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If a wire is stretched to make it 0.1% longer, its resistance will

  • Option 1)

    increase\: by \: 0.05

  • Option 2)

    increase\: by \: 0.2

  • Option 3)

    decrease\: by \: 0.2

  • Option 4)

    decrease\: by \: 0.05

 

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S solutionqc
Answered Nov 11, 2018

As we have learned

 

Ratio of resistances -

If length is given

R \alpha l^{2}

- wherein

\frac{R_{1}}{R_{2}}=\left (\frac{l_{1}}{l_{2}} \right )^{2}

 

 lA = constant \\A = \frac{K}{l}

R = \frac{\rho l}{A } = \rho \frac{l }{K/l }= \frac{\rho }{K }l^2

 

l= l_0 + (0.001 l)\\R' = \frac{\rho }{K}l_0^2 [ 1+0.001]^2 = R_0 [1+0.002]

R increases by 0.2 % 

 

 

 

 

 

 

 

 

 


Option 1)

increase\: by \: 0.05

Option 2)

increase\: by \: 0.2

Option 3)

decrease\: by \: 0.2

Option 4)

decrease\: by \: 0.05

If \Theta _{i}  is the inversion temperature , \Theta _{n} is the neutral temperature , \Theta _{c}  is the temperature of the cold junction, then

  • Option 1)

    \Theta _{i}+\Theta _{c}=\Theta _{n}

  • Option 2)

    \Theta _{i}-\Theta _{c}=2\Theta _{n}

  • Option 3)

    \frac{\Theta _{i}+\Theta _{c}}{2}=\Theta _{n}

  • Option 4)

    \Theta _{c}-\Theta _{i}=2\Theta _{n}

 

View All (1) Answers

A Avinash
Answered Jul 07, 2018

\Theta _{c}+\Theta _{i} = 2\Theta _{n}

\Rightarrow \Theta _{n} = \frac{\Theta _{c}+\Theta _{i}}{2}


Option 1)

\Theta _{i}+\Theta _{c}=\Theta _{n}

This option is incorrect.

Option 2)

\Theta _{i}-\Theta _{c}=2\Theta _{n}

This option is incorrect.

Option 3)

\frac{\Theta _{i}+\Theta _{c}}{2}=\Theta _{n}

This option is correct.

Option 4)

\Theta _{c}-\Theta _{i}=2\Theta _{n}

This option is incorrect.

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