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The resistance of a wire is ‘R’ ohm. If it is melted and stretched to ‘n’ times its original length, its new resistance will be

(A)

\(\large \frac{R}{n}\)

(B)

\(\large n^{2}R\)

(C)

\(\large \frac{R}{n^{2}}\)

(D)

nR

NEET - Medical UG-2017 Easy

Answer & Solution:

\(\large{A=area~of~cross~ section,~l=length~ of ~ wire~\\ Volume(V)=constant\\Al=A_{new}l_{new}\\ Al=A_{new}nl\\ A_{new}={\Large{{A}\over n}}\\ Resistance(R)=\rho{\Large{{l}\over A}}\\ R_{new}=\rho{\Large{{l_{new}}\over A_{new}}}=\rho{\Large{{n^2l}\over A}}=n^2R\\ }\)

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Other Questions of the same topic

Q1:

A capacitor of \(\large 2\mu F\) is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :

(A)

80%

(B)

0%

(C)

20%

(D)

75%

NEET - Medical UG-2016 Medium

Q2:

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. \(\large E_{0}\) and a resistance \(\large r_{1}\). An unknown e.m.f. E is balanced at a length \(\iota\) of the potentiometer wire. The e.m.f. E will be given by :

(A)

\(\large \frac{LE_{0}r}{\left ( r+r_{1} \right )\iota }\)

(B)

\(\large \frac{LE_{0}r}{\iota r_{2} }\)

(C)

\(\large \frac{E_{0}r}{\left ( r+r_{1} \right ) }.\frac{\iota }{L}\)

(D)

\(\large \frac{E_{0}\iota }{L }\)

NEET - Medical UG-2015 Hard

Q3:

The resistance in the two arms of the meter bridge are 5 \(\large \Omega \) and R \(\large \Omega \) , respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 \(\large \iota _{1} \) . The resistance 'R' is :

(A)

10 \(\large \Omega \)

(B)

15 \(\large \Omega \)

(C)

20 \(\large \Omega \)

(D)

25 \(\large \Omega \)

NEET - Medical UG-2014 Easy

Chemistry Biology

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