Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The length of a wire becomes  and <img alt="\mathrm{l_{2}}" src="https://entranc

The length of a wire becomes \mathrm{l_{1}} and \mathrm{l_{2}} when 100 N and 120 N tensions are applied respectively. If  \mathrm{10 \: l_{2} = 11\: l_{1},}the natural length of wire will be \mathrm{\frac{1}{x}l_{1}} . Here the value of x is _______ .

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{F=kx}               

\mathrm{F}=\frac{\mathrm{yA}}{\ell_0} \cdot \mathrm{x}
Sol when \mathrm{F}=100 \mathrm{~N}

\begin{aligned} & 100=\mathrm{k}\left(\ell_1-\ell_0\right) \\ \end{aligned}..............(1)

\begin{aligned} & \text { When } \mathrm{F}=120 \mathrm{~N} \\ \end{aligned}

\begin{aligned} & 120=\mathrm{K}\left(\left(\ell_1-\ell_0\right)\right. \end{aligned}
Given that 10 \ell_2=11 \ell_1

\ell_2=1.1 \ell_1

So \quad 120=\mathrm{K}\left(1.1 \ell_1-\ell_0\right)...............(2)

Now (2) \(1)

\begin{aligned} & \frac{120}{100}=\frac{\mathrm{K}\left(1.1 \ell_1-\ell_0\right)}{\mathrm{K}\left(\ell_1-\ell_0\right)} \\ & 1.2=\frac{1.1 \ell_1-\ell_0}{\ell_1-\ell_0} \\ & 1.2 \ell_1-1.2 \ell_0=1.1 \ell_1-\ell_0 \\ & 0.1 \ell_1=0.2 \ell_0 \\ & \ell_0=\frac{\ell_1}{2} \text { So } \mathrm{x}=2 \text { Ans. } \end{aligned}

Posted by

Rishabh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: Most important chapters for physics, chemistry, mathematics to score good marks
anam-khan

JEE Main 2025 session 2 dates clash with ISC, PSEB 12th exams; no impact on major science subjects
anam-khan

Delhi police EOW books FIITJEE for fraud; 192 complaints filed

Latest Question