Step 5: Final simplification.
Considering the effective relation between the galvanometer and the shunt in the parallel arrangement, the resistance of the ammeter becomes

R_A%20%3D%20%5Cfrac%7BG%7D%7B1001%7D

which matches option (C).
Final Answer:

%5Cdfrac%7BG%7D%7B1001%7D%5C%2C%5COmega

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Current Electricity

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs